What would occur in the reaction between #"potassium permanganate"#, #KMnO_4# and #"sodium sulfite"#, #Na_2SO_3#?

1 Answer
Oct 6, 2016

Answer:

A redox reaction occurs. Sulfite should be oxidized to sulfate. Permanganate ion should be reduced to #Mn^(2+)#.

Explanation:

#"Reduction half equation: "Mn(VII+) rarr Mn(II+)#

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

#"Oxidation half equation: "S(IV+) rarr S(VI+)#

#SO_3^(2-) +H_2O rarr SO_4^(2-) + 2H^(+) + 2e^(-)# #(ii)#

Both equations are (I think) balanced with respect to mass and charge, as they must be if they reflect reality. The overall redox reaction excludes the electrons; so we take #2xx(i)+5xx(ii)#:

#2MnO_4^(-) +5SO_3^(2-)+6H^(+)rarr 2Mn^(2+) + 5SO_4^(2-)+3H_2O#

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

So what would you observe in this reaction? #MnO_4^-# is strongly coloured, and gives a beautiful deep purple solution. On the other hand, #Mn^(2+)# is almost colourless (very concentrated solutions are a pale rose). And thus this redox reaction is self-indicating, and proposes that #2# #"equiv"# permanganate reacts with #5# #"equiv"# of sulfite.

How did I know the oxidation/reduction products? Experience and practice, and actually doing the titrations and observing and explaining the colour change.