In a copper voltameter,the mass deposited in | vedclass

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(A) The total charge $q$ passed through the voltameter is equal to the area under the current-time graph.Area $= 	ext{Area of trapezium} = \frac{1}{2

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$m/2$

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(A) The total charge $q$ passed through the voltameter is equal to the area under the current-time graph.Area $= 	ext{Area of trapezium} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$Area $= \frac{1}{2} 	imes (10 + 30) 	imes 100 	imes 10^{-3} \, 	ext{A} = \frac{1}{2} 	imes 40 	imes 0.1 = 2 \, 	ext{C}$.According to Faraday's law of electrolysis,$m = z q$,where $z$ is the $ECE$ of copper.Given that the mass deposited is $m$ for a charge $q = 2 \, 	ext{C}$,we have:$m = z 	imes 2$$z = \frac{m}{2}$.

In a copper voltameter,the mass deposited in $30\, \text{s}$ is $m\, \text{g}$. If the current-time graph is as shown in the figure,the Electrochemical Equivalent $(ECE)$ of copper is:

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