The standard reduction potentials of Cu^2+ — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Given Half-Reactions and Standard Potentials 1. $\ce{Cu^{2+}(aq) + 2e^{-} -> Cu(s)} \quad E^\circ_1 = 0.337\text{ V} \quad (n_1 = 2)$ 2. $\ce{Cu^{2+}(aq) + e^{-} -> Cu^{+}(aq)} \quad E^\circ_2 = 0.153\text{ V} \quad (n_2 = 1)$ Target: $\ce{Cu^{+}(aq) + e^{-} -> Cu(s)} \quad E^\circ_3 = ? \quad (n_3 = 1)$ Step 2 - Establish the Thermodynamic Relationship Standard electrode potentials are intensive and cannot be directly added. We use Gibbs free energy ($\Delta G^\circ = -nFE^\circ$) which is extensive and additive. Reaction 3 = Reaction 1 $-$ Reaction 2, so: $$\Delta G^\circ_3 = \Delta G^\circ_1 - \Delta G^\circ_2$$ $$-n_3 F E^\circ_3 = -n_1 F E^\circ_1 - (-n_2 F E^\circ_2)$$ $$n_3 E^\circ_3 = n_1 E^\circ_1 - n_2 E^\circ_2$$ Step 3 - Calculate the Standard Potential $$1 \times E^\circ_3 = 2 \times (0.337\text{ V}) - 1 \times (0.153\text{ V})$$ $$E^\circ_3 = 0.674\text{ V} - 0.153\text{ V} = \boxed{0.521\text{ V}}$$ Step 4 - Evaluation of Options * **Option (A)** $0.184\text{ V}$ — incorrect, this is the direct subtraction of potentials (not valid for intensive properties). * **Option (B)** $0.521\text{ V}$ — correct. * **Option (C)** $0.490\text{ V}$ — incorrect. * **Option (D)** $-0.184\text{ V}$ — incorrect. $$\text{Correct Answer: } \boxed{\text{B}}$$