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Question

Calculate the volume of gas liberated at the anode at STP during the electrolysis of a $$CuSO_{4}$$ solution by a current of $$1\ A$$ passed for $$16$$ minutes and $$5$$ seconds.


A
224 mL
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B
56 mL
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C
112 mL
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D
448 mL
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Solution

The correct option is A $$56\ mL$$
Given:-
Current, $$I=1A$$

Time, $$t$$=$$16$$ min $$5$$ sec= $$965$$ seconds

$$n=2$$

$$Cu^{+2}+2e^-\longrightarrow Cu$$

$$4OH^{-1}\longrightarrow  O_{2} + 2H_{2}O +4e^{-1}$$

$$W=\cfrac {I\times t\times M}{nF}$$

$$\Rightarrow W= \cfrac {1\times 965\times 63.5}{2\times 96500}$$

$$\Rightarrow W= 0.3175g$$

$$ 2\times63.5g\longrightarrow 22.4\  litre O_{2\ }$$

$$\Rightarrow 0.3175g \longrightarrow \cfrac {0.3175}{63.5\times2}\times 22.4$$

Volume of gas liberated during electrolysis = $$56\ ml$$

Chemistry

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