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Derive expression to calculate emf (reduction) of the following half cells at 25^@C Cu^(2+)|Cu(s) |
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Answer» SOLUTION :`Cu^(2+)+2E leftrightarrow Cu(s)` (REDUCTION) `E_(Cu^(2+),Cu)=E_(Cu^(2+),Cu)-0.0591/2 LOG""([Cu])/([Cu^(2+)])` `=E_(Cu^(2+),Cu) +0.0591/2 log[Cu^(2+)]` |
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