Question

Question #d0cfc

Answer 1

0.25 mole `O_2`

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^(+)+1e^(-)->Ag" " " " " " " " " "xi^@=0.80V`

And, water is getting oxidized (at the anode) as it is illustrated in the following oxidation equation:
`2H_2O->O_2+4H^(+)+4e^(-)" "-xi^@=-1.23V`

The overall redox reaction would then be:
Reduction: `color(blue)("[")Ag^(+)+1e^(-)->Agcolor(blue)("]"xx4)" " " " " " "xi^@=0.80V`
Oxidation: `2H_2O->O_2+4H^(+)+4e^(-)" "-xi^@=-1.23V`
Redox: `color(blue)(4)Ag^(+)+2H_2O->O_2+4H^(+)+color(blue)(4)Ag" "xi^@=-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):

`?molO_2=1cancel(molAg)xx(1molO_2)/(4cancel(molAg))=color(green)(0.25molO_2)`

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