Question

If $3$ faradays of electricity are passed through an iron (II) bromide solution, how many grams of iron metal will be deposited? (At. mass of iron $=56)$.

• A. $80g$
• B. $82g$
• C. $84g$
• D. $90g$

Answer 1

The correct option is C $84g$

Half Cell Reaction:-

$F{e}_{\left(aq\right)}^{2+}+2e^{-}\to F{e}_{\left(s\right)}$.

1 Faraday $\to$ charge on 1 mole of electron.

$\therefore$ 3 Faraday $\to$ charge on 3 mole of electron.

$\to$ 2 mole of electron produces 1 mole of $Fe$.

$\therefore$ 3 mole electron produces $\frac{3}{2}$ mole of $Fe$.

$\therefore \frac{3}{2}mole=\frac{weightofFe\left(W\right)}{AtomicmassofFe}$

$\therefore \frac{3}{2}=\frac{W}{56}$.

$\therefore W=\frac{3\times 56}{2}=84g$.
Option C is correct.