[Four-digit Integer] Given E° = 0.08 V for Fe^3+(cyt b) — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Given:} $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+})$ for cytochrome $b$ = 0.08 V; for cytochrome $c_1$ = 0.20 V. \textbf{Step 1: Identify anode and cathode.} Lower $E^\circ$ = anode (oxidation): cytochrome $b$ ($E^\circ = 0.08$ V) \[ \text{Fe}^{2+}(\text{cyt }b) \rightarrow \text{Fe}^{3+}(\text{cyt }b) + e^- \] Higher $E^\circ$ = cathode (reduction): cytochrome $c_1$ ($E^\circ = 0.20$ V) \[ \text{Fe}^{3+}(\text{cyt }c_1) + e^- \rightarrow \text{Fe}^{2+}(\text{cyt }c_1) \] \textbf{Step 2: Calculate $E^\circ_{\text{cell}}$.} \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.20 - 0.08 = 0.12\ \text{V} \] \textbf{Step 3: Calculate equilibrium constant.} $n = 1$ (one electron transferred): \[ \log K_{eq} = \frac{n \times E^\circ_{\text{cell}}}{0.06} = \frac{1 \times 0.12}{0.06} = 2 \] \[ K_{eq} = 10^2 = \boxed{100} \] Four-digit answer: \textbf{0100}