`E^(@)` for `Fe//Fe^(2+)` is `+0.44 V` and `E^(@)` for `Cu//Cu^(2+)` is `-0.32 V`. Then, in the cell

E^(@) of Fe^(2 +) //Fe = - 0.44 V, E^(@) of Cu //Cu^(2+) = -0.34 V . Then in the cell

Standard E^0 of the half cell Fe | Fe^(2+) is +0.44V and standard E^0 of half cell Cu|Cu^(2+) is -0.32V ,then

The E^(o) for half cells Fe//Fe^(2+) and Cu//Cu^(2+) are –0.44 V and +0.32 V respectively. Then which of these is true?

If e_(Fe^(2+) //Fe)^@ =- 0. 44 1V . And E_(Fe^(3+)//Fe^(2+))^(o) = 0. 771 V . The standard emf of the reaction Fe +2 Fe^(3+) rarr 3Fe^(2+) will be .

For the disproportion of copper: 2 Cu^(+) to Cu^(+2) + Cu E^(0) is :- Given E^(0) for Cu^(+2)//Cu is 0.34 V & E^(0) for Cu^(+2)//Cu^(+) is 0.15 V:

The standard reduction potential data at 25^(@)C is given below E^(@) (Fe^(3+), Fe^(2+)) = +0.77V , E^(@) (Fe^(2+), Fe) = -0.44V , E^(@) (Cu^(2+),Cu) = +0.34V , E^(@)(Cu^(+),Cu) = +0.52 V , E^(@) (O_(2)(g) +4H^(+) +4e^(-) rarr 2H_(2)O] = +1.23V E^(@) [(O_(2)(g) +2H_(2)O +4e^(-) rarr 4OH^(-))] = +0.40V , E^(@) (Cr^(3+), Cr) =- 0.74V , E^(@) (Cr^(2+),Cr) = - 0.91V , Match E^(@) of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below teh lists: {:(List-I,List-II),((P)E^(@)(Fe^(3+),Fe),(1)-0.18V),((Q)E^(@)(4H_(2)O hArr 4H^(+)+4OH^(+)),(2)-0.4V),((R)E^(@)(Cu^(2+)+Curarr2Cu^(+)),(3)-0.04V),((S)E^(@)(Cr^(3+),Cr^(2+)),(4)-0.83V):} Codes:

Copper from copper sulphate solution can be displaced by …………. The standard reducation potentials of some electrodes are given below: (a) E^(@) (Fe^(2+), Fe) = -0.44 V (b) E^(@) (Zn^(2+), Zn) = -0.76 V (c) E^(@) (Cu^(2+), Cu) = +0.34 V (d) E^(@) (H^(+), 1//2H_(2)) = +0.34 V