# Solutions 2: Electrochemistry

## Q2.1

1. The $$\mathrm{E^\circ}$$ value will be estimated to be less than 0. In Table P2, metals listed above hydrogen ($$\ce{Pb}$$ through $$\ce{Li}$$) should react with $$\ce{HCl}$$. Metals with an $$\mathrm{E^\circ}$$ above hydrogen will not. These metals must be able to displace $$\ce{H2(g)}$$ from acidic solutions.
2. The $$\mathrm{E^\circ}$$ value will be between +0.771 and +0.154. Being displaced means that the ion is forced out of solution and into metal state. The metal must be below $$\ce{Sn^4+}$$ indicating it is a stronger reducing agent (more willing to give up electrons) and below $$\ce{Fe^3+}$$.
3. The $$\mathrm{E^\circ}$$ value must be lower than 0.956, the $$\mathrm{E^\circ}$$ for $$\ce{NO3}$$. This follows the same reasoning in 1a, but this time the metal must be able to displace $$\ce{NO (g)}$$ from acidic solution. Thus, the predicted $$\mathrm{E^\circ}$$ value will be lower than 0.956.
4. The $$\mathrm{E^\circ}$$ must be lower than -2.924, meaning that the metal $$\ce{M}$$ will get oxidized and will push the $$\ce{K+}$$ ion into solution. Thus, the $$\mathrm{E^\circ}$$ will be lower than -2.924.

For review on this topic, visit the page "Electrochemistry 1: Introduction" and Using Redox Potentials to Predict the Feasibility of Reactions.

## Q2.2

1. The $$\mathrm{E^{\circ}_{cell} = E^{\circ}_{(\textrm{reduction half-cell})} - E^{\circ}_{(\textrm{oxidation half-cell})} = 0.340 - (-0.763)= 1.103}$$. Because the $$\mathrm{E^\circ_{cell}}$$ is positive, the direction of spontaneous change is in the forward reaction. Recognize that $$\ce{Cu^2+}$$ is the stronger oxidizing agent (see Table P2) and will be more likely to be reduced. Thus, the $$\ce{Cu^2+}$$ half reaction is where copper is reduced and will be the reduction half-cell. The $$\ce{Zn^2+}$$half reaction will be the oxidation half-cell because Zinc gets oxidized at the anode.
2. The $$\mathrm{E^{\circ}_{cell} = E^{\circ}_{(\textrm{reduction half-cell})} - E^{\circ}_{(\textrm{oxidation half-cell})} = 1.065 - 0.854= 0.211}$$. Because the $$\mathrm{E^\circ_{cell}}$$ is positive, the direction of spontaneous change is in the forward direction. Recognize that $$\ce{Br-}$$ is more likely to be reduced and will be the reduction half-cell.
3. The $$\mathrm{E^{\circ}_{cell} = E^{\circ}_{(\textrm{reduction half-cell})} - E^{\circ}_{(\textrm{oxidation half-cell})} = \mathrm{1.358 - 0.771= 0.587}}$$. Because the $$\mathrm{E^\circ}$$ is positive, the direction of spontaneous change is in the forward direction. Recognize that the reduction of chlorine takes place at the cathode, making it the reduction half-cell. The oxidation of iron occurs at the anode, making it the oxidation half-cell.

## Q2.3

Not given (at least here)

## Q2.4

a: The $$\mathrm{n}$$ (moles of electrons) $$\mathrm{= 2}$$

\begin{align} \mathrm{E^{\circ}_{cell}} & = \mathrm{E^{\circ}_{(reduction)}- E^{\circ}_{(oxidation)}} \\ & = \mathrm{-0.255 - (-0.257) = 0.002\;V} \end{align}

\begin{align} \mathrm{\Delta G^{\circ}} & = \mathrm{-nF E^{\circ}_{cell}}= -(\textrm{2 mol e}^- ) \times (\textrm{96,485 C/mole e}^-) \times (\textrm{0.002 V}) \\ & = \mathrm{-385.9\;J} \end{align}

b: The $$\mathrm{n}$$ (moles of electrons) $$\mathrm{= 6}$$

$$\mathrm{E^{\circ}_{cell} = E^{\circ}_{(reduction)} - E^{\circ}_{(oxidation)} = 1.087 - (-1.676) = 2.763\;V}$$

