Question

The equilibrium constant for the reaction $M(s) + 2Ag^{+}(aq) \rightarrow M^{2+}(aq) + 2Ag(s)$ is $10^{15}$. What is the Gibbs energy change ($\Delta_r G^{\ominus}$ in KJ mol$^{-1}$) for this reaction?
($\frac{RT}{F} \times 2.303 = 0.06V$; F = 96500 C mol$^{-1}$)

• A. $-86850$
• B. $-96500$
• C. $-86.85$
• D. $-96.5$

Hint:

Remember: $\Delta G^\ominus = -nFE^\ominus_{cell}$. Also $E^\ominus_{cell} = \frac{0.0591}{n} \log K_c$ (at 298K). Combining these gives the same result.

Answer 1

The Correct Option is C

Step 1: Formula for Standard Gibbs Energy Change:
The relationship between standard Gibbs energy change ($\Delta_r G^{\ominus}$) and the equilibrium constant ($K_{eq}$) is: \[ \Delta_r G^{\ominus} = -RT \ln K_{eq} \] \[ \Delta_r G^{\ominus} = -2.303 RT \log_{10} K_{eq} \]
Step 2: Using the given constant:
The problem provides the value $\frac{2.303 RT}{F} = 0.06$ V. This implies $2.303 RT = 0.06 \times F$. So, substitute this into the Gibbs energy equation: \[ \Delta_r G^{\ominus} = - (0.06 \times F) \times \log_{10} K_{eq} \]
Step 3: Calculation:
Given $K_{eq} = 10^{15}$. So, $\log_{10} K_{eq} = 15$. Given $F = 96500$ C mol$^{-1}$. \[ \Delta_r G^{\ominus} = -0.06 \times 96500 \times 15 \text{ J mol}^{-1} \] \[ \Delta_r G^{\ominus} = -0.9 \times 96500 \] \[ \Delta_r G^{\ominus} = -86850 \text{ J mol}^{-1} \]
Step 4: Unit Conversion:
The question asks for the answer in KJ mol$^{-1$}. \[ \Delta_r G^{\ominus} = \frac{-86850}{1000} \text{ kJ mol}^{-1} \] \[ \Delta_r G^{\ominus} = -86.85 \text{ kJ mol}^{-1} \]
Step 5: Final Answer:
-86.85.