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

1 Answer
Sep 16, 2017

Answer:

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.