Question

Consider the reaction, $H^{+}+I{O}_4^{-}+I^{-}\to I_2+H_{2}O$. The ratio of the coefficients of $IO{{}_4}^{-},I^{-}$ and $I_2$, respectively, are:

• A. $1:7:8$
• B. $1:2:4$
• C. $1:7:4$
• D. none of the above

Answer 1

The correct option is B $1:2:4$

The unbalanced redox equation is as follows:
$H^{+}+I{O}^{-}{}_4+I^{-}\to I_2+H_{2}O$
All atoms other than $H$ and $O$ are balanced.
The oxidation number of iodine changes from $+7$ (in $I{O}_4^{-}$) to $0$ (in $I_2$). The change in the oxidation number of iodine is $7$.
The oxidation number of iodine changes from $-1$ (in $I^{-}$) to $0$ (in $I_2$). The change in the oxidation number of iodine is $1$.
The increase in the oxidation number is balanced with a decrease in the oxidation number by multiplying $I^{-}$ and $I_2$ with $7$.
$H^{+}+I{O}_4^{-}+7I^{-}\to 4I_2+H_{2}O$
O atoms are balanced by adding $3$ water molecules on RHS.
$H^{+}+I{O}_4^{-}+7I^{-}\to 4I_2+4H_{2}O$
Hydrogen atoms are balanced by adding $7$ $H^{+}$ atoms on the LHS.
$8H^{+}+I{O}_4^{-}+7I^{-}\to 4I_2+4H_{2}O$
This is the balanced chemical equation.
Thus, the ratio of coefficients of $IO{{}_4}^{-},I^{-}$ and $I_2$ is $1:2:4$.