`100mL` of `0.3 M Fe^(3+)(aq)` ions were electrolyzed by a charge of `0.072F.` In electrolysis, metal was deposited and `O_(2)(g)` was evolved. At the end of electrolysis, it is desired to oxidize the un`-`electrolyzed metal ion.
`Fe^(3+)+e^(-) rarrFe^(2+)`
`Fe^(2+)+2e^(-)rarrFe`
The moles of `Fe^(2+)` ions left un`-`electrolyzed in the solution is
`a. 0.009" "b.0.021" "c.0.072" "d.0.042`

Explanation

`a.` mmoles of `Fe^(3+)=100xx0.3=30`
Charge `=0.072F=0.072xx10^(3)=72mF`
`i. Fe^(3+)+e^(-) rarr Fe^(2+)`
`30` mmolees of `Fe^(3+)` requires `30mF` and `30` mmoles of `Fe^(2+)` formed.
`ii. mF` left`=72-30=42mF`
`iii. Fe^(2+)+2e^(-) rarr Fe(s)`
`42mF` will electrolyze `-21 m mol` of `Fe^(2+)`
`[` Since `2e^(-)=2F` or `2mF=1 mmol` of `Fe^(2+)=1mmol `o f `Fe]`
`:.` mmoles of `Fe^(2+)` left `=30-21`
`9 mmol`
`=9xx10^(-3)=0.009 mol`