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Given below are the half -cell reactions...

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

A

`-033V`, the reaction will occur

B

`-2.69V`, the reaction will not occur

C

`-2.269V`, the reaction will occur

D

`-0.33 V`, the reaction will not occur

Text Solution

Verified by Experts

The correct Answer is:
B
`Mn^(2+)(aq)+2e rarr Mn(s), DeltaG_(1)^(0)=2xxFxx1.18J" …[1]"`
`2Mn^(3+)(aq)+2erarr2Mn^(2+)(aq),`
`DeltaG_(2)^(0)=-2xxFxx1.5J" …[2]"`
Subtracting equation (2) from (1), we get -
`3Mn^(2+)(aq)rarr Mn(s)+2Mn^(3+)`
`DeltaG_(3)^(@)=DeltaG_(1)^(@)-DeltaG_(2)^(@)" ...[3]"`
or, `-nFE^(0)=(2.36+3.02)F`
or, `-2FE^(0)=5.38F" "therefore E^(@)=-2.69V`
As `E^(0)lt0`, reaction does not take place.
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