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For Hg²⁺/Hg (E° = +0.79 V) and Cu²⁺/Cu (E° = +0.34 V), the spontaneous reaction is:
AHg + Cu²⁺ → Hg²⁺ + Cu
BNeither
CBoth
DCu + Hg²⁺ → Cu²⁺ + Hg
Answer & Solution
Correct answer: D. Cu + Hg²⁺ → Cu²⁺ + Hg
Larger E° is the stronger oxidizer at the cathode. Hg²⁺ (E° = +0.79) > Cu²⁺ (+0.34), so Hg²⁺ is the oxidizer. Cu (lower E°) gets oxidized. Net: Cu + Hg²⁺ → Cu²⁺ + Hg. E°_cell = 0.79 - 0.34 = +0.45 V (positive, spontaneous).
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