Animated Solution for Chemistry - Electrochemistry: A solution of ${\text{Ni(NO}}_3{)}_2$ is electrolysed between platinum electrodes using $0.1$ Faraday electricity. How many mole of Ni will be deposited at the cathode?

Depositing Nickel

Imagine you are looking at an electrolytic cell. We have a beaker filled with a beautiful green solution of nickel nitrate, ${\text{Ni(NO}}_3{)}_2$. Submerged in this solution are two inert platinum electrodes. When we connect a battery and pass exactly $0.1\text{ F}$ (Faraday) of electricity through this circuit, a fascinating transformation begins.

Let's focus our attention on the cathode, the negatively charged electrode. The positively charged nickel ions, ${\text{Ni}}^{2+}$, are naturally drawn towards it. Once they reach the surface of the platinum cathode, a reduction reaction takes place.

The equation for this process is:

This simple equation tells us a profound story. It says that every single nickel ion requires exactly two electrons to neutralize its charge and become a solid atom of nickel. In the language of electrochemistry, we say the n-factor for this reduction is $2$.

Now, let's scale this up from single atoms to moles. If one nickel ion needs two electrons, then one mole of nickel ions will need two moles of electrons.

We know from Faraday's laws that the charge of one mole of electrons is exactly one Faraday ($1\text{ F}$). Therefore, to deposit $1\text{ mole}$ of solid nickel, we must supply $2\text{ F}$ of electrical charge.

But the problem states we are only passing $0.1\text{ F}$ of electricity. How much nickel will that deposit? We can use a straightforward unitary method:

If $2\text{ F}$ deposits $1\text{ mole}$ of Ni, Then $0.1\text{ F}$ will deposit $\frac{1}{2}\times 0.1\text{ moles}$ of Ni.

Let's do the final math:

Alternatively, you can use the concept of equivalents. The number of equivalents of a substance discharged at an electrode is always numerically equal to the charge passed in Faradays. So, passing $0.1\text{ F}$ means $0.1\text{ equivalents}$ of ${\text{Ni}}^{2+}$ are discharged.

Since $\text{Moles}=\frac{\text{Equivalents}}{\text{n-factor}}$, we get $\frac{0.1}{2}=0.05\text{ moles}$. Both paths lead us to the same elegant truth. Exactly $0.05$ moles of nickel will beautifully plate the platinum cathode!