[DutchGirl13]

For this problem, I am supposed to find E^o for the half reaction Pd(OH)₂ +2e- ::equil::Pd_s +2OH^- given that the Ksp for Pd(OH)₂ is 3E-28 and E^o=0.915 V for the half reaction Pd²⁺+2e- ::equil::Pd_(s).
I put the Ksp and the n=2 in the Nernst equation and solved for E^o and got -0.8141 V. Next, I flipped the given half reaction to cancel the electrons and therefore changed the sign on the potential. Using the -0.8141 V for my cell voltage and -0.915 as my anode voltage, I calculated the voltage of the cathodic half reaction to be -1.73. This answer does not sit well with me. Is a negative answer okay? When I flip the known half reaction around, everything in it cancels with something in the unknown equation, and that did not happen in the sample problem we did in class. Have I done something wrong?

[Pradeep]

Yes. You have done a wrong calculation.

[FreeTheBee]

I don't think flipping and anodes are necessary. Try to write the Nernst equation for Pd²⁺  Pd(s)  and then modify it to take the given Ksp into account. Finally, think of how the standard potential is defined.