For the cell Zn \ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write down the Cell Reactions For the given galvanic cell: $$\ce{Zn(s) \mid Zn^{2+}(aq) \parallel Cu^{2+}(aq) \mid Cu(s)}$$ The cell consists of a zinc anode (where oxidation occurs) and a copper cathode (where reduction occurs). The individual half-cell reactions are: * At the Anode (Oxidation): $$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}$$ * At the Cathode (Reduction): $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$$ By adding the two half-reactions, we obtain the overall cell reaction: $$\ce{Zn(s) + Cu^{2+}(aq) <=> Zn^{2+}(aq) + Cu(s)}$$ The number of moles of electrons transferred in this balanced redox reaction is $n = 2$. Step 2 - Apply the Nernst Equation The Nernst equation relates the electromotive force (EMF) of the cell ($E_{\text{cell}}$) to its standard potential ($E^\circ_{\text{cell}}$) and the concentrations of the active ions in solution at $298\text{ K}$ ($25^\circ\text{C}$): $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{2.303 RT}{nF} \log Q$$ Where: * $E^\circ_{\text{cell}}$ is the standard cell potential. * $n = 2$ is the number of electrons transferred. * $\frac{2.303 RT}{F} \approx 0.0591\text{ V}$ at $298\text{ K}$. * $Q$ is the reaction quotient. Since the activities of pure solid phases are equal to $1$, $Q$ is defined as the ratio of the concentration of anodic metal ions to cathodic metal ions: $$Q = \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]}$$ Substituting these parameters into the Nernst equation gives the initial EMF ($E_1$): $$E_1 = E^\circ_{\text{cell}} - \frac{0.0591\text{ V}}{2} \log \left( \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]} \right)$$ Step 3 - Analyze the Effect of Doubling Ion Concentrations Let the initial concentrations of the zinc and copper ions be $[\ce{Zn^{2+}}]$ and $[\ce{Cu^{2+}}]$, respectively. If the concentration of both ions is doubled, the new concentrations are: $$[\ce{Zn^{2+}}]' = 2 [\ce{Zn^{2+}}]$$ $$[\ce{Cu^{2+}}]' = 2 [\ce{Cu^{2+}}]$$ Substituting these new values into the Nernst equation gives the new EMF ($E_2$): $$E_2 = E^\circ_{\text{cell}} - \frac{0.0591\text{ V}}{2} \log \left( \frac{[\ce{Zn^{2+}}]'}{[\ce{Cu^{2+}}]'} \right)$$ $$E_2 = E^\circ_{\text{cell}} - \frac{0.0591\text{ V}}{2} \log \left( \frac{2 [\ce{Zn^{2+}}]}{2 [\ce{Cu^{2+}}]} \right)$$ Since the factor of $2$ appears in both the numerator and the denominator, it cancels out completely: $$E_2 = E^\circ_{\text{cell}} - \frac{0.0591\text{ V}}{2} \log \left( \frac{[\ce{Zn^{2+}}]}{[\ce{Cu^{2+}}]} \right)$$ $$E_2 = E_1$$ Because the concentration ratio in the reaction quotient remains unchanged, the cell EMF remains completely unaffected. Step 4 - Explanation of Options * **Option (A) is incorrect** because the EMF does not double; the concentration ratio, which determines the concentration-dependent potential, remains constant. * **Option (B) is incorrect** because the EMF is not halved. * **Option (C) is correct** because, as mathematically derived, doubling the concentrations of both the product and reactant ions keeps their ratio unchanged, meaning the EMF remains the same. * **Option (D) is incorrect** because the EMF does not drop to zero under these conditions. $$\text{Correct Option: } \boxed{\text{C}}$$