How many grams of silver would be deposited at cathode if $0.05d{m}^3$ of oxygen measured at NTP is liberated at the anode when silver nitrate solution is electrolysed between platinum electrodes?
(Given: Atomic mass of $Ag=108,O=16$)

The correct option is C $0.95g$
Electrolysis of $AgN{O}_3$
At cathode:
$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$
At anode:
$2H_{2}O\left(l\right)\to 4H^{+}\left(aq\right)+O_2\left(g\right)+4e^{-}$
For 1 mole of $O_2$, 4 Faraday charge is released
At NTP,
$1mol\text{of}{O}_2=22.4L$
$1L=\frac{1}{22.4}\text{mole of}{O}_2$
$0.05L=\frac{0.05}{22.4}=0.0022\text{mole of}{O}_2$
For 1 mole of $O_2$, 4 Faraday charge is released
For 0.0022 mole,
$=4\times 0.0022F$
$=0.0088F$
$=0.0088\times 96500$
$\approx 850C$
$\text{Charge released, Q}=850C$
For cathode reaction,
$A{g}^{+}+e^{-}\to Ag$
On passing $1F\text{or}96500C$ of charge, one mole of $Ag$ or $108g$ of $Ag$ is produced.
$\therefore$

$850C\text{of charge gives,}$

$=\frac{850\times 108}{96500}$

$=0.95g$

Thus, $0.95g$ of silver is deposited at cathode.