19: Electron Transfer Reactions

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19.1  Electrical Fundamentals

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Exercise \(\PageIndex{1}\)

What is the relationship between potential energy (U) in joules and electric potential (E_cell​) in volts?
Note: In electrochemistry, we use E_cell to represent the electric potential across a cell, while U represents energy. This relationship describes the work required to move a charged particle across an electric potential gradient. N

Answer

The change in potential energy (ΔU) of a charge q moving through an electric potential difference E_cell is given by:

ΔU=q E_cell​

where

- ΔU = change in potential energy (joules, J)
-q = charge (coulombs, C)
-E_cell = electric potential difference (volts, V = J C⁻¹)

Engineers often write this relationship using V instead of E_cell:

ΔU=qV

If we express it in terms of work, the equation becomes:

W=qV

where W is the work done to move the charge. The energy stored or released by this process is the same quantity, expressed as U (potential energy).

 

 

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Exercise \(\PageIndex{2}\)

What is the charge of an electron?

Answer

1.6x10^-19 Coulombs, this is one of the defining SI constants, and you do not need to memorize it as a number.

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Exercise \(\PageIndex{3}\)

What is the charge of a mol of electrons

Answer

A Faraday (F)  1F=96500 C/mol e^-

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Exercise \(\PageIndex{1}\)

What is an Ampere, the unit of electrical current?

Answer

1A = 1C/sec

19.2 Oxidation-Reduction Reactions

Redox Chemistry

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Exercise \(\PageIndex{1}\)

In the following reaction,

 

Fe²⁺(aq) + Ag⁺(aq) -> Fe³⁺(aq) + Ag(s)
 

a.	Fe^2^+ is oxidized and Fe^3^+ is reduced.
b.	Fe^2^+ is oxidized and Ag^+ is reduced.
c.	Ag^+ is oxidized and Ag(s) is reduced.
d.	Ag^+ is oxidized and Fe^2^+ is reduced.
e.	Ag^+ is oxidized and Fe^3^+ is reduced.

Answer

b.	Fe^2^+ is oxidized and Ag^+ is reduced.

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Exercise \(\PageIndex{1}\)

Salts of Cu⁺ disproportionate in water to form Cu²⁺ salts and Cu metal,

2 Cu⁺ (aq) --> Cu²⁺ (aq) + Cu (s)

Which species is OXIDIZED and which species is REDUCED in this reaction, respectively?

a.  Cu⁺ (aq) and Cu⁺ (aq), respectively.      b.  Cu⁺ (aq) and Cu²⁺ (aq), respectively.
c.  Cu⁺ (aq) and Cu (s), respectively.              d.Cu²⁺ (aq) and Cu (s), respectively.
e.  This is not a reduction-oxidation reaction.

Answer

a.  Cu⁺ (aq) and Cu⁺ (aq), respectively.

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Balancing Redox Reactions in Acidic Solutions

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Exercise \(\PageIndex{2.1a}\)

Balance the reaction in an acidic solution:

\(Cu(s)+NO_{3}^{-}(aq)\rightarrow Cu^{2+}(aq)+NO_{2}(g)\)

Answer

\[Cu(s)+NO_{3}^{-}(aq)\rightarrow Cu^{2+}(aq)+NO_{2}(g)\]

\[Cu(s)\rightarrow Cu^{2+}(aq)\]

\[NO_{3}^{-}(aq)\rightarrow NO_{2}(g)\]

balanced oxidation half reaction:

\[Cu(s)\rightarrow Cu^{2+}(aq)+2e^{-}\]

balanced reduction half reaction:

\[NO_{3}^{-}(aq)+2H^{+}(aq)+e^{-}\rightarrow NO_{2}(g)+H_{2}O(l)\]

adding so electrons cancel 

\[2NO_{3}^{-}(aq)+4H^{+}(aq)+Cu(s) \rightarrow 2NO_{2}(g)+2H_{2}O(l)+Cu^{2+}(aq)\]

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Exercise \(\PageIndex{2.1b}\)

Balance the reaction in an acidic solution:

\(Mn^{2+}(aq)+BiO_{3}^{-}(aq)\rightarrow Bi^{3+}(aq)+MnO_{4}^{-}(aq)\)

Answer

\[Mn^{2+}(aq)+BiO_{3}^{-}(aq)\rightarrow Bi^{3+}(aq)+MnO_{4}^{-}(aq)\]

\[Mn^{2+}(aq)+4H_{2}O(l)\rightarrow MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\]

\[BiO_{3}^{-}(aq)+6H^{+}(aq)+2e^{-}\rightarrow Bi^{3+}(aq)+3H_{2}O(l)\]

\[2Mn^{2+}(aq)+5BiO_{3}^{-}(aq)+14H^{+}(aq)\rightarrow 5Bi^{3+}(aq)+7H_{2}O(l)+2MnO_{4}^{-}(aq)\]

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Exercise \(\PageIndex{2.1c}\)

Balance the reaction in an acidic solution:

\(Cr_{2}O_{7}^{2-}+(aq)+Cl^{-}(aq)\rightarrow Cr^{3+}(aq)+ClO_{3}^{-}(aq)\)

Answer

\[Cr_{2}O_{7}^{2-}+(aq)+Cl^{-}(aq)\rightarrow Cr^{3+}(aq)+ClO_{3}^{-}(aq)\]

balanced reduction half reaction:

\[Cr_{2}O_{7}^{2-}+(aq)+14H^{+}(aq)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l)\]

Balanced oxidation half reaction:

\[Cl^{-}(aq)+3H_{2}O(l)\rightarrow ClO_{3}^{-}(aq)+6H^{+}(aq)+6e^{-}\]

Adding half reactions so electrons lost=gained, and cancelling common species

\[Cr_{2}O_{7}^{2-}+(aq)+8H^{+}(aq)+Cl^{-}(aq)\rightarrow 2Cr^{3+}(aq)+4H_{2}O(l)+ClO_{3}^{-}(aq)\]

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Exercise \(\PageIndex{2.1d}\)

Balance the reaction in an acidic solution:

\(IO_{3}^{-}(aq)+HNO_{2}(aq)\rightarrow I^{-}(aq)+NO_{3}^{-}(aq)\)

Answer

\[IO_{3}^{-}(aq)+HNO_{2}(aq)\rightarrow I^{-}(aq)+NO_{3}^{-}(aq)\]

\[IO_{3}^{-}(aq)+6H^{+}(aq)+6e^{-}\rightarrow I^{-}(aq)+3H_{2}O(l)\]

\[HNO_{2}(aq)+H_{2}O(l)\rightarrow NO_{3}^{-}(aq)+3H^{+}(aq)+2e^{-}\]

\[IO_{3}^{-}(aq)+3HNO_{2}(aq)\rightarrow I^{-}(aq)+3NO_{3}^{-}(aq)+3H^{+}(aq)\]

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Exercise \(\PageIndex{2.1e}\)

Balance the reaction in an acidic solution:

\(Fe_{2}S(s)+Ce^{4+}(aq)\rightarrow Fe^{2+}(aq)+SO_{4}^{2-}(aq)+Ce^{3+}(aq)\)

Answer

\[
\text{Unbalanced skeletal equation:}\quad
\mathrm{Fe_2S(s)} + \mathrm{Ce^{4+}(aq)}
\;\longrightarrow\;
\mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)} + \mathrm{Ce^{3+}(aq)}
\]

Step 1: Split into half-reactions

 

\[
\text{Iron + sulfur:}\quad
\mathrm{Fe_2S(s)} \;\longrightarrow\; \mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)}
\]

\[
\text{Cerium:}\quad
\mathrm{Ce^{4+}(aq)} \;\longrightarrow\; \mathrm{Ce^{3+}(aq)}
\]

We now balance each half-reaction separately 

Step 2: Balance the first half-reaction

    a. Balance all elements except O and H

\[
\mathrm{Fe_2S(s)} \;\longrightarrow\; \mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)}
\]

Iron: there are 2 Fe atoms on the left, so we put a 2 in front of \(\mathrm{Fe^{2+}}\):

\[
\mathrm{Fe_2S(s)} \;\longrightarrow\; 2\,\mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)}
\]

Sulfur is already balanced (1 S on each side).

    b. Balance O by adding water

There are 4 O atoms on the right in \(\mathrm{SO_4^{2-}}\) and none on the left, so we add
\(4\,\mathrm{H_2O}\) to the left:

\[
\mathrm{Fe_2S(s)} + 4\,\mathrm{H_2O(l)}
\;\longrightarrow\;
2\,\mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)}
\]

    c. Balance H by adding \(\mathrm{H^+}\)

On the left there are \(4 \times 2 = 8\) H atoms in \(4\,\mathrm{H_2O}\); there are no H
atoms on the right. So we add \(8\,\mathrm{H^+}\) to the right:

\[
\mathrm{Fe_2S(s)} + 4\,\mathrm{H_2O(l)}
\;\longrightarrow\;
2\,\mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)} + 8\,\mathrm{H^+(aq)}
\]

    d Balance charge by adding electrons}

Now count charges:

- Left side: \(\mathrm{Fe_2S}\) and \(\mathrm{H_2O}\) are neutral, so total charge \(= 0\).
- Right side: \(2\,\mathrm{Fe^{2+}} \Rightarrow +4\), \(\mathrm{SO_4^{2-}} \Rightarrow -2\),
  \(8\,\mathrm{H^+} \Rightarrow +8\).

Total right-hand charge:
\[
(+4) + (-2) + (+8) = +10
\]

To bring the right-hand charge down to 0, we add \(10\) electrons to the
this is an oxidation, so electrons appear as products:

\[
\mathrm{Fe_2S(s)} + 4\,\mathrm{H_2O(l)}
\;\longrightarrow\;
2\,\mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)} + 8\,\mathrm{H^+(aq)} + 10\,e^-
\]

This completes the oxidation half-reaction.

Step 3: Balance the reduction half-reaction (we know this is reduction since the other was oxidation)

\[
\mathrm{Ce^{4+}(aq)} \;\longrightarrow\; \mathrm{Ce^{3+}(aq)}
\]

    a. Balance all elements except O and H

Cerium is already balanced (1 Ce on each side).

    b. and c.: Balance O and H

There are no O or H atoms, so nothing to do.

    d Balance charge by adding electrons

Charge on the left is \(+4\), charge on the right is \(+3\). To reduce the
charge on the left, add \(1\) electron to the left side:

\[
\mathrm{Ce^{4+}(aq)} + e^- \;\longrightarrow\; \mathrm{Ce^{3+}(aq)}
\]

Now both sides have charge \(+3\) and this is the reduction half reaction because electrons appear as reactants

Step 4: Make electrons lost = electrons gained

Oxidation half-reaction:
\[
\mathrm{Fe_2S(s)} + 4\,\mathrm{H_2O(l)}
\;\longrightarrow\;
2\,\mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)} + 8\,\mathrm{H^+(aq)} + 10\,e^-
\]

Reduction half-reaction:
\[
\mathrm{Ce^{4+}(aq)} + e^- \;\longrightarrow\; \mathrm{Ce^{3+}(aq)}
\]

To cancel electrons, multiply the reduction half-reaction by 10:

\[
10\,\mathrm{Ce^{4+}(aq)} + 10\,e^- \;\longrightarrow\; 10\,\mathrm{Ce^{3+}(aq)}
\]

Step 5: Add half-reactions and cancel electrons

Add the two half-reactions:

\[
\mathrm{Fe_2S(s)} + 4\,\mathrm{H_2O(l)} + 10\,\mathrm{Ce^{4+}(aq)} + 10\,e^-
\;\longrightarrow\;
2\,\mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)} + 8\,\mathrm{H^+(aq)}
+ 10\,\mathrm{Ce^{3+}(aq)} + 10\,e^-
\]

Cancel the \(10\,e^-\) on both sides:

\[
\mathrm{Fe_2S(s)} + 4\,\mathrm{H_2O(l)} + 10\,\mathrm{Ce^{4+}(aq)}
\;\longrightarrow\;
2\,\mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)} + 8\,\mathrm{H^+(aq)}
+ 10\,\mathrm{Ce^{3+}(aq)}
\]

\[
\boxed{
\mathrm{Fe_2S(s)} + 4\,\mathrm{H_2O(l)} + 10\,\mathrm{Ce^{4+}(aq)}
\;\longrightarrow\;
2\,\mathrm{Fe^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)} + 8\,\mathrm{H^+(aq)}
+ 10\,\mathrm{Ce^{3+}(aq)}
}
\]

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Exercise \(\PageIndex{2.1f}\)

Write a balanced chemical equation for the following reaction in an acidic solution.

