electrochemistry

Question: On passing $C$ Ampere of electricity through a electrolyte solution for $t$ second, $m$ gram metal deposits on cathode. The equivalent weight $E$ of the metal is

1) $E=\frac{C\times t}{m\times 96500}$

2) $E=\frac{C\times m}{t\times 96500}$

3) $E=\frac{96500\times m}{C\times t}$

4) $E=\frac{C\times t\times 96500}{m}$

Solution:

Explanation: No Explanation

Faraday's Laws of electrolysis

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