Question

Write the stoichiometric coefficient for the following reaction:

$xI_{2} + yOH^{-} \rightarrow IO_{3}^{-} + zI^{-} + 3H_{2}O$

x y z

• A. 6 3 5
• B. 3 2 3
• C. 3 6 5
• D. 3 3 3

Answer 1

Answer: (C)

3 6 5

Answer 2

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

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

Reduction half: $I_{2} \rightarrow I^{-}$

Oxidation half –

Adding $OH^{-},I_{2} + 12OH^{-} \rightarrow 2IO_{3}^{-} + 6H_{2}O$

Adding electrons:

$I_{2} + 12OH^{-} \rightarrow 2IO_{3}^{-} + 6H_{2}O + 10e^{-}$

Reduction half –

$I_{2} + 2e^{-} \rightarrow 2I^{-}$

Balancing $e^{-},I_{2} + 12OH^{-} \rightarrow 2IO_{3}^{-} + 6H_{2}O + 10e^{-}$

$I_{2} + 2e \rightarrow 2I^{-}\rbrack \times 5$

Adding both reactions

$6I_{2} + 12OH^{-} \rightarrow 2IO_{3}^{-} + 10I^{-} + 6H_{2}O$

Dividing by 2,

$3I_{2} + 6OH^{-} \rightarrow IO_{3}^{-} + 5I^{-} + 3H_{2}O$