 

Cr₂O₇^2-(aq) + Fe²⁺(aq) -> Cr³⁺(aq) + Fe³⁺(aq)

Answer

Balance the reaction in an acidic solution:

\[
\ce{Cr2O7^{2-}(aq) + Fe^{2+}(aq) -> Cr^{3+}(aq) + Fe^{3+}(aq)}
\]

**balanced reduction half reaction:**

\[
\ce{Cr2O7^{2-}(aq) + 14 H^{+}(aq) + 6 e^- -> 2 Cr^{3+}(aq) + 7 H2O(l)}
\]

**Balanced oxidation half reaction:**

\[
\ce{Fe^{2+}(aq) -> Fe^{3+}(aq) + e^-}
\]

**Adding half reactions so electrons lost = gained, and cancelling common species**

(Multiply the oxidation half reaction by 6 and add:)

\[
\ce{Cr2O7^{2-}(aq) + 14 H^{+}(aq) + 6 Fe^{2+}(aq) -> 2 Cr^{3+}(aq) + 7 H2O(l) + 6 Fe^{3+}(aq)}
\]
 

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Exercise \(\PageIndex{2.1g}\)

.  In acid solution, dichromate ion oxidizes ferrous iron to ferric iron as illustrated in the partial equation:

Fe²⁺ + Cr₂O₇^2- <=> Fe³⁺ + Cr³⁺

When this equation is balanced, the coefficient of Fe³⁺ is:

a. 1                  b. 2                  c. 4                  d. 6                  e. none of the above

Answer

d. 6  

\[
\mathrm{Fe^{2+} + Cr_2O_7^{2-} \rightleftharpoons Fe^{3+} + Cr^{3+}}
\]

Separate the reaction into two half-reactions.

\[
\mathrm{Fe^{2+} \rightarrow Fe^{3+}}
\]

Balance charge by adding one electron to the more positive side:

\[
\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}
\]

Now balance the dichromate half-reaction in acid solution:

\[
\mathrm{Cr_2O_7^{2-} \rightarrow Cr^{3+}}
\]

Balance chromium atoms:

\[
\mathrm{Cr_2O_7^{2-} \rightarrow 2Cr^{3+}}
\]

Balance oxygen atoms by adding water:

\[
\mathrm{Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O}
\]

Balance hydrogen atoms by adding hydrogen ions:

\[
\mathrm{14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O}
\]

Now balance charge by adding electrons. The left side has a net charge of

\[
14(+1) + (-2) = +12
\]

The right side has a net charge of

\[
2(+3) = +6
\]

Add six electrons to the left side:

\[
\mathrm{6e^- + 14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O}
\]

The iron half-reaction produces one electron, so multiply it by 6:

\[
\mathrm{6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-}
\]

Add the two half-reactions:

\[
\mathrm{6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-}
\]

\[
\mathrm{6e^- + 14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O}
\]

Cancel the electrons:

\[
\mathrm{6Fe^{2+} + 14H^+ + Cr_2O_7^{2-}
\rightarrow
6Fe^{3+} + 2Cr^{3+} + 7H_2O}
\]

Therefore, the coefficient of \(\mathrm{Fe^{3+}}\) is

\[
\boxed{6}
\]

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Balancing Redox Reactions in Basic Solutions

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Exercise \(\PageIndex{2.2a}\)

Balance the reaction in a basic solution:

\(Fe(s)+NO_{3}^{-}(aq)\rightarrow FeO_{2}^{2-}(aq)+NH_{3}(aq)\)

Answer

\[Fe(s)+NO_{3}^{-}(aq)\rightarrow FeO_{4}^{2-}(aq)+NH_{3}(aq)\]

Balance as if acidic

\[Fe(s)+2H_{2}O(l)\rightarrow FeO_{2}^{2-}(aq)+4H^{+}(aq)+2e^{-}\]

\[NO_{3}^{-}(aq)+9H^{+}(aq)+8e^{-}\rightarrow NH_{3}(aq)+3H_{2}O(l)\]

 

\[4Fe(s)+\cancel{8}5H_{2}O(l)+NO_{3}^{-}(aq)+\cancel{9H^{+}(aq)}+\cancel{8e^{-}}   \rightarrow 4FeO_{2}^{2-}(aq)+\cancel{16}7H^{+}(aq)+\cancel{8e^{-}}+ NH_{3}(aq)+\cancel{3H_{2}O(l)}  \]

\[4Fe(s)+5H_{2}O(l)+NO_{3}^{-}(aq)   \rightarrow 4FeO_{2}^{2-}(aq)+7H^{+}(aq)+ NH_{3}(aq)  \]

Add 7 hydroxide to each side to get rid of hydronium

\[4Fe(s)+5H_{2}O(l)+NO_{3}^{-}(aq) +7OH^-  \rightarrow 4FeO_{2}^{2-}(aq)+\underbrace{7H^{+}(aq) + 7OH^-}_{7 \; H_2O} + NH_{3}(aq)  \]

cancel waters

\[4Fe(s)+NO_{3}^{-}(aq)+7OH^{-}(aq)\rightarrow 4FeO_{2}^{2-}(aq)+NH_{3}(aq)+2H_{2}O(l)\]

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Exercise \(\PageIndex{2.2b}\)

Balance the reaction in a basic solution:

\(Cr\left ( OH \right )_{3}(s)+BrO^{-}(aq)\rightarrow CrO_{4}^{2-}(aq)+Br_{2}(l)\)

Answer

\[Cr\left ( OH \right )_{3}(s)+BrO^{-}(aq)\rightarrow CrO_{4}^{2-}(aq)+Br_{2}(l)\]

Balance as if acidic

\[Cr\left ( OH \right )_{3}(s)+H_{2}O(l)\rightarrow CrO_{4}^{2-}(aq)+5H^{+}(aq)+3e^{-}\]

\[2BrO^{-}(aq)+4H^{+}(aq)+2e^{-}\rightarrow Br_{2}(l)+2H_{2}O(l)\]

\[2Cr\left ( OH \right )_{3}(s)+2H_{2}O(l) + 6BrO^{-}(aq)+12H^{+}(aq) \rightarrow 2CrO_{4}^{2-}(aq)+10H^{+}(aq)+3Br_{2}(l)+6H_{2}O(l)\]

Cancel common terms

\[2Cr\left ( OH \right )_{3}(s)+ 6BrO^{-}(aq)+2H^{+}(aq) \rightarrow 2CrO_{4}^{2-}(aq)+3Br_{2}(l)+4H_{2}O(l)\]

Add hydroxide to both sides to cancel hydronium,

\[2Cr\left ( OH \right )_{3}(s)+ 6BrO^{-}(aq)+\underbrace{2H^{+}(aq) +2OH^-}_{2 \; H_2O}\rightarrow 2CrO_{4}^{2-}(aq)+3Br_{2}(l)+4H_{2}O(l) + 2OH^-\]

cancel common terms

\[2Cr\left ( OH \right )_{3}(s)+6BrO^{-}(aq)\rightarrow 2CrO_{4}^{2-}(aq)+2H_{2}O(l)+3Br_{2}(l)+2OH^{-}(aq)\]

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Exercise \(\PageIndex{2.2c}\)

Balance the reaction in a basic solution:

\(Pb\left ( OH \right )_{4}^{2-}(aq)+ClO^{-}(aq)\rightarrow PbO_{2}(s)+Cl^{-}(aq)\)

Answer

\[Pb\left ( OH \right )_{4}^{2-}(aq)+ClO^{-}(aq)\rightarrow PbO_{2}(s)+Cl^{-}(aq)\]

 

\[Pb\left ( OH \right )_{4}^{2-}(aq)\rightarrow PbO_{2}(s)+2H_{2}O(l)+2e^{-} \]

\[ClO^{-}(aq)+2H^{+}(aq)+2e^{-}\rightarrow Cl^{-}(aq)+H_{2}O(l)\]

 

\[Pb\left ( OH \right )_{4}^{2-}(aq)+ClO^{-}(aq)+2H^{+}(aq)\rightarrow PbO_{2}(s)+3H_{2}O(l)+Cl^{-}(aq)\]

\[Pb\left ( OH \right )_{4}^{2-}(aq)+ClO^{-}(aq)+\underbrace{2H^{+}(aq)+2OH^-(aq)}_{2H_2O}\rightarrow PbO_{2}(s)+3H_{2}O(l)+Cl^{-}(aq)+H_{2}O(l)+2OH^-\]

 

\[ClO^{-}(aq)+Pb\left ( OH \right )_{4}^{2-}(aq)\rightarrow PbO_{2}(s)+H_{2}O(l)+Cl^{-}(aq)+2OH^{-}(aq)\]

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Exercise \(\PageIndex{2.2d}\)

Balance the reaction in a basic solution:

\(Cl^{-}(aq)+MnO_{4}^{-}(aq)\rightarrow Cl_{2}(g)+MnO_{2}(s)\)

Answer

\[Cl^{-}(aq)+MnO_{4}^{-}(aq)\rightarrow Cl_{2}(g)+MnO_{2}(s)\]

\[2Cl^{-}(aq)\rightarrow Cl_{2}(g)+2e^{-}\]

\[MnO_{4}^{-}(aq)+ 4H^+3e^-\rightarrow MnO_{2}(s)+2H_2O\]

Combining

\[6Cl^{-}(aq)+2MnO_{4}^{-}(aq)+ 8H^+\rightarrow 3Cl_{2}(g) +2MnO_{2}(s)+4H_2O\]

Add hydroxide

\[6Cl^{-}(aq)+2MnO_{4}^{-}(aq)+ \underbrace{8H^+8OH^-}_{8H_2O}\rightarrow 3Cl_{2}(g) +2MnO_{2}(s)+4H_2O+8OH^-\]

\[6Cl^{-}(aq)+2MnO_{4}^{-}(aq)+4H_{2}O(l)\rightarrow 3Cl_{2}(g)+2MnO_{2}(s)+8OH^{-}(aq)\]

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Exercise \(\PageIndex{2.2e}\)

Balance the reaction in a basic solution:

\(IO_{3}^{-}(aq)+NO_{2}^{-}(aq)\rightarrow I^{-}(aq)+NO_{3}^{-}(aq)\)

Answer

\[I O_{3}^{-}(a q)+N O_{2}^{-}(a q) \rightarrow I^{-}(a q)+N O_{3}^{-}(a q)\]

 

\[I O_{3}^{-}(a q)+6H^++6e^- \rightarrow I^{-}(a q)+3H_2O\]

\[N O_{2}^{-}(a q)+1 H_2O\rightarrow N O_{3}^{-}(a q)+2H^++2 e^{-}\]

 

\[I O_{3}^{-}(a q)+\cancel{6H^+}+3N O_{2}^{-}(a q)+\cancel{3 H_2O}\rightarrow I^{-}(a q)+\cancel{3H_2O}+3N O_{3}^{-}(a q)+\cancel{6H^+}\]

 

\[I O_{3}^{-}(a q)+3 N O_{2}^{-}(a q) \rightarrow I^{-}(a q)+3 N O_{3}^{-}(a q)\]

 

19.3: Electrochemical Cells

Electrochemical Cell Notation

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Exercise \(\PageIndex{3.1}\)

Write the following reactions in standard electrochemical cell notation at standard state conditions.

1. \(C u(s)+C d^{2+}(a q) \rightarrow C u^{2+}(a q)+C d(s)\)
2. \(2 N a(s)+F e^{2+}(a q) \rightarrow 2 N a^{+}(a q)+F e(s)\)
3. \(3Z n(s)+2 F e^{3+}(a q) \rightarrow 3 Z n^{2+}(a q)+2 F e(s)\)
4. \(2 A l(s)+3 M n^{2+}(a q) \rightarrow 2 A l^{3+}(a q)+3 M n(s)\)
5. \(2 H^{+}(a q)+Z n(s) \rightarrow H_{2}(g)+Z n^{2+}(a q)\)
6. \(H g(l)+S n^{4+}(a q) \rightarrow H g_{2}^{2+}+Sn(s)\)

Answer a.

\[Cu(s)|Cu^{2+}(1 M)||Cd^{2+}(1 M)|Cd(s)\]

Answer b.

\[Na(s)|Na^{+}(1 M)||Fe^{2+}(1 M)|Fe(s)\]

Answer c.

\[Zn(s)|Zn^{2+}(1 M)||Fe^{3+}(1 M)|Fe(s)\]

Answer d.

\[Al(s)|A l^{3+}(1 M)||Mn^{2+}(1 M)|Mn(s)\]

Answer e.

\[Zn(s)|Zn^{2+}(1 M)||H^{+}(1 M)|H_{2}(1 atm)|Pt(s)\]

Answer f.

\[Hg(l)|Hg_{2}^{2+}(1 M)||Sn^{4+}(1 M)|Sn(s)\]

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Exercise \(\PageIndex{3.2}\)

Consider a galvanic cell based on the following overall reaction,

Fe(s) + 2Ag⁺(aq) --> Fe²⁺(aq) + 2Ag(s)

a. Write the cell in shorthand notation when the concentration of Ag+ ions is 0.050 M and the concentration of Fe⁺² ions is 1.50 M.

Answer

\[Fe(s)|Fe^{+2}(1.50M)||Ag^+(0.050M)|Ag(s)\]

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Exercise \(\PageIndex{1}\)

Which is the correct shorthand notation for the following galvanic cell reaction when [Cu²⁺(aq)] = 6.0 M and [Sn²⁺(aq)] = 0.001 M?

Cu²⁺(aq) + Sn(s) --> Cu(s) + Sn²⁺(aq), E^o = +0.48 V

a.  Cu²⁺(aq)|Sn(s)||Cu(s)| Sn²⁺(aq)                  b.  Cu²⁺(6.0)|Sn(s)||Cu(s)| Sn²⁺(0.001)
c.  Cu²⁺(6.0)|Cu(s)||Sn(s)| Sn²⁺(0.001)            d.  Cu²⁺(6.0)|Cu(s)|| Sn²⁺(0.001)|Sn(s)
e.  none of the above

Answer

none of the above

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Voltaic Cell

For many of these problems you will need to use a table of standard reduction potentials

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Exercise \(\PageIndex{3.2}\)

What is the half-reaction at the anode in the voltaic reaction?

