A solution of $$Fe_2(SO_4)_3$$ is electrolyzed for '$$x$$' min with a current of $$1.5$$ A to deposit $$0.3482$$ g of Fe. The value of $$x$$ is ______ [nearest integer]
Given : $$1 F = 96500$$ C mol$$^{-1}$$. Atomic mass of Fe $$= 56$$ g mol$$^{-1}$$

We need to find the time required to deposit 0.3482 g of Fe from a solution of $$Fe_2(SO_4)_3$$.

$$Fe^{3+} + 3e^- \to Fe$$

The n-factor for $$Fe^{3+}$$ is 3 (three electrons are required per iron atom).

$$\text{Moles of Fe} = \frac{0.3482}{56} = 0.006218 \text{ mol}$$

$$Q = n \times F \times z$$

where $$n$$ = moles of Fe, $$z$$ = number of electrons = 3, $$F$$ = 96500 C/mol

$$Q = 0.006218 \times 3 \times 96500$$

$$Q = 0.018654 \times 96500$$

$$Q = 1800.1 \text{ C}$$

$$Q = I \times t$$

$$t = \frac{Q}{I} = \frac{1800.1}{1.5} = 1200.1 \text{ s}$$

$$t = \frac{1200.1}{60} = 20.0 \text{ min}$$

The value of $$x$$ is 20 minutes.