# AP Chemistry: 9.7 Galvanic (Voltaic) and Electrolytic Cells – Exam Style questions with Answer- MCQ

### Question

The standard reduction potentials for the half-reactions related to the galvanic cell represented above are listed in the table below.

Which of the following gives the value of$$E^ {\circ}_{cell}$$for the cell?

(A) $$E^ {\circ}_{cell}$$= −1.50 V
(B) $$E^ {\circ}_{cell}$$= −0.80 V
(C) $$E^ {\circ}_{cell}$$= −0.02 V
(D) $$E^ {\circ}_{cell}$$ = +0.02 V

Ans:C

To find the standard cell potential ( $$\mathrm{E}^{\circ}$$ cell) for the given galvanic cell, we need to use the reduction potentials provided in the table and apply the formula:
$$\mathrm{E}^{\circ}$$ cell $$=\mathrm{E}^{\circ}$$ cathode $$-\mathrm{E}^{\circ}$$ anode
The cathode half-reaction is the reduction reaction with the more positive reduction potential, while the anode half-reaction is the reduction reaction with the less positive (or more negative) reduction potential.

In this case, the $$\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}$$ (s) half-reaction has a reduction potential of $$-0.76 \mathrm{~V}$$, which is more positive than the $$\mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s})$$ half-reaction with a reduction potential of $$-0.74 \mathrm{~V}$$.

Therefore, the cathode half-reaction is:
$\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(\mathrm{s}) \quad \mathrm{E}^{\circ} \text { cathode }=-0.76 \mathrm{~V}$

And the anode half-reaction is:
$\mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s}) \quad \mathrm{E}^{\circ} \text { anode }=-0.74 \mathrm{~V}$

Substituting these values into the formula:
\begin{aligned} & \mathrm{E}^{\circ} \text { cell }=\mathrm{E}^{\circ} \text { cathode }-\mathrm{E}^{\circ} \text { anode } \\ & \mathrm{E}^{\circ} \text { cell }=-0.76 \mathrm{~V}-(-0.74 \mathrm{~V}) \\ & \mathrm{E}^{\circ} \text { cell }=-0.02 \mathrm{~V} \end{aligned}

So, the correct answer is (C) $$\mathrm{E}^{\circ} \mathrm{cell}=-0.02 \mathrm{~V}$$.

Questions

Based on the information in the table above, which of the following shows the cell potential and the Gibbs free energy change for the overall reaction that occurs in a standard galvanic cell?

$$E_{cell}(V)$$    $$\bigtriangleup G ^{\circ } (\frac{Kj}{mol_{rxn}})$$

(A) +1.63                -157
(B) +1.63                -944
(C) +5.63                -543
(D) +5.63                -3262

Ans: B

To find the cell potential ($$E_{\text{cell}}^{\circ}$$) and the Gibbs free energy change ($$\Delta G^{\circ}$$) for the overall reaction,use the given half-reactions and their standard reduction potentials ($$E^{\circ}$$).

The overall reaction is the sum of the two half-reactions:

• $$\mathrm{Mg}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Mg}(s)$$ with $$E^{\circ} = -2.37$$ V
• $$\mathrm{Cr}^{3+}(aq) + 3e^{-} \rightarrow \mathrm{Cr}(s)$$ with $$E^{\circ} = -0.74$$ V

balance the number of electrons transferred in each half-reaction.

• $$3(\mathrm{Mg}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Mg}(s))$$
This gives: $$3\mathrm{Mg}^{2+}(aq) + 6e^{-} \rightarrow 3\mathrm{Mg}(s)$$
• $$2(\mathrm{Cr}^{3+}(aq) + 3e^{-} \rightarrow \mathrm{Cr}(s))$$
This gives: $$2\mathrm{Cr}^{3+}(aq) + 6e^{-} \rightarrow 2\mathrm{Cr}(s)$$

Adding these two balanced half-reactions together,

$3\mathrm{Mg}^{2+}(aq) + 2\mathrm{Cr}^{3+}(aq) \rightarrow 3\mathrm{Mg}(s) + 2\mathrm{Cr}(s)$

$E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} – E_{\text{anode}}^{\circ}$

the cathode (reduction) half-reaction is the reduction of chromium ($$\mathrm{Cr}^{3+}$$) and the anode (oxidation) half-reaction is the reduction of magnesium ($$\mathrm{Mg}^{2+}$$).

$E_{\text{cell}}^{\circ} = -(-0.74) – (-2.37) = 1.63 \text{ V}$

$\Delta G^{\circ} = -nF E_{\text{cell}}^{\circ}$

where $$n$$ is the number of moles of electrons transferred in the balanced overall reaction, and $$F$$ is the Faraday constant ($$96485 \text{ C/mol}$$).

