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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 |
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Answer» Solution :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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