\begin{align} \mathrm{\Delta G^{\circ}} & = \mathrm{-nF E^{\circ}_{cell}}= -(\textrm{6 mol e}^- ) \times (\textrm{96,485 C/mole e}^-) \times (\textrm{2.763 V}) \\ & = \mathrm{-1.60 \times 10^6\;J}\end{align}

c: The $$\mathrm{n}$$ (moles of electrons transferred) $$\mathrm{= 2}$$

$$\mathrm{E^{\circ}_{cell}= -0.005\;V}$$

\begin{align} \mathrm{\Delta G^{\circ}} & = \mathrm{-nF E^{\circ}_{cell}}= -(\textrm{2 mol e}^- ) \times (\textrm{96,485 C/mole e}^-) \times (\textrm{0.005 V}) \\ & = \mathrm{964.8\;J}\end{align}

For review on this topic, visit the page: Cell potentials and Free (Gibbs) energy

## Q2.5

1. $$\textrm{Reduction: } \mathrm{Na^+(aq)+e^- \rightarrow Na(s)} \quad \mathrm{E^{\circ}=-2.713\:V}$$
$$\textrm{Oxidation: } \mathrm{K(s) \rightarrow K^+(aq)+e^-} \quad \hspace{3mm} \mathrm{E^{\circ}=-2.924\:V}$$

\begin{align} \mathrm{E^{\circ}_{cell}} &= \mathrm{E^{\circ}_{reduction}- E^{\circ}_{oxidation}} \\ & = \mathrm{-2.713 - (-2.924)} \\ & = \mathrm{0.211\:V} \end{align}

When $$\mathrm{\Delta G^\circ < 0}$$ and $$\mathrm{E^\circ_{cell} > 0}$$, it is an exoenergonic reaction.

1. $$\textrm{Reduction: } \mathrm{2IO_3^-(aq)+12H^+(aq)+10e^- \rightarrow I_2(g)+6H_2O(l)} \quad \hspace{14 mm}\mathrm{E^{\circ}=+1.20\:V}$$
$$\textrm{Oxidation: } \mathrm{2Mn^{2+}(aq)+8H_2O(l) \rightarrow 2MnO_4^-(aq)+16H^+(aq)+10e^-} \quad \mathrm{E^{\circ}=+1.51\:V}$$

\begin{align} \mathrm{E^{\circ}_{cell}} &= \mathrm{E^{\circ}_{reduction}- E^{\circ}_{oxidation}} \\ & = \mathrm{1.20 - 1.51} \\ & = \mathrm{-0.31\:V} \end{align}

Since $$\mathrm{E^\circ_{cell} < 0}$$, $$\mathrm{\Delta G^\circ > 0}$$ and the reaction is endoenergonic

1. $$\textrm{Reduction: } \mathrm{Mg^{2+}(aq)+2e^- \rightarrow Mg(s)} \quad\hspace{104 pt} \mathrm{E^{\circ}=-2.356\:V}$$
$$\textrm{Oxidation: } \mathrm{Pb^{2+}(aq)+2H_2O(l) \rightarrow PbO_2(s)+4H^+(aq)+2e^-} \quad \mathrm{E^{\circ}=+1.455\:V}$$

\begin{align} \mathrm{E^{\circ}_{cell}} & = \mathrm{E^{\circ}_{reduction}- E^{\circ}_{oxidation}} \\ & = \mathrm{-2.356 - 1.455} \\ & = \mathrm{-3.811\:V} \end{align}

Since $$\mathrm{E^\circ_{cell} < 0}$$, $$\mathrm{\Delta G^\circ > 0}$$ and the reaction is endoenergonic

1. $$\textrm{Reduction: } \mathrm{O_2(g)+4H^+(aq)+4e^- \rightarrow 2H_2O(l)} \quad \hspace{77 pt} \mathrm{E^{\circ}=1.229\:V}$$
$$\textrm{Oxidation: } \mathrm{2Mn^{2+}(aq)+4H_2O(l) \rightarrow 2MnO_2(s)+8H^+(aq)+4e^-} \quad \mathrm{E^{\circ}=+1.23\:V}$$

\begin{align} \mathrm{E^{\circ}_{cell}} & = \mathrm{E^{\circ}_{reduction}- E^{\circ}_{oxidation}} \\ & = \mathrm{1.229-1.23} \\ & = \mathrm{-0.001\:V} \end{align}

Since $$\mathrm{E^\circ_{cell} < 0}$$, $$\mathrm{\Delta G^\circ > 0}$$ and the reaction is endoenergonic.