In the given reaction, 6 moles of electrons are transferred.

$\Delta G^{\circ} = -6 \times 96485 \text{ C/mol} \times 1.63 \text{ V} = -944 \text{ kJ/mol}$

Therefore, the correct answer is option (B):

$(B) \quad +1.63 \text{ V} \quad \text{and} \quad -944 \text{ kJ/mol}$

### Question

Pacemakers are electronic devices that help regulate the heart rate. Currently, lithium-iodine cells are commonly used to power pacemakers and have replaced zinc-mercury cells. Table 1 provides the operating cell potential, E , for each cell. Table 2 provides the standard reduction potentials for several half-reactions related to zinc-mercury and zinc-air cells.

The use of zinc-mercury cells in hearing aids has been replaced by zinc-air cells that operate using the oxidation of Zn by $$O_2$$ from the air, generating a potential of +1.60V . Table 2 provides the standard reduction potentials for the half-reactions used in zinc-mercury and zinc-air cells. Which of the following best explains the modification to the cell design that is mostly responsible for the difference in standard cell potentials for zinc-mercury and zinc-air cells?

A The greater standard cell potential of the Zn-air cell compared to that of the zinc-mercury cell most likely results from the thermodynamically more favorable reduction of $$O_2$$ compared to $$HgO$$.
B The greater standard cell potential of the Zn-air cell compared to that of the zinc-mercury cell most likely results from the greater number of moles of $$e^−$$ required to reduce $$O_2$$ compared to $$HgO$$.
C The greater standard cell potential of the Zn-air cell compared to that of the zinc-mercury cell most likely results from the thermodynamically less favorable reduction of $$O_2$$ compared to $$HgO$$.
D The greater standard cell potential of the Zn-aircell compared to that of the zinc-mercury cell most likely results from the greater number of moles of hydroxide ions required to reduce $$[Zn(OH)_4]^{2−}$$ compared to $$Zn(OH)_2$$.

Ans:A

Based on the information in table 2 and the cell potential given, the reaction that takes place in the zinc-air cell is $$2Zn+4 OH^−+O^2+2 H_2O→2[Zn(OH)_4]^{2−}$$. The relatively small difference in the standard reduction potential for Zn(OH)2 compared to that for [Zn(OH)4]2− indicates that the oxidation of zinc under both of these conditions cannot account for the difference in E°cell between the zinc-mercury and zinc-air cells. Therefore, the larger, more positive standard reduction potential for $$O_2$$ compared to that for $$HgO$$ is mostly responsible for the difference in cell potentials. This is justified by ΔG$$=−nFE$$ because a larger, more positive E° results in a larger, more negative ΔG and a reaction that is thermodynamically more favorable.

### Question

Ans:C

Replacing the $$Pb-Pb(NO_3)_2$$ half-cell with an $$Al-Al(NO_3)_3$$ half-cell, not the $$Cu-Cu(NO_3)_2$$ half-cell with a $$Ag-AgNO_3$$ half-cell, causes the greatest increase in E°cell. The replacement means that the reaction taking place is $$2Al(s)+3Cu_^{2+}(aq)→2Al_3+(aq)+3Cu(s)$$, and $$E^{\circ}_{cell}=+2.00V$$.

### Question

The galvanic cell illustrated above was constructed using a salt bridge containing $$KNO_3$$. A second cell is constructed from identical half-cells but uses $$NaNO_3$$ for the salt bridge. Which of the following best explains whether the initial potential of the second cell will be different from the initial potential of the first cell?

A The initial potential of the second cell will be lower than the initial potential of the first cell because Na has a lower molar mass than K has.

B The initial potential of the second cell will be higher than the initial potential of the first cell because Na is less electronegative than K is.

C The initial potential of the second cell will be the same as the first cell because the ions from the salt bridge are not oxidized or reduced during cell operation.

D The initial potential of the second cell will be the same as the first cell because the standard reduction potentials of Group 1 metals are very similar.

Ans:C

Based on the movement of ions in the salt bridge shown in the illustration provided for the first cell, the redox reaction that takes place is $$Cu_2^+(aq)+Pb(s)→Cu(s)+Pb_2^+(aq)$$ . The ions from the salt bridge are not part of the redox process that generates the cell potential, and as a result, the initial E° for the two cells will be the same.

### Question

Standard reduction potentials for the half-reactions associated with the electrochemical cell shown above are given in the table below.

What is the value of E° for the cell?
(A) 0.04 V
(B) 0.84 V
(C) 1.56 V
(D) 2.36 V