MCQ
Which of the following is not correct regarding the electrolytic preparation of $\ce{H_2O_2}\ ?$
- ALead is used as cathode.
- B$50\%\ \ce{H_2SO_4}$ is used.
- ✓Hydrogen is liberated at anode.
- DSulphuric acid undergoes oxidation.
✓
Answer
Correct option: C.
Hydrogen is liberated at anode.
$\ce{H_2O_2}$ can be prepared by electrolysis of $50\%$ solution of $\ce{H_2SO_4}$. In this method, hydrogen is liberated at cathode.
$\text{H}_2\text{SO}_4\rightleftharpoons2\text{H}^++\text{HSO}^-_4$
At anode $2\text{HSO}_4^-\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{H}_2\text{S}_2\text{O}_8+2\text{e}^-$
$\text{H}_2\text{S}_2\text{O}_8+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ }2\text{H}_2\text{SO}_4+\text{H}_2\text{O}_2$
At cathode $2\text{H}^++2\text{e}^-\xrightarrow{\ \ \ \ \ \ \ }\text{H}_2\uparrow$
$\text{H}_2\text{SO}_4\rightleftharpoons2\text{H}^++\text{HSO}^-_4$
At anode $2\text{HSO}_4^-\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{H}_2\text{S}_2\text{O}_8+2\text{e}^-$
$\text{H}_2\text{S}_2\text{O}_8+2\text{H}_2\text{O}\xrightarrow{\ \ \ \ \ \ \ }2\text{H}_2\text{SO}_4+\text{H}_2\text{O}_2$
At cathode $2\text{H}^++2\text{e}^-\xrightarrow{\ \ \ \ \ \ \ }\text{H}_2\uparrow$
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