\(2Cu^{+}(aq)+I_{2}(l)\rightarrow 2Cu^{2+}(aq)+2I^{-}(aq)\)

Answer

\[Cu^{+}\rightarrow Cu^{2+}+e^{-}\]

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Exercise \(\PageIndex{3.3}\)

What is the reaction occurring at the cathode from Question 19.3.2?

Answer

\[I_{2}+2e^{-}\rightarrow 2I^{-}\]

Exercise \(\PageIndex{3.4}\)

Which one can occur at the cathode of an electrochemical cell?

1. \(NO\rightarrow NO_{2}^{-}\)
2. \(Cr_{2}O_{7}^{2-}\rightarrow Cr^{7+}\)
3. \(I_{2}\rightarrow I^{-}\)
4. none of the above

Answer

c. \(I_{2}\rightarrow I^{-}\)

Exercise \(\PageIndex{3.5}\)

Which one can occur at the anode of an electrochemical cell?

1. \(Cr_{2}O_{7}^{2-}\rightarrow Cr^{7+}\)
2. \(Fe^{2+} \rightarrow Fe\)
3. \(I_{2}\rightarrow I^{-}\)
4. none of the above

Answer

a. \(Cr_{2}O_{7}^{2-}\rightarrow Cr^{7+}\)

Exercise \(\PageIndex{3.6}\)

Using standard redox potentials, determine which reaction occurs at the anode.

1. \(Al(s) \rightarrow Al^{3+}(aq)+3 e^{-}\)
2. \(ClO_{3}^{-}(aq)+6H^{+}(aq)+6 e^{-} \rightarrow Cl^{-}(aq)+3H_{2} O(l)\)

 

1. i.
2. ii.
3. neither
4. both

Answer

a. i.

Exercise \(\PageIndex{3.7}\)

Using the same information in Question 19.3.6, which electrode is consumed?

1. anode
2. cathode
3. neither
4. both

Answer

a. anode

Exercise \(\PageIndex{3.8}\)

Using the same information in Question 19.3.6, which electrode is positive?

1. anode
2. cathode
3. both
4. neither

Answer

b. cathode

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Exercise \(\PageIndex{3.9}\)

Which of the following statements about a salt bridge in a voltaic cell is TRUE?

1. 1. Free electrons flow through the salt bridge to maintain electrical neutrality in the two half-cells.
   2. The salt bridge allows the ions present in the two half-cells to mix extensively.
   3. The wire must be connected directly to the salt bridge in order for the salt bridge to be able to maintain electrical neutrality in the two half-cells.
   4. In some cases, a salt bridge functions as the anode.
   5. Ions from the electrolyte in the salt bridge flow into each half-cell to maintain electrical neutrality.

Answer

Ions from the electrolyte in the salt bridge flow into each half-cell to maintain electrical neutrality.

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Exercise \(\PageIndex{3.10}\)

Which of the following statements about a salt bridge in a voltaic cell is TRUE?

1. 1. Free electrons flow through the salt bridge to maintain electrical neutrality in the two half-cells.
   2. The salt bridge allows the ions present in the two half-cells to mix extensively.
   3. The wire must be connected directly to the salt bridge in order for the salt bridge to be able to maintain electrical neutrality in the two half-cells.
   4. In some cases, a salt bridge functions as the anode.
   5. Ions from the electrolyte in the salt bridge flow into each half-cell to maintain electrical neutrality.

Answer

Ions from the electrolyte in the salt bridge flow into each half-cell to maintain electrical neutrality.

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19.4: Electrochemical Cell Fundamentals

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Half-reaction	ϵred^0
\(I_{2}(s)+2e^{-}\rightarrow 2I^{-}(aq)\)	0.54 V
\(Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s)\)	-0.44 V
\(Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq)\)	0.77 V
\(Cr^{3+}(aq)+3e^{-}\rightarrow Cr(s)\)	-0.74 V

 

Figure \(\PageIndex{1}\): Use the following information to answer Questions 19.4.1-19.4.5

Exercise \(\PageIndex{4.1}\)

Determine the standard potential (V) for the cell reaction.

\(2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\)

Answer

\[2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\]

Anode: \[2I^{-}(aq) \rightarrow I_{2}(s)+2e^{-}\]

Cathode: \[Fe^{3+}(aq)+e^{-} \rightarrow Fe^{2+}(aq)\]

\[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.77-0.54=0.23 v\]

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Exercise \(\PageIndex{4.2}\)

Determine the standard potential (V) for the cell reaction.

\(Cr(s)+3Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)+Cr^{3+}(aq)\)

Answer

\[Cr(s)+3Fe^{3+}(aq) \rightarrow 3Fe^{2+}(a )+Cr^{3+}(aq)\]

Anode: \[Cr(s) \rightarrow Cr^{3+}(aq)+3 e^{-}\]

Cathode: \[Fe^{3+}(aq)+e^{-} \rightarrow Fe^{2+}(aq)\]

\[E^{0}=E_{\text {cathode}}^{o}-E_{\text {anode}}^{o}=0.77+0.74=1.51 v\]

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Exercise \(\PageIndex{4.3}\)

Determine the standard potential (V) for the cell reaction.

\(Fe(s)+2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)\)

Answer

\[Fe(s)+2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)\]

Anode: \[Fe(s) \rightarrow Fe^{2+}(aq)+2e^{-}\]

Cathode: \[Fe^{3+}(aq)+e^{-} \rightarrow Fe^{2+}(aq)\]

\[E^{0}=E_{\text {cathode}}^{o}-E_{\text {anode}}=0.77-(-0.44)=1.21 v\]

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Exercise \(\PageIndex{4.4}\)

Determine the standard potential (V) for the cell reaction.

\(2Cr(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq)\)

Answer

\[2Cr(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq)\]

Anode: \[Cr(s) \rightarrow Cr^{3+}(aq)+3 e^{-}\]

Cathode: \[Fe^{2+}(aq)+e^{-} \rightarrow Fe(s)\]

\[E^{0}=E_{cathode}^{o}-E_{anode}=-0.44-(-0.74)=0.30 v\]

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Exercise \(\PageIndex{4.5}\)

Determine the standard potential (V) for the cell reaction.

\(3I_{2}(s)+2Cr(s) \rightarrow 2Cr^{3+}(aq)+6l^{-}(aq)\)

Answer

\[3l_{2}(s)+2Cr(s) \rightarrow 2Cr^{3+}(aq)+6I^{-}(aq)\]

Anode: \[Cr(s) \rightarrow Cr^{3+}(aq)+3 ^{-}\]

Cathode: \[I_{2}(s)+2e^{-} \rightarrow 2I^{-}(aq)\]

\[E^{0}=E_{\text {cathode} \omega \dot{e}}^{o}-E_{\text {anode}}^{o}=0.54-(-0.74)=1.28 \mathrm{v}\]

 

19.5: Standard Electrochemical Potentials

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Exercise \(\PageIndex{5.1}\)

What is the E°_cell?

\(Cu(s)+Cd^{2+}(aq) \rightarrow Cu^{2+}(aq)+Cd(s)\)

Answer

\[E^{0}_{cell}=E^{0}_{red}\left ( cathode \right )-E^{0}_{red}\left ( anode \right )\]

\[Cu(s)|Cu^{2+}(1 M)|Cd^{2+}(1 M)|Cd(s)\]

\[E^{0}_{cell}=-0.403V-\left ( 0.337V \right )\]

\[E^{0}_{cell}=-0.740V\]

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Exercise \(\PageIndex{5.2}\)

What is the expression for Q, in the equation:

\(3Zn(s)+2Fe^{3+}(aq) \rightarrow 3Zn^{2+}(aq)+2Fe(s)\)

Answer

\[Q=\frac{\left [ Zn^{2+} \right ]^{3}}{\left [ Fe^{3+} \right ]^{2}}\]

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Exercise \(\PageIndex{5.3}\)

What is the E_cell?

\(3Zn(s)+2Au^{3+}(0.004 M) \rightarrow 3Zn^{2+}(0.0051 M)+2Au(s)\)

Answer

\[E_{cell}=E^{0}_{cell}-\frac{RT}{nF}lnQ\]

\[E^{0}_{cell}=+1.50V-\left ( -0.763V \right )=+2.26V\]

\[Q=\frac{\left [ Zn^{2+} \right ]^{3}}{\left [ Au^{3+} \right ]^{2}}=\frac{\left [ 0.0051 \right ]^{3}}{\left [ 0.004 \right ]^{2}}\]

\[n=6\]

\[E_{cell}=+2.26V-\frac{(8.314)298}{6(96,500)}ln\frac{\left [ 0.0051 \right ]^{3}}{\left [ 0.004 \right ]^{2}}=2.28V\]

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Exercise \(\PageIndex{5.4}\)

E°_cell is +2.26V, what is ∆G^o ?

\(3Zn(s)+2Au^{3+}(aq) \rightarrow 3Zn^{2+}(aq)+2Au(s)\)

Answer

\[\Delta G^{\circ}=-nFE^{\circ}\]

\[\Delta G^{\circ}=-\left(6 mol\right)\left(96,500J/V\right)(2.26 V)\]

\[\Delta{G}^{\circ}=-1.31 {MJ}\]

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Exercise \(\PageIndex{5.5}\)

E°_cell is -0.763, what is the ∆G?

\(2H^{+}(aq)+Zn(s) \rightarrow H_{2}(g)+Zn^{2+}(aq)\)

Answer

\[\Delta G^{\circ}=-nFE^{\circ}\]

\[\Delta G^{\circ}=-2 mol e^{-}\left(96,500{J}/{V}\cdot mol\,\,e^{-}\right)(-0.763 V)\]

\[\Delta{G}^{\circ}=147 {kJ}\]

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Exercise \(\PageIndex{1}\)

.  Consider the following half-reactions:

 

Fe^3+(aq) + e^- ® Fe^2+(aq)	Eº = +0.77 V
Sn^2+(aq) + 2 e^- ® Sn(s)	Eº = -0.14 V
Fe^2+(aq) + 2 e^- ® Fe(s)	Eº = -0.44 V
Al^3^+(aq) + 3 e^- ® Al(s)	Eº = -1.66 V
Mg^2^+(aq) + 2 e^- ® Mg(s)	Eº = -2.37 V

 

Which of the above metals or metal ions are able to oxidize Al(s)?

a.	Fe^3+ and Sn^2+
b.	Fe^3+, Sn^2+, and Fe^2+
c.	Fe^2+, Sn, and Fe
d.	Mg and Mg^2+
e.	Mg^2^+ only

 

Answer

b.	Fe^3+, Sn^2+, and Fe^2+

19.6: Electrochemistry and Thermodynamics

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Half-reaction	εred°
\(I_{2}(s)+2e^{-} \rightarrow 2I^{-}(aq)\)	0.54 V
\(Fe^{2+}(aq)+2e^{-} \rightarrow Fe(s)\)	-0.44 V
\(Fe^{3+}(aq)+e^{-} \rightarrow Fe^{2+}(aq)\)	0.77 V
\(Cr^{3+}(aq)+3e^{-} \rightarrow Cr(s)\)	-0.74 V

 

Figure \(\PageIndex{2}\): Use the following table to answer questions 19.6.1-19.6.5

Exercise \(\PageIndex{6.1}\)

Using the given information, determine the ΔG⁰ in J for the following cell reaction. F=96500J/V mol

\(2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\)

Answer

\[2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\]

\[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.77-0.54=0.23 v\]

\[\Delta G^{0}=-nFE^{0}=-96500 \times 2 \times 0.23=-44390{J}\]

Exercise \(\PageIndex{6.2}\)

Using the given information, determine the ΔG⁰ in J for the following cell reaction. F=96500J/V mol

\(Cr(s)+3 Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)+Cr^{3+}(aq)\)

Answer

\[Cr(s)+3 Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)+Cr^{3+}(aq)\]

\[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.77+0.74=1.51 v\]

\[\Delta G^{0}=-nFE^{0}=-96500 \times 3 \times 1.51=-437145{J}\]

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Exercise \(\PageIndex{6.3}\)

Using the given information, determine the ΔG⁰ in J for the following cell reaction. F=96500J/V mol

\(Fe(s)+2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)\)

Answer

\[Fe(s)+2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq)\]

\[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.77-(-0.44)=1.21 v\]

\[\Delta G^{0}=-nFE^{0}=-96500 \times 2 \times 1.21=-233530{J}\]

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Exercise \(\PageIndex{6.4}\)

Using the given information, determine the ΔG⁰ in J for the following cell reaction. F=96500J/V mol

\(2Cr(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq)\)

Answer

\[2Cr(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq)\]

\[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=-0.44-(-0.74)=0.30 v\]

\[\Delta G^{0}=-nFE^{0}=-96500 \times 6 \times 0.30=-173700{J}\]

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Exercise \(\PageIndex{6.5}\)

Using the given information, determine the ΔG⁰ in J for the following cell reaction. F=96500J/V mol

\(3I_{2}(s)+2Cr(s) \rightarrow 2Cr^{3+}(aq)+6I^{-}(aq)\)

Answer

\[3I_{2}(s)+2Cr(s) \rightarrow 2Cr^{3+}(aq)+6I^{-}(aq)\]

\[E^{0}=E_{cathode}^{o}-E_{anode}^{o}=0.54-(-0.74)=1.28 v\]

\[\Delta G^{0}=-nFE^{0}=-96500 \times 6 \times 1.28=-741120{J}\]

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Exercise \(\PageIndex{6.6}\)

Given the following standard reduction potentials,

 

Pb^2^+(aq) + 2 e^- ® Pb(s)	Eº = -0.126 V
PbSO4(s) + 2 e^- ® Pb(s) + SO 4^2-(aq)	Eº = -0.356 V

 

determine the K_sp for PbSO₄(s) at 25ºC.

