If a current of 10 A is passed through acidulated water for 100 s then what volume of electrolytic gas at STP will be produced?

Explanation

Quantity of electricity, `Q=10xx100=1000C.`
During electrolysis of acidulated water, `H_(2)` gas is liberated at the cathode while `O_(2)` gas is evolved at the anode.
`96500C(=1F)" of electricity = 0.5 mol of "H_(2)`.
`"So, 1000C of electricity "=(0.5xx1000)/(96500)=5.18xx10^(-3)" mol of "H_(2)`.
`"At, STP, the volume of "5.18xx10^(-5)" mol "H_(2)`
`=22.4xx5.18xx10^(-3)L="116 mL."`
The ratio of volumes of `H_(2)` to `O_(2)` evolved during electrolysis of acidulated water is `2:1`.
So, STP , the volume of `O_(2)` evolved `=(1)/(2)xx116 = 58 mL."`
and total volume of the liberated gases `=(116+58)`
`="174 mL."`