For the cell reaction: 4Br^- + + 4H^+ ⇌ 2 + 2; E° = 0.18 V. The value of (log Kc) at 298 K is [2.303 — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Number of Electrons Transferred ($n$-factor) To find the number of electrons ($n$) involved in the balanced redox reaction, we split the overall cell reaction into its oxidation and reduction half-reactions: $$\text{Overall reaction: } \ce{4Br^-(aq) + O2(g) + 4H^+(aq) <=> 2Br2(l) + 2H2O(l)}$$ 1. **Oxidation Half-Reaction (at the Anode):** Four bromide ions are oxidized to form two moles of liquid bromine, releasing four electrons: $$\ce{4Br^-(aq) -> 2Br2(l) + 4e^-}$$ 2. **Reduction Half-Reaction (at the Cathode):** Gaseous oxygen reacts with four hydrogen ions and gains four electrons to form two moles of water: $$\ce{O2(g) + 4H^+(aq) + 4e^- -> 2H2O(l)}$$ Comparing the two half-reactions, we find that the total number of electrons transferred in the balanced equation is: $$n = 4$$ Step 2 - Relate Standard Cell Potential ($E^\circ$) to the Equilibrium Constant ($K_c$) The standard electromotive force of the cell ($E^\circ$) is thermodynamically related to the equilibrium constant ($K_c$) at a given temperature $T$ by the formula: $$E^\circ = \frac{2.303 RT}{nF} \log_{10} K_c$$ We are given: * Standard cell potential ($E^\circ$) = $0.18\text{ V}$ * The Nernst pre-factor value, $\frac{2.303 RT}{F} = 0.06\text{ V}$ * Number of transferred electrons ($n$) = $4$ Step 3 - Substitute the Values and Calculate $\log_{10} K_c$ Now, we substitute these parameters directly into our relation: $$0.18\text{ V} = \frac{0.06\text{ V}}{4} \log_{10} K_c$$ Simplify the coefficient: $$0.18\text{ V} = 0.015\text{ V} \times \log_{10} K_c$$ Isolate the term $\log_{10} K_c$ to perform the division: $$\log_{10} K_c = \frac{0.18\text{ V}}{0.015\text{ V}}$$ $$\log_{10} K_c = \boxed{12}$$ Step 4 - Evaluate and Explain the Options * **Option (A) is correct:** As shown by the mathematical calculation, the value of $\log_{10} K_c$ is exactly $12$. * **Option (B) is incorrect:** This value ($6$) is obtained if the number of electrons ($n$) is incorrectly taken as $2$ instead of $4$. * **Option (C) is incorrect:** This value ($18$) represents a calculation error, such as multiplying instead of dividing or choosing an incorrect $n$-value. * **Option (D) is incorrect:** This value ($3$) is obtained if we multiply the potential by the coefficient in an incorrect arrangement: $\frac{0.18 \times 0.06}{4}$ or other arithmetic errors. $$\text{Correct Option: } \boxed{\text{A}}$$