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

1.22 g

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0.44 g

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0.95 g

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0.33 g

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Solution

The correct option is C $0.95 g$
Electrolysis of $A g N O_{3}$
At cathode:
$A g^{+} (a q) + e^{-} \to A g (s)$
At anode:
$2 H_{2} O (l) \to 4 H^{+} (a q) + O_{2} (g) + 4 e^{-}$
For 1 mole of $O_{2}$ , 4 Faraday charge is released
At NTP,
$1 m o l of O_{2} = 22.4 L$
$1 L = \frac{1}{22.4} mole of O_{2}$
$0.05 L = \frac{0.05}{22.4} = 0.0022 mole of O_{2}$
For 1 mole of $O_{2}$ , 4 Faraday charge is released
For 0.0022 mole,
$= 4 \times 0.0022 F$
$= 0.0088 F$
$= 0.0088 \times 96500$
$\approx 850 C$
$Charge released, Q = 850 C$
For cathode reaction,
$A g^{+} + e^{-} \to A g$
On passing $1 F or 96500 C$ of charge, one mole of $A g$ or $108 g$ of $A g$ is produced.
$∴$

$850 C of charge gives,$

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

$= 0.95 g$

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

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How many grams of silver would be deposited at cathode if 0.05 dm3 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)

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