A) 0.0836 L
B) 0.0432 L
C) 0.1672 L
D) 0.836 L
Correct Answer: D
Solution :
0.836L Quantity of electricity passed \[=\left( 4A \right)\times \left( 30\times 60s \right)=7200C\] \[{{\operatorname{H}}_{2}}O\to {{\operatorname{H}}_{2}}+\frac{1}{2}{{O}_{2}}\] \[\operatorname{i}.e. 2{{H}^{+}}+2{{e}^{-}}\to {{\operatorname{H}}_{2}}\] \[2 \times 96500C liberate {{H}_{2}}=22.42 at STP\] \[\therefore 7200C liberate = \frac{22.4}{2\times 96500} \times 7200 = 0.836 L\]
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