`Hg_(2)CI_(2)` is produced by the electrolytic reduction of `Hg^(2+)` ion in presence of `CI^(-)` ion is `2Hg^(2+) +2CI^(-) +2e^(Theta) rarr Hg_(2)CI_(2)`. Calculate the current required to have a rate production of 44g per hour of `Hg_(2)CI_(2)`.
[Atomic weight of `Hg = 200.6]`:-

Options

A. 5 ampere
B. 4 amapere
C. 6.5 ampere
D. 3.5 ampere

Answer

A

Explanation

`w = Z It w = (E)/(96500) xx I xx t`
`(w)/(t) = (200.6 xx 2+71)/(2 xx 96500) xx 1 (44)/(3600) =(236.1)/(96500) xx1`
`I = 5 Amp`.