1.

In passing the electricity through the acidulated H_(2)O , 5.6 dm^(3) of O_(2) liberated at anode . The volume of H_(2) liberated at cathode will be

Answer»

`5.6 DM^(3)`
`22.4 dm^(3)`
`11.2 dm^(3)`
`44.8 dm^(3)`

SOLUTION :By passing 1 F , 1 g equivalent of `H_(2)` (1 g `-= 11.2 dm^(3)`) and 1 g equivalent of `O_(2)` ( 8 g `-= 5.6 dm^(3))` liberated.


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