If the mass deposited in 30 s is m,what will | vedclass

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(B) The charge $Q$ passed through the voltameter is equal to the area under the $i-t$ graph.$Q = 	ext{Area of the trapezium} = \frac{1}{2} 	imes (

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$\frac{m}{2}$

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(B) The charge $Q$ passed through the voltameter is equal to the area under the $i-t$ graph.$Q = 	ext{Area of the trapezium} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$$Q = \frac{1}{2} 	imes (10 + 30) 	imes 100 	imes 10^{-3} \, A \cdot s = 2 \, C$.According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = z \cdot Q$,where $z$ is the $E.C.E$ (Electrochemical Equivalent).Therefore,$z = \frac{m}{Q} = \frac{m}{2}$.

If the mass deposited in $30 \ s$ is $m$,what will be the $E.C.E$ of the copper voltameter?

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