In the Hall-Heroult cell, a current of 100 A is passed for 9650 s with 100% current efficiency. Al is deposited at the cathode (Al^3+ + 3e- -> Al), and the graphite anode is oxidised only to CO2 (C + 2O^2- -> CO2 + 4e-). With F = 96500 C/mol, Al=27, C=12, the masses of Al deposited and carbon consumed are respectively

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In the Hall-Heroult cell, a current of $100 A$ is passed for $9650 s$ with $100%$ current efficiency. Al is deposited at the cathode ( $Al X^{3 +} + 3 e X^{-} Al$ ), and the graphite anode is oxidised only to $CO X_{2}$ ( $C + 2 O X^{2 -} CO X_{2} + 4 e X^{-}$ ). With $F = 96500 C/mol$ , $Al = 27$ , $C = 12$ , the masses of Al deposited and carbon consumed are respectively

Q = 100 \times 9650 = 9.65 \times 1 0^{5} C

, so

n (e^{-}) = 10 mol

. Al:

10/3 mol \times 27 = 90 g

. Carbon: 4 mol

e^{-}

per mol C, so

10/4 = 2.5 mol

= 30 g

. (If CO were formed, 2

e^{-}

per C would give

60 g