Consider the disproportionation reaction of in hot, concentrated alkali: What is the total number of moles of electrons transferred when exactly 3 moles of undergo this reaction?

Question

Consider the disproportionation reaction of $I_2$ in hot, concentrated alkali: 
 $3I_2 + 6OH^- \rightarrow 5I^- + IO_3^- + 3H_2O$ 
 What is the total number of moles of electrons transferred when exactly 3 moles of $I_2$ undergo this reaction?

Prashant Kotian · Accepted Answer

Correct answer: 5. In the reactant , there are 6 Iodine atoms (O.S. = 0). 5 of these atoms reduce to (-1), gaining a total of 5 moles of electrons. 1 of these atoms oxidizes to (+5), losing a total of 5 moles of electrons. Thus, the net electron transfer is 5 moles.

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