Saved Bookmarks
| 1. |
Using the standard electrode potentials given in thepredict if the reaction between the following if feasible : (a) Fe^(3+) (a) and I^(-)(aq), (b)ag^(+)(aq)and Cu(s), (c )Fe^(3+)(aq)and Cu (s) (d) Ag(s)and Fe^(3+)(aq),(e )Br_(2)(aq)and Fe^(2+)(aq) |
|
Answer» Solution :(a) The possible reaction between `Fe^(3+) (aq)` and `I^(-)(aq)` is `2Fe^(3+)(aq)+2I^(-)(aq)rarr2Fe^(2+)(aq)+I_(2)(s)` The above redox reaction can be split into the following two half reaction one involving oxidation and the other redcution wiritng electrode potential for each half RECTION from OVERALL reaction `2Fe^(3+)(aq) +2I^(-)(aq)rarr2Fe^(2+)+I_(2)(s)` since the EMF for the above reaction `Ag^(+)` (aq) and Cu (s) is `Cu(s)+2Ag^(+)(aq)rarrCu^(2+)(aq)+2ag(s)` The above redox reaction can be split into the following two half reaction writing electrode potential for each half reaction form we have oxidation`Cu(s)rarrCu^(2)(aq)+2e^(-), E^(@)=-0.34 V` Reduction `Ag^(+)(aq)+E^(-)rarrAg(s)xx2, E^(@)=+0.80 V` overall reaction `Cu(s)+2Ag^(+)(aq)rarrCu^(2+) (aq)+2 (s),E^(@)=+0.46V` since the EMF of the above reaction comes out to be postive therefore the above reaction is feasible (c ) suppose the reaction between `Fe^(3+)(aq)rarr Cu^(aq)+2Ag(s),E^(@)=+0.46 V` since the EMF of the above reaction comes oiut to be postive therefore to the following equatin `Cu(s)+2Fe^(3+)(a)rarr3Cu(aq)+2Fe^(2+)(aq)` the above reaction can be split into the following two half reaction wirting elecrode potential for each half reaction from we have oxidation `Cu(s)rarrCu^(2+)(aq)+2e^(-) , E^(@)=-0.34 V` Reduction `Fe^(3+)(aq)+E^(-)rarrFe^(2+)(aq)xx2 , E^(2)=+0.77V` overall reaction `Cu(s)_+2Fe^(3+)(aq)rarrCu^(2+)+2Fe^(2+)(aq),E^(@)=+0.43 V` since the EMF of the reaction is postive therefore the above reaction is feasible alternatively if the raction between 1 `Fe^(3+)` (aQ) and Cu (s) occurs according to the followng equation The EMF of the reaction comes out to be -ve -0.36 V(-0.34 V -0.034 V)and hence this rection is not feasible (d) suppose the reaction between Ag(s) and`Fe^(3+)` (Aq) occurs accoding to the following eqation The above reaction can be split intothe following two half rection writing electrode potential for each half reaction from we have oxidation `Ag(s)rarrAg^(+)(aq)+e^(-), E^(@)=-0.80 V` Reduction `Fe^(3+)(aq)+E^(-)rarrFe^(2+)(aq), E^(@)=+0.77 V` since the EMF of the reaction is negative therefore the above reaction is not feasible Alternatively the rection between Ag (s) and `Fe^(3+)` (aQ) MAY occur according to the following equation `3Ag(s)+Fe^(3+)(aq)rarr3Ag^(+)(aq)+Fe(s)` on similar lines we can calcualte the EMF of this reaction comes to be even followng equation `Br_(2)(aq)+2Fe^(2+)(aq)rarr2Br^(-)(aq)+2Fe^(2+)(aq)` the above reaction can be split in to the followng two half rection writing electrode potential for each half reaction from the we have oxidation `Fe^(2+)(aq)rarrFe^(3+)(aq)+e^(-)xx2,E^(@)=-0.77 V` Reduction : `Br_(2)(aq)+2e^(-)rarrBr^(-)rarr2Br^(-)(aq),E^(@)=+1.09 V` overall reaction `2Fe^(2+)(aq)+Br_(2)(aq)rarr2Fe^(3+)(aq)+2 Br^(-)(aq)'E^(@)=+0.32 V` since the EMF for the above reaction is positive therefore this reaction is feasible |
|