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A 0.20 M KOH solution is electrolysed fo...

A 0.20 M KOH solution is electrolysed for 1.5h using a current of 8.00A . The number of moles of `O_(2)` produced at anode is

A

0.48

B

0.224

C

0.112

D

`2.24xx10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`4OH^(-)rarr2H_(2)O+O_(2)+4e^(-)`
1 mol 4 mol
Thus, 1 mol, of `O_(2)` is produced from `4xx96500C`
`1=8.00A,t=1.5xx60xx60s`
`q=1xxt=8xx1.5xx60xx60C `
`=43200C`
`4 xx96500C` produces `O_(2)=1` mol
`43200C` will produce `O_(2)`
`=(43200)/(4xx96500)=0.112` mol
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