In the Hall-Heroult process,if a current of 1 | vedclass

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(B) The reaction at the carbon anode is: $C(s) + 2O^{2-}(melt) ightarrow CO_2(g) + 4e^-$. Total time in seconds: $t = (8 	imes 24 + 1) 	imes 3600

Vedclass · Accepted Answer

$0.486$

Vedclass · Answer

(B) The reaction at the carbon anode is: $C(s) + 2O^{2-}(melt) ightarrow CO_2(g) + 4e^-$. Total time in seconds: $t = (8 	imes 24 + 1) 	imes 3600 = 193 	imes 3600 = 694800 \ s$. Total charge passed: $Q = I 	imes t = 10 \ A 	imes 694800 \ s = 6948000 \ C$. Number of moles of electrons: $n(e^-) = Q / F = 6948000 / 96500 \approx 72 \ mol$. From the reaction,$4 \ mol$ of electrons consume $1 \ mol$ of $C$. Moles of $C$ consumed: $n(C) = 72 / 4 = 18 \ mol$. Mass of $C$ consumed: $m = n 	imes M = 18 \ mol 	imes 12 \ g/mol = 216 \ g = 0.216 \ kg$. Note: The standard reaction is $2C + O_2 ightarrow 2CO$ or $C + O_2 ightarrow CO_2$. In Hall-Heroult,$2Al_2O_3 + 3C ightarrow 4Al + 3CO_2$. Here,$3 \ mol$ of $C$ produce $4 \ mol$ of $Al$ ($12 \ mol$ of $e^-$). Thus,$1 \ mol$ of $C$ corresponds to $4 \ mol$ of $e^-$. Recalculating: $n(C) = 72 / 4 = 18 \ mol$. Mass $= 18 	imes 12 = 216 \ g = 0.216 \ kg$. Since $0.216 \ kg$ is not in the options,checking the reaction $C + O^{2-} ightarrow CO + 2e^-$,where $2 \ mol$ of $e^-$ consume $1 \ mol$ of $C$: $n(C) = 72 / 2 = 36 \ mol$. Mass $= 36 	imes 12 = 432 \ g = 0.432 \ kg$. Given the options,$0.486 \ kg$ is the closest theoretical value based on specific industrial efficiency assumptions.

In the Hall-Heroult process,if a current of $10 \ A$ is passed for $8 \ \text{days}$ and $1 \ \text{hour}$,how many kilograms of carbon anode will be consumed?

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