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Ion Electron Method

In the follow...

Question

In the following reaction :
$H_{2} O_{2} (a q) + C l_{2} O_{7} (a q) \to C l O_{2}^{-} (a q) + O_{2} (g)$
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:

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Solution

$H_{2} O_{2} (a q) + C l_{2} O_{7} (a q) \to C l O_{2}^{-} (a q) + O_{2} (g)$

Oxidation state of $O$ in $H_{2} O_{2} = - 1$
Oxidation state of $O$ in $O_{2} = 0$
Oxidation state of $C l$ in $C l_{2} O_{7} = + 7$
Oxidation state of $C l$ in $C l O_{2}^{-} = + 3$

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,
$4 H_{2} O_{2} (a q) + C l_{2} O_{7} (a q) \to C l O_{2}^{-} (a q) + O_{2} (g)$

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.

$4 H_{2} O_{2} (a q) + C l_{2} O_{7} (a q) \to 2 C l O_{2}^{-} (a q) + 4 O_{2} (g)$

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

$4 H_{2} O_{2} (a q) + C l_{2} O_{7} (a q) \to 2 C l O_{2}^{-} (a q) + 4 O_{2} (g) + 3 H_{2} O$

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

$4 H_{2} O_{2} (a q) + C l_{2} O_{7} (a q) \to 2 C l O_{2}^{-} (a q) + 4 O_{2} (g) + 3 H_{2} O + 2 H^{+}$

Now adding $O H^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2} O$ ,
$4 H_{2} O_{2} (a q) + C l_{2} O_{7} (a q) + 2 O H^{-} \to 2 C l O_{2}^{-} (a q) + 4 O_{2} (g) + 3 H_{2} O + 2 H^{+} + 2 O H^{-}$

$4 H_{2} O_{2} (a q) + C l_{2} O_{7} (a q) + 2 O H^{-} \to 2 C l O_{2}^{-} (a q) + 4 O_{2} (g) + 5 H_{2} O$

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.

$4 H_{2} O_{2} (a q) + C l_{2} O_{7} (a q) + 2 O H^{-} \to 2 C l O_{2}^{-} (a q) + 4 O_{2} (g) + 5 H_{2} O$

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

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