Potassium chlorate is prepared by the electrolysis of KCl in basic solution
$$6\text{OH}^- + \text{Cl}^- \to \text{ClO}_3^- + 3\text{H}_2\text{O} + 6e^-$$. If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce $$10\,\text{g}$$ of $$\text{KClO}_3$$ using a current of $$2\,\text{A}$$ is........ (Given: F = 96,500 C/mol; molar mass of $$\text{KClO}_3$$ = 122 g mol$$^{-1}$$)

We have to prepare potassium chlorate, $$\text{KClO}_3$$, by electrolysis. The given anodic half-reaction is

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

This equation tells us, first of all, how many electrons are needed. One mole of $$\text{ClO}_3^-$$ (and hence one mole of $$\text{KClO}_3$$) is produced by the loss of $$6$$ moles of electrons. So the electron requirement is

$$n_{e^-} = 6 \; \text{mol e}^- \; \text{per mol of } \text{KClO}_3$$

Now let us find how many moles of $$\text{KClO}_3$$ we want to make. The mass required is $$10\;\text{g}$$ and the molar mass is $$122\;\text{g mol}^{-1}$$, so

$$n_{\text{KClO}_3} \;=\; \frac{10\;\text{g}}{122\;\text{g mol}^{-1}} = 0.08197\;\text{mol}$$

Next, we convert this amount of product into the corresponding amount of electrons. Multiplying by the factor of $$6$$ from the balanced half-reaction we get

$$n_{e^-} = 0.08197\;\text{mol} \times 6 = 0.4918\;\text{mol e}^-$$

Faraday’s law of electrolysis states that the total charge required is given by

$$Q = n_{e^-}\,F$$

where $$F = 96\,500\;\text{C mol}^{-1}$$ is the Faraday constant. Substituting the value of $$n_{e^-}$$ we obtain

$$Q_{\text{ideal}} = 0.4918\;\text{mol}\times 96\,500\;\text{C mol}^{-1} = 4.7427 \times 10^{4}\;\text{C}$$

However, we are told that only $$60\%$$ of the current actually does the desired work. The current efficiency is therefore $$0.60$$, meaning the real charge that has to pass through the cell is larger than the ideal charge. We divide by the efficiency:

$$Q_{\text{actual}} = \frac{Q_{\text{ideal}}}{0.60} = \frac{4.7427 \times 10^{4}\;\text{C}}{0.60} = 7.9045 \times 10^{4}\;\text{C}$$

Now let us translate this charge into the time required when the current is $$2\;\text{A}$$. The basic relation is

$$Q = I\,t$$

so

$$t = \frac{Q_{\text{actual}}}{I} = \frac{7.9045 \times 10^{4}\;\text{C}}{2\;\text{A}} = 3.9523 \times 10^{4}\;\text{s}$$

To express this time in hours we divide by $$3600\;\text{s h}^{-1}$$, i.e.

$$t = \frac{3.9523 \times 10^{4}\;\text{s}}{3600\;\text{s h}^{-1}} = 10.978\;\text{h}$$

Finally, we round to the nearest whole hour:

$$t \approx 11\;\text{h}$$

Hence, the correct answer is Option 11.