Question

An electrolytic cell consists of a pair of inert metallic electrodes in a solution buffered to $\mathbf{pH}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}$and containing nickel sulphate $\left({\mathrm{NiSo}}_4\right)$ at a concentration of 1.00 M. A current of 2.00 A is passed through the cell for 10.0 hours.

(a)What product is formed at the cathode?

(b)What is the mass of this product?

(c)If the pH is changed to $\mathbf{pH}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}$, what product willform at the cathode?

Short answer

1. Nickel is formed at the cathode.
2. The mass of the product is 21.9 g.
3. If the pH changes to 1.0, the formed product will be hydrogen gas.

Step-by-step solution

Finding the product

Usually, reduction occurs at the cathode, and oxidation occurs at the anode.

At cathode, ${\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Ni}\left(\mathrm{s}\right)$

The half-cell potential, $\mathrm{E}\left(\left.{\mathrm{Ni}}^{2+}\right|\mathrm{Ni}\right)={\mathrm{E}}^0\left(\left.{\mathrm{Ni}}^{2+}\right|\mathrm{Ni}\right)-\frac{0.0592}{\mathrm{n}}{\log }_{10}\mathrm{Q}$, ${\mathrm{E}}^0$ is the cell potential under standard conditions, n is the number of transferred electrons, and Q is the reaction quotient.

Given the concentration of nickel sulfate, ${\mathrm{M}}_{{\mathrm{NiSO}}_4}=1.00\mathrm{M}$

The reduction potential of ${\mathrm{Ni}}^{2+}$is -0.23 V. The number of transferred electrons, $\mathrm{n}=2$Let us substitute these values in the equation, we get

$\begin{array}{c}\mathrm{E}\left(\left.{\mathrm{Ni}}^{2+}\right|\mathrm{Ni}\right)={\mathrm{E}}^0\left(\left.{\mathrm{Ni}}^{2+}\right|\mathrm{Ni}\right)-\frac{0.0592}{\mathrm{n}}{\log }_{10}\mathrm{Q}\\ =-0.23\mathrm{V}-\frac{0.0592}{2}{\log }_{10}\frac{1}{\left[{\mathrm{Ni}}^{2+}\right]}\\ =-0.23\mathrm{V}-\frac{0.0592}{2}{\log }_{10}\frac{1}{1}\\ =-0.23\mathrm{V}\end{array}$

Since the cell potential is greater than -0.414 V, the product formed is Ni.

Step 2: Calculation of the total charge

We need to find the total charge to calculate the mass of nickel. Total charge is calculated by the following formula:

The total charge, $\mathrm{Q}=\mathrm{It}$, I is the current and t is the time.

Given current,$\mathrm{I}=2.00\mathrm{A}$

Time,

$\begin{array}{c}\mathrm{t}=10\mathrm{h}\\ =10\mathrm{h}\times 60\times 60\\ =3.6\times {10}^4\mathrm{s}\end{array}$

Substitute the values and we get

$\begin{array}{c}\mathrm{Q}=2.00\mathrm{A}\times 3.6\times {10}^4\mathrm{s}\\ =7.2\times {10}^4\mathrm{As}\end{array}$

The total charge of Nickel is $7.2\times {10}^4\mathrm{As}$.

Calculation of the mass of nickel

The following equation calculates mass of nickel:

Mass of nickel, ${\mathrm{m}}_{\mathrm{Ni}}=\frac{\mathrm{MQ}}{\mathrm{nF}}$, where M is the molar mass of nickel, F is the Faraday constant, $\left(\mathrm{F}=96485.34{\mathrm{Cmol}}^{-1}\right)$

$\begin{array}{c}{\mathrm{m}}_{\mathrm{Ni}}=\frac{58.69\times 7.2\times {10}^4\mathrm{As}}{2\times 96485.34{\mathrm{Cmol}}^{-1}}\\ =21.8980\mathrm{g}~21.9\mathrm{g}\end{array}$

The mass of nickel is 21.9 g.

Finding the product after changing pH

At the cathode, when the pH is 1.00, then ${\mathrm{H}}^{+}$ is discharged. So, the product formed after changing pH is hydrogen gas.