How is #"Mohr's salt"# oxidized by #"potassium permanganate"# to give #Fe^(3+)# and #Mn^(2+)#? How is #"hydrogen peroxide"# oxidized by the same reagent to give #"dioxygen gas"#?

1 Answer
Mar 23, 2017

Answer:

Well, in the first scenario, #"ferrous ion"# is oxidized to #"ferric ion"#:

Explanation:

You used the so-called #"Mohr's salt"#, which is a stable and convenient source of #"Fe(II) salt"#, i.e. #[(NH_4)_2Fe(SO_4)_2]*6H_2O#. We can ignore the ammonium and sulfate shrubbery, and represent the redox transition:

#Fe^(2+) rarr Fe^(3+) + e^-# #(i)#

Note that the ammonium ions, and the sulfate ions are simply along for the ride. They do not take place in the redox transition, and so may be eliminated from consideration. Meanwhile the electron acceptor, was the OXIDANT, the permanganate ion. This is reduced from #Mn(VII+)# to #Mn(II+)#, according to the following stoichiometry:

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

And #5xx(i) + (ii)=#

#MnO_4^(-) +5Fe^(2+) +8H^(+)rarr Mn^(2+) +5Fe^(3+) + 4H_2O# #(iii)#

And what we would we observe in the reaction? The macroscopic colour change observed would be from deeply purple (the permanganate colour) to almost colourless #Mn^(2+)# ion. The colour change is important as it allows you to vizualize the stoichiometric endpoint.

And in the second scenario, #"hydrogen peroxide"# is oxidized to #"dioxygen gas"#; i.e. #2xxO^(-I)rarrO_2^(0)#:

#H_2O_2 rarr O_2(g) +2H^+ +2e^-# #(iv)#

And again we cross-multiply: #2xx(ii)+5xx(iv):#

#2MnO_4^(-) +5H_2O_2+6H^(+)rarr 2Mn^(2+) +8H_2O+5O_2(g)#

Which (I think) is balanced with respect to mass and charge.

Again, you can very accurately determine the stoichiometric endpoints by means of the pronounced colour change.

If I have missed the point of your question, would you clarify?