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[chem geek]

Thiosulfate is supposed to reduce chlorine directly and not form combined chlorines. There are many possible reactions, but one of them is the following:

Na2S2O3•5H2O --> 2Na+ + S2O3(2-) + 5H2O

2S2O3(2-) + HOCl --> S4O6(2-) + OH- + Cl-

There are other reactions that take the thiosulfate all the way to sulfate or even sulfur, but I believe these are less likely. None of them form combined chlorine -- all produce chloride ion. The oxidation potentials are as follows:

2S2O3(2-) --> S4O6(2-) + 2e- ..... Eo = -0.08V

S2O3(2-) + 6OH- --> 2SO3(2-) + 3H2O + 4e- ..... Eo = +0.571V
2x( SO3(2-) + 2OH- --> SO4(2-) + H2O + 2e- ) ..... Eo = +0.93V
--------------------------------------------------------------------------
S2O3(2-) + 10OH- --> 2SO4(2-) + 5H2O + 8e- ..... Eo = +1.501V

The above reaction would result in the following with chlorine:

S2O3(2-) + 6OH- + 4HOCl --> 2SO4(2-) + 5H2O + 4Cl-
or equivalently
S2O3(2-) + 4HOCl + H2O --> 2SO4(2-) + 6H+ + 4Cl-
Thiosulfate Ion + Hypochlorous Acid + Water --> Sulfate Ion + Hydrogen Ion + Chloride Ion
or
S2O3(2-) + 4OCl- + H2O --> 2SO4(2-) + 2H+ + 4Cl-

which is very acidic. My experience from using thiosulfate drops in high chlorine (but < 10 ppm so as to not significantly affect the pH test directly) water and its affect on the pH test is that the reaction is alkaline, not acidic, so I believe these latter reactions don't occur. If we calculate actual oxidation potential for the 8 electron reaction then RT/nF = 8.314472*300/(8*96485.3415) = 0.00323 while ln(10^(14-7.5)^10) = -149.7 so the actual E = +1.501 - 0.00323 * 172.7 = 1.502 - 0.483 = +1.019V. This is still a very likely reaction so perhaps the rate of the reaction is slow (high activation energy?).

Richard