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Cu^(2+)+2e^(-) rarr Cu. For this, graph ...

`Cu^(2+)+2e^(-) rarr Cu`. For this, graph between `E_(red)` versus `ln[Cu^(2+)]` is a straight line of intercept `0.34V`, then the electrode oxidation potential of the half cell `Cu|Cu^(2+)(0.1M)` will be

A

`-0.34+(0.0591)/(2)V`

B

`0.34+0.0591V`

C

`0.34V`

D

`-0.34V`

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The correct Answer is:
A
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