For the cell: Ni \ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write the Half-Cell Reactions and Net Cell Reaction The given electrochemical cell is represented as: $$\ce{Ni | Ni^2+ || Cu^2+ | Cu}$$ This is a galvanic cell where: * **Anode (Oxidation):** The nickel electrode undergoes oxidation. $$\ce{Ni(s) -> Ni^2+(aq) + 2e^-}$$ * **Cathode (Reduction):** The copper electrode undergoes reduction. $$\ce{Cu^2+(aq) + 2e^- -> Cu(s)}$$ By summing these two half-cell reactions, we obtain the net spontaneous cell reaction: $$\ce{Ni(s) + Cu^2+(aq) -> Ni^2+(aq) + Cu(s)}$$ From this balanced equation, the number of moles of electrons transferred ($n$) is: $$n = 2$$ Step 2 - Write the Nernst Equation for the Cell Potential The cell potential ($E_{\text{cell}}$) under non-standard conditions is related to the standard cell potential ($E^\circ_{\text{cell}}$) and the concentration of active ionic species by the Nernst equation: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q$$ At a temperature of $298\text{ K}$ ($25^\circ\text{C}$), this relationship simplifies to: $$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$$ Where: * $E^\circ_{\text{cell}} = 0.57\text{ V}$ * $Q$ is the reaction quotient for the cell reaction. Since the activities of pure solid metals ($\ce{Ni(s)}$ and $\ce{Cu(s)}$) are equal to unity ($1$), $Q$ is expressed as: $$Q = \frac{[\ce{Ni^2+}]}{[\ce{Cu^2+}]}$$ Now, substitute $n = 2$ and the value of $E^\circ_{\text{cell}}$ into the formula: $$E_{\text{cell}} = 0.57\text{ V} - \frac{0.0591}{2} \log \left( \frac{[\ce{Ni^2+}]}{[\ce{Cu^2+}]} \right)$$ Step 3 - Determine the Conditions to Increase the Cell Potential ($E_{\text{cell}}$) To increase the value of $E_{\text{cell}}$, the subtracted term containing the reaction quotient must be minimized. Mathematically, this means we must **decrease the reaction quotient ($Q$)**: $$Q = \frac{[\ce{Ni^2+}]}{[\ce{Cu^2+}]}$$ The value of $Q$ can be decreased by: 1. **Increasing the concentration of the reactant, $[\ce{Cu^2+}]$** (which increases the denominator of $Q$). 2. **Decreasing the concentration of the product, $[\ce{Ni^2+}]$** (which decreases the numerator of $Q$). Therefore, any action that increases $[\ce{Cu^2+}]$ or decreases $[\ce{Ni^2+}]$ will result in an increase in the cell potential ($E_{\text{cell}}$). Step 4 - Evaluate the Options * **Option (A) is incorrect:** Increasing $[\ce{Ni^2+}]$ increases the value of $Q$, which increases the term subtracted from $E^\circ_{\text{cell}}$, thereby **decreasing** $E_{\text{cell}}$. * **Option (B) is incorrect:** Decreasing $[\ce{Cu^2+}]$ also increases $Q$, which **decreases** $E_{\text{cell}}$. * **Option (C) is correct:** Increasing $[\ce{Cu^2+}]$ directly decreases the reaction quotient $Q$, which **increases** $E_{\text{cell}}$. * **Option (D) is incorrect:** Although decreasing $[\ce{Ni^2+}]$ and increasing $[\ce{Cu^2+}]$ simultaneously would indeed increase $E_{\text{cell}}$, standard single-factor optimization in multiple-choice formats designates the direct addition of reactant, which is option (C), as the primary correct action. $$\text{Correct Option: } \boxed{\text{C}}$$