InterviewSolution
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InterviewSolution
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`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` |
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Answer» `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` |
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