1. Identify the cell reaction

Given cell: $\text{Pt}(s)|H_2(g)|H^+(aq)\;||\;Ag^+(aq)|Ag(s)$

Left electrode is the anode and right electrode is the cathode.

• Anode (oxidation): $H_2 \rightarrow 2H^+ + 2e^-$

• Cathode (reduction): $Ag^+ + e^- \rightarrow Ag$

Multiply the silver half-reaction by 2: $2Ag^+ + 2e^- \rightarrow 2Ag$

Overall reaction: $H_2 + 2Ag^+ \rightarrow 2H^+ + 2Ag$

So, the number of electrons transferred is: $n=2$

2. Use the relation between $\Delta G^\circ$ and $E^\circ_{cell}$

The formula is: $\Delta G^\circ = -nFE^\circ_{cell}$

Given:

• $n=2$
• $F=96500\;\text{C mol}^{-1}$ (approximately)
• $E^\circ_{cell}=0.5332\;\text{V}$

Thus, $\Delta G^\circ = -2 \times 96500 \times 0.5332$

3. Calculate

First, $96500 \times 0.5332 = 51453.8$

Then, $\Delta G^\circ = -2 \times 51453.8 = -102907.6\;\text{J mol}^{-1}$

Convert to kJ mol$^{-1}$: $\Delta G^\circ = -102.9\;\text{kJ mol}^{-1}$

Nearest integer: $\boxed{-103}$

4. About $\Delta_f G^\circ$

The question writes $\Delta_f G^\circ$, but from the electrochemical data given, what is directly obtained is the standard Gibbs free energy change of the cell reaction: $\Delta G^\circ \approx -103\;\text{kJ mol}^{-1}$

If the exam/key expects magnitude only, it may be written as $103$.

5. Comparison with stored answer

Stored correct answer: 51OR103

Our derived value is $103$ in magnitude (or $-103$ with sign for $\Delta G^\circ$). Hence it matches one of the stored acceptable values.