A quantity of 6.539 × 10^-2 g of metallic zinc is added to 100 ml of saturated solution of AgCl. The — Electrochemistry Chemistry Question

💡 Solution & Explanation

**Step 1: Write the cell reaction.** Zinc displaces silver from silver ion solution: $$\ce{Zn(s) + 2Ag+(aq) -> Zn^{2+}(aq) + 2Ag(s)}$$ This is the reaction that establishes equilibrium after the zinc is added. **Step 2: Identify the standard cell potential.** - **Cathode (reduction):** $\ce{Ag+(aq) + e- -> Ag(s)}, \quad E^\circ = +0.80\ \text{V}$ - **Anode (oxidation):** $\ce{Zn(s) -> Zn^{2+}(aq) + 2e-}, \quad E^\circ_{\text{red}} = -0.76\ \text{V}$ $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode(red)}} = 0.80 - (-0.76) = 1.56\ \text{V}$$ **Step 3: Calculate log K (= log K_eq).** Number of electrons transferred: $n = 2$. At 25°C: $$\log K = \frac{n \times E^\circ}{0.0591} = \frac{2 \times 1.56}{0.0591} = \frac{3.12}{0.0591} = 52.79 \approx \mathbf{52.88}$$ (Using 0.059 V as given in the passage: $\log K = 3.12/0.059 = 52.88$) $$\boxed{\text{Answer: B — } \log K = 52.88}$$