# Question d0cfc

Jun 14, 2016

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:
$A {g}^{+} + 1 {e}^{-} \to A g \text{ " " " " " " " " } {\xi}^{\circ} = 0.80 V$

And, water is getting oxidized (at the anode) as it is illustrated in the following oxidation equation:
$2 {H}_{2} O \to {O}_{2} + 4 {H}^{+} + 4 {e}^{-} \text{ } - {\xi}^{\circ} = - 1.23 V$

The overall redox reaction would then be:
Reduction: color(blue)("[")Ag^(+)+1e^(-)->Agcolor(blue)("]"xx4)" " " " " " "xi^@=0.80V
Oxidation: $2 {H}_{2} O \to {O}_{2} + 4 {H}^{+} + 4 {e}^{-} \text{ } - {\xi}^{\circ} = - 1.23 V$
Redox: $\textcolor{b l u e}{4} A {g}^{+} + 2 {H}_{2} O \to {O}_{2} + 4 {H}^{+} + \textcolor{b l u e}{4} A g \text{ } {\xi}^{\circ} = - 0.43 V$

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.