Question #21c09

1 Answer
Jan 7, 2016

Answer:

Thiosulfate ion, #S_2O_3^(2-)# is an interesting customer in the context of redox chemistry. The average oxidation number of sulfur in the anion is #II^+#. Do you agree? Why?

Explanation:

In fact, one of the sulfurs has replaced an oxygen. In #SO_4^(2-)#, sulfate anion, the oxidation number of sulfur is #VI^+#. We could designate the central sulfur in thiosulfate as #VI^+# also, and the terminal sulfur has the SAME oxidation state as oxygen, i.e. #II^-#. The average oxidation state of the sulfurs in thiosulfate is of course still #II^+#.

Anyway, to your problem:

Oxidation:

#S_2O_3^(2-) + 3H_2O(l) rarr 2SO_3^(2-) + 6H^+ + 4e^-# #(i)#

Reduction:

#S_2O_3^(2-) + 6H^+ +4 e^(-) rarr 2Sdarr + 3H_2O# #(ii)#

Both (i) and (ii) are balanced with respect to mass and charge (as they must be!), and, conveniently, we don't have to cross multiply because 4 electrons are involved in each reaction.)

Overall #(i)+(ii)#:

#2S_2O_3^(2-) rarr 2SO_3^(2-) + 2Sdarr#

So, what do you observe? Treatment of thiosulfate with strong acid should result in a fine white precipitate of elemental sulfur (reduction), and also the formation of (acrid) #H_2SO_3# (oxidation); clearly a disproportionation reaction has occurred.

Do I win £5-00?