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E^(@) values of Mg^(2+)//Mg is -2.37V, ...

`E^(@)` values of `Mg^(2+)//Mg` is `-2.37V,` of `Zn^(2+)//Zn` is `-0.76V` and `Fe^(2+)// Fe` is - `0.44V`
Which of the statement is correct ?

A

Zn will reduce `Fe^(2+)`

B

Zn will reducing `Mg^(2+)`

C

Mg oxidises Fe

D

Zn oxidises Fe

Text Solution

Verified by Experts

The correct Answer is:
A
`Zn+ Fe^(2+) rightarrow Zn^(2+) + Fe`
`E_(Cell)^(@) = E_(RP)^(@)("Cathode")-E_(RP)^(@)("anode")= - 0.44- (-0.76) = + 0.32V Rightarrow (+) Ve` therefore will be spontaneous
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