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Question

# Find the emf of the following cell. (Take RT ln 10F=0.6 Pt, H2(g)/H+(aq)//H+(aq)/H2(g), Pt10atm 10−9M 10−3M 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 VCell is equivalent to Pt, H2(g)/H+(aq)//H+(aq)/H2(g), Pt10atm 10−9M 10−3M 0.1atm oxidation half cell: H2(g)10 atm→2H+10−9 M+2e− reduction half cell: 2H+(aq.)10−3 M+2e−→H2(g)0.1 atm overall reaction: 2H+(aq.)10−3 M+H2(g)10 atm→H2(g)0.1 atm+2H+10−9 M apply nearst equation for n=2 E0cell=0−0.062log(10−9)2×0.110×(10−3)2=−0.062 log 10−14=0.42V

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