PROBLEM: One half-cell in a voltaic cell is constructed from a silver wire dipped into a 0.25 M solution of AgNO3. The other half-cell consists of a zinc electrode in a 0.010 M solution of Zn(NO3)2. Calculate the cell potential. From the table of standard reduction potentials: Zn2+(aq) + 2 e–  Zn(s) E° = –0.76 V Ag+(aq) + e–  Ag(s) E° = +0.80 V More positive reduction potential is the cathode First balance the reaction and calculate the E°. Zn(s)  Zn2+(aq) + 2 e– E° = +0.76 V 2[Ag+(aq) + e–  Ag(s)] E° = +0.80 V Zn(s) + 2 Ag+(aq)  Zn2+(aq) + 2 Ag(s) Eºcell = Eºcathode + Eºanode = (0.80 V) + (+0.763 V) = +1.56 V Next use the Nerst equation to calculate the voltage under non-standard conditions