E°Cu2+/ cu = + 0.34 V and E°Ag+/Ag = + 0.80 V.Cu+2Ag+→ Cu2++2Ag(s)Applying the nernst equation E = E°-0.0592logCu2+Ag+2E° = E°cathode - E°anode = 0.80 - 0.34 = 0.46 VE = 0.46 - 0.0592log0.1301.0×10-4 = 0.46-0.0592×1.30×107 = 0.46 - 0.21 = 0.25 V.
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