How many coloumbs are required for the oxidation of 1 mole of `H_(2)O "to" O_(2)`

Explanation

`{:(H_(2)O rarr2H^(+) + 1//2 O_(2)+2e^(-)),(1 mol " "2 mol):}`
(Here, oxidation number of oxygen changes from -2 in `H_(2)O` to 0 in `O_(2)`. Therefore, 2 moles of electrons are lost during the oxidation of 1 mol of water.)
The charge Q on n mol of electrons is given by Q=nF
Thus, quantity of electricity required for oxidation of 1 mol of `H_(2)O "to" O_(2)`,
`Q=2 mol xx96500 "C mol"^(-1)=193000 C =1.93xx10^(5)C`.