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Question

E0Zn2+/Zn=0.76V The EMF of the cell

Zn/Zn2+(1M)||HCl(pH=2)|H2(1atm),Pt is ::


A
0.878V
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B
0.642V
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C
0.875V
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D
0.700V
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Solution

The correct option is B 0.642V
Given
log[H+]=2
[H]+=102

2e+Zn2+Zn E=0.76

Zn+2H+Zn2++H2

Ecell=(0.76)0.0592log[Zn2+][H+]2

=0.760.0592×2pH

=0.760.0592×2×2

=0.642V

Option B is correct.

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