InterviewSolution
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InterviewSolution
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`E_(cell)=0.78` volt for the following cell.`underset(Fe_((s))|Fe_((aq))^(2+))||underset((0.01M)(Cu_((aq))^(2+)|Cu_((s)))` `E_(Fe//Fe^(2+)(aq))^(@)=0.44V,E_(Cu//Cu^(2+)(aq))^(@)=-0.34V`A. x cannot be predictedB. x=0.01MC. `xgt0.01M`D. `xlt0.01M` |
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Answer» Correct Answer - B If a solution contains two or more catios, the cation to be deposited on the cathode is the one which has higher reduction potential. Here the reaction will be `Fe+Cu^(2+)toFe^(2+)+Cu` `E_(cell)=E_(cell)^(o)-(0.0591)/(n)"log"([Fe^(2+)])/([Cu^(2+)])` `0.78=0.78-(0.591)/(n)"log"[(Fe^(2+))/(Cu^(2+))]` or `0="log"(x)/(0.01)implies(x)/(0.01)=1impliesx=0.01`. |
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