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Given below are the half cell reactions ...

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

A

`-2.69` V , the reaction will not occur

B

`-2.96` V , the reaction will occur

C

`-0.33` V , the reaction will not occur

D

`-0.33` V, the reaction will occur.

Text Solution

Verified by Experts

The correct Answer is:
A
(i) `Mb^(2+) + 2e^(-) rarr Mn, E^(@) = -1.18 V`
`DeltaG_(1)^(@)= -2 F(-1.18)=2.36 V`...(i)
(ii) `Mn^(3+) +e^(-) rarr Mn^(2+), E^(@)= +1.51 V`
`DeltaG_(2)^(@) = -F(1.51) = -1.51F` ....(ii)
Eq. (i) `-2 xx `eq. (ii)
`3Mn^(2+) rarr Mn + 2Mn^(3+)`
`DeltaG_(3)^(@) = DeltaG_(1)^(@)-2DeltaG_(1)^(@)`
`=[2.36-2-(1.51)]F`
`=(2.36 + 3.02)F=5.38 F`
But `DeltaG_(3)^(@)= -2FE^(@)`
`5.38 = -2FE^(@) :. E^(@)= -2.69V`
Since `E^(@)` value is negative the reaction will not occur.
So, option (A) is correct.
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