The EMF of the cell: Zn \ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Cell Reactions and the Electron Transfer ($n$-factor) The given electrochemical cell is represented as: $$\ce{Zn \mid Zn^{2+}(0.01\text{ M}) \parallel Fe^{2+}(0.001\text{ M}) \mid Fe}$$ To analyze this cell, we break it down into its respective half-cell reactions: * **Anode (Oxidation Half-Cell, Left Side):** $$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}$$ * **Cathode (Reduction Half-Cell, Right Side):** $$\ce{Fe^{2+}(aq) + 2e^- -> Fe(s)}$$ By combining these two half-reactions, we obtain the balanced overall cell reaction: $$\ce{Zn(s) + Fe^{2+}(aq) -> Zn^{2+}(aq) + Fe(s)}$$ From this balanced equation, we can determine the number of moles of electrons transferred ($n$-factor): $$n = 2$$ Step 2 - Write the Nernst Equation and Calculate the Reaction Quotient ($Q$) The Nernst equation relates the actual cell potential ($E_{\text{cell}}$) to the standard cell potential ($E^\circ_{\text{cell}}$) at $298\text{ K}$: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log Q$$ For a two-electron process at $298\text{ K}$, we approximate the term $\frac{2.303 RT}{F} \approx 0.059\text{ V}$ (which gives $\frac{0.059\text{ V}}{2} = 0.0295\text{ V}$): $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059\text{ V}}{2} \log Q$$ The reaction quotient ($Q$) for the cell reaction involves only the concentrations of the aqueous ions, as the activities of pure solid metals are equal to $1$: $$Q = \frac{[\ce{Zn^{2+}}]}{[\ce{Fe^{2+}}]}$$ Substituting the given concentrations ($[\ce{Zn^{2+}}] = 0.01\text{ M}$ and $[\ce{Fe^{2+}}] = 0.001\text{ M}$): $$Q = \frac{0.01\text{ M}}{0.001\text{ M}} = 10$$ Step 3 - Calculate the Standard Cell Potential ($E^\circ_{\text{cell}}$) Now, substitute the given actual cell potential ($E_{\text{cell}} = 0.2905\text{ V}$) and $Q = 10$ into the Nernst equation: $$0.2905\text{ V} = E^\circ_{\text{cell}} - 0.0295\text{ V} \times \log(10)$$ Since the base-10 logarithm of $10$ is exactly $1$ ($\log(10) = 1$): $$0.2905\text{ V} = E^\circ_{\text{cell}} - 0.0295\text{ V}$$ $$E^\circ_{\text{cell}} = 0.2905\text{ V} + 0.0295\text{ V}$$ $$E^\circ_{\text{cell}} = 0.32\text{ V}$$ Step 4 - Calculate the Equilibrium Constant ($K_{\text{eq}}$) At chemical equilibrium, the actual cell potential becomes zero ($E_{\text{cell}} = 0$), and the reaction quotient ($Q$) becomes equal to the equilibrium constant ($K_{\text{eq}}$). The standard cell potential is related to the equilibrium constant by: $$E^\circ_{\text{cell}} = \frac{2.303 RT}{nF} \log K_{\text{eq}}$$ Substituting the values $E^\circ_{\text{cell}} = 0.32\text{ V}$, $n = 2$, and the standard Nernst factor: $$0.32\text{ V} = \frac{0.059\text{ V}}{2} \log K_{\text{eq}}$$ $$0.32\text{ V} = 0.0295\text{ V} \log K_{\text{eq}}$$ Isolating $\log K_{\text{eq}}$: $$\log K_{\text{eq}} = \frac{0.32}{0.0295}$$ Taking the base-10 antilogarithm of both sides yields the final expression for the equilibrium constant: $$K_{\text{eq}} = 10^{\frac{0.32}{0.0295}}$$ Step 5 - Evaluate and Explain the Options Let us compare our derived value with the options from the printed curriculum reference: * **Option (A) is incorrect:** This option uses the natural exponential base $e$ instead of base $10$. The standard relationship with $0.059\text{ V}$ is formulated using the common base-10 logarithm. * **Option (B) is correct:** As mathematically shown, the equilibrium constant is exactly $10^{\frac{0.32}{0.0295}}$. * **Option (C) is incorrect:** This value of $0.26$ in the exponent is incorrect and would arise from a subtraction error in calculating the standard cell potential. * **Option (D) is incorrect:** This uses the un-halved factor $0.0591$ in the denominator, neglecting that two electrons are transferred in the redox process. $$\text{Correct Option: } \boxed{\text{B}}$$