Question

An electrolysis cell contains a solution of 0.1M$\mathit{N}\mathit{i}\mathit{S}{\mathit{O}}_{\mathbf{4}}$. The anode and cathode are both strips of Pt. foil. Another electrolysis cell contains the same solution, but the electrolysis are strips of Ni foil. To each case a current of 0.10A flows through the cell for 10 hours.

1. Write a balanced equation for the chemical reaction that occurs at the anode in each cell.
2. Calculate the mass in grams of the product formed at the anode in each cell. (The product may be a gas, a solid or an ionic species in solution)

Short answer

a. When pt. foil strip act as anode and ${\mathbf{\text{OH}}}^{\mathbf{\text{-}}}$oxidised at anode according to the following equation.

$O{H}^{-}\to O_2\left(g\right)+2H_{2}O+4e^{-}\left(Anode\right)$

When ‘Ni’ foil strip act as anode, Nioxidised at anode according to the following equation.

$Ni\left(s\right)\to N{i}^{2+}+2e^{-}\left(Anode\right)$

b. When ‘pt.’ foil strip act as anode 0.296 gm${\mathbf{\text{O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$ produced at anode. When ‘Ni’ foil strip acts as anode, 1.086gm of ${\mathbf{\text{Ni}}}^{\mathbf{\text{2 +}}}$ions are produced at anode.

Step-by-step solution

Step-1: Number of ionic species in electrolytes solution

In 0.1M$NiS{O}_4$solution, the species that form ions in the solution due to dissociation of $NiS{O}_4$and $H_{2}O$

$$\begin{gathered} NiS{O}_4\to N{i}^{2+}+S{O}_4^{2-}............Equation\left(1\right) \\ {H}_{2}O\rightleftharpoons H^{+}+O{H}^{-}..............Equation\left(2\right) \end{gathered}$$

From above equation we concluded that the ionic species present in the solution are$\mathit{N}{\mathit{i}}^{\mathbf{2}\mathbf{+}}\mathbf{,}\mathit{S}{\mathit{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{,}{\mathit{H}}^{\mathbf{+}}\mathbf{\&}\mathit{O}{\mathit{H}}^{\mathbf{-}}$

Note:

When there are multiple number of electrodes present in a solution, species having highest oxidation potential will oxidized at the anode.

Step-2: Chemical reaction when anode is ‘Pt.’ foil

In this case ‘Pt’ foil strip act as anode, which is aninert electrode and do not participate in the electrochemical reaction.

So, available species for oxidation at anode are$S{O}_4^{2-}$and$O{H}^{-}$ as they can easily loose electrons standard reduction potential of$S{O}_4^{2-}$is greater than $O{H}^{-}$.

The standard reduction potential is converted to standard oxidation potential by using following relation

Standard oxidation potential = - (Standard reduction potential)

Hence, oxidation potential of$S{O}_4^{2-}$is less than $O{H}^{-}$ so, $\mathit{O}{\mathit{H}}^{\mathbf{-}}$oxidised at anode

Step-3: Balanced equation of anode when ‘Pt.’ is used

$4O{H}^{-}\to O_2\left(g\right)+2H_{2}O+4e^{-}\left(Anode\right).........Equation\left(4\right)$

Hence, $\mathit{O}{\mathit{H}}^{\mathbf{-}}$isoxidised at anodewhen ‘Pt.’ foilis used.

Step-4: Chemical reaction when anode is ‘Ni’ foil

‘Ni’ is an active electrode and it participate in electrochemical reaction. So, the ions available for oxidation at anode are$S{O}_4^{2-}$,$O{H}^{-}$and Ni

Standard reduction potential of the ions is i.e.,$S{O}_4^{2-}>O{H}^{-}>Ni$

Standard oxidation potential of the ions is i.e.,$S{O}_4^{2-}<O{H}^{-}<Ni$

From this we concluded that ‘Ni” oxidised at anode

Step-5: Balanced equation of anode when ‘Ni’ foil is used

$Ni\left(s\right)\to N{i}^{2+}+2e^{-}\left(Anode\right)...........Equation\left(5\right)$

Hence, when ‘Ni’ foil strip used at anode, ‘Ni’ is oxidised at the anode.

Step-6: Mass of product when Pt. foil act as anode

Calculate the number of moles of electron used when 0.1A current flow through the cell for 10 hours by using following formula:

$n=\frac{It}{96485Cmo{l}^{-1}}..........Equation\left(6\right)$

n = number of moles of electron

I = current

t = time

Putting,

$$\begin{gathered} I=0.1A=0.1C{s}^{-1} \\ t=10hours=36000s \\ n=\frac{\left(0.1c{s}^{-1}\right)\left(36000s\right)}{96485cmo{l}^{-1}}=0.037mole \end{gathered}$$

According to equation (4)

For 4 moles of $e^{-}$produced, the mass of $O_2\left(g\right)$ is:

Mass of $O_2\left(g\right)$= $\left(0.037mol{e}^{-1}\right)\times \left(\frac{1}{4}\frac{mol{O}_2}{mol{e}^{-1}}\right)\times \left(32gmo{l}^{-1}\right)$of

$=0.296g$of$O_2\left(g\right)$

Hence, $0.296g$ of $O_2\left(g\right)$ is produced at the anode

Step-7: Mass of product when ‘Ni’ foil act as anode

The number of moles of $e^{-}$ transferred when 0.1A current flows through cell for 10 hours are 0.03 mole as calculated in previous step.

According to equation (5)

2 moles $e^{-}$ are produced with 1 mole of $N{i}^{2+}$

Mass of$N{i}^{2+}$$=\left(0.037mol{e}^{-1}\right)\times \left(\frac{1}{4}\frac{molN{i}^{2+}}{mol{e}^{-1}}\right)\times \left(58.69gmo{l}^{-1}\right)$

$=1.086g N{i}^{2+}$

Hence, $1.086g$of $N{i}^{2+}$produced at anode.