## Q2.6

1. Reduction: $$\mathrm{Fe^{3+} (aq) + e^- \rightarrow Fe^{2+}(aq)}$$ at the cathode (reduction occurs at the cathode)

Oxidation: $$\mathrm{Ag(s) \rightarrow Ag^+(aq) + e^-}$$ at the anode

$$\mathrm{E^{\circ}_{cell} = (0.771 - 0.800) = -0.029\;V}$$

1. Reduction: $$\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)}$$ at the cathode, where reduction occurs

Oxidation: $$\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}$$ at the anode

$$\mathrm{E^{\circ}_{cell} = 0.340 - (-0.763) = 1.103\;V}$$

1. Reduction: $$\mathrm{Cu^{2+}(aq) +2e^- \rightarrow Cu(s)}$$ at the cathode, where reduction occurs

Oxidation: $$\mathrm{Cd(s) \rightarrow Cd^{2+} + 2e^-}$$ at the anode

$$\mathrm{E^{\circ}_{cell}= E^{\circ}_{(right)} - E^{\circ}_{(left)} = 0.340 + 0.403 = 0.743}$$

1. Balance the equation, multiplying the iron half equation by 2 to equal the electrons

Reduction: $$\mathrm{Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)}$$ at the cathode

Oxidation: $$\mathrm{2(Fe^{2+}(aq) \rightarrow 2Fe^{3+}(aq) +e^-)}$$ at the anode

$$\mathrm{E^{\circ}_{cell} = 1.358 - (-0.440) = 1.798\;V}$$

For review on this topic, visit the page "Electrochemistry 2: Galvanic cells and electrodes".

not given

not given

## Q2.9

1. $$\mathrm{Al(s)| Al^{3+}(0.36\:M)||Sn^{4+}(0.086\:M),\,Sn^{2+}(0.54\:M)|Pt}$$

$$\mathrm{E^{\circ}_{cell} = 0.154 - (-1.676) = 1.83}$$

Using the Nernst equation $$\mathrm{E_{cell} = E^{\circ}_{cell}-\dfrac{0.0592}{n}\times\log Q}$$, we simply need to set up the $$\mathrm{Q}$$ and establish that our $$\mathrm{n}$$ (moles of electrons transferred) is 6.

Thus, the $$\mathrm{Q} = \dfrac{(0.54)(0.36)^2}{(0.086)}$$ and plugging in the found $$\mathrm{E^\circ_{cell}}$$ gives us an $$\mathrm{E_{cell}}$$ value of 1.827 V

1. $$\mathrm{Sn(s)|Sn^{2+}(0.01\:M)||Pb^{2+}(0.700\:M)|Pb(s)}$$

$$\mathrm{E^{\circ}_{cell} = -0.125 - (-0.137) = 0.012\:V}$$

Using the Nernst equation $$\mathrm{E_{cell} = E^{\circ}_{cell} - \dfrac{0.0592}{n}\times \log Q}$$, we now need to establish our $$\mathrm{Q}$$. The overall equation would be $$\mathrm{Pb^{2+}(aq) + Sn(s) \rightarrow Sn^{2+}(aq) + Pb(s)}$$

Thus, the $$\mathrm{Q}$$ would be $$\dfrac{0.01}{0.700}$$ and the $$\mathrm{E_{cell}}$$ calculates out to 0.067V

For review on this topic, visit the page "Electrochemistry 4: The Nernst Equation".

## Q2.10

To push a reaction forward spontaneously, the $$\mathrm{E_{cell}}$$ must be greater than 0. We find the $$\mathrm{E^\circ_{cell}}$$ = 0.460 V.

Next, plug in the values given to the Nernst equation. The number of moles of electrons being transferred is 2. The equation will be set up as:

$0 > 0.460 - \dfrac{0.0592}{2} \times \log\dfrac{2}{x^2}$

Remember to account for coefficients in setting up the $$\mathrm{Q}$$. Now solve for x $$\ce{[Ag+]}$$ and we find that $$\mathrm{x = 2.4 \times 10^{-8}\, M}$$.

For review on this topic, visit the page "Electrochemistry 4: The Nernst Equation".

## Q2.11

Set up the Nernst equation with the values appropriate for this equation. The moles of electrons transferred (n) are 2. The $$\mathrm{E^\circ_{cell}}$$ for this would be 0.

Thus, the Nernst equation sets up as $$\mathrm{0.0567 = E^{\circ}_{cell} - \dfrac{0.0592}{n}\times \log Q}$$

$\mathrm{0.0567 = E^{\circ}_{cell} - \dfrac{0.0592}{n}\times \log \dfrac{x}{0.100}}$

Solve for the value of x algebraically; this will yield an X value of 0.001215.

