If current (I) of an electrolytic cell is dou | vedclass

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(D) According to Faraday's first law of electrolysis,the mass $(w)$ deposited is given by $w = Z 	imes I 	imes t$,where $Z$ is the electrochemical e

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remains same

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(D) According to Faraday's first law of electrolysis,the mass $(w)$ deposited is given by $w = Z 	imes I 	imes t$,where $Z$ is the electrochemical equivalent.Since $Z = \frac{	ext{eq. wt.}}{96500}$,we can write $\frac{w}{	ext{eq. wt.}} = \frac{I 	imes t}{96500}$.Let the initial current be $I_1 = I$ and time be $t_1 = t$. The initial ratio is $R_1 = \frac{I 	imes t}{96500}$.If the current is doubled $(I_2 = 2I)$ and time is halved $(t_2 = \frac{t}{2})$,the new ratio is $R_2 = \frac{(2I) 	imes (t/2)}{96500} = \frac{I 	imes t}{96500}$.Therefore,$R_1 = R_2$,which means the ratio remains the same.

If current $(I)$ of an electrolytic cell is doubled and time $(t)$ for electrolysis is halved,then the $\frac{\text{mass}}{\text{eq. wt.}}$ ratio becomes :-

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