AP Chemistry 9.7 Coupled Reactions- MCQs - Exam Style Questions

1. Determine Cathode and Anode:
A galvanic cell is spontaneous, so \(E^{\circ}_{cell}\) must be positive. The half-reaction with the more positive (less negative) reduction potential will be the cathode (reduction).

• \(Cr^{3+}\) Reduction: \(E^{\circ} = -0.74\) V   (More positive, so this is the Cathode)
• \(Mg^{2+}\) Reduction: \(E^{\circ} = -2.37\) V   (More negative, so this will be the Anode)

2. Calculate \(E^{\circ}_{cell}\):
The anode half-reaction must be reversed (oxidation).
\(E^{\circ}_{cell} = E^{\circ}_{cathode} – E^{\circ}_{anode}\)
\(E^{\circ}_{cell} = (-0.74 \text{ V}) – (-2.37 \text{ V})\)
\(E^{\circ}_{cell} = -0.74 + 2.37 = +1.63 \text{ V}\)
This result eliminates options (C) and (D).

3. Determine Moles of Electrons (n):
To calculate \(\Delta G^{\circ}\), we need the total number of electrons transferred (\(n\)). We must balance the electrons in the half-reactions:

• Cathode: \(Cr^{3+}(aq) + 3 e^{-} \rightarrow Cr(s)\)   (multiply by 2)
• Anode: \(Mg(s) \rightarrow Mg^{2+}(aq) + 2 e^{-}\)   (multiply by 3)

This gives:
\(2~Cr^{3+} + 6 e^{-} \rightarrow 2~Cr\)
\(3~Mg \rightarrow 3~Mg^{2+} + 6 e^{-}\)
The least common multiple of electrons is \(6\). Therefore, \(n = 6\).

4. Calculate \(\Delta G^{\circ}\):
Use the formula \(\Delta G^{\circ} = -nFE^{\circ}_{cell}\), where \(F\) is the Faraday constant (\(\approx 96,485\) J/(V·mol)).
\(\Delta G^{\circ} = -(6 \text{ mol}) \times (96,485 \frac{\text{J}}{\text{V·mol}}) \times (1.63 \text{ V})\)
\(\Delta G^{\circ} = -943,925.55\) J/mol\(_{rxn}\)
\(\Delta G^{\circ} \approx -944\) kJ/mol\(_{rxn}\)

This matches option (B).
✅ Answer: (B)