How do we quantitatively represent the oxidation of sodium thiosulfate, #S_2O_3^(2-)#, by dichromate ion, #Cr_2O_7^(2-)#?

1 Answer
Jan 3, 2016

Answer:

This is a somewhat complicated redox equation. You should end with #Cr^(3+)# and sulfate ions.

Explanation:

#Cr(VI)# is reduced to #Cr(III)#:

#Cr_2O_7^(2-) +14H^(+)+ 6e^(-)rarr 2Cr^(3+) + 7H_2O(l)# #(i)#

Thiolsulfate is oxidized to sulfate:

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

Both equations (I think) are balanced with respect to mass and charge (as they must be!).

So I cross multiply to give the overall redox equation (I wish to remove the electrons!) and cancel common reagents:

#4xx(i)+3xx(ii)#:

#4Cr_2O_7^(2-) +26H^(+)+ 3S_2O_3^(2-) rarr 8Cr^(3+) + 6SO_4^(2-) + 13H_2O(l)#

So at the end of your titration, you should have #Cr^(3+)#, and sulfate ions; acidium ions will be in excess with these reactions. What should signal the endpoint in this redox reaction?

Note that in thiosulfate, I have always found it useful to regard the terminal sulfur as the precise analogue of oxygen, which has the #-II# oxidation state, whereas the central sulfur has a #VI^+# oxidation state, as in sulfate. The average sulfur oxidation state in thiosulfate is still #II^+#.