Number of electron transferred in each case when potassium permanganate acts as oxidizing agent give MnO2, Mn²+ Mn(OH) 3 and MnO4²-?

Sep 16, 2017

Well you have to write the individual reduction half equations.....

Explanation:

Permanganate ion is reduced to a lower oxidation state manganese ion, and the deep red colour of permanganate dissipates to give a green colour in the case of manganate ion, a brown solid in the the case of the neutral oxide, or colourless in the case of $M {n}^{2 +}$

$1.$ $M n \left(+ V I I\right) \rightarrow M n \left(+ I V\right) :$

${\underbrace{M n {O}_{4}^{-}}}_{\text{deep red}} + 3 {e}^{-} + 4 {H}^{+} \rightarrow M n {O}_{2} + 2 {H}_{2} O$

This normally occurs in basic conditions so we add $4 \times H {O}^{-}$ to each side.....

$M n {O}_{4}^{-} + 3 {e}^{-} + 2 {H}_{2} O \rightarrow {\underbrace{M n {O}_{2} \left(s\right)}}_{\text{brown black solid}} + 4 H {O}^{-}$

$2.$ $M n \left(+ V I I\right) \rightarrow M n \left(+ I I\right) :$

$M n {O}_{4}^{-} + 5 {e}^{-} + 8 {H}^{+} \rightarrow {\underbrace{M {n}^{2 +}}}_{\text{almost colourless}} + 4 {H}_{2} O$

$3.$ $M n \left(+ V I I\right) \rightarrow M n \left(+ I I I\right) :$

$M n {O}_{4}^{-} + 4 {e}^{-} + 5 {H}^{+} \rightarrow M n {\left(O H\right)}_{3} + {H}_{2} O$

And again we add $5 \times H {O}^{-}$ to each side because basic conditions are required.....

$M n {O}_{4}^{-} + 4 {e}^{-} + 4 {H}_{2} O \rightarrow M n {\left(O H\right)}_{3} \left(s\right) \downarrow + 5 H {O}^{-}$

$4.$ $M n \left(+ V I I\right) \rightarrow M n \left(+ V I\right) :$

${\underbrace{M n {O}_{4}^{-}}}_{\text{purple" +e^(-)rarr underbrace(MnO_4^(2-))_"green}}$

Are charge and mass balanced in each scenario? All care taken, but no responsibility admitted.