The electrode potential (in millivolts) of in a solution buffered at pH = 3 and which is also satura — Electrochemistry Chemistry Question
Question
The electrode potential (in millivolts) of $2Ag(s) + S^{2-}(aq) \rightarrow Ag_2S(s) + 2e^-$ in a solution buffered at pH = 3 and which is also saturated with 0.1 M – $H_2S$, is (Given: for $H_2S$, $K_{a1} = 10^{-8}$; $K_{a2} = 10^{-13}$, $K_{sp} \text{ of } Ag_2S = 4 \times 10^{-48}$, $E^\circ_{Ag \| Ag^+} = 0.80 \text{ V}$, $\log 2 = 0.3$, $2.303RT/F = 0.06$)
💡 Solution & Explanation
For $H_2S \rightleftharpoons 2H^+ + S^{2-}$, $K_{overall} = K_{a1}K_{a2} = 10^{-21}$. $[S^{2-}] = (K_{overall} [H_2S]) / [H^+]^2 = (10^{-21} \times 0.1) / (10^{-3})^2 = 10^{-16}$ M. For saturated $Ag_2S$, $[Ag^+] = \sqrt{K_{sp}/[S^{2-}]} = \sqrt{4 \times 10^{-48} / 10^{-16}} = 2 \times 10^{-16}$ M. For the oxidation half-reaction $Ag \rightarrow Ag^+ + e^-$, $E = -E^\circ_{red} - 0.06 \log[Ag^+] = -0.80 - 0.06 \log(2 \times 10^{-16}) = -0.80 - 0.06(0.3 - 16) = -0.80 + 0.942 = +0.142$ V. In millivolts, this is 142 mV. Therefore, correct answer is 0142.