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Using the standart electrode potential find out the pair between which redox reaction is not feasible E^(@) values : Fe^(3+)//Fe^(2+)=+0.77,I_(2)//I^(-)=+0.54,Cu^(2+)//Cu=+0.34,Ag^(+)//Ag=+0.80V |
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Answer» SOLUTION :CALCULATE the `E_("cell")^(@)` of the four redox reaction If `E_("cell")^(@)` of a reaction is -ve that reaction will not occur (a) `2Fe^(3+)+2e^(-)rarr2 Fe^(2+),E^(@)=+0.77 V` `2 I^(-)rarrI_(2)+2e^(-),E^(@)=-0.54V` `2Fe^(3+)+2I^(-) rarr 2Fe^(2+)+I_(2),E_("Cell")^(@) =+0.23 v` This reaction is feasible SINCE `E_("cell")^(@)=+0.23V` (b)`CurarrCu^(2+)+2e^(-),E^(@)=-0.34 V` `2Ag^(+)+2e^(-)rarr2A,E^(@)=+0.80 V` `CU+2Ag^(+)rarr2Cu^(2+)+2agE^(@)=+0.46V` This reaction is feasible since `E^(@)` is +ve (c ) `2Fe^(3+)+2e^(-)rarr2Fe^(2+),E^(@)=+0.77V` `CurarrCu^(2+)+2e^(-),E^(@)=-0.34V` `2Fe^(3+)+Curarr2Fe^(2+)+Cu^(2+),E^(@)=+0.43 V` This reaction is Feasible since `E_("cell")^(@)` is +ve (d) `AgrarrAg^(+)+E^(-)E^(2)=-0.80 V` `Fe^(3+)+E^(-)rarrFe^(2+),E^(@)=-0.03 V` This reaction is not feasible since `E_(cell)^(@)` is -ve thus option (d) is correct |
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