An excess of liquid mercury is added to an acidified solution of . It is found that 10% of remains a — Electrochemistry Chemistry Question
Question
An excess of liquid mercury is added to an acidified solution of $10^{-3} \text{ M } - Fe^{3+}$. It is found that 10% of $Fe^{3+}$ remains at equilibrium at 25°C. The value of $E^\circ_{Hg_2^{2+}\|Hg}$ (in mV), assuming that the only reaction that occurs is: $2Hg + 2Fe^{3+} \rightarrow Hg_2^{2+} + 2Fe^{2+}$, is (Given: $E^\circ_{Fe^{3+}\|Fe^{2+}} = 0.7724 \text{ V}$, $\log 2 = 0.3, \log 3 = 0.48, 2.303RT/F = 0.06$)
💡 Solution & Explanation
Initial $[Fe^{3+}] = 10^{-3}$ M. At equilibrium, 10% remains, so $[Fe^{3+}] = 10^{-4}$ M. Moles reacted = $9 \times 10^{-4}$ M. Thus, $[Fe^{2+}] = 9 \times 10^{-4}$ M. By stoichiometry, $[Hg_2^{2+}] = 4.5 \times 10^{-4}$ M. $K_{eq} = ([Hg_2^{2+}][Fe^{2+}]^2) / [Fe^{3+}]^2 = (4.5 \times 10^{-4} \times (9 \times 10^{-4})^2) / (10^{-4})^2 = 3.645 \times 10^{-2}$. $E^\circ_{cell} = (0.06/2) \log K_{eq} = 0.03 \log(0.03645) = 0.03 \times (-1.438) = -0.0431$ V. $E^\circ_{cell} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{Hg_2^{2+}/Hg} \Rightarrow -0.0431 = 0.7724 - E^\circ_{Hg_2^{2+}/Hg}$. $E^\circ_{Hg_2^{2+}/Hg} = 0.7724 + 0.0431 = 0.8155$ V = 815.5 mV. Rounded to integer, 815. Therefore, correct answer is 0815.