We know that when we are trying to recognize iodate, we use sodium bisulphite to the solution and with the help of starch paper recognize the evolved $\ce{I2}$.But when I studied the books I found actually Iodide is formed? What's the solution to this controversy?


1 Answer 1


Not much controversy here, IMO. There is more than one reaction involved, with slightly more complex kinetics.

As so often, Wikipedia already knows, so I will just quote the relevant part of the following article: http://en.wikipedia.org/wiki/Iodine_clock_reaction

Iodate variation

An alternative protocol uses a solution of iodate ion (for instance potassium iodate) to which an acidified solution (again with sulfuric acid) of sodium bisulfite is added.

In this protocol, iodide ion is generated by the following slow reaction between the iodate and bisulfite:

$$\ce{ IO3^{−} + 3 HSO3^{−} -> I^{−} + 3 HSO4^{−}}$$

This is the rate determining step. The iodate in excess will oxidize the iodide generated above to form iodine:

$$\ce{ IO3^{−} + 5 I^{−} + 6 H+ -> 3 I2 + 3 H2O}$$

However, the iodine is reduced immediately back to iodide by the bisulfite:

$$\ce{ I2 + HSO3^{−} + H2O → 2 I^{−} + HSO4^{−} + 2 H+}$$

When the bisulfite is fully consumed, the iodine will survive (i.e., no reduction by the bisulfite) to form the dark blue complex with starch.

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