|Ag Anode Cathode # At anode, oxidation occur H2→2H++2e− \# At cathode, reduction occur 2Ag++2e−→2Ag Adding equation (1) and (2), we get n=2, where n= cancelled out electron Now, ∆G∘=−nFEcello =−2×96500×0.5332 =−102907.6 =−102.9kJ∕mol =−103kJ∕mol