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Question

For the cell, Pt|Cl2(g, 0.4 bar)|Cl−(aq, 0.1 M)||Cl−(aq, 0.01 M)|Cl2(g, 0.2 bar)|Pt. The cell potential at 298 K is :

A
0.051 V
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B
0.051 V
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C
0.102 V
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D
0.0255 V
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Solution

The correct option is C 0.051 V
Anode: 2Cl(aq)Cl2(g)+2e
Cathode: Cl2(g)+2e2Cl(aq)
Cell reaction: 2Cl(aq)+Cl2(g)2Cl(aq)+Cl2(g)
Ecell=0.062log[Cl]2RHS.(PCl2)LHS[Cl]2LHS.(PCl2)RHS (Ecell=0)
=0.062log(0.01)2(0.1)2×(0.40.2)0.051V

Hence option (A) is correct.

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