Question

Calculate $\Delta G^{\circ}$ for the cell: $Sn_{(s)}|Sn_{(1M)}^{2+}||Ag_{(1M)}^{+}|Ag_{(s)}$ at $25^{\circ}C (E_{cell}^{\circ}=0.90~V)$

• A. -173.7 kJ
• B. -225.3 kJ
• C. -100.2 kJ
• D. -290.8 kJ

Hint:

Logic Tip: A positive cell potential ($E_{cell}^{\circ}>0$) always results in a negative $\Delta G^{\circ}$, indicating a spontaneous reaction. Also, remember that $1 \text{ Volt} \times 1 \text{ Coulomb} = 1 \text{ Joule}$.

Answer 1

The Correct Option is A

Concept:
The standard Gibbs free energy change ($\Delta G^{\circ}$) for an electrochemical cell is related to the standard cell potential ($E_{cell}^{\circ}$) by the equation: $$\Delta G^{\circ} = -nFE_{cell}^{\circ}$$ where $n$ is the number of moles of electrons transferred in the balanced redox reaction, and $F$ is Faraday's constant ($1 F \approx 96500 \text{ C mol}^{-1}$).
Step 1: Determine the number of electrons transferred (n).
From the given cell representation: $Sn_{(s)}|Sn_{(1M)}^{2+}||Ag_{(1M)}^{+}|Ag_{(s)}$, we can write the half-reactions: Oxidation (Anode): $Sn_{(s)} \rightarrow Sn^{2+} + 2e^-$ Reduction (Cathode): $Ag^+ + e^- \rightarrow Ag_{(s)}$ To balance the electrons, we multiply the reduction half-reaction by 2: Overall reaction: $Sn_{(s)} + 2Ag^+ \rightarrow Sn^{2+} + 2Ag_{(s)}$ Thus, the number of moles of electrons transferred is $n = 2$.
Step 2: Calculate $\Delta G^{\circ}$.
Substitute the values into the formula ($n = 2$, $F = 96500 \text{ C mol}^{-1}$, $E_{cell}^{\circ} = 0.90 \text{ V}$): $$\Delta G^{\circ} = -nFE_{cell}^{\circ}$$ $$\Delta G^{\circ} = -2 \times 96500 \times 0.90$$ $$\Delta G^{\circ} = -193000 \times 0.90$$ $$\Delta G^{\circ} = -173700 \text{ J}$$
Step 3: Convert the result to kilojoules.
To convert Joules to kilojoules, divide by 1000: $$\Delta G^{\circ} = \frac{-173700}{1000} \text{ kJ}$$ $$\Delta G^{\circ} = -173.7 \text{ kJ}$$