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  4. What will be the proportion of moles of metal (Cu: Ni :Ag) at cathode according to the second law of Faraday?

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Q. What will be the proportion of moles of metal $ (Cu: Ni :Ag) $ at cathode according to the second law of Faraday?

ManipalManipal 2008Electrochemistry

Solution:

According to Faraday's second law of electrolysis "when the same quantity of electricity is passed through solutions of different electrolytes connected in series, the weights of substances produced at the electrodes are directly proportional to their equivalent weights".
If $M$ stands for atomic weight of substances
Equivalent weight of $Cu =\frac{M}{2}$
Equivalent weight of $Ni =\frac{M}{2}$
Equivalent weight of $Ag =\frac{M}{1}$
So, the proportion of moles of metals
$\frac{M}{2}: \frac{M}{2}: \frac{M}{1} $
$\therefore 1: 1: 2$