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The `0.1M` copper sulphate solution in which copper electrode is dipped at `25^(@)C`. Calculate the electrode potential of copper electrode. [Given : `E^(0) Cu^(+2)//Cu = 0.34V]`

Answer» `Cu^(+2) +2e^(-) rarr Cu`
`E_(red) = E_(red)^(0) - (0.0591)/(n) log .(["Product"])/(["Reactant"])`
Here `n = 2`
So `E = 0.34 -(0.0591)/(2) log 10`
`= 0.34 - 0.03 = 0.31` volts


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