When `4.0 A` of current is passed through a `1.0L , 0.10M Fe^(3+)(aq)` solution for `1.0 ` hour, it is partly reduced to `Fe(s)` and partly of `Fe^(2+)(aq)`. The correct statements `(s)` is `(` are `):`

Options

A. `0.10 mol` of electrons are required to convert all `Fe^(3+)` to `Fe^(2+)`
B. `0.025 mol` of `Fe(s)` will be deposited.
C. `0.075 mol` of iron remains as `Fe^(2+)`.
D. `0.050 mol` of iron remains as `Fe^(2+)`

Answer

a,b,c

Explanation

Number of Faradays `=(4xx1xx3600)/(96500)=0.15`
Initially, moles of `Fe^(3+)=0.1xx1 =0.1`
First, `Fe^(3+)` will get reduced to `Fe^(2+)` .
`Fe^(3+)+e^(-) rarr Fe^(2+)`
`1F-=1 mol Fe^(3+)` deposited
`implies0.15 F -=0.15 mol Fe^(3+)` deposited `gt Fe^(3+) ` available.
Thus, `1 mol Fe^(3+)-=1F`
`implies 0.1 mol Fe^(3+)-=0.1 F` electricity is used `-= 0.1 mol Fe^(2+)` produced
`implies0.15 -0.1 =0.05 F ` electricity left for the reduction of `Fe^(2+)` .
`Fe^(2+) + 2e^(-) rarr Fe`
`2F-=1 mol Fe^(2+)`
`implies0.05 F -=(0.05)/(2) =0.25 mol Fe^(2+)` reduced `-= 0.025 mol Fe` deposited .
`implies Fe^(2+)` left `=0.1-0.025=0.075 mol`