Question

During the disproportionation of iodine to iodide and iodate ions, the ratio of iodate and iodide ions formed in alkaline medium is:

• A. $1:5$
• B. $5:1$
• C. $3:1$
• D. $1:3$

Answer 1

Answer: (A)

Reduction half-cell
$(i)I_2\times 2I^{-}+I{O}^{3-}$
$(ii)I_2+2e^{-}\times 2I^{-}...(1)$
Oxidation half-cell
$(i)I_2\to 2I^{-}+I{O}^{3-}$
$(ii)I_2\to 2I{O}_{3-}+10e^{-}$
$(iii)I_2+12^{-}OH\to 2I{O}^{3-}3-+10e^{-}$
$(iv)I_2+12^{-}OH\to 2I{O}^{3-}+10e^{-}+6H_{2}O....(2)$ Multiply eqn. $(1)\times 5+$ eqn. (2)
$6I_2+I{2}^{-}OH\to 1O{I}^{-}+2I{O}_3^{-}+6H_{2}O$
Ratio of $\frac{I{O}_3^{-}}{I^{-}}=\frac{2}{10}=1.5$