Question

Give the electrode reactions for formation of (i) lead metal and bromine vapours from molten $PB{ r }_{ 2 }$ using inert electrodes (ii) ${ H }_{ 2 }$ and ${ O }_{ 2 }$ gas [2:1] from acidified water using inert Pt electrodes.

Answer 1

Hint: In electrolysis, a non-spontaneous process is carried out by supplying the required electrical energy at both the cathode and the anode. Reduction takes place at the cathode while oxidation takes place at the anode.

Complete answer:
In electrolysis, a non-spontaneous process is carried out by supplying the required electrical energy at both the cathode and the anode. When molten $PB{ r }_{ 2 }$ is electrolysed, two products are obtained: bromine gas at the anode and lead metal at the cathode. For this electrolysis, generally graphite electrodes are used that are inert in nature.
The positively charged lead ions ${ Pb }^{ 2+ }$ are attracted towards the negatively charged cathode and get reduced to give the lead metal. The reaction is shown below:

$\begin{matrix} Reaction\quad at\quad \\ cathode \end{matrix}:\begin{matrix} { Pb }^{ 2+ }(l) \\ Lead\quad ion \end{matrix}+\begin{matrix} { 2e }^{ - } \\ electrons \end{matrix}\rightarrow \begin{matrix} Pb(s) \\ Lead\quad metal \end{matrix}$
The negatively charged bromide ions get attracted to the positively charged anode and are oxidised to bromine vapours. The reaction is shown below:

$\begin{matrix} Reaction\quad at\quad \\ anode \end{matrix}:\begin{matrix} { 2Br }^{ - }(l) \\ Bromide\quad ions \end{matrix}\rightarrow \begin{matrix} { Br }_{ 2 }(g) \\ Bromine\quad vapours \end{matrix}+\begin{matrix} { 2e }^{ - } \\ electrons \end{matrix}$

(ii) Acidified water is actually a solution of dilute sulphuric acid. When this solution is electrolysed, two gases: oxygen and hydrogen are produced at the inert electrodes in the ratio 1:2 respectively. The hydrogen ions are reduced at the cathode and hydrogen gas is obtained. Similarly the water molecules are oxidised at the anode and oxygen gas is obtained. The reactions are given below:

$\begin{matrix} Reaction\quad at\quad \\ anode \end{matrix}:\begin{matrix} { 2H }_{ 2 }O(l) \\ water \end{matrix}\rightarrow \begin{matrix} { 4H }^{ + }(aq) \\ Hydrogen\quad ions \end{matrix}+\begin{matrix} { 4e }^{ - } \\ electrons \end{matrix}+\begin{matrix} { O }_{ 2 }(g) \\ Oxygen\quad gas \end{matrix}$

$\begin{matrix} Reaction\quad at\quad \\ cathode \end{matrix}:\begin{matrix} { 2H }^{ + }(aq) \\ Hydrogen\quad ions \end{matrix}+\begin{matrix} { 2e }^{ - } \\ electrons \end{matrix}\rightarrow \begin{matrix} { H }_{ 2 }(g) \\ Hydrogen\quad gas \end{matrix}]\times 2$
Hence the overall reaction is:

$2{ H }_{ 2 }O(l)\rightarrow 2{ H }_{ 2 }(g)+{ O }_{ 2 }(g)$
Hence both the questions are explained.

Note: For the electrolysis of water, acid (sulphuric acid) is added to it because water contains very small amount of hydrogen and hydroxide ions due to which it is a poor conductor of electricity but by adding sulphuric acid to water the concentration of hydrogen ions increases along with the addition of sulphate ions which makes water a good conductor of electricity and the electrolysis process can be carried out easily.