19: Electron Transfer Reactions

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19.1  Electrical Fundamentals

 

19.2 Oxidation-Reduction Reactions

Balancing Redox Reactions in Acidic Solutions

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 

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

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

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

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

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

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

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

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

Exercise \(\PageIndex{2.1d}\)

Balance the reaction in an acidic solution:

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

Answer

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

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

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

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

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

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

\[Fe_{2}S(s)+4H_{2}O(l)\rightarrow 2Fe^{2+}(aq)+SO_{4}^{2-}(aq)+8H^{+}(aq)+10e^{-}\]

\[Ce^{4+}(aq)+e^{-}\rightarrow Ce^{3+}(aq)\]

\[Fe_{2}S(s)+4H_{2}O(l)+10Ce^{4+}(aq)\rightarrow 2Fe^{2+}(aq)+SO_{4}^{2-}(aq)+8H^{+}(aq)+10Ce^{3+}(aq)\]

 

Balancing Redox Reactions in Basic Solutions

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

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

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

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

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

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

 

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

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

 

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

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

 

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

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

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

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

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

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.

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

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.

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

 

Voltaic Cell

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

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

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

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

 

 

19.4: Electrochemical Cell Fundamentals

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

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

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

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

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

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

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

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

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

Exercise \(\PageIndex{5.1}\)

What is the E°_cell?

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

Answer

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

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

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

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

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

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

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

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

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

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

\[n=6\]

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

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

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

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

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

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

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

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

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

 

19.6: Electrochemistry and Thermodynamics

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 

19.7: Electrochemical Cells under Nonstandard Conditions

The Nernst Equation

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

\[E=E^{0}-\frac{0.0592}{n} \log Q=1.21-\frac{0.0592}{2} \log \frac{0.01^{3}}{0.02^{2}}=1.29v\]

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

\[E=E^{0}-\frac{0.0592}{n} \log Q=0.30-\frac{0.0592}{6} \log \frac{0.005^{2}}{0.01^{3}}=0.31v\]

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

\[E=E^{0}-\frac{0.0592}{n} \log Q\]

\[1.50=1.21-\frac{0.0592}{2} \log \frac{x^{3}}{0.02^{2}}\]

\[\log \frac{x^{3}}{0.02^{2}}=-9.797\]

\[3 \log x-2 \log 0.02=-9.797\]

\[x=4.0 \times 10^{-5} {M}\]

 

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.

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

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

 

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

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

 

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

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

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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\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}\]
 

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.

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

 

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

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

 

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

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

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}\

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\

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}\]

 

19.9: Electrolysis

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

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

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

 

General Questions

Exercise \(\PageIndex{1}\)

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

Exercise \(\PageIndex{2}\)

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 \(\PageIndex{3}\)

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.

Exercise \(\PageIndex{4}\)

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 \(\PageIndex{5}\)

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 \(\PageIndex{6}\)

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

Answer

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

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

Exercise \(\PageIndex{7}\)

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 \(\PageIndex{8}\)

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 )

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

Exercise \(\PageIndex{9}\)

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

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

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

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

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

Exercise \(\PageIndex{10}\)

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

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

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

Exercise \(\PageIndex{11}\)

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

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

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

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

Exercise \(\PageIndex{12}\)

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 \(\PageIndex{13}\)

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⁺

Exercise \(\PageIndex{14}\)

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

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

Exercise \(\PageIndex{15}\)

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 \(\PageIndex{16}\)

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

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

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

Exercise \(\PageIndex{17}\)

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 \(\PageIndex{18}\)

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

Answer

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

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

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

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

Exercise \(\PageIndex{19}\)

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

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

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

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

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

Exercise \(\PageIndex{20}\)

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

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

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

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

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

Exercise \(\PageIndex{21}\)

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 \(\PageIndex{22}\)

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

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

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

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

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

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

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

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

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

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

\[E_{cell}=1.34V\]

Exercise \(\PageIndex{23}\)

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 \(\PageIndex{24}\)

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

Exercise \(\PageIndex{25}\)

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

Exercise \(\PageIndex{26}\)

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 \(\PageIndex{27}\)

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. \(Cu^{2+}+2e^{-} \rightarrow Cu\)

Exercise \(\PageIndex{28}\)

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 \(\PageIndex{29}\)

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 \(\PageIndex{30}\)

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 \\

\[=6.47 \,g\]

Exercise \(\PageIndex{31}\)

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 \(\PageIndex{32}\)

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

Exercise \(\PageIndex{33}\)

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 \(\PageIndex{34}\)

Which of the following species is the best reducing agent?

1. Cd
2. Mg²⁺
3. I^-
4. Sn
5. F^-

Answer

a. Cd