For Hg²⁺/Hg (E° = +0.79 V) and Cu²⁺/Cu (E° = +0.34 V), the spontaneous…

Question

For Hg²⁺/Hg (E° = +0.79 V) and Cu²⁺/Cu (E° = +0.34 V), the spontaneous reaction is:

Accepted 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).

Answer

Hg + Cu²⁺ → Hg²⁺ + Cu

Answer

Neither

Answer

Both

For Hg²⁺/Hg (E° = +0.79 V) and Cu²⁺/Cu (E° = +0.34 V), the spontaneous reaction is:

Keep practising Unit 9B1: Electrochemical Cells and Standard Potentials