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What is the emf of the cell when [Ni2+] = 0.281 M and [Zn2+] = 0.895 M?

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The voltaic cell uses the following reaction and operates at 298 K.

Zn(s) + Ni2+(aq) -----> Zn2+(aq) + Ni(s)

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  1. We know that the following relationship exists...

    Eo = (RT/nF)lnK,

    where R= 8.314JK-1mol-1

    T = temperature = 298K

    n = number of electrons transferred = 2

    F = 96500 Jmol-1

    K = [products]/[reactants] = 3.1851

    Plugging everything in to solve for Eo, you should get +0.0149V

    [Answer: see above]

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