`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+)` ionB. `Cu^(2+)` oxidises FeC. Cu reduces `Fe^(2+)` ionD. `Cu^(2+)` reduces Fe

`E^(@)` for `Fe//Fe^(2+)` is `+0.44 V` and `E^(@)` for `Cu//Cu^(2+)` i

1.	`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+)` ionB. `Cu^(2+)` oxidises FeC. Cu reduces `Fe^(2+)` ionD. `Cu^(2+)` reduces Fe
Answer» Correct Answer - B `E_(OP)^(@)` of `Fe gt E_(OP)^(@)` of Cu [ `because E_(RP (Fe))^(@) lt E_(RP (Cu))^(@)]` Thus Fe gets oxidised of `Fe to Fe^(2+) + 2e^(-)`, `Cu^(2+) + 2e^(-) to Cu`

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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,A. Cu oxidises `Fe^(2+)` ionB. `Cu^(2+)` oxidises FeC. Cu reduces `Fe^(2+)` ionD. `Cu^(2+)` reduces Fe

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