The standard EMF for the cell reaction: Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s) is 1.10 V at 25°C. Th — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Cell Reaction and Number of Electrons Transferred ($n$) The given spontaneous cell reaction is: $$\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$$ This redox reaction consists of the following two half-reactions: * **Oxidation Half-Reaction (at the Anode):** $$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}$$ * **Reduction Half-Reaction (at the Cathode):** $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$$ Comparing the two half-reactions, the number of moles of electrons transferred ($n$) in the balanced overall cell reaction is: $$n = 2$$ Step 2 - Formulate the Nernst Equation for the Cell The actual electromotive force ($E_{\text{cell}}$) of the cell under non-standard concentration conditions is related to the standard electromotive force ($E^\circ_{\text{cell}}$) at $25^\circ\text{C}$ ($298.15\text{ K}$) by the Nernst equation: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log_{10} Q$$ At $25^\circ\text{C}$, the constant term is $\frac{2.303 RT}{F} \approx 0.0591\text{ V}$. Substituting this constant and $n = 2$ into the equation yields: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591\text{ V}}{2} \log_{10} Q$$ Where $Q$ is the reaction quotient, defined as the ratio of the active masses of the products to the reactants. Since the activities of pure solids ($\ce{Zn(s)}$ and $\ce{Cu(s)}$) are defined as unity ($1$): $$Q = \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]}$$ Therefore, the Nernst equation for this specific cell simplifies to: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591\text{ V}}{2} \log_{10}\left(\frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]}\right)$$ Step 3 - Substitute the Given Values and Calculate the Actual EMF We are given the following values: * Standard cell potential ($E^\circ_{\text{cell}}$) = $1.10\text{ V}$ * Concentration of zinc ions ($[\ce{Zn^{2+}}]$) = $0.1\text{ M}$ * Concentration of copper ions ($[\ce{Cu^{2+}}]$) = $0.1\text{ M}$ Substituting these values with their units into our formulated equation: $$E_{\text{cell}} = 1.10\text{ V} - \frac{0.0591\text{ V}}{2} \log_{10}\left(\frac{0.1\text{ M}}{0.1\text{ M}}\right)$$ $$E_{\text{cell}} = 1.10\text{ V} - \frac{0.0591\text{ V}}{2} \log_{10}(1)$$ Since the base-10 logarithm of $1$ is exactly zero ($\log_{10}(1) = 0$): $$E_{\text{cell}} = 1.10\text{ V} - \frac{0.0591\text{ V}}{2} \times 0$$ $$E_{\text{cell}} = 1.10\text{ V} - 0\text{ V}$$ $$E_{\text{cell}} = \boxed{1.10\text{ V}}$$ Step 4 - Evaluate and Explain the Options * **Option (A) is correct:** As calculated, because the concentration of the anodic product is equal to the concentration of the cathodic reactant, the reaction quotient is exactly $1$. This causes the concentration-dependent correction term in the Nernst equation to become zero. Consequently, the actual cell EMF is identical to the standard cell EMF, which is $1.10\text{ V}$. * **Option (B) is incorrect:** This value ($1.041\text{ V}$) is obtained if one incorrectly performs the logarithm calculation or assumes a concentration ratio other than $1$. * **Option (C) is incorrect:** This negative potential ($-1.10\text{ V}$) represents a cell where the reverse reaction would be spontaneous, which does not apply to this spontaneous forward process. * **Option (D) is incorrect:** This is a mathematically incorrect and negative potential. $$\text{Correct Option: } \boxed{\text{A}}$$