Question

A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively:
A.${{\text{H}}_{\text{2}}}{\text{,}}{{\text{O}}_{\text{2}}}$
B.${{\text{O}}_{\text{2}}}{\text{,}}{{\text{H}}_{\text{2}}}$
C.${{\text{O}}_{\text{2}}}{\text{,}}{{\text{N}}_{\text{2}}}$
D.${{\text{O}}_{\text{2}}}{\text{,S}}{{\text{O}}_{\text{2}}}$

Answer 1

Hint: To answer this question, you should recall the electrochemical series to get the idea of the reduction potential as it helps to determine which ions are released at different electrodes. The basic concept involved is that reduction takes place at cathode and oxidation takes place at the anode.

Complete step by step answer:
An aqueous solution of sodium sulphate contains the following ions: \[{\text{N}}{{\text{a}}^{\text{ + }}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, O}}{{\text{H}}^{\text{ - }}}{\text{, S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}\]. To calculate which ions undergo reaction, you should know the electrochemical series.
The cation \[\left( {{\text{N}}{{\text{a}}^{\text{ + }}}{\text{, }}{{\text{H}}^{\text{ + }}}} \right)\] which has a lower discharge potential will be discharged at cathode and the anion \[\left( {{\text{O}}{{\text{H}}^{\text{ - }}}{\text{, S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}} \right)\] which has a lower discharge potential will be discharged at anode.
At the cathode, reduction of cation will take place.
Hydrogen ions having lower discharge (or higher reduction) potential than sodium ions will be liberated at the cathode.
Thus, we can write reaction at cathode as:
\[{\text{2}}{{\text{H}}^{\text{ + }}}{\text{ + 2}}{{\text{e}}^{\text{ - }}} \to {{\text{H}}_{\text{2}}} \uparrow \]

At the anode, oxidation of anion will take place.
Hydroxide ions having a lower discharge potential than sulphate ions will be liberated at the anode.
Thus, we can write reaction at anode as:
\[{\text{4O}}{{\text{H}}^ - } \to {\text{2}}{{\text{H}}_{\text{2}}}{\text{O + }}{{\text{O}}_{\text{2}}}{\text{ + 4}}{{\text{e}}^{\text{ - }}}\]
Hence, \[{{\text{H}}_{\text{2}}}\] and \[{{\text{O}}_{\text{2}}}\] will be liberated.

So, the correct answer is Option A .

Note:
We should know that at the electrodes, electrons are absorbed or released by the atoms and ions. There is gain or loss of electrons which are released and passed into the electrolyte. The uncharged atoms separate from the electrolyte upon exchanging electrons. This is called discharging and this concept helps determine the release of ions on different electrodes.
The increasing order of discharge of few cations is:
\[{{\text{K}}^{\text{ + }}}{\text{, C}}{{\text{a}}^{{\text{2 + }}}}{\text{, N}}{{\text{a}}^{\text{ + }}}{\text{, M}}{{\text{g}}^{{\text{2 + }}}}{\text{, A}}{{\text{l}}^{{\text{3 + }}}}{\text{, Z}}{{\text{n}}^{{\text{2 + }}}}{\text{, F}}{{\text{e}}^{{\text{2 + }}}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, C}}{{\text{u}}^{{\text{2 + }}}}{\text{, A}}{{\text{g}}^{\text{ + }}}{\text{, A}}{{\text{u}}^{{\text{3 + }}}}\]
The increasing order of discharge of some of the anions is:
\[{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{, N}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{, O}}{{\text{H}}^{\text{ - }}}{\text{, C}}{{\text{l}}^{\text{ - }}}{\text{, B}}{{\text{r}}^{\text{ - }}}{\text{, }}{{\text{I}}^{\text{ - }}}\]