# Electrolysis

The process of chemical decomposition of the electrolyte by the passage of electricity through its molten or dissolved state is termed as electrolysis. It is the chemical change (Redox reaction occurring at electrode on passing current). The device in which this process is carried out is called electrolytic cell.

In this process first of all electrolyte is ionised. The cations deposit on cathode where reduction takes place. The anions deposit on anode where oxidation takes place. The conversion of ions in the neutral particles at the respective electrodes is referred to as primary change. The product obtained on primary change may be collected as such or may undergo secondary change to form final products. The reactions taking place in these changes are called electrochemical reactions.

It is interesting to note that liberation of a particular ion depends upon the nature of electrode, discharged potential and concentration of ions in a solution. According to the preferential discharge theory, if more than one type of cations or anions or both are present in the electrolyte, then the ion is discharged which requires least energy or discharged potential.

The decreasing order of discharged potential or the increasing order of deposition of some ions is given below:

Cations: $Na^+, ca^+, Mg^{++}, Al^{3+}, Zn^{++}, Fe^{++}, H^+, Cu^{++}, Hg^{++}, Ag+, Ag^{3+}$ etc.

Anions: $SO^{-2}_4 , NO^-_3 , OH^- , Cl^-, Br^-, I^-$ etc.

To explain the above, we consider the following examples:

1. Electrolysis of molten $PbBr_2$: Molten $PbBr_2$ consists of $Pb^{++}$  and $Br^-$ ions which deposit on cathode and anode respectively on passing electricity. The changes may be represented as:

$PbBr_2 \leftrightharpoons Pb^{++} + 2Br^-$

At cathode $Pb^{++} + 2e^- \to Pb$

At anode $Br^- e^- \to Br$ (Primary change)

$Br + Br \to Br_2$ (Primary change)

2. Electrolysis of aqueous solution of NaCl: Aqueous NaCl consists of the following ions in solution:

$NaCl(aq) + H_2O (l) \leftrightharpoons Na^+(aq) + H^+(aq) + Cl^-(aq) + OH^-(aq)$

Thus aqueous solution of NaCl consists of

$Na^+ \text{and} \hspace{3mm}cations H^+ , \text{Cations} , Cl^- \text{and} \hspace{3mm} OH^-$ anions.

On passing electricity $H^+$ions deposit at the cathode $H^+$ in preference and produce $H_2$ gas. Thus$H_2O$ molecules reduce at cathode as follows:

$2H_2O + 2e^-\to 2OH^-(aq) + H_2(g)$

On the other hand $Cl^-$ ions deposit at the anode in preference and produce $Cl_2$ gas as:

$Cl^- \to Cl + e^+$ (Primary change)

$Cl + Cl \to Cl_2(g)$ (Secondary change)

Thus overall reaction taking place on electrolysis of aqueous solution of NaCl may be given as:

$2Na^+ + 2Cl^- + 2H_2O \overset{Electricity}{\rightarrow} 2Na^+ +2OH^- + H_2(g) ++ Cl_2(g)$

I.e.

$Na^+ \text{and} OH^-$ ions remain in the solution and the solution on evaporation gives crystals of NaOH.

• Due to over voltage concept in place of $O_2$ here $Cl_2$ is liberated at anode, i.e., for$O_2$ additional voltage is needed here.

3. Electrolysis of aqueous solution of $CuSO_4$ using Pt electrodes (Inert electrodes): Aqueous $CuSO_4$ consists following ions:

$CuSO_4 + H2O \leftrightharpoons + Cuy^{++} + H^+ + OH^- + SO^{-2}_4$

On passing electricity $Cu^{++}$ ions deposit at the cathode in preference due to higher electrode potential than ion.

I.e.

$Cu^{++} + 2e^- \to Cu$

On the other hand OH‘ ions deposit at the anode in preference and produce $O_2$ gas.

$4OH^- \to 2H_2O + O_2 + 4e^-$ (At anode)

Thus overall reaction taking place on electrolysis of aqueous solution of $CuSO_4$ using Pt-electrodes may be given as:

$2Cu^{++} + 2SO^{-2}_4 + 2H_2O \overset{Electricity}{\rightarrow} 2Cu +O_2 + 4H^+ + 2SO^{-2}_4$

4. Electrolysis of aqueous solution of $CuSO_4$ using Cu electrodes: Thus ionisation of CuSO4 may be given as:

$CuSO_4 + H_2O \leftrightharpoons Cu^{++} + H^+ + OH^- + SO^{-2}_4$

It is the concept of electroplating that Cu-metal is liberated at cathode. Here blue colour of the solution gets faded due to electrolysis.

On passing electricity $Cu^{++}$ions deposit at the cathode as follows.

$Cu^{++} + 2e^- \to Cu$ (At cathode)

On the other hand, anode itself undergoes loss of electrons to form $Cu^{++}$ ions which go into the solution.

$Cu \to Cu^{++} + 2e^-$

Thus the net concentration of $Cu^{++}$ ions remains the same.

Here Cu is transferred from the anode to cathode so the solution concentration does not change.

• This concept develops the concept of refining used for purification of metals.

5.       Electrolysis of $CuCl_2$using inert electrodes:

At anode $2Cl^- (aq.) \to Cl_2(g) + 2e^-$

At anode $Cu^{+2}(aq.) \to 2e^- + Cu(s)$

Net reaction $Cu^{2+}(aq.) + 2Cl^-(aq.) \to Cu(s) + Cl_2(g)$

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