On carrying out the electrolysis of acidified water, the volume of hydrogen liberated at `STP` condition is `22.4L`. The volume of oxygen liberated is

Options

A. `2.8 cm^(3)`
B. `5.6 cm^(3)`
C. `8.4 cm^(3)`
D. `11.2 cm^(3)`

Answer

A

Explanation

`H_(2)O to (1)/(2) O_(2) + 2e^(-)`
i.e., `O^(2-) to (1)/(2) O_(2) + 2e^(-)`
i.e., 1 mole of `H_(2)O` require = `2 xx 96500 C`
= `1.93 xx 10^(5) C `