Question

Write the stoichiometric coefficient for the following reaction:
$x{I}_2+yO{H}^{-}+I{O}_3^{-}+z{I}^{-}+3H_{2}O$

	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)

$I_2+O{H}^{-}\to I{O}_3^{-}+I^{-}+H_{2}O$
Oxidation half: $I_2+O{H}^{-}\to I{O}_3^{-}+H_{2}O$

Reduction half : $I_2\to I^{-}$
Oxidation half-
Adding $O{H}^{-},I_2+12O{H}^{-}\to 2I{O}_3^{-}+6H_{2}O$
Adding electrons:
$I_2+12O{H}^{-}\to 2I{O}_3^{-}+6H_{2}O+10e^{-}$
Reduction half -
$I_2+2e^{-}\to 2I^{-}$
Balancing $e^{-}$,
$I_2+12O{H}^{-}\to 2I{O}_3^{-}+6H_{2}O+10e^{-}$
$I_2+2e\to 2I^{-}]\times 5$
Adding both reactions
$6I_2+12O{H}^{-}\to 2I{O}_3^{-}+10I^{-}+6H_{2}O$
Dividing by $2$,
$3I_2+6O{H}^{-}\to I{O}_3^{-}+5I^{-}+6H_{2}O$