PbSO₄(s) -> Pb²⁺(aq)+ SO ₄^2-(aq)

Answer

\[
\begin{aligned}
\text{Given:}\quad 
&\ce{Pb^{2+}(aq) + 2e^- -> Pb(s)} \qquad E^\circ_1 = -0.126~\text{V} \\
&\ce{PbSO4(s) + 2e^- -> Pb(s) + SO4^{2-}(aq)} \qquad E^\circ_2 = -0.356~\text{V}
\end{aligned}
\]

We want \(K_\text{sp}\) for \(\ce{PbSO4(s) -> Pb^{2+}(aq) + SO4^{2-}(aq)}\).
Reverse the first half-reaction to make oxidation and add to the second:

\[
\begin{aligned}
\ce{Pb(s) -> Pb^{2+}(aq) + 2e^-} &\qquad E^\circ_{\text{ox},1} = -E^\circ_1 = +0.126~\text{V}\\
\ce{PbSO4(s) + 2e^- -> Pb(s) + SO4^{2-}(aq)} &\qquad E^\circ_2 = -0.356~\text{V}
\end{aligned}
\]

Adding (electrons and \(\ce{Pb(s)}\) cancel) gives the target dissolution reaction:
\[
\ce{PbSO4(s) -> Pb^{2+}(aq) + SO4^{2-}(aq)}
\]
with overall
\[
E^\circ_\text{rxn} = E^\circ_{\text{ox},1} + E^\circ_2 = (+0.126) + (-0.356) = -0.230~\text{V}.
\]

Relate \(E^\circ\) to the equilibrium constant \(K\) via
\[
\Delta G^\circ = -n F E^\circ \quad\text{and}\quad \Delta G^\circ = -RT\ln K
\;\Rightarrow\; \ln K = \frac{n F E^\circ}{RT}.
\]
Here \(n=2\), \(F=96485.33212~\text{C mol}^{-1}\), \(R=8.314462618~\text{J mol}^{-1}\text{K}^{-1}\), \(T=298.15~\text{K}\), and \(E^\circ=-0.230~\text{V}\). Substituting (no intermediate rounding):

\[
\ln K = \frac{(2)(96485.33212)(-0.230)}{(8.314462618)(298.15)} = -17.90400247\dots
\]
\[
K = e^{-17.90400247\dots} = 1.6764496879\dots\times 10^{-8}.
\]

Therefore,
\[
\boxed{K_\text{sp}(\ce{PbSO4},\,25^\circ\text{C}) \approx 1.68\times 10^{-8}}
\]
\[
\text{(equivalently, }pK_\text{sp} = -\log_{10}K \approx 7.776\text{).}
\]
 

Once we got E^o we could have solved this in one step

\(K \;=\; e^{-\Delta G^\circ/RT} \;=\; e^{\,n F E^\circ/RT} \;=\; \exp\!\left(\dfrac{(2)(96485.33212)\,(-0.230~\mathrm{V})}{(8.314462618~\mathrm{J\,mol^{-1}K^{-1}})(298.15~\mathrm{K})}\right) \;=\; 1.67\times 10^{-8}\)
 

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Exercise \(\PageIndex{6.7}\)

Use the Standard Reduction Potentials given below to calculate K_f for Zn(NH₃)₄²⁺ at 25^oC.

Zn²⁺(aq) + 4NH₃(aq) <==> Zn(NH₃)₄²⁺(aq), K_f = ??

half reaction	E^o, V
Zn(NH3)4^2+(aq) + 2e^- <==> Zn(s) + 4NH3(aq)	-1.04
Zn^2+(aq) + 2e^- <==> Zn(s)	-0.76

 

Answer

\[
\mathrm{Zn^{2+}(aq) + 4NH_3(aq) \rightleftharpoons Zn(NH_3)_4^{2+}(aq)}
\]

We are given two reduction half-reactions:

\[
\mathrm{Zn(NH_3)_4^{2+}(aq) + 2e^- \rightleftharpoons Zn(s) + 4NH_3(aq)}
\]

\[
E^\circ = -1.04 \, \text{V}
\]

and

\[
\mathrm{Zn^{2+}(aq) + 2e^- \rightleftharpoons Zn(s)}
\]

\[
E^\circ = -0.76 \, \text{V}
\]

To obtain the formation reaction for the complex ion, reverse the first half-reaction:

\[
\mathrm{Zn(s) + 4NH_3(aq) \rightleftharpoons Zn(NH_3)_4^{2+}(aq) + 2e^-}
\]

When a half-reaction is reversed, the sign of \(E^\circ\) changes:

\[
E^\circ_{\text{ox}} = +1.04 \, \text{V}
\]

Now add this to the reduction of zinc ion:

\[
\mathrm{Zn^{2+}(aq) + 2e^- \rightleftharpoons Zn(s)}
\]

\[
E^\circ_{\text{red}} = -0.76 \, \text{V}
\]

Adding the two half-reactions gives:

\[
\mathrm{Zn^{2+}(aq) + 4NH_3(aq)
\rightleftharpoons
Zn(NH_3)_4^{2+}(aq)}
\]

The cell potential is

\[
E^\circ_{\text{cell}} = E^\circ_{\text{red}} + E^\circ_{\text{ox}}
\]

\[
E^\circ_{\text{cell}} = (-0.76 \, \text{V}) + (+1.04 \, \text{V})
\]

\[
E^\circ_{\text{cell}} = +0.28 \, \text{V}
\]

Now use the relationship between \(E^\circ\) and \(K\):

\[
\Delta G^\circ = -nFE^\circ
\]

and

\[
\Delta G^\circ = -RT \ln K
\]

Therefore,

\[
RT \ln K = nFE^\circ
\]

\[
\ln K = \frac{nFE^\circ}{RT}
\]

For this reaction,

\[
n = 2
\]

\[
\ln K_f =
\frac{(2)(96485 \, \text{C/mol})(0.28 \, \text{V})}
{(8.314 \, \text{J/mol K})(298.15 \, \text{K})}
\]

\[
\ln K_f = 21.8
\]

\[
K_f = e^{21.8}
\]

\[
K_f = 2.9 \times 10^9
\]

\[
\boxed{K_f = 2.9 \times 10^9}
\]

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19.7: Electrochemical Cells under Nonstandard Conditions

The Nernst Equation

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Exercise \(\PageIndex{7.1}\)

Determine the cell voltage E at 25°C, when [Fe²⁺]=[I^-]=0.01M, [Fe³⁺]=0.02M.

\(2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s), \quad E^{\circ}=0.23 v\)

Answer

\[E=E^{0}-\frac{0.0592}{n} \log Q=0.23-\frac{0.0592}{2} \log \frac{0.01^{2}}{0.01^{2} \times 0.02^{2}}=0.13 v\]

Exercise \(\PageIndex{7.2}\)

Determine the cell voltage E at 25°C, when [Fe²⁺]=0.01M, [Fe³⁺]=0.02M

\(Fe(s)+2F e^{3+}(aq) \rightarrow 3 Fe^{2+}(aq), \quad E^{\circ}=1.21 v\)

Answer

\[
\text{Reaction:}\quad
\text{Fe}(s) + 2\,\text{Fe}^{3+}(aq) \rightarrow 3\,\text{Fe}^{2+}(aq)
\]
\[
E^\circ = 1.21~\text{V}, \quad [\text{Fe}^{2+}] = 0.010~\text{M}, \quad [\text{Fe}^{3+}] = 0.020~\text{M}, \quad T = 298~\text{K}
\]

\[
\text{Nernst equation (natural log form):}\quad
E = E^\circ - \frac{RT}{nF} \ln Q
\]

First, determine \(n\), the number of electrons transferred.

Oxidation half-reaction:
\[
\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^-
\]

Reduction half-reaction:
\[
\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq)
\quad (\times 2)
\]

Thus, the total number of electrons transferred is
\[
n = 2
\]

Next, write the reaction quotient \(Q\). Solids do not appear in \(Q\):

\[
Q = \frac{[\text{Fe}^{2+}]^3}{[\text{Fe}^{3+}]^2}
  = \frac{(0.010)^3}{(0.020)^2}
\]

Now substitute into the Nernst equation (using \(R = 8.314~\text{J mol}^{-1}\text{K}^{-1}\),
\(F = 96485~\text{C mol}^{-1}\)):

\[
E = 1.21
    - \frac{(8.314)(298)}{(2)(96485)}
      \ln\!\left(\frac{(0.010)^3}{(0.020)^2}\right)
\]

Evaluating the expression:

\[
E \approx 1.29~\text{V}
\]

\[
\boxed{E \approx 1.29~\text{V}}
\]
 

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Exercise \(\PageIndex{7.3}\)

Determine the cell voltage E at 25°C, when [Fe²⁺]=0.01M, [Cr³⁺]=0.005M

\(2{Cr}(s)+3Fe^{2+}(aq) \rightarrow 3Fe(s)+2Cr^{3+}(aq), \quad E^{\circ}=0.30 v\)

Answer

\[
\text{Given reaction:}\quad
2\text{Cr}(s) + 3\text{Fe}^{2+}(aq) \rightarrow 3\text{Fe}(s) + 2\text{Cr}^{3+}(aq)
\]
\[
E^\circ = 0.30~\text{V}, \quad [\text{Fe}^{2+}] = 0.010~\text{M}, \quad [\text{Cr}^{3+}] = 0.0050~\text{M}, \quad T = 298~\text{K}
\]

\[
\text{Nernst equation (natural log form):}\quad
E = E^\circ - \frac{RT}{nF} \ln Q
\]

First, determine \(n\), the number of electrons transferred.

\[
\text{Cr}(s) \rightarrow \text{Cr}^{3+} + 3e^- \quad (\times 2)
\]
\[
\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}(s) \quad (\times 3)
\]
\[
n = 6 \text{ electrons}
\]

Next, write the reaction quotient \(Q\). Solids do not appear in \(Q\):

\[
Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3}
     = \frac{(0.0050)^2}{(0.010)^3}
     = 25
\]

Now substitute into the Nernst equation (using \(R = 8.314~\text{J mol}^{-1}\text{K}^{-1}\),
\(F = 96485~\text{C mol}^{-1}\), \(T = 298~\text{K}\)):

\[
E = 0.30
    - \frac{(8.314)(298)}{(6)(96485)}
      \ln(25)
\]

\[
E = 0.30
    - \left(\frac{8.314 \times 298}{6 \times 96485}\right)\ln(25)
    = 0.30
    - (0.004282\ldots)(3.2189\ldots)
\]

\[
E \approx 0.30 - 0.0138 = 0.286~\text{V}
\]

\[
\boxed{E \approx 0.29~\text{V}}
\]
 

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Exercise \(\PageIndex{7.4}\)

Calculate the equilibrium constant for the reaction at 25°C. R=8.314J/Kmol, F=96500 J/Vmol

\(2I^{-}(aq)+2Fe^{3+}(aq) \rightarrow 2Fe^{2+}(aq)+I_{2}(s)\)

Answer

\[K_{eq}=e^{-\frac{\Delta G^{\circ}}{RT}}=e^{\frac{nFE^{0}}{RT}}=e^{\frac{2(9650)(0.23)}{8.314(288.15)}}=5.9\times 10^{7}\]

or at 25⁰C,

\[K_{eq}=10^{\frac{nE^{0}}{0.0592}}=10^{\frac{2 \times 0.23}{0.0592}}=5.9 \times 10^{7}\]

Exercise \(\PageIndex{7.5}\)

For the reaction, if the E_cell is 1.50v and [Fe³⁺]=0.02M, determine the concentration of Fe²⁺.

