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E^(@) for Fe//Fe^(2+) is +0.44 V and E^(...

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

A

`Cu` oxidises `Fe^(2+)` ion

B

`Cu^(2+)` oxidises iron

C

`Cu` reduces `Fe^(2+)` ion

D

`Cu^(2+)` reduces `Fe`

Text Solution

Verified by Experts

The correct Answer is:
B
`underset("Oxidation")(Fe)+underset("Reduction")(Cu^(2+)) rarr Fe^(2+) +Cu`
`E_("cell")^(@)=E_(Fe//Fe^(2+))^(@)+E_(Cu^(2+)//Cu)^(@)`
`=0.44+0.32`
`=0.76 V`
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