Cu2+(aq) + 2 e- → Cu(s) E° = 0.34 V Cr3+(aq) + e- → Cr2+(aq) E° = -0.41 V According to the half-reac — Electrochemistry Chemistry Question

STEPS:

1. Compare the standard reduction potentials ($E^\circ$): The standard reduction potential for the reduction of copper(II) ions is $E^\circ = 0.34\text{ V}$. The standard reduction potential for the reduction of chromium(III) ions is $E^\circ = -0.41\text{ V}$.
2. Identify the cathode (reduction) and anode (oxidation): In a spontaneous redox reaction under standard conditions, the species with the more positive standard reduction potential is more easily reduced and will run in the forward direction (as the reduction half-reaction):
$$\text{Cu}^{2+}(aq) + 2 e^- \rightarrow \text{Cu}(s) \quad E^\circ_{\text{red}} = 0.34\text{ V}$$
The species with the less positive (more negative) reduction potential will be forced to undergo oxidation, meaning its half-reaction must be reversed:
$$\text{Cr}^{2+}(aq) \rightarrow \text{Cr}^{3+}(aq) + e^- \quad E^\circ_{\text{ox}} = +0.41\text{ V}$$
3. Balance the electrons transferred: To combine these half-reactions into a balanced net ionic equation, the number of electrons gained in reduction must equal the number of electrons lost in oxidation. Multiply the chromium oxidation half-reaction by 2:
$$2\text{ Cr}^{2+}(aq) \rightarrow 2\text{ Cr}^{3+}(aq) + 2 e^-$$
4. Combine the half-reactions: Add the reduction and balanced oxidation equations together to cancel out the electrons:
$$\text{Cu}^{2+}(aq) + 2\text{ Cr}^{2+}(aq) \rightarrow \text{Cu}(s) + 2\text{ Cr}^{3+}(aq)$$
5. Calculate the standard cell potential ($E^\circ_{\text{cell}}$):
$$E^\circ_{\text{cell}} = E^\circ_{\text{red}} + E^\circ_{\text{ox}} = 0.34\text{ V} + 0.41\text{ V} = +0.75\text{ V}$$
Because $E^\circ_{\text{cell}}$ is positive ($E^\circ_{\text{cell}} > 0$), this reaction is thermodynamically favorable (spontaneous) under standard conditions. This perfectly matches Option B.

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WHY_OTHERS_WRONG:

• A is incorrect: This reaction represents the reduction of both $\text{Cu}^{2+}$ and $\text{Cr}^{3+}$ simultaneously without a corresponding oxidation partner. Additionally, the transfer of electrons is not balanced.
• C is incorrect: This represents the exact reverse of the spontaneous reaction. Its standard cell potential would be $E^\circ_{\text{cell}} = -0.75\text{ V}$, which is thermodynamically unfavorable (non-spontaneous) under standard conditions.
• D is incorrect: This represents a non-spontaneous reverse reaction that is also unbalanced with respect to electron transfer.
• E is incorrect: This equation is not balanced for mass or charge (there are $2\text{ Cu}$ atoms on the reactant side but only $1\text{ Cr}$ ion changing oxidation state without balancing the overall loss/gain of electrons).

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