\(Fe(s)+2F e^{3+}(aq) \rightarrow 3 Fe^{2+}(aq)\)

Answer

\[
\text{Overall reaction:}\quad
\mathrm{Fe(s)} + 2\,\mathrm{Fe^{3+}(aq)} \;\longrightarrow\; 3\,\mathrm{Fe^{2+}(aq)}
\]

\[
E = 1.50~\text{V}, \quad E^\circ = 1.21~\text{V}, \quad [\mathrm{Fe^{3+}}] = 0.020~\text{M}, \quad T = 298~\text{K}
\]

\[
\textbf{Half-reactions (to make the electron flow explicit):}
\]

Oxidation (anode): metallic iron is oxidized to ferrous iron
\[
\mathrm{Fe(s)} \;\longrightarrow\; \mathrm{Fe^{2+}(aq)} + 2\,e^-
\]

Reduction (cathode): ferric iron is reduced to ferrous iron
\[
\mathrm{Fe^{3+}(aq)} + e^- \;\longrightarrow\; \mathrm{Fe^{2+}(aq)}
\quad (\times 2)
\]

So, the total number of electrons transferred is
\[
n = 2
\]

\[
\textbf{Reaction quotient } Q:
\]
Solids do not appear in \(Q\), so for
\[
\mathrm{Fe(s)} + 2\,\mathrm{Fe^{3+}(aq)} \;\longrightarrow\; 3\,\mathrm{Fe^{2+}(aq)}
\]
we have
\[
Q = \frac{[\mathrm{Fe^{2+}}]^3}{[\mathrm{Fe^{3+}}]^2}
\]

\[
\textbf{Nernst equation (ln form):}
\]
\[
E = E^\circ - \frac{RT}{nF}\,\ln Q
\]

Substitute \(Q = \dfrac{[\mathrm{Fe^{2+}}]^3}{[\mathrm{Fe^{3+}}]^2}\):
\[
E = E^\circ - \frac{RT}{nF}\,\ln\!\left( \frac{[\mathrm{Fe^{2+}}]^3}{[\mathrm{Fe^{3+}}]^2} \right)
\]

Rearrange to solve for the logarithm:
\[
E - E^\circ = -\frac{RT}{nF}\,\ln\!\left( \frac{[\mathrm{Fe^{2+}}]^3}{[\mathrm{Fe^{3+}}]^2} \right)
\]
\[
\ln\!\left( \frac{[\mathrm{Fe^{2+}}]^3}{[\mathrm{Fe^{3+}}]^2} \right)
= -\,\frac{nF}{RT}\,(E - E^\circ)
\]

Exponentiate both sides:
\[
\frac{[\mathrm{Fe^{2+}}]^3}{[\mathrm{Fe^{3+}}]^2}
= \exp\!\left[-\,\frac{nF}{RT}\,(E - E^\circ)\right]
\]

Therefore,
\[
[\mathrm{Fe^{2+}}]^3
= [\mathrm{Fe^{3+}}]^2 \,
  \exp\!\left[-\,\frac{nF}{RT}\,(E - E^\circ)\right]
\]

\[
[\mathrm{Fe^{2+}}]
= \left\{
    [\mathrm{Fe^{3+}}]^2 \,
    \exp\!\left[-\,\frac{nF}{RT}\,(E - E^\circ)\right]
  \right\}^{1/3}
\]

Now substitute numerical values
(\(R = 8.314~\mathrm{J\,mol^{-1}K^{-1}},\, T = 298~\mathrm{K},\,
F = 96485~\mathrm{C\,mol^{-1}},\, n = 2,\,
E = 1.50~\mathrm{V},\, E^\circ = 1.21~\mathrm{V},\,
[\mathrm{Fe^{3+}}] = 0.020~\mathrm{M}\)):

\[
[\mathrm{Fe^{2+}}]
=
\left\{
    (0.020)^2
    \exp\!\left[
      -\,\frac{(2)(96485)}{(8.314)(298)}
      \,(1.50 - 1.21)
    \right]
\right\}^{1/3}
\]

Evaluating the expression (keeping full calculator precision in the intermediate steps) gives
\[
[\mathrm{Fe^{2+}}] \approx 4.0 \times 10^{-5}~\mathrm{M}
\]

\[
\boxed{[\mathrm{Fe^{2+}}] \approx 4.0 \times 10^{-5}~\text{M}}
\]

 

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Exercise \(\PageIndex{7.6}\)

Consider a galvanic cell based on the following overall reaction,

Fe(s) + 2Ag⁺(aq) --> Fe²⁺(aq) + 2Ag(s)

.  Calculate the cell potential (in V) for this reaction at 25^oC when the concentration of Ag⁺ ions is 0.050 M and the concentration of Fe²⁺ ions is 1.50 M.

half reaction	E^o, V
Fe^2^+(aq) + 2e^- <==> Fe(s)	-0.44
Ag^+(aq)  e^- <==> Ag(s)	+0.80

 

Answer

Expert Solution

1.  Calculate E⁰

\[
E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode}
= 0.80 - (-0.44) = 1.24~\text{V}
\]

 

2. Balance Eq and calculate n:
\[
\mathrm{Fe(s)} \longrightarrow \mathrm{Fe^{2+}(aq)} + 2e^-
\]
\[
2\,\mathrm{Ag^+(aq)} + 2e^- \longrightarrow 2\,\mathrm{Ag(s)}
\]

Overall cell reaction:
\[
\mathrm{Fe(s)} + 2\,\mathrm{Ag^+(aq)} \longrightarrow
\mathrm{Fe^{2+}(aq)} + 2\,\mathrm{Ag(s)}
\]

n=2

3. Solve Nernst equation

For this reaction,
\[
E = E^\circ_\text{cell} - \frac{RT}{nF}
      \ln Q
= E^\circ_\text{cell} - \frac{RT}{nF}
      \ln\!\left(
        \frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Ag^+}]^2}
      \right)
\]

Substitute \(E^\circ_\text{cell} = 1.24~\text{V}\), \(n = 2\),
\([\mathrm{Fe^{2+}}] = 1.50~\text{M}\),
\([\mathrm{Ag^+}] = 0.050~\text{M}\),
\(R = 8.314~\mathrm{J\,mol^{-1}K^{-1}}\),
\(F = 96485~\mathrm{C\,mol^{-1}}\):

\[
E
= 1.24
- \frac{(8.314)(298)}{(2)(96485)}
  \ln\!\left(
    \frac{1.50}{(0.050)^2}
  \right)
\]

Evaluate:

\[
E \approx 1.16~\text{V}
\]

\[
\boxed{E \approx 1.16~\text{V at }25^\circ\mathrm{C}}
\]

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Exercise \(\PageIndex{7.7}\)

Calculate the cell potential (in V) at 25^oC for the following overall cell reaction,

Cu²⁺(aq) + Sn(s) --> Cu(s) + Sn²⁺(aq), E^o = +0.48 V

when [Cu²⁺(aq)] = 6.0 M and [Sn²⁺(aq)] = 0.001 M.

Answer

\[
\mathrm{Cu^{2+}(aq) + Sn(s) \rightarrow Cu(s) + Sn^{2+}(aq)}
\]

Use the Nernst equation at \(25^\circ \mathrm{C}\):

\[
E = E^\circ - \frac{0.0592}{n}\log Q
\]

For this reaction, 2 electrons are transferred:

\[
n = 2
\]

The reaction quotient is written using only aqueous species. Pure solids are not included.

\[
Q = \frac{[\mathrm{Sn^{2+}}]}{[\mathrm{Cu^{2+}}]}
\]

Substitute the given concentrations:

\[
Q = \frac{0.001}{6.0}
\]

\[
Q = 1.67 \times 10^{-4}
\]

Now substitute into the Nernst equation:

\[
E = 0.48 - \frac{0.0592}{2}\log(1.67 \times 10^{-4})
\]

\[
E = 0.48 - (0.0296)(-3.78)
\]

\[
E = 0.48 + 0.112
\]

\[
E = 0.592 \, \text{V}
\]

\[
\boxed{E = 0.59 \, \text{V}}
\]

Because \(Q < 1\), the reaction has more reactant ion than product ion relative to standard conditions, so the cell potential is larger than \(E^\circ\).

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Exercise \(\PageIndex{1}\)

Consider the following standard reduction potentials,

half reaction	E^o, V
Fe^3^+(aq) + e^- <==> Fe^2^+(aq)	0.77
H2O2(aq) + 2e^- <==> 2OH^-(aq)	0.88

For the voltaic cell reaction at 298K below, calculate the Fe²⁺ concentration (in M) that would be needed to produce a cell potential equal to 0.16 V at 25^oC when (OH^-) = 0.10 M, (Fe³⁺) = 0.50 M and (H₂O₂) = 0.35 M.

2Fe²⁺(aq) + H₂O₂(aq) --> 2Fe³⁺(aq) + 2OH^-(aq)

Answer

\[
\mathrm{2Fe^{2+}(aq) + H_2O_2(aq) \rightarrow 2Fe^{3+}(aq) + 2OH^-(aq)}
\]

First calculate the standard cell potential.

The cathode reaction is the reduction of hydrogen peroxide:

\[
\mathrm{H_2O_2(aq) + 2e^- \rightarrow 2OH^-(aq)}
\]

\[
E^\circ_{\text{cathode}} = 0.88 \, \text{V}
\]

The anode reaction is the oxidation of iron(II):

\[
\mathrm{Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^-}
\]

The given reduction potential for the iron half-reaction is

\[
\mathrm{Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq)}
\]

\[
E^\circ_{\text{anode, as reduction}} = 0.77 \, \text{V}
\]

Therefore,

\[
E^\circ_{\text{cell}}
=
E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
\]

\[
E^\circ_{\text{cell}} = 0.88 - 0.77
\]

\[
E^\circ_{\text{cell}} = 0.11 \, \text{V}
\]

Now use the natural log form of the Nernst equation:

\[
E = E^\circ - \frac{RT}{nF}\ln Q
\]

For this reaction,

\[
n = 2
\]

The reaction quotient is

\[
Q =
\frac{[\mathrm{Fe^{3+}}]^2[\mathrm{OH^-}]^2}
{[\mathrm{Fe^{2+}}]^2[\mathrm{H_2O_2}]}
\]

Substitute the desired cell potential:

\[
0.16
=
0.11
-
\frac{(8.314 \, \text{J/mol K})(298.15 \, \text{K})}
{(2)(96485 \, \text{C/mol})}
\ln Q
\]

\[
0.16
=
0.11
-
(0.01285)\ln Q
\]

\[
0.05 = -(0.01285)\ln Q
\]

\[
\ln Q = -3.89
\]

\[
Q = e^{-3.89}
\]

\[
Q = 2.04 \times 10^{-2}
\]

Now substitute into the reaction quotient:

\[
2.04 \times 10^{-2}
=
\frac{(0.50)^2(0.10)^2}
{[\mathrm{Fe^{2+}}]^2(0.35)}
\]

\[
2.04 \times 10^{-2}
=
\frac{(0.25)(0.010)}
{0.35[\mathrm{Fe^{2+}}]^2}
\]

\[
2.04 \times 10^{-2}
=
\frac{0.0025}
{0.35[\mathrm{Fe^{2+}}]^2}
\]

Solving for \([\mathrm{Fe^{2+}}]^2\):

\[
[\mathrm{Fe^{2+}}]^2
=
\frac{0.0025}
{(0.35)(2.04 \times 10^{-2})}
\]

\[
[\mathrm{Fe^{2+}}]^2 = 0.350
\]

\[
[\mathrm{Fe^{2+}}] = 0.592 \, \text{M}
\]

\[
\boxed{[\mathrm{Fe^{2+}}] = 0.59 \, \text{M}}
\]

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Galvanic Cell Potentials

Cell 1:

\(Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s) \quad -0.44\,volts\)
\(Mg^{2+}(aq)+2e^{-}\rightarrow MG(s) \quad -2.37\,volts\)

Cell 2:

\(Ga^{3+}(aq)+3e^{-}\rightarrow Ga(s) \quad -0.53\,volts\)
\(Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s) \quad -1.18\,volts\)

Cell 3:

\(Au^{3+}(aq)+3e^{-}\rightarrow Au(s) \quad 1.50\,volts\)
\(Sr^{2+}(aq)+2e^{-}\rightarrow Sr(s) \quad -2.89\,volts\)

Figure \(\PageIndex{1}\): Use the following reactions to answer questions 19.7.6-19.7.11.

Exercise \(\PageIndex{7.6}\)

Find the following values for cell 1.

1. What is the E°_cell for cell 1?
2. What is the ∆G°_rxn for cell 1?

Answer a.

\[Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s) \quad -0.44\,volts\]

\[Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s) \quad -2.37\,volts\]

 

\[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

\[E^{0}_{cell}=-0.44V+2.37V=1.93V\]

Answer b.

\[Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s) \quad -0.44\,volts\]

\[Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s) \quad -2.37\,volts\]

 

\[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

\[E^{0}_{cell}=-0.44V+2.37V=1.93V\]

 

\[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

\[\Delta G^{0}_{rxn}=-(2\,mol\,e^{-})(96500C/mol\,e^{-})(1.93V)=-372490CV\]

Highlight here for hypothes.is annotations and indicate question number in your annotation.

Exercise \(\PageIndex{7.7}\)

What is the E_cell for cell 1(0.4M Mg²⁺ and 1.4M Fe²⁺)?

Answer

\[Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s) \quad -0.44\,volts\]

\[Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s) \quad -2.37\,volts\]

 

\[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

\[E^{0}_{cell}=-0.44V+2.37V=1.93V\]

 

\[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

\[\Delta G^{0}_{rxn}=-(2\,mol\,e^{-})(96500C/mol\,e^{-})(1.93V)=-372490CV\]

 

\[E_{cell}={E}_

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^{\circ}-\left(\frac

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\right) \ln {Q}\]

\[E_{cell}=1.93 {V}-\left(\frac{8.314 {J} / {K} \cdot {m} {ol}(298.15 {K})}{2 {mol} {e}^{-}\left(96500 {C} / {m} {ol} {e}^{-}\right)}\right) \ln \left(\frac{0.4 {M} {Mg}^{2+}}{1.4 {M} {Fe}^{2+}}\right)\]

\[{E}_

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=1.95 {V}\]
 

Highlight here for hypothes.is annotations and indicate question number in your annotation.

Exercise \(\PageIndex{7.8}\)

Find the following values for cell 2.

1. What is the E°_cell for cell 2?
2. What is the ∆G°_rxn for cell 2?

Answer a.

\[Ga^{3+}(aq)+3e^{-}\rightarrow Ga(s) \quad -0.53\,volts\]
\[Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s) \quad -1.18\,volts\]

 

\[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

\[E^{0}_{cell}=-0.53V+1.18V=0.65V\]

Answer b.

\[Ga^{3+}(aq)+3e^{-}\rightarrow Ga(s) \quad -0.53\,volts\]
\[Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s) \quad -1.18\,volts\]

 

\[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

\[E^{0}_{cell}=-0.53V+1.18V=0.65V\]

 

\[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

\[\Delta G^{0}_{rxn}=-(6\,mol\,e^{-})(96500C/mol\,e^{-})(0.65V)=-376350CV\]

Highlight here for hypothes.is annotations and indicate question number in your annotation.

