1.

Half-cell reactions are given below : Mn^(2+)+2e^(-) to Mn,E_(0)=-1.18V 2(Mn^(3+)+e^(-) to Mn^(2+)),E_(0)=+1.51V then what is E_(0) for 3Mn^(2+) to Mn+2Mn^(3+) ?

Answer»

`-0.33V`, no REACTION
`-0.33V`, reaction can be possible
`-2.69V`, no reaction
`-2.69V`, no reaction

Solution :For given reactioin :
`underset(+3)(3MN^(2+)) to underset(0)(M)n+underset(+3)(2MN)^(3+)`
So, half OXIDATION : `2Mn^(2+) to 2Mn^(3+)+2e^(-)("Anode")`
`UNDERLINE("and half reduction : "Mn^(2+)+2e^(-) to Mn" (Cathode)")`
Total reaction : `3Mn^(2+) to Mn+2Mn^(3+)`
So `E_("reaction")^(@)=E_("reduction")^(@)("Cathode")-E_("red")^(@)("Anode")`
`=E_(Mn^(2+)|Mn)^(@)-E_(Mn^(3+)|Mn^(2+))^(@)`
`=(-1.18-1.51)`
`=-2.69V`
`E_("Reaction")^(@) lt 0` so no reaction is possible and here, `E_(("Reaction"))^(@)=-2.68V=-2.69V`.


Discussion

No Comment Found

Related InterviewSolutions