During the electrolysis of acidified water, `O_(2)` gas is formed at the anode. To produce `O_(2)` gas at the anode at the rate of `0.224 Ml` per second at `STP`, the current passed is

Options

A. `0.224A`
B. `2.24 A`
C. `9.64 A`
D. `3.86 A`

Answer

d

Explanation

First method
At anode `(` oxidation `):`
`H_(2)O rarr 2H^(o+)+2e^(-)+(1)/(2) O_(2)(g)`
`2F=2xx96500C=(1)/(2) mol O_(2)at STP. `
`=(22.4)/(2)=11.2L of O_(2)`
`11.2xx10^(3) mL =2 xx 96500C`
`0.224 mL of O_(2)` per second `=(96500xx0.224mL)/(11.2 xx 10^(3)mL)`
`=3.86 A`
Second methode
`1F=96500C=1 Eq of O_(2)=(22.4 )/(4) =5.6 L`
`5.6 xx 10^(3)mL =96500C`
`0.224 mL ` per second of `O_(2)=(96500xx0.224)/(5.6xx10^(3)mL)`
`=3.86A`