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Given below are the half-cell reactions ...

Given below are the half-cell reactions :
`Mn^(2+)+2e^(-) rarr Mn, E^(@)=-1.18 V`
`2(Mn^(3+)+e^(-) rarr Mn^(2+)), E^(@)=+1.51 V`
The `E^(@)` for `3 Mn^(2+) rarr Mn+2Mn^(3+)` will be :

A

`-0.33" V "`, the reaction will occur

B

`-2.69" V "`, the reaction will not occur

C

`-2.69" V "`, the reaction will occur

D

`-0.33" V "`, the reaction will not occur

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `Mn^(2)+2e^(-) to Mn,E^(@)=-1.18" V "`
`(2Mn^(2)+2e^(-) to 2Mn^(3+)+2e^(-),E^(@)=-1.51" V ")/(3Mn^(2+)(aq) to Mn+2Mn^(3+),E^(@)=-(1.18+1.51V)=-2.69" V )`
Since `E^(@)` is negative, the reaction will not accur.
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