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Question

At pH=2, EQuinhydrone will be:

(EoQuihydrone=1.30V)

1067625_9f90f2319e9b4e0893931807ead38b21.png

A
1.36V
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B
1.30V
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C
1.42V
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D
1.20V
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Solution

The correct option is A 1.36V
Given: PH=2
Given: E0=1.30VPH=log[H+1][H+1]=102
Applying Nernst Equation,
E=E00.0592log[H+1]E=1.300.0592log[102]E=1.300.0592×(2)E=1.36V

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