Question

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

• A. x - ( 6 ), y - ( 3 ), z - (5)
• B. x - ( 3 ), y - ( 2), z - (3)
• C. x - ( 3 ), y - (6 ), z - (5)
• D. x - ( 3 ), y - ( 3 ), z - (3)

Answer 1

The correct option is C x - ( 3 ), y - (6 ), z - (5)

Given reaction:

$x{I}_2+yO{H}^{-}\to I{O}_3^{-}+z{I}^{-}+3H_{2}O$

in a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.

Balancing a chemical reaction as:

Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:

$\stackrel{0}{I_2}+O{H}^{-}\to \stackrel{+5}{I}{O}_3^{-}+\stackrel{-1}{I^{-}}+3H_{2}O$

Step 2: Identify the atoms that are oxidized and those that are reduced as:

Reduction: $\stackrel{0}{I_2}\to \stackrel{-1}{I^{-}}$

Oxidation: $\stackrel{0}{I_2}\to \stackrel{+5}{I}{O}_3^{-}$

Step 3: oxidation-number change is:

Reduction: $\stackrel{0}{I_2}\to 2\stackrel{-1}{I^{-}}$: Gain of 2 electron

Oxidation: : $\stackrel{0}{I_2}\to 2\stackrel{+5}{I}{O}_3^{-}$: Loss of total 10 electrons

Step 4: Balance the total change in oxidation number as:

Reduction: : $\stackrel{0}{I_2}\to 2\stackrel{-1}{I^{-}}\times 5$: Gain of 10 electron total

Oxidation: : $\stackrel{0}{I_2}\to 2\stackrel{+5}{I}{O}_3^{-}\times 1$: Loss of 10 electron

Reduction: : $5\stackrel{0}{I_2}\to 10\stackrel{-1}{I^{-}}$: Gain of 10 electron total

Oxidation: : $\stackrel{0}{I_2}\to 2\stackrel{+5}{I}{O}_3^{-}$: Loss of 10 electron

Step 5: Balance O atoms in oxidation reaction by adding $H_{2}O$ and then balance H by $H^{+}$ as:

Oxidation: : $\stackrel{0}{I_2}+6H_{2}O\to 2\stackrel{+5}{I}{O}_3^{-}+12H^{+}$

Step 6: For base catalysed reaction add $O{H}^{-}$ to both side to neutralize $H^{+}$ as:

Oxidation: : $\stackrel{0}{I_2}+6H_{2}O+12O{H}^{-}\to 2\stackrel{+5}{I}{O}_3^{-}+12H^{+}+12O{H}^{-}$

or,

Reduction: : $5\stackrel{0}{I_2}\to 10\stackrel{-1}{I^{-}}$

Oxidation: : $\stackrel{0}{I_2}+12O{H}^{-}\to 2\stackrel{+5}{I}{O}_3^{-}+6H_{2}O$

Thus overall reaction is:

$5\stackrel{0}{I_2}+\stackrel{0}{I_2}+12O{H}^{-}\to 10\stackrel{-1}{I^{-}}+2\stackrel{+5}{I}{O}_3^{-}+6H_{2}O$

OR $6\stackrel{0}{I_2}+12O{H}^{-}\to 10\stackrel{-1}{I^{-}}+2\stackrel{+5}{I}{O}_3^{-}+6H_{2}O$

or $3\stackrel{0}{I_2}+6O{H}^{-}\to 5\stackrel{-1}{I^{-}}+\stackrel{+5}{I}{O}_3^{-}+3H_{2}O$

compairing the above balanced reaction with given reaction we get:

$x{I}_2+yO{H}^{-}\to I{O}_3^{-}+z{I}^{-}+3H_{2}O$

thus $x=3,y=6,z=5$