Question

A solution of $Ni{(N{O_3})_2}$ is electrolysed between platinum electrodes using a current of $5$ amperes for $20$ minutes. What mass of $Ni$ is deposited at the cathode?

Answer 1

First we should find the charge passed through, with the help of the values of current and time given. Then we can write the equation for this reaction and get the mass of nickel deposited when the number of moles of electrons in the reaction are used, and then compare this with our actual charge passed to get the final result.
Formulas used: $Q = It$
Where $Q$ is the charge, $I$ is the current and $t$ is the time in seconds.

Complete answer:
We know that charge passed through a solution is the product of the current and the time for which the current is passed. That is:
$Q = It$
Where $Q$ is the charge, $I$ is the current and $t$ is the time in seconds.
Here, it is given that $I = 5A$ and $t = 20\min = 20 \times 60 = 1200s$. Substituting this, we get:
$Q = 5 \times 1200 = 6000C$
Now let us take a look at the reaction involved:
The nickel ion, which is in $+ 2$ oxidation state, is being reduced to solid nickel by accepting $2$ electrons. Thus:
$N{i^{2 + }}(aq.) + 2{e^ - } \to Ni(s)$
As we can see, two moles of electrons are needed to produce one mole of nickel, whose molar mass is $58.7g/mol$. As we know, one mole of electrons carries a charge of $96487C$, otherwise known as ‘Faraday’. Therefore, two moles will carry a charge of $(96487 \times 2) = 1,92,974C$. Thus, if we take mass deposited by passing $6000C$ as $x$:

Charge passed	Mass of nickel produced
$1,92,974C$	$58.7g/mol$
$6000C$	$x$

Using cross -multiplication rule, we have:
$x = \dfrac{{58.7 \times 6000}}{{1,92,974}}$
On solving this, we get:
$x = 1.825g/mol$
Thus, the mass of nickel produced is $1.825g$.

Note:
The equations we have used here are derived from the Faraday’s laws of electrolysis, which gives us a quantitative relationship between the charge passed and the mass of the element produced. Note that electrolysis is the process of splitting up aqueous solutions of ionic compounds through the passage of electricity. Here, as the nickel is being reduced, it will get deposited at the cathode.