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

Question

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

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

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 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.

4 views Answered 2023-02-05 15:29