Electrolysis of acidulated water yields `H_(2)` at the cathode and `O_(2)` at the anode. In an experiment, the total volume of liberate `H_(2)` and `O_(2)`, gases during the electrolysis was found to be 16.8 mL at STP. What amount of electricity in coulomb was used in the experiment?

Explanation

In electrolysis of water, the ratio of the volume of liberated `H_(2)` to `O_(2)=2:1`. In 16.8 mL of gas mixture at STP, the volume of `H_(2)=(16.8xx2)/(3)="11.2 mL and that of "O_(2)`
`=(16.8xx1)/(3)="5.6 mL"`
Let the quantity of electricity used during electrolysis be Q coulomb. When 96500 C of electricity is passed, the amount of `H_(2)` evolved = 1.008 g.
So, Q coulombs of electricity will liberate `(1.008xxQ)/(96500)g` of `H_(2)`.
`"11.2 mL of "H_(2)" at STP has a mss of "(2.016xx11.2)/(22400)g`
`=1.008 xx10_(g)^(-3)`.
Therefore, `(1.008xxQ)/(96500)=1.008xx10^(-3)" or, "Q=96.5C.`