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

Answer

A, B, C

Explanation

Number of faradays `= (4xx1xx36000 /( 96500) = 0.15`
Initialy, moles of `Fe^(3+)= 0.1 xx1 = 0.1`
First , `Fe^(3+)` will get reduced to `Fe^(2+)`
. `Fe^(3+)+ e^(-) rightarrow Fe^(2+)`
1F `equiv` 1 mole `Fe^(3+)` deposited `rightarrow 0.15F equiv 0.15 ("mole") Fe^(3+)` deposited > `Fe^(3+)` Available
Thus , 1mole `Fe^(3+) equiv 1F`
`Rightarrow 0.1 ("mole")Fe^(3+) equiv 0.1F` electricity is used `equiv 0.025` mole `Fe^(2+)`produced
`Fe^(2) + 2e6(-) rightarrow Fe`
2F equiv 1` mole Fe^(2+)`
`Rightarrow 0.05 F equiv (0.025)` mole `Fe^(2+)` reduced `equiv0.025` mole Fe deposited
`rightarrow Fe^(2+)` left `=` 0.1 - 0.025 `=` 0.075m moles