The solution of , in which copper rod is immersed, is diluted to 10 times. The reduction electrode p — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write down the Reduction Half-Reaction When a copper rod is immersed in an aqueous copper sulfate ($\ce{CuSO4}$) solution, the reduction half-reaction established at the copper electrode surface is: $$\ce{Cu^{2+}(aq) + 2e^- <=> Cu(s)}$$ The number of moles of electrons transferred in this half-reaction is $n = 2$. Step 2 - Apply the Nernst Equation According to the Nernst equation at $298\text{ K}$ ($25^\circ\text{C}$), the reduction potential ($E$) of the copper electrode is given by: $$E = E^\circ - \frac{2.303 RT}{nF} \log Q$$ Where: * $E^\circ$ is the standard reduction potential of the $\ce{Cu^{2+}/Cu}$ electrode. * $n = 2$ 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, which is expressed as: $$Q = \frac{1}{[\ce{Cu^{2+}}]}$$ Substituting these values into the Nernst equation: $$E = E^\circ - \frac{0.0591\text{ V}}{2} \log \left(\frac{1}{[\ce{Cu^{2+}}]}\right)$$ Using the logarithmic property $\log(x^{-1}) = -\log(x)$, we can rewrite this as: $$E = E^\circ + \frac{0.0591\text{ V}}{2} \log [\ce{Cu^{2+}}]$$ $$E = E^\circ + 0.02955\text{ V} \times \log [\ce{Cu^{2+}}]$$ Step 3 - Analyze the Effect of Dilution Let the initial concentration of copper ions in the solution be $C_1 = [\ce{Cu^{2+}}]$. The initial reduction potential ($E_1$) is: $$E_1 = E^\circ + 0.02955\text{ V} \times \log C_1$$ When the solution is diluted to 10 times, the volume increases tenfold, meaning the concentration of the dissolved copper ions decreases to one-tenth of its original value: $$C_2 = \frac{C_1}{10} = 0.1 C_1$$ The new reduction potential ($E_2$) after dilution is: $$E_2 = E^\circ + 0.02955\text{ V} \times \log \left( \frac{C_1}{10} \right)$$ Step 4 - Calculate the Change in Reduction Potential We can find the change in the electrode potential ($\Delta E = E_2 - E_1$) by subtracting the initial potential from the final potential: $$\Delta E = \left( E^\circ + 0.02955\text{ V} \times \log\left(\frac{C_1}{10}\right) \right) - \left( E^\circ + 0.02955\text{ V} \times \log C_1 \right)$$ $$\Delta E = 0.02955\text{ V} \times \left[ \log\left(\frac{C_1}{10}\right) - \log C_1 \right]$$ Using the logarithmic identity $\log(a) - \log(b) = \log\left(\frac{a}{b}\right)$: $$\Delta E = 0.02955\text{ V} \times \log\left( \frac{C_1/10}{C_1} \right)$$ $$\Delta E = 0.02955\text{ V} \times \log\left(\frac{1}{10}\right)$$ $$\Delta E = 0.02955\text{ V} \times \log(10^{-1})$$ $$\Delta E = 0.02955\text{ V} \times (-1) = \boxed{-0.0295\text{ V}}$$ The negative sign indicates a decrease in the potential value. Thus, the reduction electrode potential decreases by $0.0295\text{ V}$. Step 5 - Explanation of Options * **Option (A) is incorrect** because it states that the potential increases. Dilution lowers the concentration of active species ($\ce{Cu^{2+}}$) that undergo reduction, shifting the equilibrium to favor oxidation (the reverse direction) according to Le Chatelier's principle, which decreases the reduction potential. * **Option (B) is correct** because our calculation shows a change of $-0.0295\text{ V}$, representing a decrease of $0.0295\text{ V}$. * **Option (C) is incorrect** because it represents an increase of $0.059\text{ V}$, which does not divide the Nernst coefficient by the number of electrons transferred ($n = 2$). * **Option (D) is incorrect** because it represents a decrease of $0.059\text{ V}$, which would only occur if the reaction involved a single electron transfer ($n = 1$). $$\text{Correct Option: } \boxed{\text{B}}$$