17. The solubility product () of can be found using the concentration cell . If the EMF is 0.164 V a — Electrochemistry Chemistry Question

💡 Solution & Explanation

$E = -0.059 \log([Ag^+]_{anode} / [Ag^+]_{cathode})$. $0.164 = 0.059 \log(0.1 / x) \implies \log(0.1/x) = 2.78 \implies 0.1/x = 602.5 \implies x = [Ag^+]_{anode} = 1.66 \times 10^{-4} M$. $K_{sp} = [Ag^+]^2[CrO_4^{2-}] = (2S)^2(S)$. Since $[Ag^+]=2S$, $S = 0.83 \times 10^{-4}$. $K_{sp} = 4(0.83 \times 10^{-4})^3 = 2.28 \times 10^{-12}$.