Question

In the Hall–Héroult process for the electrolytic production of aluminium, ${\mathbf{\text{Al}}}^{\mathbf{\text{3+}}}$ions from${\mathbf{\text{Al}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{3}}}$dissolved in molten cryolite$\left({\text{Na}}_{\text{3}}{\text{AlF}}_{\text{6}}\right)$are reduced to$\mathbf{\text{Al(l)}}$while carbon (graphite) is oxidized to${\mathbf{\text{CO}}}_{\mathbf{\text{2}}}$by reaction with oxide ions.

(a) Write balanced equations for the half-reactions at the anode and at the cathode, and for the overall cell reaction.

(b) If a current of$\mathbf{\text{50,000A}}$ is passed through the cell for a period of$\mathbf{\text{24}}$hours, what mass of aluminium will be recovered?

Short answer

1. Balanced equation for the half-cell reaction at the anode –

${\text{C}}_{\text{-}}{\text{(s)+2O}}^{\text{2-}}\text{(aq)}\to {\text{CO}}_{\text{2(g)}}{\text{+4e}}^{\text{-}}$

Balanced equation for the half-cell reaction at the cathode –

${{\text{Al}}^{\text{3+}}}_{\text{(aq)}}{\text{+3e}}^{\text{-}}\to {\text{Al}}_{\text{(l)}}$

The overall cell reaction is –

${\text{3C}}_{\text{-}}{\text{(s)+6O}}_{\text{(aq)}}^{\text{2-}}{{\text{+4Al}}^{\text{3+}}}_{\text{(aq)}}\to {\text{3CO}}_{\text{2(g)}}{\text{+4Al}}_{\text{(l)}}$

(b) If a current of$\text{50,000A}$ is passed through the cell for a period of$\text{24}$ hours, of aluminium will be recovered.

Step-by-step solution

Concept Introduction

An oxidation process in which electrons are lost or a reduction reaction in which electrons are gained is known as a half-cell reaction. The processes take place in an electrochemical cell, where electrons are lost at the anode via oxidation and consumed at the cathode via reduction.

Balanced Equation for half-cell Reaction

The anode half-cell reaction for the oxidation –

${\text{C}}_{\text{-}}{\text{(s)+2O}}^{\text{2-}}\text{(aq)}\to {\text{CO}}_{\text{2(g)}}{\text{+4e}}^{\text{-}}$

The cathode half-cell reaction for the reduction –

${{\text{Al}}^{\text{3+}}}_{\text{(aq)}}{\text{+3e}}^{\text{-}}\to {\text{Al}}_{\text{(l)}}$

Multiplying the first reaction with 3 and the second one with 4 to satisfy number of electrons on both sides –

$\begin{array}{c}{\text{C}}_{\text{-}}{\text{(s)+2O}}^{\text{2-}}\text{(aq)}\to {\text{CO}}_{\text{2(g)}}{\text{+4e}}^{\text{-}}\text{/3}\\ {\text{Al}}_{\text{(aq)}}^{\text{3+}}{\text{+3e}}^{\text{-}}\to {\text{Al}}_{\text{(l)}}\text{/4}\end{array}$

Now knowing these half-reactions, write the overall reaction –

${\text{3C}}_{\text{-}}{\text{(s)+6O}}_{\text{(aq)}}^{\text{2-}}{{\text{+4Al}}^{\text{3+}}}_{\text{(aq)}}\to {\text{3CO}}_{\text{2(g)}}{\text{+4Al}}_{\text{(l)}}$

Therefore, the overall reaction obtained is ${\text{3C}}_{\text{-}}{\text{(s)+6O}}_{\text{(aq)}}^{\text{2-}}{{\text{+4Al}}^{\text{3+}}}_{\text{(aq)}}\to {\text{3CO}}_{\text{2(g)}}{\text{+4Al}}_{\text{(l)}}$.

Mass of Aluminium

The amount of current passed:$\text{I=50,000A}$.

The time period:$\begin{array}{c}{\text{t=24h=24h×3600=86400s}}_{}^{}\\ {\text{1A=1Cs}}^{\text{-}}\text{1h=3600s}\end{array}$.

To calculate$\text{n}$(number of moles) use the formula –

$\text{n=}\frac{\text{I×t}}{\text{z×F}}$

Plug the given data in knowing that$\text{F}$is constant${\text{96485Cmol}}^{\text{-}}$–

$\text{n=}\frac{{\text{50Cs}}^{\text{-}}\text{×86400s}}{{\text{3×96485Cmol}}^{\text{-}}}$

Divide and multiply these numbers and it is obtained –

.$\text{n=14.92mol}$

As it is determined number of moles for$\text{Al}$, plug this data in to calculate mass –

$\text{m(Al)=n×M}$

Molar mass for$\text{Al}$is$\text{26.98g/mol}$–

$\text{m(Al)=14.92mol×26.98g/mol}$

Multiply these numbers and it is obtained –

$\text{m(Al)=403g}$

Therefore, the mass of aluminium obtained is $\text{m(Al)=403g}$.