[Single-digit Integer] If [Fe^3+] at equilibrium, when potassium iodide is added to a solution of Fe — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Redox Half-Reactions and the Overall Cell Reaction When potassium iodide ($\ce{KI}$) is added to a solution of iron(III) ions ($\ce{Fe^3+}$), a spontaneous redox reaction occurs where $\ce{Fe^3+}$ oxidizes iodide ions ($\ce{I^-}$) to molecular iodine ($\ce{I2}$), while being reduced to iron(II) ions ($\ce{Fe^2+}$). The individual half-reactions and their standard reduction potentials ($E^\circ$) are: * **Reduction half-reaction (at the cathode):** $$\ce{Fe^3+(aq) + e^- -> Fe^2+(aq)} \quad E^\circ_{\ce{Fe^3+/Fe^2+}} = 0.77\text{ V}$$ * **Oxidation half-reaction (at the anode):** $$\ce{2I^-(aq) -> I2(s) + 2e^-} \quad E^\circ_{\ce{I2/I^-}} = 0.53\text{ V}$$ To balance the electrons transferred, we multiply the reduction half-reaction by $2$: $$\ce{2Fe^3+(aq) + 2e^- -> 2Fe^2+(aq)}$$ Adding the balanced half-reactions gives the overall cell reaction: $$\ce{2Fe^3+(aq) + 2I^-(aq) -> 2Fe^2+(aq) + I2(s)}$$ Here, the number of moles of electrons transferred per mole of the balanced equation is $n = 2$. Step 2 - Calculate the Standard Cell Potential ($E^\circ_{\text{cell}}$) The standard potential for this electrochemical cell is calculated using the formula: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ Substitute the given values into the formula: $$E^\circ_{\text{cell}} = 0.77\text{ V} - 0.53\text{ V}$$ $$E^\circ_{\text{cell}} = 0.24\text{ V}$$ Step 3 - Calculate the Equilibrium Constant ($K_{\text{eq}}$) At chemical equilibrium, the cell potential ($E_{\text{cell}}$) is exactly zero. The relationship between the standard cell potential ($E^\circ_{\text{cell}}$) and the equilibrium constant ($K_{\text{eq}}$) is given by the Nernst equation: $$E^\circ_{\text{cell}} = \frac{2.303 RT}{nF} \log_{10} K_{\text{eq}}$$ Substituting the slope factor $\frac{2.303 RT}{F} = 0.06\text{ V}$ and $n = 2$: $$E^\circ_{\text{cell}} = \frac{0.06\text{ V}}{2} \log_{10} K_{\text{eq}}$$ $$0.24\text{ V} = 0.03\text{ V} \times \log_{10} K_{\text{eq}}$$ Solving for $\log_{10} K_{\text{eq}}$: $$\log_{10} K_{\text{eq}} = \frac{0.24\text{ V}}{0.03\text{ V}}$$ $$\log_{10} K_{\text{eq}} = 8$$ $$K_{\text{eq}} = 10^8$$ Step 4 - Determine the Equilibrium Concentrations of the Species The expression for the equilibrium constant ($K_{\text{eq}}$) of the reaction is: $$K_{\text{eq}} = \frac{[\ce{Fe^2+}]^2}{[\ce{Fe^3+}]^2 [\ce{I^-}]^2}$$ Since the equilibrium constant is extremely large ($K_{\text{eq}} = 10^8$), the forward redox reaction proceeds virtually to completion. * The initial concentration of $\ce{Fe^3+}$ is $0.50\text{ M}$. * At completion, virtually all of the $\ce{Fe^3+}$ is converted to $\ce{Fe^2+}$: $$[\ce{Fe^2+}] \approx 0.50\text{ M}$$ * The iodide ion concentration is maintained at a high excess such that its equilibrium concentration is: $$[\ce{I^-}] = 1.0\text{ M}$$ Step 5 - Calculate the Equilibrium Concentration of \ce{Fe^3+} and Solve for $x$ We rearrange the equilibrium expression to solve for $[\ce{Fe^3+}]$: $$[\ce{Fe^3+}]^2 = \frac{[\ce{Fe^2+}]^2}{K_{\text{eq}} \times [\ce{I^-}]^2}$$ Substitute the values: $$[\ce{Fe^3+}]^2 = \frac{(0.50\text{ M})^2}{10^8 \times (1.0\text{ M})^2}$$ $$[\ce{Fe^3+}]^2 = \frac{0.25\text{ M}^2}{10^8}$$ $$[\ce{Fe^3+}]^2 = 2.5 \times 10^{-9}\text{ M}^2 = 25 \times 10^{-10}\text{ M}^2$$ Taking the square root of both sides: $$[\ce{Fe^3+}] = \sqrt{25 \times 10^{-10}\text{ M}^2}$$ $$[\ce{Fe^3+}] = 5 \times 10^{-5}\text{ M}$$ Comparing this result with the given expression $[\ce{Fe^3+}] = x \times 10^{-5}\text{ M}$, we find: $$x = \boxed{5}$$