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

Question

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

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. `-0.33 V`, the reaction will not occur
B. `-0.33 V`, the reaction wil occur
C. `-2.69 V`, the reaction will not occur
D. `-2.69 V`, the reaction will occur

Monjur Alahi

6 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 - C
`{:(Mn^(2+)+2e^(-) rarr Mn,E_(Mn^(2+)//Mn)^(@)=-1.18 V),(2Mn^(2+) rarr 2Mn^(3+)+2e^(-),E_(Mn^(3+)//Mn^(2+))^(@)=+1.51 V),(bar(3Mn^(2+) rarr 2Mn^(3+)+Mn),):}`
`E_("Redox process")^(@)=E_(Mn^(2+)//Mn^(2+))^(@)`
`=-1.81-1.51=-2.69 V`

6 views Answered 2023-02-05 15:29