Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.
6 OH$$^-$$ + Cl$$^- \to$$ ClO$$_3^-$$ + 3H$$_2$$O + 6e$$^-$$
A current of xA has to be passed for 10 h to produce 10.0 g of potassium chlorate. The value of x is ___. (Nearest integer)
(Molar mass of KClO$$_3$$ = 122.6 g mol$$^{-1}$$, F = 96500 C)

The electrolytic reaction for preparation of potassium chlorate is:

$$6\text{OH}^- + \text{Cl}^- \rightarrow \text{ClO}_3^- + 3\text{H}_2\text{O} + 6e^-$$

This shows that 6 electrons are transferred per mole of $$\text{KClO}_3$$ produced. First, calculate the moles of $$\text{KClO}_3$$ needed:

$$n = \frac{10.0}{122.6} = 0.08157 \text{ mol}$$

The total charge required:

$$Q = n \times 6 \times F = 0.08157 \times 6 \times 96500 = 47,227 \text{ C}$$

The current required to deliver this charge in $$t = 10 \text{ h} = 10 \times 3600 = 36000 \text{ s}$$:

$$I = \frac{Q}{t} = \frac{47227}{36000} \approx 1.31 \text{ A}$$

Rounding to the nearest integer, the value of $$x$$ is $$\mathbf{1}$$ A.