[Four-digit Integer] An excess of liquid mercury is added to an acidified solution of 10^-3 M - Fe^3 — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Redox Reaction and Half-Cell Processes When liquid mercury ($\ce{Hg}$) is added to an acidified solution containing iron(III) ions ($\ce{Fe^3+}$), a spontaneous redox reaction takes place. The liquid mercury is oxidized to mercurous ions ($\ce{Hg2^2+}$), while the ferric ions ($\ce{Fe^3+}$) are reduced to ferrous ions ($\ce{Fe^2+}$). We write the balanced half-reactions as follows: * **Anode (Oxidation half-reaction):** $$2\ce{Hg(l) -> Hg2^{2+}(aq) + 2e^-}$$ * **Cathode (Reduction half-reaction):** $$\ce{Fe^{3+}(aq) + e^- -> Fe^{2+}(aq)}$$ To balance the number of electrons transferred in the redox process, we multiply the cathode half-reaction by $2$: $$2\ce{Fe^{3+}(aq) + 2e^- -> 2Fe^{2+}(aq)}$$ Adding the oxidation and reduction half-reactions yields the overall balanced equilibrium reaction: $$2\ce{Hg(l) + 2\ce{Fe^{3+}}(aq) <=> Hg2^{2+}(aq) + 2\ce{Fe^{2+}}(aq)}$$ This confirms that the number of moles of electrons transferred in the balanced overall reaction is: $$n = 2$$ Step 2 - Determine the Equilibrium Concentrations We are given: * Initial concentration of $\ce{Fe^3+}$: $[\ce{Fe^3+}]_0 = 10^{-3}\text{ M}$ * At equilibrium, $10\%$ of the initial $\ce{Fe^3+}$ remains unreacted at $25^\circ\text{C}$. We calculate the equilibrium concentration of $\ce{Fe^3+}$: $$[\ce{Fe^3+}]_{\text{eq}} = 10\% \times 10^{-3}\text{ M} = 0.10 \times 10^{-3}\text{ M} = 10^{-4}\text{ M}$$ The concentration of $\ce{Fe^3+}$ that reacted to form products is: $$\Delta[\ce{Fe^3+}] = [\ce{Fe^3+}]_0 - [\ce{Fe^3+}]_{\text{eq}}$$ $$\Delta[\ce{Fe^3+}] = 10^{-3}\text{ M} - 10^{-4}\text{ M} = 9 \times 10^{-4}\text{ M}$$ Using the stoichiometry of the balanced overall equation: * The concentration of $\ce{Fe^2+}$ produced is equal to the concentration of reacted $\ce{Fe^3+}$: $$[\ce{Fe^2+}]_{\text{eq}} = \Delta[\ce{Fe^3+}] = 9 \times 10^{-4}\text{ M}$$ * The concentration of $\ce{Hg2^2+}$ produced is equal to half of the concentration of reacted $\ce{Fe^3+}$: $$[\ce{Hg2^2+}]_{\text{eq}} = \frac{1}{2}\Delta[\ce{Fe^3+}] = \frac{9 \times 10^{-4}\text{ M}}{2} = 4.5 \times 10^{-4}\text{ M}$$ Step 3 - Calculate the Equilibrium Constant ($K_{\text{eq}}$) The equilibrium constant expression for the reaction is: $$K_{\text{eq}} = \frac{[\ce{Hg2^2+}][\ce{Fe^2+}]^2}{[\ce{Fe^3+}]^2}$$ Substituting the equilibrium values into the formula: $$K_{\text{eq}} = \frac{(4.5 \times 10^{-4}\text{ M}) \times (9 \times 10^{-4}\text{ M})^2}{(10^{-4}\text{ M})^2}$$ $$K_{\text{eq}} = \frac{4.5 \times 10^{-4} \times 81 \times 10^{-8}}{10^{-8}}$$ $$K_{\text{eq}} = 4.5 \times 81 \times 10^{-4}$$ $$K_{\text{eq}} = 364.5 \times 10^{-4} = 3.645 \times 10^{-2}$$ Step 4 - Calculate the Logarithm of the Equilibrium Constant We can calculate the value of $\log_{10} K_{\text{eq}}$ in two ways: * **Method A (Analytical derivation