This value is not the answer; it is how much $$\ce{Pb^2+}$$ needs to be saturated in the equation. To find the $$\mathrm{K_{sp}}$$, note the stoichiometric values for $$\ce{[Pb]}$$ and $$\mathrm{[I^-]^2}$$. The $$\ce{Pb}$$ and $$\ce{I}$$ concentrations have been found. Now plug in your x value and $$\mathrm{K_{sp} = [Pb][I^-]^2}$$and $$\mathrm{K_{sp} =1.79 \times 10^{-9}}$$.

For review on this topic, visit the page "Electrochemistry 4: The Nernst Equation".

## Q2.12

a) ox: $$\ce{[Ag \rightarrow Ag+ + e-]\times 2}$$; red: $$\ce{Cu^2+ + 2 e- \rightarrow Cu}$$

balanced: $$\ce{2 Ag(s) + Cu^2+ (aq) \rightarrow 2 Ag+ (aq) + Cu(s)}$$

Eo = Eoox(Ag) + Eored(Cu2+) = -0.80 V + 0.34 V = -0.46 V. Not spontaneous.

b) ox: $$\ce{[Ni(s) \rightarrow Ni^2+ + 2 e-] \times 5}$$; red: $$\ce{[5 e- + 8 H+ + MnO_4- \rightarrow Mn^2+ + 4 H2O] \times 2}$$

balanced: $$\ce{5 Ni(s) + 16 H+ (aq) + 2 MnO_4- (aq) \rightarrow 5 Ni^2+ (aq) + 2 Mn^2+ (aq) + 8 H2O}$$

Eo = Eoox(Ni) + Eored(MnO4-) = +0.23 V + 1.51 V = +1.74 V. Spontaneous.

c) ox: $$\ce{[2 H2O + Mn^2+ \rightarrow MnO2 + 4 H+ + 2 e-] \times 3}$$

red: $$\ce{[3 e- + 4 H+ + NO3- \rightarrow NO + 2 H2O] \times 2}$$

balanced: $$\ce{2 H2O + 3 Mn^2+ (aq) + 2 NO3- (aq) \rightarrow 3 MnO2(s) + 4 H+ (aq) + 2 NO(g)}$$

Eo = Eoox(Mn2+) + Eored(NO3-) = -1.21 V + 0.96 V = -0.25 V. Not spontaneous.

## Q2.13

1. Eo = Eoox(Fe) + Eored(Fe3+) = +0.44 V + 0.77 V = +1.21 V. Spontaneous.
2. Eo = Eoox(I-) + Eored(Zn2+) = - 0.54 V - 0.76 V = -1.30 V. Not spontaneous.

## Q2.14

Eoox(Br-) = -1.09 V. An ion will oxidize Br- if it has an Eored > +1.09 V.

1. $$\mathrm{E^\circ_{red}(Pb^{2+}) = -0.13\, V \Rightarrow}$$ No.
2. $$\mathrm{E^\circ_{red}(H^+) = 0 \Rightarrow}$$ No.
3. $$\mathrm{E^\circ_{red}(Au^{3+}) = +1.50\, V \Rightarrow}$$ Yes.
4. $$\mathrm{E^\circ_{red}(MnO_4^-) = +1.51\, V \Rightarrow}$$ Yes.

## Q2.15

1. Eoox(Al) = +1.66 V; Eoox(Pb) = +0.13V. Al has a greater tendency to be oxidized. Thus, for spontaneous reaction, Al will be oxidized and Pb will be reduced: $$\ce{2 Al(s) + 3 Pb^2+ (aq) \rightarrow 2 Al^3+ (aq) + 3 Pb(s)}$$
2. -
3. lead (Pb); since $$\ce{Pb^2+ \rightarrow Pb}$$

## Q2.16

1. Iron (Fe) is the strongest reducing agent because it is always oxidized. It reduces all of the other metals.
2. Gold (Au) is the weakest reducing agent because it is never oxidized.
3. Using the voltages of the other metals with Pb.
Half-reaction Eored (volts)
Au3+ + 3 e- ↔ Au + 0.80 v
Pb2+ + 2 e- ↔ Pb 0.00 v
Ni2+ + 2 e- ↔ Ni - 0.10 v
Fe2+ + 2 e- ↔ Fe - 0.25 v