Question

Sample of $Al_2O_3$ in a molten fluoride bath is electrolysed using a current of 1.55 amperes using graphite rods. Oxygen liberated at anode forms x g $CO_2$ in 2 hours. Find x (report to nearest integer), if current efficiency is 80%

Answer 1

1

Answer 2

The problem involves the electrolysis of $Al_2O_3$ using graphite anodes, where oxygen liberated reacts with the anode to form $CO_2$. We need to calculate the mass of $CO_2$ formed considering the current efficiency.

1. Anode Reaction: During the electrolysis of molten $Al_2O_3$ using graphite (carbon) anodes, oxygen ions ($O^{2-}$) from $Al_2O_3$ migrate to the anode and react with carbon to form carbon dioxide. The reaction at the anode is: $C(s) + 2O^{2-}(melt) \rightarrow CO_2(g) + 4e^-$ This equation shows that 4 moles of electrons are required to produce 1 mole of $CO_2$.

2. Calculate Total Charge Passed (Q): Given current (I) = 1.55 amperes Given time (t) = 2 hours Convert time to seconds: $t = 2 \text{ hours} \times 60 \text{ min/hour} \times 60 \text{ s/min} = 7200 \text{ s}$ The total charge passed is given by $Q = I \times t$: $Q = 1.55 \text{ A} \times 7200 \text{ s} = 11160 \text{ C}$

3. Calculate Actual Charge Used for Reaction ($Q_{actual}$): Given current efficiency = 80% = 0.80 $Q_{actual} = Q \times \text{efficiency}$ $Q_{actual} = 11160 \text{ C} \times 0.80 = 8928 \text{ C}$

4. Calculate Moles of Electrons ($n_e$): Faraday's constant (F) = 96485 C/mol (approximately 96500 C/mol for calculations) $n_e = \frac{Q_{actual}}{F}$ Using F = 96485 C/mol: $n_e = \frac{8928 \text{ C}}{96485 \text{ C/mol}} \approx 0.09253 \text{ mol}$

5. Calculate Moles of $CO_2$ Produced ($n_{CO_2}$): From the anode reaction ($C + 2O^{2-} \rightarrow CO_2 + 4e^-$), 4 moles of electrons produce 1 mole of $CO_2$. $n_{CO_2} = \frac{n_e}{4}$ $n_{CO_2} = \frac{0.09253 \text{ mol}}{4} \approx 0.02313 \text{ mol}$

6. Calculate Mass of $CO_2$ (x): Molar mass of $CO_2$ ($M_{CO_2}$) = Atomic mass of C + 2 $\times$ Atomic mass of O $M_{CO_2} = 12.011 \text{ g/mol} + 2 \times 15.999 \text{ g/mol} = 44.01 \text{ g/mol}$ Mass of $CO_2$ ($x$) = $n_{CO_2} \times M_{CO_2}$ $x = 0.02313 \text{ mol} \times 44.01 \text{ g/mol} \approx 1.018 \text{ g}$

7. Report to Nearest Integer: Rounding 1.018 g to the nearest integer gives 1 g.