
During an electrolysis of conc. \[{H_2}S{O_4}\], perdisulphuric acid \[\left( {{H_2}{S_2}{O_8}} \right)\] and \[{O_2}\] form in equimolar amount . The amount of \[{H_2}\] that will form simultaneously will be:
\[\left( {2{H_2}S{O_4} \to {H_2}{S_2}{O_8} + 2{H^ + } + 2{e^ - }} \right)\]
A. Thrice that of \[{O_2}\] in moles
B. Twice that of \[{O_2}\] in moles
C. Equal to that of \[{O_2}\] in moles
D. Half of that of \[{O_2}\] in moles
Answer
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Hint: As we know that, electric current is just a flowing stream of electrons, called charged particles. Electrolysis is yet another electrochemical reaction that absorbs electric energy.
Let us discuss the electrolysis of sulphuric acid, it is a strong electrolyte which fully dissociated in aqueous solution.
${H_2}S{O_4} \to 2{H^ + } + S{O_4}^{2 - }$
Water is a weak electrolyte and is only slightly dissociated.${H_2}O \to {H^ + } + O{H^ - }$.
${H_2}O \to {H^ + } + O{H^ - }$
The hydrogen ions move into the cathode during electrolysis, and are discharged there. And in case of electrolysis in sulphuric acid, it gives at anode per disulphuric acid and hydrogen ion. We can see it by the given reaction. Two reactions are given at the anode and at the cathode.
$\begin{array}{l}
Anode\;\left\{ \begin{array}{l}
2{H_2}S{O_4} \to {H_2}{S_2}{O_8} + 2{H^ + } + 2{e^ - }\\
2{H_2}O \to {O_2} + 4{H^ + } + 4{e^\_}
\end{array} \right\}\\
\underline {Cathode\;\left\{ {2{H_2}O \to {H_2} + 2O{H^ - } - 2{e^\_}} \right\} \times 3}
\end{array}$
$2{H_2}S{O_4} + 8{H_2}O \to {H_2}{S_2}{O_8} + {O_2} + 3{H_2} + 6{H^ + } + 6O{H^\_}$
We see here that here the ratio of ${O_2}$ and ${H_2}$ is $1:3$.
Therefore, the correct answer is A.
Additional Information:
When we talk about the application of electrolysis, two large scale applications of electrolysis come to our mind. The first is the electrolysis of salt water to produce sodium hydroxide and chlorine gas, both of which are huge articles of commerce. The cost of electricity dominates the economics of this process. The second large scale application is the manufacture of the alkali metals from their molten chloride or electroplating metal objects with a protective layer of a different metal.
Note:
One of the most important uses of electrolysis is the production of aluminium from its oxide. And also copper is purified by electrolysis.
Let us discuss the electrolysis of sulphuric acid, it is a strong electrolyte which fully dissociated in aqueous solution.
${H_2}S{O_4} \to 2{H^ + } + S{O_4}^{2 - }$
Water is a weak electrolyte and is only slightly dissociated.${H_2}O \to {H^ + } + O{H^ - }$.
${H_2}O \to {H^ + } + O{H^ - }$
The hydrogen ions move into the cathode during electrolysis, and are discharged there. And in case of electrolysis in sulphuric acid, it gives at anode per disulphuric acid and hydrogen ion. We can see it by the given reaction. Two reactions are given at the anode and at the cathode.
$\begin{array}{l}
Anode\;\left\{ \begin{array}{l}
2{H_2}S{O_4} \to {H_2}{S_2}{O_8} + 2{H^ + } + 2{e^ - }\\
2{H_2}O \to {O_2} + 4{H^ + } + 4{e^\_}
\end{array} \right\}\\
\underline {Cathode\;\left\{ {2{H_2}O \to {H_2} + 2O{H^ - } - 2{e^\_}} \right\} \times 3}
\end{array}$
$2{H_2}S{O_4} + 8{H_2}O \to {H_2}{S_2}{O_8} + {O_2} + 3{H_2} + 6{H^ + } + 6O{H^\_}$
We see here that here the ratio of ${O_2}$ and ${H_2}$ is $1:3$.
Therefore, the correct answer is A.
Additional Information:
When we talk about the application of electrolysis, two large scale applications of electrolysis come to our mind. The first is the electrolysis of salt water to produce sodium hydroxide and chlorine gas, both of which are huge articles of commerce. The cost of electricity dominates the economics of this process. The second large scale application is the manufacture of the alkali metals from their molten chloride or electroplating metal objects with a protective layer of a different metal.
Note:
One of the most important uses of electrolysis is the production of aluminium from its oxide. And also copper is purified by electrolysis.
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