For a reaction A(s) + 2B^+ -> A^2+ + B(s); Kc has been found to be 10^12. The EMF of the cell is — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the Redox Reaction and Determine the Electron Transfer ($n$-factor) We are given the following chemical reaction: $$\ce{A(s) + 2B^+(aq) -> A^{2+}(aq) + 2B(s)}$$ Let us break this redox reaction down into its individual oxidation and reduction half-reactions: 1. **Oxidation half-reaction (at the anode):** $$\ce{A(s) -> A^{2+}(aq) + 2e^-}$$ 2. **Reduction half-reaction (at the cathode):** $$\ce{2B^+(aq) + 2e^- -> 2B(s)}$$ By balancing the charges and the atoms, we find that the total number of moles of electrons transferred ($n$-factor) in the balanced chemical equation is: $$n = 2$$ Step 2 - State the Relation between Standard EMF ($E^\circ_{\text{cell}}$) and the Equilibrium Constant ($K_c$) The Nernst equation for a general cell reaction is: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} Q$$ When the cell reaction reaches a state of dynamic chemical equilibrium, the actual cell potential ($E_{\text{cell}}$) drops to zero ($E_{\text{cell}} = 0\text{ V}$), and the reaction quotient ($Q$) becomes equal to the equilibrium constant ($K_c$): $$0 = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} K_c$$ $$E^\circ_{\text{cell}} = \frac{2.303 RT}{nF} \log_{10} K_c$$ At standard temperature $25^\circ\text{C}$ ($298.15\text{ K}$), the Nernst slope factor $\frac{2.303 RT}{F}$ is approximately constant: $$\frac{2.303 RT}{F} \approx 0.0591\text{ V}$$ Substituting this constant into our equilibrium equation yields: $$E^\circ_{\text{cell}} = \frac{0.0591\text{ V}}{n} \log_{10} K_c$$ *(Note: Although the question statement asks for the "EMF of the cell", at equilibrium, the actual operating EMF $E_{\text{cell}}$ is zero. The question is asking for the standard EMF, $E^\circ_{\text{cell}}$, of the cell).* Step 3 - Substitute the Given Values and Calculate $E^\circ_{\text{cell}}$ We are given: * Equilibrium constant ($K_c$) = $10^{12}$ * Number of electrons ($n$) = $2$ Substitute these values into the formula: $$E^\circ_{\text{cell}} = \frac{0.0591\text{ V}}{2} \log_{10}\left(10^{12}\right)$$ Using the logarithmic identity $\log_{10}(10^x) = x$: $$\log_{10}\left(10^{12}\right) = 12$$ Now compute the standard EMF: $$E^\circ_{\text{cell}} = \frac{0.0591\text{ V}}{2} \times 12$$ $$E^\circ_{\text{cell}} = 0.02955\text{ V} \times 12$$ $$E^\circ_{\text{cell}} = \boxed{0.354\text{ V}}$$ Step 4 - Evaluate and Explain the Options * **Option (A) is correct:** As shown by the calculation, the standard EMF of the cell is exactly $0.354\text{ V}$. * **Option (B) is incorrect:** This value ($0.708\text{ V}$) is exactly double the correct value, which is obtained if one forgets to divide the Nernst factor by $n = 2$ in the denominator. * **Option (C) is incorrect:** This value ($0.534\text{ V}$) is mathematically incorrect and does not satisfy the Nernst equilibrium relationship for $n=2$ and $K_c=10^{12}$. * **Option (D) is incorrect:** This value ($0.453\text{ V}$) is mathematically incorrect and is likely a rearrangement of the correct digits in an incorrect order. $$\text{Correct Option: } \boxed{\text{A}}$$