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
B
56 mL
C
112 mL
D
448 mL

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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