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How cud i solve this problem.can somebody plz help me.thnx

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The Ksp of Zn(OH)2 is 1.8*10^-14.Find Ecell for the follwing half-reaction.

Zn(OH)2(s) 2e- = Zn(s) 2OH-(aq)

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  1. the Nernst equation:

    E = Eo - (0.0592 / n)(log K)

    at equilibrium becomes:

    0 = Eo - (0.0592 / n)(log K)

    Eo =-+ (0.0592/2e-)(log 1.8e-14)

    Eo =  (0.0592/2e-)(-13.74)

    Eo =  - 0.407

    your answer is: Eo (2 sig figs)= - 0.41 volts

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