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E^(@) of Fe^(2 +) //Fe = - 0.44 V, E^(@)...

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

A

`Cu^(2+)` Oxidizes Fe

B

`Fe^(2+)` oxidizes Cu

C

Cu Reduces `Fe^(2+)`

D

Fe reduces `Cu^(2+)`

Text Solution

Verified by Experts

The correct Answer is:
D
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E^(@) for Fe//Fe^(2+) is +0.44 V and E^(@) for Cu//Cu^(2+) is -0.32 V . Then, in the cell,

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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

Copper from copper sulphate solution can be displacesd by. (The standard reduction potentials of some electrodes are given below): E^(o)(Fe^(2+)//Fe) = - 0.44 V, E^(o) (Zn^(2+)//Zn) = - 0.76 V E^(o) (Cu^(2+)//Cu) = + 0.34 V: E^(o)(Cr^(3+)//Cr) = - 0.74 V E^(o) (H^(+0)/H_(2)) = 0.00V

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