How is #"thiosulfate anion"#, #S_2O_3^(2-)#, oxidized by #"permanganate anion"#, #MnO_4^(-)#?

1 Answer
Jan 27, 2016

Answer:

#5S_2O_3^(2-) + 8MnO_4^(-) + 14H^(+) rarr 8Mn^(2+) + 10SO_4^(2-) + 7H_2O(l)#

Explanation:

Thiosulfate is oxidized to sulfate #S(II)rarrS(VI)#:

#S_2O_3^(2-) + 5H_2O rarr 2SO_4^(2-) + 10H^(+) + 8e^-# #(i)#

Permanganate (#Mn(VII)#) is reduced to #Mn^(2+)#:

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

If we take #5xx(i)+8xx(ii)# (we do this cross-multiplication so that electrons, #e^ -#, do not appear in the reaction), we get (finally!):

#8MnO_4^(-) + 5S_2O_3^(2-) +14H^(+)rarr 8Mn^(2+) +10SO_4^(2-)+7H_2O(l)#

This equation (I think) is balanced with respect to mass and charge (I did a double take when I realized there were 47 oxygens on each side!). The reaction is also self-indicating: deep purple permanganate ion is reduced to (almost) colourless #Mn^(2+)#. I leave it to you to give the complete reaction (i.e. where #MnO_4^-# appears as #KMnO_4# and thiosulfate appears as #Na_2S_2O_3#).

Note that I designated the AVERAGE oxidation state of sulfur in thiosulfate as #+II#. I could have treated the terminal sulfur as the analogue of oxygen in sulfate, i.e. a #-II# oxidation state, and the central sulfur as the analogue of the central sulfur in sulfate, i.e. #+VI#. The average oxidation state of #S# of course is the same, #+II# in each case.