If I take the electrodes as nickel and silver, I've according to the nernst equation that

$\ce{Ni(s) + 2Ag+(aq) -> Ni^2+(aq) + Ag(s)}$

Now $E=E_{0} - \frac{RT}{2F} \ln\left(\frac{[\ce{Ni^{2+}}]}{[\ce{Ag+}]^2} \right)$

But if I double the stoichiometric coefficients, the value inside the logarithm would square up. So if I write the same reaction but with different coefficients, I'll get different values of potential of the cell?


1 Answer 1


No, you won't get a different value. The number of electrons transfered in that case is 4, so it will be $E=E_{0} - \frac{RT}{4F} \ln\left(\frac{[\ce{Ni^{2+}}]^{2}}{[\ce{Ag+}]^{4}} \right)$.


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