Discussion Forum

 Given below  are the half  -cell reactions

$Mn^{2+}+e^{-}\rightarrow Mn;E^{0}=-1.18eV$

$2(Mn^{3+}+e^{-}\rightarrow Mn^{2+});E^{0}=+1.51eV$ the E° for    $3Mn^{2+}\rightarrow Mn+2 Mn^{3+}$ will be

Answer:

Explanation:

 Standard electrode potential  of reaction  [E°]  can be calculated as

   $E_{cell}^{0}=E_{R}-E_{P}$

where,  $E_{R}$   = SRP of reactant

           $E_{P}$   = SRP of product

   If      $E_{cell}^{0}$   = +ve  , then reaction is  spontaneous  otherwise non-spontaneous

$Mn^{3+}$      $\underrightarrow{E_{1}^{0}=1.51V}$        $Mn^{2+}$                

  $Mn^{2+}$       $\underrightarrow{E_{2}^{0}=1.181V}$     Mn

   For   $Mn^{2+}$ disproportionation

             E° =-1.51 V  - 1.18 v   =-2.69 v <  0