If at equilibrium, when potassium iodide is added to a solution of initially at 0.50 M until M, is M — Electrochemistry Chemistry Question
Question
If $[Fe^{3+}]$ at equilibrium, when potassium iodide is added to a solution of $Fe^{3+}$ initially at 0.50 M until $[I^-] = 1.0$ M, is $x \times 10^{-5}$ M, the value of $x$ is (Given $E^\circ_{Fe^{3+}\|Fe^{2+}} = 0.77$ V, $E^\circ_{I_2\|I^-} = 0.53$ V, $2.303RT/F = 0.06$)
💡 Solution & Explanation
The redox reaction is $2Fe^{3+} + 2I^- \rightleftharpoons 2Fe^{2+} + I_2(s)$. $E^\circ_{cell} = 0.77 - 0.53 = 0.24$ V. Equilibrium constant $K_{eq}$: $\log K_{eq} = (n E^\circ) / 0.06 = (2 \times 0.24) / 0.06 = 8 \Rightarrow K_{eq} = 10^8$. Since $K_{eq}$ is large, almost all initial $Fe^{3+}$ reacts. Initial $[Fe^{3+}] = 0.50$ M; at equilibrium, $[Fe^{2+}] \approx 0.50$ M. The activity of solid $I_2$ is 1, and equilibrium $[I^-] = 1.0$ M. $K_{eq} = [Fe^{2+}]^2 / ([Fe^{3+}]^2 [I^-]^2) \Rightarrow 10^8 = (0.50)^2 / ([Fe^{3+}]^2 \times 1.0^2) = 0.25 / [Fe^{3+}]^2$. Thus, $[Fe^{3+}]^2 = 25 \times 10^{-10} \Rightarrow [Fe^{3+}] = 5 \times 10^{-5}$ M. Therefore $x = 5$. Therefore, correct answer is 5.