Two electrolytic cells containing silver nitrate solution and dilute sulphuric acid solution were connected in series. A steady current of 2.5 amp was passed through them till 1.078 g of silver was deposited. [Ag=107.8 g `mol^(-1)`,1 F=96,500 C]
(i) How much electricity was consumed ?
(ii) What was the weight of oxygen gas liberated ?

Explanation

`O_(2)` gas liberated at anode of first cell and second cell according to the equation :
`2 H_(2) O (l) to 4H^(+) (aq) + 4 e^(-) + O_(2) (g)`
`4 xx 96500 C` electricity = 1 mol of `O_2` at each cell
`= 32` g of `O_2` at each cell
965 C of electricity = `(32)/(4 xx 96500) xx 965 = 0.08 g O_2` at each cell
Total amount of `O_2` at anode of both cells = `2 xx 0.08 = 0.16` g .