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The EMF of the cell Ni|Ni^(2+) (0.01M) ...

The EMF of the cell `Ni|Ni^(2+)` (0.01M)
`Cl^(-)(0.01M)//Cl_(2)`, pt is__ V if the SRP of nickel and chlorine electrodes are `-0.25` and `+1.36 V` respectively

A

`+1.61`

B

`-1.61`

C

`+1.79`

D

`-1.79`

Text Solution

Verified by Experts

The correct Answer is:
C
`E_("cell") = E_("cell")^(@) -(0.059)/2 log [Ni^(2+)][Cl^(-)]`
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