`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 ironC. `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 ironC. `Cu` reduces `Fe^(2+)` ionD. `Cu^(2+)` reduces `Fe`
Answer» Correct Answer - 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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`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 ironC. `Cu` reduces `Fe^(2+)` ionD. `Cu^(2+)` reduces `Fe`

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