The current strength required to displace $0.1 g$ of $H_{2}$ in $10 s e c$ is:

9.65

amp

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1.988

amp

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198

amp

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965

amp

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Solution

The correct option is D $965$ amp
According to Faraday's law of Electrolysis
Mass deposited $(ω) = \frac{E q u i v a l e n t m a s s (z) \times C u r r e n t (A m p) \times t i m e s (s e c), t}{F a r a d a y c o n s t a n t} 0.1 g = \frac{1 g \times I \times 10 s e c}{96500} I = 965 A$
Thus $965$ Amp current is required to displace $0.1 g$ of $H_{2}$

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