Question

The change in potential of the half-cell Cu\textsuperscript{2+|Cu, when aqueous Cu\textsuperscript{2+} solution is diluted 100 times at 298 K? \(\left( \frac{2.303 RT}{F} = 0.06 \right)\)}

• A. increases by 120 mV
• B. decreases by 120 mV
• C. increases by 60 mV
• D. decreases by 60 mV
• E. no change

Hint:

Dilution decreases the concentration of reactants ($Cu^{2+}$), which pushes the equilibrium to the left, thereby decreasing the reduction potential.

Answer 1

The Correct Option is D

Concept: The electrode potential ($E$) of a half-cell depends on the concentration of ions as described by the Nernst Equation.
• Half-cell Reaction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$.
• Nernst Equation: $E = E^0 - \frac{2.303 RT}{nF} \log \frac{1}{[Cu^{2+}]}$.

Step 1: Calculate the potential difference ($\Delta E$). Let initial concentration be $C_1$. Diluting 100 times means final concentration $C_2 = C_1 / 100$. \[ E_1 = E^0 - \frac{0.06}{2} \log \frac{1}{C_1} \] \[ E_2 = E^0 - \frac{0.06}{2} \log \frac{1}{(C_1 / 100)} = E^0 - 0.03 \log \frac{100}{C_1} \]

Step 2: Find the change $E_2 - E_1$. \[ \Delta E = E_2 - E_1 = -0.03 \left( \log \frac{100}{C_1} - \log \frac{1}{C_1} \right) \] \[ \Delta E = -0.03 \log \left( \frac{100/C_1}{1/C_1} \right) = -0.03 \log(100) \] \[ \Delta E = -0.03 \times 2 = -0.06 \text{ V} = -60 \text{ mV} \] The negative sign indicates a decrease.