If a current of 2A flowing for 1 hour liberates a certain amount of ions, then a current of 8A flowing for the same time will liberate ions

sixteen times

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six times

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eight times

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four times

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Solution

The correct option is B four times
Given, $I_{1} = 2 A$ $I_{2} = 8 A$
$t = t_{1} = t_{2} = 1 m i n$
The mass of a substance liberated at an electrode due to passage of current I is given by
$m = Z I t$
Hence, the mass of a substance liberated at an electrode due to passage of current $I_{1}$ is
$m_{1} = Z I_{1} t_{1}$ ............(1)
and the mass of a substance liberated at an electrode due to passage of current $I_{2}$ is
$m_{2} = Z I_{2} t_{2}$ ............(2)
dividing (2) by (1 ) we get,

$\frac{m_{2}}{m_{1}} = \frac{Z I_{2} t_{2}}{Z I_{1} t_{1}}$

$\frac{m_{2}}{m_{1}} = \frac{8}{2}$
$\Rightarrow m_{2} = 4 m_{1}$

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