The E_cell for Ag(s) \| AgI (satd) \|\| Ag+ (0.10 M) \| Ag (s) is +0.413 V. What is the value of K_s — Electrochemistry Chemistry Question

💡 Solution & Explanation

For the concentration cell, $E = \frac{0.0591}{1} \log \frac{[Ag^+]_c}{[Ag^+]_a}$. $0.413 = 0.0591 \log \frac{0.10}{[Ag^+]_a}$, yielding $\log \frac{0.10}{[Ag^+]_a} \approx 7$. This gives $[Ag^+]_a = 10^{-8}$ M. In a saturated AgI solution in pure water, $[Ag^+] = [I^-] = 10^{-8}$ M. Therefore, $K_{sp} = [Ag^+][I^-] = (10^{-8})(10^{-8}) = 1.0 \times 10^{-16}$.