During the industrial extraction of aluminium using the Hall-Héroult process, a steady cur… — JAMB UTME

During the industrial extraction of aluminium using the Hall-Héroult process, a steady current of $96.5\text{ A}$ is passed through an electrolytic cell containing molten alumina ( $Al_2O_3$ ) dissolved in molten cryolite for $5.0\text{ hours}$ . What is the mass of pure aluminium, in grams, deposited at the cathode?

(Take $1\text{ Faraday} = 96,500\text{ C mol}^{-1}$ , Relative atomic mass: $\text{Al} = 27$ )

Answer: 162 g

Answer

The mass of pure aluminium deposited at the cathode is

162\text{ g}

Using

Q = I \times t

, a current of

96.5\text{ A}

for

5.0\text{ hours}

(

18,000\text{ s}

) yields a total charge of

1,737,000\text{ C}

, which equals

18\text{ Faradays}

(

18\text{ moles of electrons}

). Since the reduction of

\text{Al}^{3+}

\text{Al}

requires 3 electrons per atom (

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}

18\text{ moles}

of electrons liberate

6\text{ moles}

of aluminium metal. Multiplying by the molar mass of aluminium (

27\text{ g mol}^{-1}

) gives

162\text{ g}

Step-by-Step Solution

Convert time to seconds and calculate total electric charge transferred

Q = 96.5\text{ A} \times (5.0 \times 3600\text{ s}) = 1,737,000\text{ C}

Electric charge is defined as current multiplied by time in seconds (

Q = I \times t

Calculate the moles of electrons transferred using Faraday's constant

Moles of

e^- = \frac{1,737,000\text{ C}}{96,500\text{ C mol}^{-1}} = 18\text{ moles}

One Faraday (

96,500\text{ C}

) corresponds to the charge carried by one mole of electrons.

Relate moles of electrons to moles of aluminium deposited using the half-equation

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}_{(s)}

, so

3\text{ mol } e^-

produces

1\text{ mol Al}

. Moles of

\text{Al} = \frac{18}{3} = 6\text{ moles}

Aluminium ion

\text{Al}^{3+}

requires three electrons for reduction to metallic aluminium.

Multiply moles of aluminium by its relative atomic mass

\text{Mass} = 6\text{ mol} \times 27\text{ g mol}^{-1} = 162\text{ g}

Mass equals molar amount multiplied by molar mass.

Key Concept

Quantitative application of Faraday's laws of electrolysis in the extraction of metals

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