# Solutions 2: Electrochemistry

- Page ID
- 10540

## Q2.1

- 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.
- 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 \(\ce{Fe^3+}\).
- 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. - 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

- 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. - 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. - 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

- \(\textrm{Reduction: } \mathrm{Na^+(aq)+e^- \rightarrow Na(s)} \quad \mathrm{E^{\circ}=-2.713\:V}\)

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

\(\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.

- \(\textrm{Reduction: } \mathrm{2IO_3^-(aq)+12H^+(aq)+10e^- \rightarrow I_2(g)+6H_2O(l)} \quad \hspace{14 mm}\mathrm{E^{\circ}=+1.20\:V}\)

\(\textrm{Oxidation: } \mathrm{2Mn^{2+}(aq)+8H_2O(l) \rightarrow 2MnO_4^-(aq)+16H^+(aq)+10e^-} \quad \mathrm{E^{\circ}=+1.51\:V} \)

\(\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

- \(\textrm{Reduction: } \mathrm{Mg^{2+}(aq)+2e^- \rightarrow Mg(s)} \quad\hspace{104 pt} \mathrm{E^{\circ}=-2.356\:V}\)

\(\textrm{Oxidation: } \mathrm{Pb^{2+}(aq)+2H_2O(l) \rightarrow PbO_2(s)+4H^+(aq)+2e^-} \quad \mathrm{E^{\circ}=+1.455\:V} \)

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

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

- \(\textrm{Reduction: } \mathrm{O_2(g)+4H^+(aq)+4e^- \rightarrow 2H_2O(l)} \quad \hspace{77 pt} \mathrm{E^{\circ}=1.229\:V}\)

\(\textrm{Oxidation: } \mathrm{2Mn^{2+}(aq)+4H_2O(l) \rightarrow 2MnO_2(s)+8H^+(aq)+4e^-} \quad \mathrm{E^{\circ}=+1.23\:V}\)

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

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

## Q2.6

- 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

The equation is already balanced.

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

- 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

Reaction is already balanced

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

- 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

Reaction is already balanced

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

- 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". *

## Q2.7

not given

## Q2.8

not given

## Q2.9

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

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

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

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

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

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

E^{o} = E^{o}_{ox}(Ag) + E^{o}_{red}(Cu^{2}^{+}) = -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: \(\ce{5 Ni(s) + 16 H+ (aq) + 2 MnO_4- (aq) \rightarrow 5 Ni^2+ (aq) + 2 Mn^2+ (aq) + 8 H2O}\)

E^{o} = E^{o}_{ox}(Ni) + E^{o}_{red}(MnO_{4}^{-}) = +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: \(\ce{2 H2O + 3 Mn^2+ (aq) + 2 NO3- (aq) \rightarrow 3 MnO2(s) + 4 H+ (aq) + 2 NO(g)}\)

E^{o} = E^{o}_{ox}(Mn^{2}^{+}) + E^{o}_{red}(NO_{3}^{-}) = -1.21 V + 0.96 V = -0.25 V. **Not** spontaneous.

## Q2.13

- E
^{o}= E^{o}_{ox}(Fe) + E^{o}_{red}(Fe^{3}^{+}) = +0.44 V + 0.77 V = +1.21 V. Spontaneous. - E
^{o}= E^{o}_{ox}(I^{-}) + E^{o}_{red}(Zn^{2}^{+}) = - 0.54 V - 0.76 V = -1.30 V.**Not**spontaneous.

## Q2.14

E^{o}_{ox}(Br^{-}) = -1.09 V. An ion will oxidize Br^{-} if it has an E^{o}_{red} > +1.09 V.

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

Answer: Au^{3}^{+} and MnO_{4}^{-}

## Q2.15

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

## Q2.16

- Iron (Fe) is the strongest reducing agent because it is always oxidized. It reduces all of the other metals.
- Gold (Au) is the weakest reducing agent because it is never oxidized.
- Using the voltages of the other metals with Pb.

Half-reaction | E^{o}_{red} (volts) |
---|---|

Au^{3}^{+} + 3 e^{-} ↔ Au |
+ 0.80 v |

Pb^{2}^{+} + 2 e^{-} ↔ Pb |
0.00 v |

Ni^{2}^{+} + 2 e^{-} ↔ Ni |
- 0.10 v |

Fe^{2}^{+} + 2 e^{-} ↔ Fe |
- 0.25 v |