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
$(i) I_{2} \times 2I^{-} + IO^{3-}$
$(ii) I_{2} + 2e^{-} \times 2I^{-} ...(1)$
Oxidation half-cell
$(i) I_{2} \to 2I^{-} + IO^{3-}$
$(ii) I_{2} \to 2IO_{3-} + 10e^{-}$
$(iii) I_{2} + 12 ^{-}OH \to 2IO^{3-} 3- + 10 e^{-}$
$(iv) I_{2} + 12 ^{-}O H \to 2IO^{3-} + 10 e^{-} + 6H_{2}O.... (2)$ Multiply eqn. $(1) \times 5 +$ eqn. (2)
$6I_{2} + I2^{-}OH \to 1OI^{-} + 2IO_{3}^{-} + 6H_{2}O$
Ratio of $\frac{IO_{3}^{-}}{I^{-}}=\frac{2}{10}=1.5$