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Question

Zn(s)+2H+(aq)Zn2+(aq)+H2(g)
The above equation has an equilibrium constant Keq=1025. What is the value of standard EMF for the reaction.
(Given : 2.303RTF=0.0591 V)

A
0.739V
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B
1.478 V
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C
0.739 V
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D
1.478 V
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Solution

The correct option is A 0.739V
Zn(s)+2H+(aq)Zn2+(aq)+H2(g)
The reaction is involves 2 electron change.
n=2
We know
E=RTnF×2.303log(k)E=0.05912×log1025=0.739 V

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