ElectrochemistryHard

Question

The correct cell diagram for the following reaction and Eo for the cell is

2AgBr(s) + H2(g) → 2Ag(s) + 2H+ + 2Br−; $E_{AgBr|Ag|Br^{-}}^{o} = + 0.10\text{ V}$

Options

A.(Pt) H2|H+||Br−|AgBr|Br2 (Pt); E° = 0.10 V
B.(Pt) H2|H+||Br−|AgBr|Br2 (Pt); E° = −0.10 V
C.(Pt) Br2|AgBr|Br−||H+|H2 (Pt); E° = 0.10 V
D.(Pt) Br2|AgBr|Br−||H+|H2 (Pt); E° = −0.10 V

Solution

Reduction occur at right electrode.

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