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Calculate the emf of the cell in which t...

Calculate the emf of the cell in which the following reaction takes place :
`Ni(s) + 2Ag^(+) (0.002 M) to Ni^(2+) (0.160 M) + 2Ag (s)`
Given that: `E_("cell")^(@) = 1.05` V

Text Solution

Verified by Experts

Applying Nernst equation and substituting the values, we get
`E_("cell") = E_("cell")^(@) -(0.0591)/n log ([Ni^(2+)])/([Ag^(+)])^(2) = 1.05 - 0.0591/2 (4.6021) = 1.05 - 0.14 V = 0.91 V`
Thus emf of the cell = 0.91 V.
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