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Cu^(2+) + 2e^(-) rightarrow Cu. For the ...

`Cu^(2+) + 2e^(-) rightarrow Cu.` For the this, graph between`E_(red)` versus In `([CU^(2+)])` is a straight line of intercept `0.34V` then electrode oxidation potential of the half - cell `(Cu) //(Cu^(2+)(0.1M)` will be: (at 298 Kelvin )

A

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

B

`0.34+0.0591V`

C

`0.34V`

D

`-0.34V`

Text Solution

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The correct Answer is:
A
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