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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+)` ion reduces Fe .

Text Solution

Verified by Experts

The correct Answer is:
B

Higher o.p `to` anode `to` oxidation take place .
`therefore` Lower o.p `to` cathode `to` Reduction take place .
`Fe + Cu^(2+) to Fe^(2+) + Cu`
`E_(cell)^(@) = E_(op(Fe))^(@) + E_(op(Cu))^(@)`
=`0.44 V + (0.32 V) ( therefore E_(Cu|Cu^(2+))^(@) = - E_(Cu^(2+)|Cu)^(@) )`
`=0.76 V`
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