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Calculate the value of the solubility product (Ksp) for CdS given the following standard reduction potentials:

CdS + 2e- --> Cd + S-2 E° = -1.21V
Cd+2 + 2e- --> Cd E° = -0.402V
CdS + Cd --> Cd + Cd+2 + S-2 E° = -0.808
What I tried was:
E = E° - (0.0591/n)log(K)
where E is zero since it is in equilibrium, and n is 2 because there are 2moles of electrons per mole of Cd.

I reversed the net equation because I was looking for the Ksp of CdS.

When I did this, I got 4.53*10^28, but the actual answer was 3.9*10^28.

Please help!
Thanks a lot.

I hope the error was from rounding up. :)

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