Question

In the following reaction :
$H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to Cl{O}_2^{-}\left(aq\right)+O_2\left(g\right)$
Moles of $O{H}^{-}$ and $H_{2}O$ respectively present in a balanced equation in a basic medium are x and y. Calculate the value of x+y:

Answer 1

$H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to Cl{O}_2^{-}\left(aq\right)+O_2\left(g\right)$

Oxidation state of $O$ in $H_2{O}_2=-1$
Oxidation state of $O$ in $O_2=0$
Oxidation state of $Cl$ in $C{l}_2{O}_7=+7$
Oxidation state of $Cl$ in $Cl{O}_2^{-}=+3$

Clearly, $H_2{O}_2$ is undergoing oxidation and $C{l}_2{O}_7$ is undergoing reduction.
using the formula of n-factor given above,
$n_f$ of $H_2{O}_2=2$
$n_f$ of $C{l}_2{O}_7=8$
simplest ratio of $n_f$ of $H_2{O}_2:C{l}_2{O}_7$ is $1:4$

Cross multiplying these with $n_f$ of each other.
we get,
$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to Cl{O}_2^{-}\left(aq\right)+O_2\left(g\right)$

Balancing the main elements on both sides, generally oxygen is not balanced in this step but we will balance as it is undergoing oxidation in this reaction.

$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)$

Adding the $H_{2}O$ to balance the oxygen,


$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+3H_{2}O$

Adding $H^{+}$ to balance hydrogen,

$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+3H_{2}O+2H^{+}$

Now adding $O{H}^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2}O$,
$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)+2O{H}^{-}\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+3H_{2}O+2H^{+}+2O{H}^{-}$

$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)+2O{H}^{-}\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+5H_{2}O$

This is the final balanced equation.

We can also see that charge on both sides is -2. which also indicates that the equation is balanced now.

$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)+2O{H}^{-}\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+5H_{2}O$

So, 2, 5 moles of $O{H}^{-}$ and $H_{2}O$ respectively are present in balanced equation.
Hence, 5+2=7