using exact algebraic relations):** We can express $3.645 \times 10^{-2}$ as a fraction: $$K_{\text{eq}} = 3.645 \times 10^{-2} = \frac{364.5}{10,000} = \frac{729}{20,000} = \frac{3^6}{2 \times 10^4}$$ Taking the logarithm: $$\log_{10} K_{\text{eq}} = \log_{10}\left(\frac{3^6}{2 \times 10^4}\right)$$ $$\log_{10} K_{\text{eq}} = 6\log_{10}(3) - \log_{10}(2) - 4$$ Substituting the given values $\log_{10}(2) = 0.3$ and $\log_{10}(3) = 0.48$: $$\log_{10} K_{\text{eq}} = 6(0.48) - 0.3 - 4$$ $$\log_{10} K_{\text{eq}} = 2.88 - 0.3 - 4 = -1.42$$ * **Method B (Standard high-precision log approximation):** Using standard logarithmic values ($\log_{10}(3) \approx 0.47712$ and $\log_{10}(2) \approx 0.30103$): $$\log_{10} K_{\text{eq}} \approx 6(0.47712) - 0.30103 - 4 = -1.438 \approx -1.44$$ Both values yield the same integer result when standard potentials are rounded. Step 5 - Calculate the Standard Cell Potential ($E^\circ_{\text{cell}}$) The Nernst equation for the cell at $25^\circ\text{C}$ is: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} Q$$ At chemical equilibrium, the cell potential is zero ($E_{\text{cell}} = 0\text{ V}$) and the reaction quotient equals the equilibrium constant ($Q = K_{\text{eq}}$): $$0 = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} K_{\text{eq}}$$ $$E^\circ_{\text{cell}} = \frac{2.303 RT}{nF} \log_{10} K_{\text{eq}}$$ Substituting the slope factor $\frac{2.303 RT}{F} = 0.06\text{ V}$ and $n = 2$: $$E^\circ_{\text{cell}} = \frac{0.06\text{ V}}{2} \log_{10} K_{\text{eq}}$$ $$E^\circ_{\text{cell}} = 0.03\text{ V} \times \log_{10} K_{\text{eq}}$$ * **Using the value from Method A ($\log_{10} K_{\text{eq}} = -1.42$):** $$E^\circ_{\text{cell}} = 0.03\text{ V} \times (-1.42) = -0.0426\text{ V}$$ * **Using the value from Method B ($\log_{10} K_{\text{eq}} = -1.44$):** $$E^\circ_{\text{cell}} = 0.03\text{ V} \times (-1.44) = -0.0432\text{ V}$$ Step 6 - Calculate the Standard Reduction Potential of the Mercury Electrode ($E^\circ_{\ce{Hg2^2+}|\ce{Hg}}$) The standard potential of our cell is given by: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$E^\circ_{\text{cell}} = E^\circ_{\ce{Fe^3+}|\ce{Fe^2+}} - E^\circ_{\ce{Hg2^2+}|\ce{Hg}}$$ Rearranging the equation to solve for $E^\circ_{\ce{Hg2^2+}|\ce{Hg}}$: $$E^\circ_{\ce{Hg2^2+}|\ce{Hg}} = E^\circ_{\ce{Fe^3+}|\ce{Fe^2+}} - E^\circ_{\text{cell}}$$ Given that $E^\circ_{\ce{Fe^3+}|\ce{Fe^2+}} = 0.7724\text{ V}$: * **With $E^\circ_{\text{cell}} = -0.0426\text{ V}$:** $$E^\circ_{\ce{Hg2^2+}|\ce{Hg}} = 0.7724\text{ V} - (-0.0426\text{ V}) = 0.7724\text{ V} + 0.0426\text{ V} = 0.8150\text{ V} = 815\text{ mV}$$ * **With $E^\circ_{\text{cell}} = -0.0432\text{ V}$:** $$E^\circ_{\ce{Hg2^2+}|\ce{Hg}} = 0.7724\text{ V} - (-0.0432\text{ V}) = 0.7724\text{ V} + 0.0432\text{ V} = 0.8156\text{ V} \approx 815\text{ mV}$$ Both calculations lead to the same four-digit integer millivolt value of $815\text{ mV}$. $$E^\circ_{\ce{Hg2^2+}|\ce{Hg}} = \boxed{0815}$$