The reduction potential at 25°C for electrode is +0.718 V. If V and V, the ratio of molar concentrat — Electrochemistry Chemistry Question
Question
The reduction potential at 25°C for $Fe^{3+}\|Fe^{2+}$ electrode is +0.718 V. If $E^\circ_{Fe^{2+}\|Fe} = -0.44$ V and $E^\circ_{Fe^{3+}\|Fe} = -0.04$ V, the ratio of molar concentrations of $Fe^{2+}$ to $Fe^{3+}$ ions in solution is ($2.303RT/F = 0.06$, $\log 5 = 0.7$)
💡 Solution & Explanation
First, calculate $E^\circ$ for $Fe^{3+} \rightarrow Fe^{2+}$ using $\Delta G^\circ$: $3 \times E^\circ_{Fe^{3+}/Fe} = 1 \times E^\circ_{Fe^{3+}/Fe^{2+}} + 2 \times E^\circ_{Fe^{2+}/Fe}$. So, $3(-0.04) = E^\circ_{Fe^{3+}/Fe^{2+}} + 2(-0.44) \Rightarrow -0.12 = E^\circ_{Fe^{3+}/Fe^{2+}} - 0.88 \Rightarrow E^\circ_{Fe^{3+}/Fe^{2+}} = 0.76$ V. Nernst equation: $E = E^\circ - 0.06 \log([Fe^{2+}]/[Fe^{3+}])$. $0.718 = 0.76 - 0.06 \log([Fe^{2+}]/[Fe^{3+}])$. $0.042 = 0.06 \log([Fe^{2+}]/[Fe^{3+}])$. $\log([Fe^{2+}]/[Fe^{3+}]) = 0.042 / 0.06 = 0.7$. Since $\log 5 = 0.7$, the ratio is 5. Therefore, correct answer is 5.