Exercise \(\PageIndex{7.9}\)

What is the E_cell for cell 2(0.5M Mn²⁺ and 1.5M Ga³⁺)?

Answer

\[Ga^{3+}(aq)+3e^{-}\rightarrow Ga(s) \quad -0.53\,volts\]
\[Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s) \quad -1.18\,volts\]

 

\[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

\[E^{0}_{cell}=-0.53V+1.18V=0.65V\]

 

\[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

\[\Delta G^{0}_{rxn}=-(6\,mol\,e^{-})(96500C/mol\,e^{-})(0.65V)=-376350CV\]

 

\[E_{cell}={E}_

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\right) \ln {Q}\]

\[E_{cell}=0.65 {V}-\left(\frac{8.314 {J} / {K} \cdot {mol}(298.15 {K})}{6 {mol} {e}^{-}\left(96500 {C} / {m} {ol} {e}^{-}\right)}\right) \ln \left(\frac{0.5 {M} {Mn}^{2+}}{1.5 {M} Gae}{3+}}\right)\]

\[{E}_

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=0.65 {V}\

Highlight here for hypothes.is annotations and indicate question number in your annotation.

Exercise \(\PageIndex{7.10}\)

Find the following values for cell 3.

1. What is the E°_cell for cell 3?
2. What is the ∆G°_rxn for cell 3?

Answer a.

\(Au^{3+}(aq)+3e^{-}\rightarrow Au(s) \quad 1.50\,volts\)
\(Sr^{2+}(aq)+2e^{-}\rightarrow Sr(s) \quad -2.89\,volts\)

 

\[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

\[E^{0}_{cell}=1.50V+2.89V=4.39V\]

Answer b.

\(Au^{3+}(aq)+3e^{-}\rightarrow Au(s) \quad 1.50\,volts\)
\(Sr^{2+}(aq)+2e^{-}\rightarrow Sr(s) \quad -2.89\,volts\)

 

\[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

\[E^{0}_{cell}=1.50V+2.89V=4.39V\]

 

\[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

\[\Delta G^{0}_{rxn}=-(6\,mol\,e^{-})(96500C/mol\,e^{-})(4.39V)=-2541810CV\

Highlight here for hypothes.is annotations and indicate question number in your annotation.

Exercise \(\PageIndex{7.11}\)

What is the E_cell for cell 3(1.2M Sr²⁺ and 0.6M Au³⁺)?

Answer

\(Au^{3+}(aq)+3e^{-}\rightarrow Au(s) \quad 1.50\,volts\)
\(Sr^{2+}(aq)+2e^{-}\rightarrow Sr(s) \quad -2.89\,volts\)

 

\[E^{0}_{cell}=E^{0}_{cathode}+E^{0}_{anode}\]

\[E^{0}_{cell}=1.50V+2.89V=4.39V\]

 

\[\Delta G^{0}_{rxn}=-nFE^{0}_{cell}\]

\[\Delta G^{0}_{rxn}=-(6\,mol\,e^{-})(96500C/mol\,e^{-})(4.39V)=-2541810CV\

 

\[E_{cell}={E}_

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\right) \ln {Q}\]

\[E_{cell}=4.39 {V}-\left(\frac{8.314 {J} / {K} \cdot {m} {ol}(298.15 {K})}{6 {mol} {e}^{-}\left(96500 {C} / {m} {ol} {e}^{-}\right)}\right) \ln \left(\frac{1.2 {M} {Sr}^{2+}}{0.6{M} {Au}^{3+}}\right)\]

\[{E}_

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=4.39 {V}\]

Highlight here for hypothes.is annotations and indicate question number in your annotation.

 

Exercise \(\PageIndex{7.12}\)

Consider the following standard reduction potentials, [NOTE: do this problem last as it is probably the hardest of the exam]

half reaction	E^o, V
Fe^3^+(aq) + e^- <==> Fe^2^+(aq)	0.77
H2O2(aq) + 2e^- <==> 2OH^-(aq)	0.88

For the voltaic cell reaction at 298K below, calculate the Fe²⁺ concentration (in M) that would be needed to produce a cell potential equal to 0.16 V at 25^oC when (OH^-) = 0.10 M, (Fe³⁺) = 0.50 M and (H₂O₂) = 0.35 M.

2Fe²⁺(aq) + H₂O₂(aq) --> 2Fe³⁺(aq) + 2OH^-(aq)

Answer

\[
\begin{aligned}
&\text{Half-reactions (as reductions):}\\
&\ce{Fe^{3+} + e^- -> Fe^{2+}} \quad E^\circ = 0.77~\text{V}
\qquad
\ce{H2O2 + 2e^- -> 2OH^-} \quad E^\circ = 0.88~\text{V}
\\[4pt]
&\text{Overall (as written): } \ce{2Fe^{2+} + H2O2 -> 2Fe^{3+} + 2OH^-},\quad
E^\circ_{\text{cell}} = 0.88 - 0.77 = 0.11~\text{V},\quad n=2.
\\[6pt]
&\text{Nernst at } 298.15~\text{K:}\quad
E = E^\circ - \frac{RT}{nF}\ln Q,\quad
Q=\frac{[\ce{Fe^{3+}}]^2[\ce{OH^-}]^2}{[\ce{Fe^{2+}}]^2[\ce{H2O2}]}
\\[4pt]
&0.16 = 0.11 - \frac{(8.314462618)(298.15)}{(2)(96485.33212)}\,
\ln\!\left(\frac{(0.50)^2(0.10)^2}{[\ce{Fe^{2+}}]^2(0.35)}\right)
\\[6pt]
&\Rightarrow\;
\ln Q = \frac{(0.11-0.16)(2)(96485.33212)}{(8.314462618)(298.15)}
= -3.892174449\ldots
\;\Rightarrow\;
Q = e^{-3.892174449\ldots} = 0.02040093696\ldots
\\[6pt]
&Q=\frac{(0.50)^2(0.10)^2}{[\ce{Fe^{2+}}]^2(0.35)}
\;\Rightarrow\;
[\ce{Fe^{2+}}]^2
=\frac{(0.50)^2(0.10)^2}{(0.35)\,Q}
=\frac{0.25\times 0.01}{0.35\times 0.02040093696\ldots}
=0.3501239749\ldots
\\[6pt]
&\boxed{[\ce{Fe^{2+}}] = \sqrt{0.3501239749\ldots} = 0.591712747\ldots \approx 0.592~\text{M}}
\end{aligned}
\]

 

In two lines:
 

\[
E = E^\circ - \frac{RT}{nF}\ln\!\left(\frac{[\ce{Fe^{3+}}]^2[\ce{OH^-}]^2}{[\ce{Fe^{2+}}]^2[\ce{H2O2}]}\right)
\]

Now solve the above for ferrous iron ([Fe⁺²]

 

\[
[\ce{Fe^{2+}}] 
= \sqrt{\frac{[\ce{Fe^{3+}}]^2[\ce{OH^-}]^2}{[\ce{H2O2}]\,e^{-\,\frac{(E^\circ - E)(nF)}{RT}}}}
= \sqrt{\frac{(0.50)^2(0.10)^2}{(0.35)\,e^{-\,\frac{(0.11-0.16)(2)(96485)}{(8.314)(298)}}}}
= \sqrt{0.3501}
= \boxed{0.59~\text{M}}
\]

 

 

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Exercise \(\PageIndex{7.12}\)

Calculate E for the following electrochemical cell at 25ºC

 

Pt(s) | H₂(g, 1.00 atm) | H⁺(aq, 1.00 M) || Cu²⁺(aq, 0.250 M) | Cu(s)

 

given the following standard reduction potentials.

 

Cu^2^+(aq) + 2 e^- -> Cu(s)	Eº = 0.337 V
2 H^+(aq) + 2 e^- -> H2(g)	Eº = 0.000 V

Answer

\[
\begin{aligned}
&\text{Half-reactions (as reductions):}\quad 
\ce{Cu^{2+} + 2e^- -> Cu(s)}\ (E^\circ=0.337~\text{V}),\quad
\ce{2H^+ + 2e^- -> H2(g)}\ (E^\circ=0.000~\text{V})\\
&\text{Cathode: }\ce{Cu^{2+}/Cu},\ \text{Anode: }\ce{H2/H^+}\ \Rightarrow\ 
E^\circ_{\text{cell}} = 0.337~\text{V},\ n=2.\\[4pt]
&\text{Overall: }\ce{H2(g) + Cu^{2+}(aq) -> 2H^+(aq) + Cu(s)}\\[6pt]
&\text{Nernst at }T=298.15~\text{K}:\quad 
E = E^\circ - \frac{RT}{nF}\ln Q,\qquad 
Q=\frac{a_{\ce{H^+}}^2}{a_{\ce{H2}}\,a_{\ce{Cu^{2+}}}}
\approx \frac{[\ce{H^+}]^2}{P_{\ce{H2}}[\ce{Cu^{2+}}]}\\[4pt]
&\Rightarrow\ 
E = 0.337 - \frac{(8.314462618)(298.15)}{(2)(96485.33212)}
\ln\!\left(\frac{(1.00)^2}{(1.00)(0.250)}\right)\\[6pt]
&\boxed{E = 0.3191912612\ldots~\text{V} \;\approx\; 0.319~\text{V}}
\end{aligned}
\]

ddd d

19.9: Electrolysis

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Exercise \(\PageIndex{9.1}\)

 How many minutes will it take to plate out 2.50g of chromium metal from Cr³⁺ solution using a current of 32.0 amps?

Answer

\[Cr^{3+}(aq)+3 e^{-} \rightarrow Cr(s)\]

\[ \left ( \frac{sec}{32.0 \; coulombs} \right )\left ( \frac{min}{60sec} \right )\left ( \frac{96500 \; coulombs}{mol \; e^-} \right )\left ( \frac{3 \; mol \;e^-}{mol \; Cr} \right )\left ( \frac{mol \; Cr}{52.00g} \right )2.5g \; Cr=7.25 \; min\]

 

Exercise \(\PageIndex{9.2}\)

What current (in A) is needed to plated out 1.50g of chromium metal from Cr³⁺ solution in 30sec?

Answer

\[\frac{1.50 g}{52 g / m o l}=2.88 \times 10^{-2} {mol}\]

\[e^{-}\, needed : 2.88 \times 10^{-2} {mol} \times 3=8.65 \times 10^{-2} {mol}\]

\[8.65 \times 10^{-2} {mol} \times 96500 {C} / {mol}=8.35 \times 10^{3} {C}\]

\[\frac{8.35 \times 10^{3} {C}}{30 {sec}}=2.78 \times 10^{2} {A}\]

Exercise \(\PageIndex{9.3}\)

How many grams of copper can be obtained if a current of 10amps passes through a solution of copper (II) sulfate for 20min?

Answer

\[10 A \times 20 \times 60 \sec =1.2 \times 10^{4} C \]

\[\frac{1.2 \times 10^{4} C}{96500 C / m o l}=0.124 m o l \]

\[\frac{0.124 m o l}{2} \times 63.55 g / m o l=3.948\]

Exercise \(\PageIndex{9.4}\)

 In an electrolytic cell, 0.05g of copper has produced from the copper (II) sulfate solution. How many grams of potassium would be plated out if the same current was applied through molten KCl?

Answer

\[\frac{0.05 g}{63.55 g / m o l} \times 2 \times 39 g / m o l=0.06 g\]

Exercise \(\PageIndex{9.5}\)

Calculate the number of kilowatt-hrs of electricity required to produce 1.00kg of Al by electrolysis of Al³⁺ if 5.00V of emf is applied.

Answer

\[\frac{1000 g}{27.0 g / m o l} \times 3 \times 96500 C / m o l=1.072 \times 10^{7} C\]

\[\frac{1.072 \times 10^{7} C \times 5.00 v}{3.6 \times 10^{6}}=14.89 k W h\]

Exercise \(\PageIndex{9.6}\)

How many minutes does it take to plate out 2.50 g chromium from a Cr⁺³ solution using a current of 32.0 amps

Answer

\[
\begin{aligned}
t
&=\; (2.50~\mathrm{g\ Cr})
\left(\frac{1~\mathrm{mol\ Cr}}{51.9961~\mathrm{g\ Cr}}\right)
\left(\frac{3~\mathrm{mol\ e^-}}{1~\mathrm{mol\ Cr}}\right)
\left(\frac{96485.33212~\mathrm{C}}{1~\mathrm{mol\ e^-}}\right)
\left(\frac{1~\mathrm{s}}{32.0~\mathrm{C}}\right)
\left(\frac{1~\mathrm{min}}{60~\mathrm{s}}\right) \\[4pt]
&=\; \boxed{7.26~\text{min}}
\end{aligned}
\]

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Exercise \(\PageIndex{1}\)

Which of the following is TRUE for an electrolytic cell when no voltage is applied?

a.  G > 0 ; E < 0                  b.  G < 0 ; E < 0                  c.  G = 0 ; E > 0
d.  G = 0 ; E = 0                  e.  G < 0 ; E > 0

Answer

a.  G > 0 ; E < 0   

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Exercise \(\PageIndex{1}\)

An electrolysis plant processes sea water to obtain Mg(s). If the plant operates its electrolytic cells at a total current of 1.00 x 10⁴ amps for 12.2 hours, how many grams of Mg(s) will be produced from sea water containing Mg²⁺?

a.  18.6 kg       b. 240 kg         c. 109 kg         d. 940 g           e. none of the above

Answer

e. none of the above

\[
\mathrm{Mg^{2+}(aq) + 2e^- \rightarrow Mg(s)}
\]

Each mole of \(\mathrm{Mg(s)}\) requires 2 moles of electrons.

