How long would it take to reduce completely `80 mL` of `0.1 M Fe_(2)(SO_(4))_(3)` to `FeSO_(4)` by passing a current of `2` ampere?

Explanation

`2e+(Fe^(3+))_(2) rarr 2Fe^(2+)`
`:.` Valence factore of `Fe_(2)(SO_(4))_(3) = 2`
Meq. of `Fe(SO_(4))_(3)` to be reeduced `= 80 xx 0.1 xx 2 = 16`
`:.` Eq. of `Fe_(2)(SO_(4))_(3)` or `((w)/(E)) = (16)/(1000)`
Now ` (w)/(E) = (i.t)/(96500)`
`:. t = (w xx 96500)/(E xx i) xx (96500 xx 16)/(1000 xx 2) = 772 sec`.