Statement I: In the Daniel cell, if concentrations of Cu^2+ and Zn^2+ ions are doubled, the EMF of c — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Evaluate Statement I (The Daniell Cell Potential and Concentration Variations) The overall cell reaction for the Daniell cell is: $$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$ This redox reaction involves the transfer of $2\text{ moles}$ of electrons ($n = 2$). At $298\text{ K}$, the cell potential ($E_{\text{cell}}$) is expressed using the Nernst equation as: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592\text{ V}}{2} \log_{10} \left(\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}\right)$$ Let the initial concentrations of both zinc and copper ions be $[\ce{Zn^2+}]$ and $[\ce{Cu^2+}]$. If the concentrations of both ionic species are doubled, the new concentrations are $2[\ce{Zn^2+}]$ and $2[\ce{Cu^2+}]$. Substituting these doubled values into the Nernst equation: $$E'_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592\text{ V}}{2} \log_{10} \left(\frac{2[\ce{Zn^2+}]}{2[\ce{Cu^2+}]}\right)$$ $$E'_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592\text{ V}}{2} \log_{10} \left(\frac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]}\right) = E_{\text{cell}}$$ Because the common factor of $2$ cancels out completely in the reaction quotient, the ratio of product concentration to reactant concentration remains unchanged. Thus, the EMF of the cell is unaffected. Therefore, **Statement I is correct**. Step 2 - Evaluate Statement II (Concentration Dependence of Single Electrode Potential) Consider a general metal-metal ion half-cell reaction (reduction): $$\ce{M^{z+}(aq) + z e^- -> M(s)}$$ The non-standard single electrode potential ($E_{\ce{M^{z+}/M}}$) at $298\text{ K}$ is given by the Nernst equation: $$E_{\ce{M^{z+}/M}} = E^\circ_{\ce{M^{z+}/M}} - \frac{0.0592\text{ V}}{z} \log_{10} \left(\frac{1}{[\ce{M^{z+}}]}\right)$$ $$E_{\ce{M^{z+}/M}} = E^\circ_{\ce{M^{z+}/M}} + \frac{0.0592\text{ V}}{z} \log_{10} [\ce{M^{z+}}]$$ This equation shows that the electrode potential is a logarithmic function of the ion concentration, not a linear function. If we double the concentration of the metal ion ($[\ce{M^{z+}}]_{\text{new}} = 2[\ce{M^{z+}}]$), the new electrode potential ($E'_{\ce{M^{z+}/M}}$) is: $$E'_{\ce{M^{z+}/M}} = E^\circ_{\ce{M^{z+}/M}} + \frac{0.0592\text{ V}}{z} \log_{10} (2[\ce{M^{z+}}])$$ $$E'_{\ce{M^{z+}/M}} = E^\circ_{\ce{M^{z+}/M}} + \frac{0.0592\text{ V}}{z} \log_{10} [\ce{M^{z+}}] + \frac{0.0592\text{ V}}{z} \log_{10} 2$$ $$E'_{\ce{M^{z+}/M}} = E_{\ce{M^{z+}/M}} + \frac{0.0592\text{ V} \times 0.301}{z} \approx E_{\ce{M^{z+}/M}} + \frac{0.0178\text{ V}}{z}$$ Instead of doubling, the electrode potential merely increases by a constant additive value of approximately $\frac{17.8}{z}\text{ mV}$ at $298\text{ K}$. Therefore, **Statement II is incorrect**. Step 3 - Evaluate the Options * **Option (A) is incorrect:** This option assumes both statements are correct, but Statement II is incorrect. * **Option (B) is incorrect:** This option assumes both statements are correct, but Statement II is incorrect. * **Option (C) is correct:** Statement I is correct, and Statement II is incorrect. * **Option (D) is incorrect:** Statement I is correct, not incorrect. $$\text{Correct Option: } \boxed{C}$$