Do reduction and oxidation occur together?

1 Answer
Aug 28, 2017

Answer:

Well, yes.....we conceive of #"oxidation"# as the loss of electrons.....

Explanation:

And we conceive of #"reduction"# as the gain of electrons.

Since charge, as well as mass, is CONSERVED in every chemical reaction, for every oxidation there must be a corresponding reduction, and we can assign #"oxidation numbers"# as arbitrary numbers with which we can assess #"conceptual electron transfer....."#

Now ammonia can oxidized to nitrate.......a formal oxidation of #N(-III)# to #N(+V)#, an EIGHT electron transfer.....

And thus for oxidation we write.....

#NH_3 +3H_2O rarrNO_3^(-)+9H^(+) +8e^(-)# #(i)#

Mass is balanced, and charge is balanced....so this is a reasonable representation of chemical change. Now the electrons are presumed to GO somewhere on oxidation; i.e. they cause a CORRESPONDING reduction of some other reagent (which of course is the OXIDIZING agent). A typical oxidizing agent is permanganate ion, #MnO_4^(-)#, which is REDUCED from #Mn(VII+)# to colourless #Mn^(2+)#, a 5 electron reduction.....

#MnO_4^(-) +8H^+ + 5e^(-)rarr Mn^(2+) + 4H_2O # #(ii)#

Charge and mass are balanced as required. Are they?

To complete the entire redox reaction, we cross multiply to remove the electrons.....#5xx(i) + 8xx(ii)#:

#8MnO_4^(-) +64H^+ +5NH_3 +15H_2O+40e^(-)rarr 8Mn^(2+) + 32H_2O +5NO_3^(-)+45H^(+) +40e^(-)#

And then we cancel out the common reagents......

#8MnO_4^(-) +cancel(64)19H^+ +5NH_3 +cancel(15H_2O)+cancel(40e^(-))rarr 8Mn^(2+) + cancel(32)17H_2O +5NO_3^(-)+cancel(45H^(+)) +cancel(40e^(-))#

To give finally.......

#8MnO_4^(-) +19H^+ +5NH_3 rarr 8Mn^(2+) +5NO_3^(-)+ 17H_2O#

Which, despite the whack coefficients, is balanced with respect to mass and charge. And we would see the deep purple colour of permanganate dissipate to give COLOURLESS #Mn^(2+)#.