Question #d0cfc

1 Answer
Jun 14, 2016

0.25 mole O2

Explanation:

We should determine which reactions are happening on every electrode.

Note that silver is getting reduced (at the cathode) as it is illustrated in the following reduction equation:
Ag++1eAg ξ=0.80V

And, water is getting oxidized (at the anode) as it is illustrated in the following oxidation equation:
2H2OO2+4H++4e ξ=1.23V

The overall redox reaction would then be:
Reduction: [Ag++1eAg]×4 ξ=0.80V
Oxidation: 2H2OO2+4H++4e ξ=1.23V
Redox: 4Ag++2H2OO2+4H++4Ag ξ=0.43V

The redox equation shows that for every 4 moles of silver reduced, 2 moles of water are oxidized to produce 1 mole of oxygen.

Therefore, the number of mole of oxygen produced when 1 mole of silver is reduced (precipitated):

?molO2=1molAg×1molO24molAg=0.25molO2

Here is a video that explains electrolysis in general:
Electrochemistry | Electrolysis, Electrolytic Cell & Electroplating.