Marshell's acid is prepared by the electrolytic oxidation of `H_(2)SO_(4)` as
`2H_(2)SO_(4) rarr H_(2)S_(2)O_(8)+2H^(o+)+2e^(-)`
Oxygen and hydrogen are byproducts. In such electrolysis `2.24L` of `H_(2)` and `0.56L` of `O_(2)` were product at `STP`. The weight of `H_(2)S_(2)O_(8)` fromed is

Options

A. `9.7g`
B. `19.4g`
C. `14.55g`
D. `29.1g`

Answer

a

Explanation

`H_(2)SO_(4) rarr H^(o+)+HSO_(4)^(c-)`
These two reactions are competing at anode `:`
`2HSO_(4)^(c-) rarr H_(2)S_(2)O_(8)+2e^(-)`
Anode reaction `:(` Oxidation of `H_(2)O)`
`[{:(2overset(c-)(O)H,rarr,H_(2)O,+,(1)/(2)O_(2),+,2e^(-)),(,,,or,,,),(H_(2)O,rarr,2H^(o+),+,(1)/(2)O_(2),+,2e^(-)):}]`
Cathode reaction `:`

Since one reaction at cathode and two reactions at anode are taking place, therefore, the equivalent of `H_(2)(g)` produced at cathode should be equal to the equivalent of `O_(2)` produced and the equivalent of `H_(2)S_(2)O_(8)` formed.
`:.` Equivalent of `H_(2)=` Equivalent of `O_(2)+` Equivalent of `H_(2)S_(2)O_(8)`
`(2.24xx2)/(22.4)=(0.56xx4)/(22.4)+(W)/(194//2)`
`(Mw of H_(2)S_(2)O_(8)=194,Ew=(194)/(2))`
After solving, we get `W=9.7g`