The equivalent weight of Na2S2O3 in the reaction is
2Na2S2O3+ I2gives Na2S4O6 + 2NaI

The equivalent weight of oxidising or reducing agent is = molecular weight / number of electron per mole transfer during the reaction.
In above mentioned reaction there is transfer of 2 electrons per 2 moles of Na2S2O3 as follows :
2S2O32- → S4O62- + 2e-
So the equivalent weight of Na2S2O3 is same as molecular weight = 158.11 g mol-1‚Äč.


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