Compute the cell potential at 25 ∘C.?
Oxidation:Reduction:Pb(s)→ Pb2+(aq,0.12M)+2e−,E∘=−0.13V
MnO4−(aq,1.55M)+4H+(aq,2.0M)+3e−→MnO2(s)+2H2O(l),E∘=1.68V
I tried 1.81-(0.059/6)(log(.12^3)/(1.55^2)(2.0^8) but this was wrong.
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your rxns are
.. Pb(s) ---> Pb(2+) + 2e's
.. Mn(7+) + 3e's --> Mn(4+)
balancing e's and combining gives this.
.. 3 Pb(s) + 2 Mn(7+) + 6e's --> 3 Pb(2+) + 6e's + 2 MnO2(s)
why do you have [H+]^8 in your equation?