First calculate the total charge passed through the electrolytic cells:

\[
q = It
\]

The current is

\[
I = 1.00 \times 10^4 \, \text{A}
\]

and the time is

\[
12.2 \, \text{h} \times \frac{3600 \, \text{s}}{1 \, \text{h}}
= 43920 \, \text{s}
\]

Therefore,

\[
q = (1.00 \times 10^4 \, \text{C/s})(43920 \, \text{s})
\]

\[
q = 4.392 \times 10^8 \, \text{C}
\]

Use Faraday's constant to convert charge to moles of electrons:

\[
1 \, \text{mol } e^- = 96485 \, \text{C}
\]

\[
\text{mol } e^- =
4.392 \times 10^8 \, \text{C}
\times
\frac{1 \, \text{mol } e^-}{96485 \, \text{C}}
\]

\[
\text{mol } e^- = 4.55 \times 10^3 \, \text{mol } e^-
\]

Now convert moles of electrons to moles of magnesium:

\[
4.55 \times 10^3 \, \text{mol } e^-
\times
\frac{1 \, \text{mol Mg}}{2 \, \text{mol } e^-}
=
2.28 \times 10^3 \, \text{mol Mg}
\]

Convert moles of magnesium to grams:

\[
2.28 \times 10^3 \, \text{mol Mg}
\times
\frac{24.31 \, \text{g Mg}}{1 \, \text{mol Mg}}
\]

\[
= 5.53 \times 10^4 \, \text{g Mg}
\]

\[
\boxed{5.53 \times 10^4 \, \text{g Mg}}
\]

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General Questions

Exercise \(\PageIndex{10.1}\)

In the voltaic cell that is represented as

Cd (S)| Cd²⁺(aq)||  Fe³⁺ (aq)| Fe²⁺(aq)Pt(s)

the electron flow will be from

1. Pt to Cd²⁺
2. Pt to Cd
3. Fe²⁺ to Cd²⁺
4. Cd²⁺ to Fe²⁺
5. Cd to Pt

Answer

e. Cd to Pt

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Exercise \(\PageIndex{10.2}\)

In the voltaic cell that is represented as

Zn(s) | Zn²⁺(1.0 M) || Cu²⁺(1.0 M) | Cu(s)

Which of the following statements is false?

1. The mass of the zinc electrode decreases during discharge.
2. The copper electrode is the anode.
3. Electrons flow through the external circuit from the zinc electrode to the copper electrode.
4. Reduction occurs at the copper electrode during discharge.
5. The concentration of Cu²⁺ decreases during discharge.

Answer

b. The copper electrode is the anode.

Exercise \(\PageIndex{10.3}\)

For a galvanic cell using Fe|Fe²⁺(1.0 M) and Pb|Pb²⁺(1.0 M) half-cells, which of the following statements is correct?

\(Fe^{2+}+2e^{-}\rightarrow Fe \quad -0.44V\)
\(Pb^{2+}+2e^{-}\rightarrow Pb \quad -0.13V\)

1. The mass of the iron electrode decreases during discharge.
2. Electrons leave the lead electrode to pass through the external circuit during discharge.
3. The concentration of Pb²⁺ increases during discharge.
4. The iron electrode is the cathode.
5. When the cell has completely discharged (to zero voltage), the concentration of Pb²⁺ is zero.

Answer

a. The mass of the iron electrode decreases during discharge.

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Exercise \(\PageIndex{10.4}\)

What is the cell reaction of the voltaic cell Cr(s)|Cr³⁺(aq)|| Cl₂(g)|Cl^-(aq)|Pt?

Answer

\[2Cr(s)+3Cl_{2}(g)\rightleftharpoons 2Cr^{3+}(aq)+6Cl^{-}(aq)\]

Exercise \(\PageIndex{10.5}\)

When Au is obtained by electrolysis from NaAuCl₄, what is the minimum number of coulombs required to produce 1.00 mol of gold?

Answer

gold is in the +3 oxidation state

\[Na(+1) \quad Au(+3) \quad 4CN(-1) \]

\[Au^{3+}+3e^{-}\rightarrow Au(s) \]

\[3mol \; e^-*96500 \frac{coulomb}{mole \; e^-}=2.90*10^{5}\,coulombs\]

Exercise \(\PageIndex{10.6}\)

How many faradays are required to convert a mole of Cr₂O₇^2- to Cr³⁺?

Answer

\[Cr_{2}O_{7}^{2}+14H^{+}+6e^{-} \rightarrow 2Cr^{3+}+7H_{2}O\]

\[1e^{-}=1F \therefore 6e^{-}=6F\]

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Exercise \(\PageIndex{10.7}\)

Which of the following cell reactions would require the use of an inert electrode?

1. \(Zn(s)+2MnO_{2}(s)+2NH_{4}^{+}(aq) \rightarrow Zn^{2+}(aq)+Mn_{2}O_{3}(s)+2NH_{3}(aq)+H_{2}O(l)\)
2. \(Zn(s)+2Ag^{+}(aq) \rightarrow Zn^{2+}(aq)+2Ag(s)\)
3. \(3Cu(s)+2Au^{3+}(aq) \rightarrow 3Cu^{2+}(aq)+2Au(s)\)
4. \(Cl_{2}(g)+2I^{-}(aq) \rightarrow 2Cl^{-}(aq)+I_{2}(s)\)

 

1. 1 only
2. 1 and 3 only
3. 2 and 3 only
4. 1 and 4 only
5. 3 and 4 only

Answer

d. 1 and 4 only

Exercise \(\PageIndex{10.8}\)

How many moles of electrons are produced form a current of 15.0 A in 1.00 hr?

Answer

 1 \; hr\frac{15 \; coulomb}{sec}\left ( \frac{mol \; e^-}{96,500 \; coulomb} \right )\left ( \frac{60 min}{hr} \right )\left ( \frac{60 sec}{min} \right )

\[x=0.560\,moles\,e^{-}\]

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Exercise \(\PageIndex{10.9}\)

The following has a potential of 0.92 V:

\(2Hg^{2+}(aq)+H_{2}(g)\rightarrow 2H^{+}(aq)+Hg_{2}^{2+}(aq)\)

If the concentration of the ions were 1.0 molar and the pressure of H₂ were 1.0 atmosphere,  then what would be the E° for the half-reaction?

\(2Hg^{2+}(aq)+2e^{-}\rightarrow Hg_{2}^{2+}\)

Answer

\[2Hg^{2+}(aq)+H_{2}(g)\rightarrow 2H^{+}(aq)+Hg_{2}^{2+}(aq)\]

\[E_{cell}=E^{0}_{cell}-\frac{0.0591}{n}log\frac{\left [ Hg_{2}^{2+} \right ]\left [ H^{+} \right ]^{2}}{\left ( P_{H_{2}} \right )\left [ Hg^{2+} \right ]} \]

\[0.92=E^{0}_{cell}-\frac{0.0591}{2}log\frac{\left [ 1.0 \right ]\left [ 1 \right ]^{2}}{\left ( 1 \right )\left [ 1.0 \right ]} \]

\[E^{0}_{cell}=0.92V\]

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Exercise \(\PageIndex{10.10}\)

A cell with the potential of 0.74 V has the cell reaction

\(2Cr+6H^{+}\rightarrow 2Cr^{3+}+3H_{2}\)

If the concentrations of the ions were 1.0 molar and the pressure of H₂ were 1.0 atmosphere, then what would be the E° of the half-reaction?

\(Cr^{3+}+3e^{-}\rightarrow Cr\)

Answer

Cell reaction: \(2Cr+6H^{+} \longrightarrow 2Cr^{3+}+3H_{2} \)

At anode: \(2Cr \longrightarrow 2Cr^{3+}+6e^{-} \)

At cathode: \(6H^{+}+6e^{-} \longrightarrow 3H_{2}\)

\[E_{cell}=E_{cell}^{0}-\frac{0.0591}{n}logQ \]

\[0.74V=\left (E_{H^{+}/H_{2}}^{0}-E_{Cr^{3+}/Cr}^{0}  \right )-\frac{0.0591}{6}log\frac{\left [ Cr^{3+} \right ]^{2}\left ( P_{H_{2}} \right )^{3}}{\left [ H^{+} \right ]^{6}}\]

\[0.74V=\left (E_{H^{+}/H_{2}}^{0}-E_{Cr^{3+}/Cr}^{0}  \right )-\frac{0.0591}{6}log\frac{\left [ 1.0 \right ]^{2}\left ( 1.0 \right )^{3}}{\left [ 1.0 \right ]^{6}} \]

\[0.74V=\left (E_{H^{+}/H_{2}}^{0}-E_{Cr^{3+}/Cr}^{0}  \right )-\frac{0.0591}{6}log(1) \]

noting log1=0

\[0.74V=0.00V-E_{Cr^{3+}/Cr}^{0} \]

\[E_{Cr^{3+}/Cr}^{0}=-0.74V \]

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Exercise \(\PageIndex{10.11}\)

A standard cell that consisted of a strip of zinc dipped into 1.0 M zinc ion and a strip of copper dipped into 1.0 M copper ion and which was connected by a salt bridge had a potential of 1.10 V

\(Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu\)

If a potential of 0.60 V were assigned to the half-reaction

\(Cu^{2+}+2e^{-}\rightarrow Cu\)

instead of 0.34 V, which is the potential given in a standard reduction potential table, the potential for the reaction would be

\(Zn^{2+}+2e^{-}\rightarrow Zn\)

Answer

\[E_{cell}=E_{cathode}^{0}-E_{anode}^{0} \]

\[E_{cell}=0.60V-1.10 \]

\[E_{cell}=-0.50V\]

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Exercise \(\PageIndex{10.12}\)

Consider the following electrode potentials:

\(Mg^{2+}+ 2e^{-} \rightarrow Mg \quad E^{0}=-2.37V\)

\(V^{2+}+2e^{-} \rightarrow V \quad E^{0}=-1.18V\)

\(Cu^{2+}+e^{-} \rightarrow Cu^{+} \quad E^{0}=0.15V\)
 

Which one of the reactions below will proceed spontaneously from left to right?

1. \(Mg^{2+}+V \rightarrow V^{2+}+Mg\)
2. \(Mg^{2+}+2Cu^{+} \rightarrow 2Cu^{2+}+Mg\)
3. \(V^{2+}+2Cu^{+} \rightarrow V+Cu^{2+}\)
4. \(V+2Cu^{2+} \rightarrow V^{2+}+2Cu^{+}\)
5. none of these

Answer

d. \(V+2Cu^{2+} \rightarrow V^{2+}+2Cu^{+}\)

Exercise \(\PageIndex{10.13}\)

Consider the following standard reduction potentials:

\(2H^{+}(aq)+2e^{-} \rightarrow H_{2}(g) \quad E^{0}=0.00V\)

\(Sn^{2+}(aq)+2e^{-} \rightarrow Sn(s) \quad E^{0}=-0.15V\)

\(Cd^{2+}(aq)+2e^{-} \rightarrow Cd(s) \quad E^{0}=-0.40V\)

which pair of substances react spontaneously

1. Sn²⁺ with Cd²⁺
2. Sn with Cd²⁺
3. Sn²⁺ with H⁺
4. Cd with Sn
5. Cd with Sn²⁺

Answer

e. Cd with Sn²⁺

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Exercise \(\PageIndex{10.14}\)

Calculate the maximum electrical work obtained when 7.10 grams of Cl₂ gas are consumed in the reaction Cd(s) + Cl₂(g) ⇌ Cd²⁺(aq) +2 Cl^-(aq). (E_cell=1.76V)

Answer

\[W_{\max }=-n \times E_{cell} \times F\]

where F=96,500kJ/V mole of electron

E_cell=1.76V(given)

gram of Cl₂=7.10g molar mass of Cl₂=70.906 g/mole

than mole of Cl₂=7.10/70.906=0.10013 mole

Half reaction of cell as

\[Cl_{2}+2e^{-}\rightarrow 2Cl^{-}\]

1 mole Cl₂ required 2 mole electrons

than 0.10013 mole Cl₂ required, mole of electron (n)=0.10013 * 2=0.20026 mole e^-

Now put all value

\[W_{max }=-\left(0.20026 \,mole \,e^{-}\right) \times(1.76V) \times\left(96.500 kJ / \mole \,e^{-} \,V\right)=34.0121584\,kJ\]

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Exercise \(\PageIndex{10.15}\)

Calculate E° for the cell reaction 2Cr + 3Sn⁴⁺ → 3Sn²⁺ + 2Cr³⁺.

\(Cr^{3+}+3e^{-} \rightarrow Cr \quad E^{0}=-0.74V\)

\(Sn^{4+}+2e^{-} \rightarrow Sn^{2+} \quad E^{0}=0.15V\)

Answer

\[E^{0}_{cell}=E_{cathode}^{0}-E_{anode}^{0}\]

\[E^{0}_{cell}=0.15-(-0.74)\]

\[E^{0}_{cell}=0.89V\]

Exercise \(\PageIndex{10.16}\)

At 25°C, calculate the voltage of the cell

\(2Ag^{+}(aq)(0.16 M)+Cu(s)\rightarrow Cu^{2+}(aq)(0.015 M)+2Ag(s)\)

if E°_cell = 0.460 V.

