Question

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

Answer 1

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 `Mn^(2+)`

`1.` `Mn(+VII)rarrMn(+IV):`

`underbrace(MnO_4^(-))_"deep red" +3e^(-)+4H^+ rarrMnO_2+2H_2O`

This normally occurs in basic conditions so we add `4xxHO^-` to each side.....

`MnO_4^(-) +3e^(-) +2H_2Orarrunderbrace(MnO_2(s))_"brown black solid" +4HO^-`

`2.` `Mn(+VII)rarrMn(+II):`

`MnO_4^(-) +5e^(-)+8H^+ rarr underbrace(Mn^(2+))_"almost colourless"+4H_2O`

`3.` `Mn(+VII)rarrMn(+III):`

`MnO_4^(-) +4e^(-)+5H^+ rarr Mn(OH)_3+H_2O`

And again we add `5xxHO^-` to each side because basic conditions are required.....

`MnO_4^(-) +4e^(-)+4H_2O rarr Mn(OH)_3(s)darr+5HO^-`

`4.` `Mn(+VII)rarrMn(+VI):`

`underbrace(MnO_4^(-))_"purple" +e^(-)rarr underbrace(MnO_4^(2-))_"green"`

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