Question

Consider the disproportionation of Iodine to iodide and iodate ions. Using Oxidation method calculate the ratio of iodate and iodide ions formed in alkaline medium :

$I_2\to I{O}_3^{-}+I^{-}$

• A. $5:1$
• B. $1:5$
• C. $3:1$
• D. $1:3$

Answer 1

The correct option is B $1:5$

Consider the redox reaction:
$I_2\to I{O}_3^{-}+I^{-}$
Oxidation state of I in $I_2=+0$
Oxidation state of I in $I{O}_3^{-}=+5$
Oxidation state of I in $I^{-}-=-1$
Cleary, $I_2$ is undergoing oxidation in $I{O}_3^{-}$ and in $I^{-}$ it is undergoing reduction.
Formula used for the n-factor calculation :
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atoms}$
using the formula of n-factor given above,
$n_f$ of $I{O}_3^{-}=5$
$n_f$ of $I^{-}=1$
Cross multiplying these with $n_f$ of each other.
we get,
$I_2\to I{O}_3^{-}+5I^{-}$

Balancing the I on both sides,
$3I_2\to I{O}_3^{-}+5I^{-}$
Adding the $H_{2}O$ to balance the oxygen,
As the LHS of the equation contains 0 oxygens while RHS contains 3 oxygens, adding 3 $H_{2}O$ to LHS.
$3I_2+3H_{2}O\to I{O}_3^{-}+5I^{-}$
adding $H^{+}$ to balance hydrogen,
As LHS of the equation contains 6 hydrogens while RHS contains 0 hydrogens, so adding 6$H^{+}$ to RHS.
$3I_2+3H_{2}O\to I{O}_3^{-}+5I^{-}+6H^{+}$
Now adding $O{H}^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2}O$,
$3I_2+3H_{2}O+6O{H}^{-}\to I{O}_3^{-}+5I^{-}+6H^{+}+O{H}^{-}$
$3I_2+3H_{2}O+6O{H}^{-}\to I{O}_3^{-}+5I^{-}+6H_{2}O$
Eliminating the unrequired common $H_{2}O$ which is present on the both sides,
$3I_2+6O{H}^{-}\to I{O}_3^{-}+5I^{-}+3H_{2}O$
This is the final balanced equation.
We can also see that charge on both sides is -6. which also indicates that the equation is balanced now.
So, the required ratio is $1:5.$