In a copper voltameter experiment,current is | vedclass

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(A) According to Faraday's first law of electrolysis,the mass of substance deposited $(m)$ is given by $m = Z \cdot I \cdot t$,where $Z$ is the electr

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(A) According to Faraday's first law of electrolysis,the mass of substance deposited $(m)$ is given by $m = Z \cdot I \cdot t$,where $Z$ is the electrochemical equivalent,$I$ is the current,and $t$ is the time.In the first case,$m_1 = Z \cdot I \cdot t$.In the second case,the current becomes $I' = \frac{I}{4}$ and the time becomes $t' = 4t$.Substituting these values,$m_2 = Z \cdot (\frac{I}{4}) \cdot (4t) = Z \cdot I \cdot t$.Therefore,$m_2 = m_1$,which means the amount of copper deposited remains the same.

In a copper voltameter experiment,current is decreased to one-fourth of the initial value but it is passed for four times the earlier duration. The amount of copper deposited will be:

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