A) \[1.93A\]
B) \[9.65A\]
C) \[19.3A\]
D) \[0.965A\]
Correct Answer: B
Solution :
No. of moles of \[{{H}_{2}}=\frac{1.12}{22400}\] No. of equivalence of hydrogen \[=\frac{1.12\times 2}{22400}={{10}^{-4}}\] No. of Faradays required \[={{10}^{-4}}\] \[\therefore \]Current to be passed in one second \[=96500\times {{10}^{-4}}\] \[=9.65A\]You need to login to perform this action.
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