Question

Calculate the number of coulombs required to deposit $6.75g\,of\,Al$ when the electrode reaction is:
$A{l^{3 + }} + 3{e^ - } \to Al$

Answer 1

Here we need to know the Faraday’s first law of electrolysis. It states that the amount of chemical reaction that occurs at any electrode during electrolysis by a current is directly proportional to the amount of electricity that is passed through the electrolyte.

Complete step by step answer:
In an electrolytic cell we use an external source of voltage to carry out a chemical reaction. Metals like sodium, magnesium, aluminium are produced industrially on a large scale by the electrochemical reduction of their respective cations since no suitable chemical reducing agents are available to carry out this job.
According to Faraday we can give the relation between amount of electricity passing through an electrolytic cell and the current by the equation-
$Q = It$
where $Q$ is the amount of electricity passing through the cell in $coulombs$, $I$is the current in $amperes$ and $t$ is the time in ${\text{seconds}}$ .
Now let us look at our reaction:
$A{l^{3 + }} + 3{e^ - } \to Al$
Here we require three mole of electrons for the reduction of aluminium ions. We know that charge on one electron is $1.6 \times {10^{ - 19}}coulomb$ . So the charge on one mole of electron will be-
$ch\arg e = {N_A} \times 1.6 \times {10^{ - 19}} = 6.023 \times {10^{ - 19}} \times 1.6 \times {10^{ - 19}} = 96487C$
This quantity is called Faraday and is represented by the symbol $F$.
So we can say that one mole of electrons has a charge of $96487C$. Mostly for approximate calculation we take $1F = 96500C$ .
Now for the reduction of $A{l^{3 + }}$ ion we require three mole of electrons or $3Faraday$.
$3F = 3 \times 96500C$
The given electrode reaction states that for deposition of one mole of aluminium or $27g\,of\,Al$ , we require a charge of $3 \times 96500C$ .
So for depositing $6.75g\,of\,Al$ we would require charge of-
Charge required $= {{6.75 \times 3 \times 96500}}{{27}} = 72375C$
So a charge of $72375C$ is required to deposit $6.75g\,Al$ .

Note: The amount of electricity required for the oxidation or reduction depends on the stoichiometry of the electrode half-reaction. Aluminium is produced by the electrolysis of aluminium oxide in the presence of cryolite. Remember charge is expressed in $coulomb(C)$ , current in $amperes(A)$ .
$1F = 96500C$