Answer

\[E_{cell}=E^{0}-(0.596/n)log(A_{red}/A_{oxd})\]

[A_red=conc. of reductant, A_oxd=conc. of oxidant]

\[E_{cell}=0.460-(0.0596 / 2) \log (0.015 / 0.16)\]

[n=no. of electrons that occur in reaction, since copper is reduced to 2, change in oxidation state =2]

\[E_{cell}=0.46-(0.0298)*(-1.028028724) \]

\[E_{cell}=0.46+0.030635255 \]

\[E_{cell}=0.491V\]

 

note you can use this from of the Nerst Eq. because you are at 298K

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\(Cr^{3+}+3e^{-}\rightarrow Cr \quad -0.74V\)
\(Pb^{2+}+2e^{-}\rightarrow Pb \quad -0.13V\)

Figure \(\PageIndex{4}\): Use the following information for questions 19.17-19.18

Exercise \(\PageIndex{10.17}\)

What is the standard cell potential for the reaction?

\(2Cr+3Pb^{2+}\rightarrow 3Pb+2Cr^{3+}\)

Answer

\[E_{cell}=E_{cathode}^{0}-E_{anode}^{0} \]

\[E_{cell}=-0.13V-(-0.74V) \]

\[E_{cell}=0.61V\]

Exercise \(\PageIndex{10.18}\)

Calculate the Gibbs free energy change for the reaction above the initial concentration of Cr³⁺ and Pb²⁺ are 1.00M.

Answer

\[\Delta G=-nFE^{0} \]

\[\Delta G=-(6) \times (96,500) \times (0.61) \]

\[\Delta G=-353190\,J \]

\[\Delta G=-353\,kJ\]

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Exercise \(\PageIndex{10.19}\)

What is the value of the reaction quotient, Q, for the cell that is constructed from the two half-reactions?

\(Zn^{2+}+2e^{-}\rightarrow Zn \quad -0.76V\)

\(Ag^{+}+e^{-}\rightarrow Ag \quad 0.80V\)
 

when the Zn²⁺ concentration is 0.0100 M and the Ag⁺ concentration is 1.25M?

Answer

\[Q= \frac{\left [ products \right ]}{\left [ reactants \right ]} \]

\[Q=\frac{\left[Zn^{2+}\right]}{\left[Ag^{+}\right]^{2}} \]

\[Q=\frac{[0.0100]}{[1.25]^{2}} \]

\[Q=6.40 \times 10^{-3}\]

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Exercise \(\PageIndex{10.20}\)

Calculate ∆G° (in joules) for the reaction 2AlI₃(aq) ⇌ 2Al(s) + 3I₂(s).

\(Al^{3+}+3e^{-} \rightleftharpoons Al \quad E^{0}=-1.66V \)

\(I_{2}(s)+2e^{-} \rightleftharpoons 2I^{1} \quad E^{0}=0.54V\)
 

Answer

\[2Al^{3+}+6e^{-}\rightleftharpoons 2Al(s) \quad E^{0}=-1.66V \]

\[6I^{-}\rightleftharpoons 3I_{2}(s)+6e^{-} \quad E^{0}=-0.54V \]

\[E^{0}=E_{cat}+E_{ano}=-1.66+(-0.54)=-2.20V \]

\[\Delta G=-nFE^{0}=-(6) \times (96,500) \times (-2.20)=1.3*10^{6}\]

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Exercise \(\PageIndex{10.21}\)

What is the reduction potential for the half-reaction Al³⁺(aq) + 3e^- → Al(s) + at 25°C is [Al⁺] = 0.10 M and E° = -1.66 V?

Answer

\[E_{Al^{3+}/Al}=E^{0}_{Al^{3+}/Al}-\frac{RT}{nF}ln\frac{1}{\left [ Al^{3+} \right ]} \]

\[E_{Al^{3+}/Al}=E^{0}_{Al^{3+}/Al}-\frac{0.0591}{n}log\frac{1}{\left [ Al^{3+} \right ]} \]

\[E_{Al^{3+}/Al}=-1.66-\frac{0.0591}{3}log\frac{1}{0.10} \]

\[E_{Al^{3+}/Al}=-1.66-0.0197 \]

\[E_{Al^{3+}/Al}=-1.68\]

Exercise \(\PageIndex{10.22}\)

What is the emf at 25°C for the following cell?

\(Cr\left|Cr^{3+}(0.010 M) \| Ag^{+}(0.00010 M)\right| Ag\)

\(Cr^{3+}+3e^{-} \rightleftharpoons Cr \quad E^{0}=-0.74 V\)

\(Ag^{+}+e^{-} \rightleftharpoons Ag\quad E^{0}=0.80V\)
 

Answer

\[Cr^{3+}(aq)+3e^{-} \rightarrow Cr(s) \quad E^{0}=-0.74V \]

\[Ag^{+}(a q)+e^{-} \rightarrow Ag(s)  \quad E^{0}=+0.80V \]

\[3Ag^{+}(aq)+Cr(s) \rightarrow Ag(s)+Cr^{3+}(aq) \]

\[E_{0}=E_{cathode}+E_{anode}=+0.80V+0.74V=1.54V \]

\[E_{cell}=E_{0}-\frac{0.0591}{n} \log \frac{[\text { products }]}{\left[\text { reactants }\right]} \]

\[E_{cell}=1.54-\frac{0.0591}{3} \log \frac{\left[Cr^{3+}\right]}{\left[Ag^{+}\right]^{3}} \]

\[E_{cell}=1.54-\frac{0.0591}{3} \log \frac{0.01}{(0.0001)^{3}} \]

\[E_{cell}=1.54-\frac{0.0591}{3} \times 10 \]

\[E_{cell}=1.54-0.197\]

\[E_{cell}=1.34V\]

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Exercise \(\PageIndex{10.23}\)

What is the Cu²⁺ concentration at 25°C in the cell Zn(s) | Zn²⁺ (1.0 M) || Cu²⁺(aq) | Cu(s)? The cell emf is 1.03 V. The standard cell emf is 1.10 V.

Answer

\[E_{cell}=E_{0}-\frac{0.0591}{n} \log \frac{[Zn^{2+}]}{\left[Cu^{2+}\right]} \]

\[1.03=1.10-\frac{0.0591}{2} \log \frac{1.0}{\left[Cu^{2+}\right]} \]

\[-0.07=-\frac{0.0591}{2} \log \frac{1.0}{\left[Cu^{2+}\right]} \]

\[2.369=\log \frac{1.0}{\left[Cu^{2+}\right]} \]

\[233.81= \frac{1.0}{\left[Cu^{2+}\right]} \]

\[[Cu^{2+}]=0.004M\]

Exercise \(\PageIndex{10.24}\)

Molten magnesium chloride is electrolyzed using insert electrodes and reactions represented by the following equations

\(2Cl^{-} \rightarrow Cl_{2}+2e^{-}\)

\(Mg^{2+}+2e^{-} \rightarrow Mg\)

concerning this electrolysis which of the following statements is true

1. oxidation occurs at the cathode
2. Mg²⁺ ions are reduced at the anode
3. electrons pass through the metallic part of the circuit from Mg²⁺ ions to the Cl^- ion
4. Cl^- ions are oxidizing agents
5. the cations and the electrolyte undergo reduction

Answer

e. the cations and the electrolyte undergo reduction

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Exercise \(\PageIndex{10.25}\)

A piece of iron half immerse in a sodium chloride solution will corrode more rapidly than a piece of iron half immersed in pure water because

1. the sodium iron oxidized the iron atoms
2. the chloride ions oxidized the iron atoms
3. the chloride ions form a precipitate with iron
4. the chloride ions increase the pH of the solution
5. the sodium ions and chloride ions carry a current through the solution

Answer

e. the sodium ions and chloride ions carry a current through the solution

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Exercise \(\PageIndex{10.26}\)

A reaction is spontaneous win

1. ∆G° is negative or E° is positive
2. ∆G° is negative or E° is negative
3. ∆G° is positive or E° is negative
4. ∆G° is positive or E° is positive
5. ∆G° is negative, ∆H° is negative, and E° is negative

Answer

a. ∆G° is negative or E° is positive

Exercise \(\PageIndex{10.27}\)

What reaction occurs at the cathode during electrolysis of aqueous CuSO₄?

1. \(2H_{2}O+2e^{-} \rightarrow H_{2}+2OH^{-}\)
2. \(Cu \rightarrow Cu^{2+}+2e^{- \)
3. \(2H_{2}O \rightarrow O_{2}+4H^{+}+4e^{-}\)
4. \(2H^{+}+2e^{-} \rightarrow H_{2}\)
5. \(Cu^{2+}+2e^{-} \rightarrow Cu\)

Answer

e. \(Cu^{2+}+2e^{-} \rightarrow Cu\)

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Exercise \(\PageIndex{10.28}\)

For a certain oxidation-reduction reaction, he is positive. this means that

1. ∆G° is negative and K is less than 1
2. ∆G° is negative and K is greater than 1
3. ∆G° is zero and K is greater than 1
4. ∆G° is positive and K is greater than 1
5. ∆G° is positive and K is less than 1

Answer

b. ∆G° is negative and K is greater than 1

Exercise \(\PageIndex{10.29}\)

Cathodic protection results when

1. iron is attached to a more active metal
2. iron is amalgamated with Mercury
3. iron is tin plated for use as a tin can
4. iron is painted to protect it from corrosion
5. iron is made amphoteric

Answer

a. iron is attached to a more active metal

Exercise \(\PageIndex{10.30}\)

What mass of chromium could be deposited by electrolysis of an aqueous solution fo Cr₂(SO₄)₂ for 60.0 minutes using constant current of 10.0 amperes? (One faraday = 96,500 coulombs.)

Answer

\[Cr^{3+}+3e^{-}=Cr \]

\[Charge = current + time=10A *\left(60^{+} 60\right)s=36000C \]

No of moles of electron flowed

\[=36000 / 96500 =0.373 \,moles \]

As 1 mole of Cr³⁺ is reduced by 3 moles of electrons, so 0.124 moles of Cr³⁺ is reduced by 0.373 moles of electrons. 

Mass of Cr deposited 

\[=0.124 * molar mass of \,Cr \]

\[=0.124 * 52.01 \,g \\

\[=6.47 \,g\]

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Exercise \(\PageIndex{10.31}\)

In a galvanic cell, the cathode is always

1. the positive electrode
2. the negative electrode
3. the positive electrode and the negative electrode, respectively
4. the electrode at which some species gain electrons
5. the electrode at which some species lose electrons

Answer

d. the electrode at which some species gain electrons

Exercise \(\PageIndex{10.32}\)

The voltage of a voltaic cell of zinc and copper at 25°C in standard state conditions would be

\(Zn^{2+}+2e^{-} \rightarrow Zn \quad E^{0}=-0.7V \)

\(Cu^{2+}+2e^{-} \rightarrow Cu \quad E^{0}=0.34V\)
 

1. between 0.76 and 1.10V
2. between 0.34 and 0.76 V
3. between 0.00 and 0.76 speak
4. less than 0.42 V
5. 1.10 V

Answer

e. greater than 1.10 V

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Exercise \(\PageIndex{10.33}\)

On the basis afford going standard electrode potentials, determine which of the following is the strongest oxidizing agent.

\(Zn^{2+}+2e^{-} \rightarrow Zn \quad E^{0}=-0.76V \)

\(Cu^{2+}+2e^{-} \rightarrow Cu \quad E^{0}=+0.34V \)

\(Cr_{2}O_{7}^{2-}+14H^{+}+6e^{-} \rightarrow 2Cr^{3+}+7H_{2}O \quad E^{0}=+1.33V\)
 

1. Zn²⁺
2. Zn
3. Cu
4. Cr₂O₇^2-
5. Cr³⁺

Answer

d. Cr₂O₇^2-

Exercise \(\PageIndex{10.34}\)

Which of the following species is the best reducing agent?

1. Cd
2. Mg²⁺
3. I^-
4. Sn
5. F^-

Answer

a. Cd

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Exercise \(\PageIndex{10.35}\)

Which cell notation describes the following reaction?

2Cr(NO₃)₃(aq) + 3Zn(s)  ->  3Zn(NO₃)₂(aq)  + 2Cr(s)

a.  Cr⁺³(aq)|Cr(s)|| Zn(s)|Zn⁺² (s)                  b.  Zn(s)|Zn⁺²(aq)||Cr⁺³(aq)|Cr(s)

c.  2Cr⁺³(aq)|2Cr(s)|| 3Zn(s)|3Zn⁺² (s)           d.  3Zn(s)|3Zn⁺²(aq)||2Cr⁺³(aq)|2Cr(s)

e. none of the above

Answer

Zn(s)|Zn⁺²(aq)||Cr⁺³(aq)|Cr(s)

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Exercise \(\PageIndex{1}\)

Consider the following standard reduction potentials,
 

half reaction	E^o, V
Ni^2+(aq) + 2e^- <==> Ni(s)	-0.23
Co^2+(aq) + 2e^- <==> Co(s)	-0.28
Fe^2+(aq) + 2e^- <==> Fe(s)	-0.41
Cr^3+(aq) + 3e^- <==> Cr(s)	-0.74
Mn^2+(aq) + 2e^- <==> Mn(s)	-1.03

Which of the following metals could be used to protect iron from rusting?

a.  Ni only                   b.  Ni and Co               c.  Fe only          d.  Mn only              e.  Mn and Cr

Answer

e.  Mn and Cr

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