If the voltage of the cell is -0.1288V at 25Celsius when [Pb2+]=0.1403M and the pressure of H2=1atm, what is the pH

Question

Pb(s)+2H+(aq)<-> Pb2+(aq) +H2(g) E0 = 0.1260V

Answer 1

Ecell = Eocell - (0.05916/2)log(Pb^2+)*PH2/(H^+)^2
Substitute and solve for H^+, then convert to pH.

Answer 2

Thank you DrBob222