Chemistry - Electrochemistry Question with Solution | TestHub

Volume of $O_2$ displaced at STP = $5600$ mL = $5.6$ L.

Moles of $O_2$ = $\frac{5.6}{22.4}=0.25$ mol.

The reaction for the liberation of $O_2$ is:
$2H_{2}O\to O_2+4H^{+}+4e^{-}$

Number of moles of electrons (Faradays) required for $0.25$ mol of $O_2$ = $0.25\times 4=1$ F.

The reaction for the deposition of silver is:
$A{g}^{+}+e^{-}\to Ag$

Since $1$ F of electricity deposits $1$ mole of Ag, the mass of Ag deposited is:
$1\times 108=108.0$ g.

Answer: $108.0\text{g}$