`A 500mL` sample of a `0.1M Cr^(3+)` is electrolyzed with a current of `96.5A`. If the remaining `[Cr^(3+)]` is `0.04M` then the duration of process is:-

Answer

`90sec`

Explanation

Initial moles of `Cr^(3+) = 0.5 xx 0.1 rArr 0.05`
Final moles of `Cr^(3+) = 0.04 xx 0.5 rArr 0.02`
Moles of `Cr^(3+)` reduced `Ϸ 0.05 -0.02 rArr 0.03`
`:.` Number of equivalents of `Cr^(3+)` reduced `=(1 xxt)/(96500)`
`=(t xx 96.5)/(96500)`
`3 xx 0.03 =(t xx 96.5)/(96500), t = 90` sec.