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$Al_2O_3$ is reduced by electrolysis at low potentials and high currents. If $4.0\times10^4$ amperes of current is passed through molten $Al_2O_3$ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, at. mass of Al = 27 g $mol^{-1}$)
A
$1.3\times10^4$ g
B
$9.0\times10^3$ g
C
$8.1\times10^4$ g
D
$2.4\times10^5$ g
Detailed Solution
Charge passed: $Q = It = 4.0\times10^4\times6\times3600 = 8.64\times10^8$ C
Moles of electrons = $\frac{Q}{F} = \frac{8.64\times10^8}{96500} = 8953$ mol
Cathode reaction: $Al^{3+} + 3e^- \rightarrow Al$, so 3 mol of electrons deposit 1 mol (27 g) of Al.
Moles of Al = $\frac{8953}{3} = 2984$ mol
Mass of Al = $2984\times27 = 8.06\times10^4$ g
$\approx 8.1\times10^4$ g
Moles of electrons = $\frac{Q}{F} = \frac{8.64\times10^8}{96500} = 8953$ mol
Cathode reaction: $Al^{3+} + 3e^- \rightarrow Al$, so 3 mol of electrons deposit 1 mol (27 g) of Al.
Moles of Al = $\frac{8953}{3} = 2984$ mol
Mass of Al = $2984\times27 = 8.06\times10^4$ g
$\approx 8.1\times10^4$ g