During the electrolysis of conc `H_(2)SO_(4)`, it was found that `H_(2)S_(2)O_(8)` and `O_(2)` liberated in a molar ratio of `3:1`. How many moles of `H_(2)` were found of moles of `H_(2)S_(2)O_(8)` ?
`(` Express your answer as `:3xx mol e s of H_(2),` integer answer is between 0 and 5`0`

Answer

5

Explanation

Let `x` mol of `O_(2)` of liberated and `3x` mol of `H_(2)S_(2)O)8)` is formed. Reactions at cathode `(` reduction `):`
`2H_(2)O+2e^(-) rarr H_(2)+2overset(c-)(O)H`
Reactions at anode `(` oxidation `):`
`i. 2H_(2)O rarr O_(2)+4H^(o+)+4e^(-)[{:(1 mol,O_(2),-=,4F,),(x mol,O_(2),=,4xF,):}]`
`ii. 2SO_(4)^(2-)rarr S_(2)O_(8)^(2-)+2e^(-)[{:(1 mol,S_(2)O_(8)^(2-),=,2F,),(3x mol,S_(2)O_(8)^(2-),=,6xF,):}]`
Total Faraday at anode `=(4x+6x)F=10xF.`
Total Faradays at cathode `=` Total Faradays at anode
`:. 2F` at cathode `-=1 mol of H_(2)`
`10xF` at cathode`-=(1)/(2F)xx10xF=5x mol of H_(2)`.
`Ratio=("Moles of "H_(2)" at cathode" )/("Moles of "H_(2)S_(2)O_(8) at anode )=(5x)/(3x)=(5)/(3)`
Number of moles of `H_(2)=3xx(5)/(3)=5`
Alternatively
Molar ratio of `H_(2)S_(2)O_(8):O_(2)`
`(n` factor `=2) (n` factor `=4)`
`=3:1`
Equivalent ratio`=3xx2:1xx4=6:4`
Total equivalent of `H_(2)S_(2)O_(8)` and `O_(2)` at anode `=6+4=10Eq`.
So total equivalent of `H_(2)` at cathode `=10`
`:.` moles of `H_(2)(n` factor `=2)=(10)/(2)=5` moles