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Question

Find the emf of the following cell. (Take RT ln 10F=0.06
Pt, H2(g)/H+(aq)//H+(aq)/H2(g), Pt10atm 109M 103M 0.1atm

A
0.24 V
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B
0.48 V
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C
0.42 V
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D
0.84 V
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Solution

The correct option is C 0.42 V
Cell is equivalent to Pt, H2(g)/H+(aq)//H+(aq)/H2(g), Pt10atm 109M 103M 0.1atm
oxidation half cell:
H2(g)10 atm2H+109 M+2e
reduction half cell:
2H+(aq.)103 M+2eH2(g)0.1 atm
overall reaction:
2H+(aq.)103 M+H2(g)10 atmH2(g)0.1 atm+2H+109 M
apply nearst equation for n=2
E0cell=00.062log(109)2×0.110×(103)2=0.062 log 1014=0.42V

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