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Using the standard electrode potential, ...

Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
`E^(θ)` 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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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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