How much electricity is required in coulomb for the oxidation of
(i) 1 mol of `H_(2)O` to `O_(2)` ?
(ii) 1 mol of FeO to `Fe_(2)O_(3)` ?

Explanation

(i) Calculation for required electricity for production of `O_(2)` from the oxidation of 1 mol
`H_(2)O` : According to following reaction `1/2` mol of `O_(2)` is produced from 1 mol of `H_(2)O`.
`H_(2)O_((l)) to (1)/(2)O_(2(g))+2H^(+)+2e^(-)`
1 mol `H_(2)O to 1/2` mol of `O_(2)`, in this 2 mole `2e^(-)` is used.
So, to produce `O_(2)` from 1 mol `H_(2)O` 2F electricity is required.
So, `2F=2xx96500=193000` coulomb electricity.
(ii) Calculation for required electricity for production of `Fe_(2)O_(3)` from oxidation of 1 mol of `FeO`:
1 mol of FeO gives following reaction to obtain `Fe_(2)O_(3)`.
`underset((+2))(Fe)O to (1)/(2) underset((+3))(Fe_(2))O_(3)+underset(1mol)(e^(-))+H^(+)`
So, oxidation of 1mol FeO gives `1/2` mole of `Fe_(2)O_(3)` and in this reaction 1 mol electron =1F=96500 coloumb electricity is used.