Cells A and B contain aqueous solutions of fe | vedclass

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(D) According to Faraday's second law of electrolysis, the mass $(m)$ of a substance deposited is given by $m = \frac{M \cdot Q}{n \cdot F}$, where $M

Vedclass · Accepted Answer

$3:2$

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(D) According to Faraday's second law of electrolysis, the mass $(m)$ of a substance deposited is given by $m = \frac{M \cdot Q}{n \cdot F}$, where $M$ is the molar mass, $Q$ is the quantity of electricity, $n$ is the valency factor, and $F$ is Faraday's constant.Since $Q$ and $F$ are constant, $m \propto \frac{M}{n}$.For cell $A$ $(FeCl_2)$, the iron is in the $Fe^{2+}$ state, so $n_A = 2$.For cell $B$ $(FeCl_3)$, the iron is in the $Fe^{3+}$ state, so $n_B = 3$.The ratio of the mass of iron deposited is $\frac{m_A}{m_B} = \frac{M/n_A}{M/n_B} = \frac{n_B}{n_A} = \frac{3}{2} = 3:2$.

Cells $A$ and $B$ contain aqueous solutions of ferrous chloride $(FeCl_2)$ and ferric chloride $(FeCl_3)$ respectively. If the same quantity of electricity is passed through both cells, what is the ratio of the mass of iron deposited in cell $A$ to that in cell $B$?

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