Ag^+(aq) + e^- <==> Ag(s)..............E^0 = 0.80V
2H^+ (aq) + 2e^- <==> H2(g)..........E^0= 0.00V
Ni^2+ (aq) + 2e^- <==> Ni(s)...........E^0 = -0.23V
What is the cell voltage for the spontaneous reaction (galvanic cell) between the silver and nickel half reacts? Could someone help me with this problem.
Ni(s) ==> Ni^2+ + 2e ............E = 0.23 v
2Ag^+ + 2e ==> 2Ag(s)............E = 0.80 v
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Ni(s) + 2Ag^+ ==> Ni^2+ + 2Ag(s) E = 1.03 v
Note that I changed the sign of the Ni given in the problem because I wrote it as an oxidation and not a reductiion. Also note that I multiplied the Ag half cell by 2 in order to make the electrons balance (and cancel). I did not multiply the voltage when I did that for the Ni half cell.