InterviewSolution
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InterviewSolution
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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 `):`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+)` |
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Answer» Correct Answer - a,b,c 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` |
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