1.

The ratio of volumes of H_(2) and O_(2) liberated on electrolysis of water is ____

Answer»

`1 : 2`
`1 : 3`
`2 : 1`
`3 : 1`

SOLUTION :`H_(2)O rarr H^(+) + OH^(-)`
Cathode : `4OH^(-) rarr 2H_(2)O + O_(2) + 4E^(-)`
ANODE : `4H^(+) + 4e^(-) rarr 2H_(2)`
`:.` The ratio of volume of `H_(2) and O_(2) "is " 2 : 1`


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