The standard reduction potentials in acidic conditions are 0.77 V and 0.53 V, respectively, for Fe^3 — Electrochemistry Chemistry Question

💡 Solution & Explanation

### Step 1 - Identify the Electrode Couples and Standard Reduction Potentials First, we identify the two redox couples involved in the given electrochemical reaction under acidic conditions: * The iron couple: $\ce{Fe^3+/Fe^2+}$ * The triiodide/iodide couple: $\ce{I3^-/I^-}$ The standard reduction potentials are given as: $$E^\circ_{\ce{Fe^3+/Fe^2+}} = 0.77 \text{ V}$$ $$E^\circ_{\ce{I3^-/I^-}} = 0.53 \text{ V}$$ ### Step 2 - Identify Cathode and Anode and Calculate Standard Cell Potential ($E^\circ_{\text{cell}}$) The overall equilibrium reaction is: $$2\ce{Fe^3+} + 3\ce{I^-} \rightleftharpoons 2\ce{Fe^2+} + \ce{I3^-}$$ By splitting this overall reaction into two half-reactions: 1. **At the Cathode (Reduction half-reaction):** $$2\ce{Fe^3+} + 2\ce{e^-} \rightarrow 2\ce{Fe^2+}$$ $$E^\circ_{\text{cathode}} = E^\circ_{\ce{Fe^3+/Fe^2+}} = 0.77 \text{ V}$$ 2. **At the Anode (Oxidation half-reaction):** $$3\ce{I^-} \rightarrow \ce{I3^-} + 2\ce{e^-}$$ $$E^\circ_{\text{anode}} = E^\circ_{\ce{I3^-/I^-}} = 0.53 \text{ V}$$ Using these potentials, we find the standard cell potential ($E^\circ_{\text{cell}}$): $$\text{Formula: } E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ $$\text{Substitution: } E^\circ_{\text{cell}} = 0.77 \text{ V} - 0.53 \text{ V}$$ $$\text{Calculation: } E^\circ_{\text{cell}} = 0.24 \text{ V}$$ ### Step 3 - Determine the Number of Electrons Transferred ($n$) From the balanced half-reactions in Step 2, $2$ moles of electrons are transferred per mole of reaction. Thus: $$n = 2$$ ### Step 4 - Calculate the Equilibrium Constant ($K$) At chemical equilibrium, the cell potential ($E_{\text{cell}}$) becomes zero. The standard cell potential ($E^\circ_{\text{cell}}$) is related to the equilibrium constant ($K$) by the Nernst equation: $$\text{Formula: } E^\circ_{\text{cell}} = \frac{2.303 RT}{nF} \log_{10} K$$ Given that $\frac{2.303 RT}{F} = 0.06$, the equation simplifies to: $$E^\circ_{\text{cell}} = \frac{0.06}{n} \log_{10} K$$ Substitute the values of $E^\circ_{\text{cell}}$ and $n$: $$\text{Substitution: } 0.24 = \frac{0.06}{2} \log_{10} K$$ $$\text{Calculation: } 0.24 = 0.03 \log_{10} K$$ $$\log_{10} K = \frac{0.24}{0.03} = 8$$ $$K = 10^8$$ $$\text{Final Answer: } K = \boxed{10^8}$$ ### Step 5 - Evaluation and Explanation of Options The options printed on Page 8.6 of the textbook are: * **(A) $2 \times 10^8$:** Incorrect. The calculation yields exactly $10^8$. * **(B) $10^8$:** **Correct.** This matches our derived value of the equilibrium constant $K = 10^8$. * **(C) $10^4$:** Incorrect. This would correspond to a standard cell potential of $0.12 \text{ V}$. * **(D) $10^{-8}$:** Incorrect. This negative exponent represents a non-spontaneous reaction at standard state where $E^\circ_{\text{cell}} < 0$. *(Note: While some test platforms or databases may contain a typographical error listing "C" as the correct answer key, the textbook's official printed answer key on page 8.13 under "Galvanic Cell" correctly lists option (b) as the correct answer, which perfectly aligns with the mathematical solution.)*