2Ag^+ + C6H12O6 + → 2Ag(s) + C6H12O7 + 2H^+. Find ln K of this reaction. — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the half-reactions.** The overall reaction: $$\ce{2Ag+(aq) + C6H12O6(aq) + H2O(l) -> 2Ag(s) + C6H12O7(aq) + 2H+(aq)}$$ **Cathode (reduction):** $$\ce{Ag+(aq) + e- -> Ag(s)}, \quad E^\circ = +0.80\ \text{V}$$ **Anode (oxidation):** $$\ce{C6H12O6(aq) + H2O(l) -> C6H12O7(aq) + 2H+(aq) + 2e-}$$ The standard reduction potential for the gluconolactone/glucose couple is $E^\circ = +0.05\ \text{V}$. **Step 2: Calculate the standard cell potential.** $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode(red)}} = 0.80 - 0.05 = 0.75\ \text{V}$$ **Step 3: Calculate ln K using the thermodynamic relation.** At 25°C, $\frac{RT}{F} = 0.02568\ \text{V}$ and $n = 2$ electrons: $$\ln K = \frac{nFE^\circ}{RT} = \frac{n \times E^\circ}{0.02568} = \frac{2 \times 0.75}{0.02568}$$ Using $\frac{F}{RT} = 38.92\ \text{V}^{-1}$ at 298 K: $$\ln K = n \times E^\circ \times \frac{F}{RT} = 2 \times 0.75 \times 38.92 = 58.38$$ $$\boxed{\text{Answer: B — } \ln K = 58.38}$$