Question

${Al}_2{O}_3$ is reduced by electrolysis at low potential and high currents. If $4.0\times {10}^4$ amperes of current is passed through molten ${Al}_2{O}_3$ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, At. mass of $Al=27g{mol}^{-1}$)

• A. $1.3\times {10}^{4}g$
• B. $9.0\times {10}^{3}g$
• C. $8.1\times {10}^{4}g$
• D. $2.4\times {10}^{5}g$

Answer 1

Answer: (C)

$8.1\times {10}^{4}g$

$A{l}_2{O}_3\left(l\right)\to 2A{l}^{3+}\left(l\right)+3O^{2-}\left(l\right)$
Reduction equation of $A{l}^{3+}:$
$A{l}^{3+}\left(aq\right)+3e^{-}\to Al\left(s\right)$
As,

Ampere = Coulomb / sec
Time unit should be second.
$6hr\times 60\frac{min}{hr}\times 60\frac{sec}{min}=21600sec$
Coulomb = Ampere x second $=40000\times 21600=8.64\times {10}^8$
1 mol $e^{-}$ carries a charge of 1 faraday, or 96485 coulombs.
Then the mole of $e^{-}$ used in the electrolysis will be:
$\frac{8.64\times {10}^8\text{coulombs}}{96485-\text{coulombs/mol}}-of-e^{-}=8954.8\text{mol of}-e^{-}$
According to the reduction equation, 3 moles of e- are needed to reduce 1 mol of Al^3+ to Al metal.
Then , the mole of Al to be deposited at the cathode will be;
$\frac{8954.8\text{mol e-}}{\frac{3-\text{mol of e-}}{\text{mol Al}}}=2985\text{mol Al}$
Molar mass of $Al:26.98g/mol$
Mass of Al produced$:2985mol\times 26.98g/mol=80535g$
$Al=8.1\times {10}^{4}g$