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Question

The potential of the cell containing two hydrogen electrodes as represented below, is:

Pt,H2(g)|H+(106M)||H+(104M)|H2(g),Pt at 298 K

A
0.118V
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B
0.0591V
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C
0.118V
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D
0.0591V
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Solution

The correct option is C 0.118V
At cathode: H+c+e1/2H2 Eo=0V
At anode: 1/2H2H+a+e Eo=0V
_____________________
Cell reaction H+cH+a Eocell=0

For concentration Eocell is 0, as both the compartment have same half cell.

Using Nernst Equation

[H+]a= concentration of H+ at anode; [H+]c= concentration of H+ at cathode

Ecell=Eocell0.05911log[H+]a[H+]c

Ecell=0.0591log[106][104]

Ecell=0.0591log102

Ecell=2×0.0591=0.1182V

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