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  4. On passing 3A of electricity for 50 min 1.8 g metal deposits. The equivalent mass of metal is

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Q. On passing $3A$ of electricity for $50$ min $1.8\, g$ metal deposits. The equivalent mass of metal is

ManipalManipal 2012Electrochemistry

Solution:

$ {{W}_{\text{metal}}}=\frac{\text{Eit}}{96500} $
$ =\frac{E\times 3\times 50\times 60}{96500} $
$ E=\frac{96500\times w}{3\times 50\times 60} $
$ E=\frac{96500\times 1.8}{3\times 50\times 60}=19.3 $