Question

Write the stoichiometric coefficient for the following reaction:
$xI_2 + yOH^- + IO_3^- + zI^- + 3H_2O$

	x	y	z
(a)	6	3	5
(b)	3	2	3
(c)	3	6	5
(d)	3	3	3

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (C)

$I_{2} + OH ^{-} \to IO_{3}^{-} + I^{-} + H_{2}O$
Oxidation half: $I_{2} + OH ^{-} \to IO_{3}^{-} + H_{2}O$

Reduction half : $I_{2} \to I^{-}$
Oxidation half-
Adding $OH^{-}, I_{2} + 12OH^{-}\to 2IO_{3}^{-} + 6H_{2}O$
Adding electrons:
$I_{2} + 12OH^{-} \to 2IO_{3}^{-} + 6H_{2}O + 10e^{-}$
Reduction half -
$I_{2} + 2e^{-} \to 2I^{-}$
Balancing $e^{-}$,
$I_{2} + 12OH^{-} \to 2IO_{3}^{-} + 6H_{2}O + 10e^{-}$
$I_{2} + 2e \to 2I^{-}] \times 5$
Adding both reactions
$6I_{2} + 12OH^{-}\to 2IO_{3}^{-} + 10I^{-}+ 6H_{2}O$
Dividing by $2$,
$3I_{2} + 6OH^{-}\to IO_{3}^{-} + 5I^{-}+ 6H_{2}O$