Question

Hydrogen peroxide oxidizes ${[Fe{(CN)_6}]^{4 - }}$ to ${[Fe{(CN)_6}]^{3 - }}$ in acidic medium but reduces ${[Fe{(CN)_6}]^{3 - }}$to ${[Fe{(CN)_6}]^{3 - }}$ in alkaline medium. The other products formed are, respectively
A.${H_2}O$ and $({H_2}O\, + \,{O_2})$
B.${H_2}O$ and $({H_2}O\, + \,O{H^ - })$
C.$({H_2}O\, + \,{O_2})$ and ${H_2}O$
D.$({H_2}O\, + \,{O_2})$ and $({H_2}O\, + \,O{H^ - })$

Answer 1

Hint: In this question we will write the reaction in both the cases and then we will check the oxidation state of the elements so that it matches with the given condition. Then the additional products in the reactions will be our answer. Increase in oxidation state is oxidation and decrease in oxidation state is known as reduction.

Complete step by step solution:
In the question we are given that hydrogen peroxide oxidizes ${[Fe{(CN)_6}]^{4 - }}$ to ${[Fe{(CN)_6}]^{3 - }}$. We will check the reactions for this. The reaction is as follows:
${[Fe{(CN)_6}]^{4 - }}\, + \,{H_2}{O_2}\, \to \,{[Fe{(CN)_6}]^{3 - }}\, + \,{H_2}O$
The oxidation number of oxygen atoms in hydrogen peroxide is -1. Oxidation number of oxygen atoms of water is -2.
On the other hand the compound's oxidation state changes from -4 to -3.
So we see that as oxygen atoms undergo reduction. So hydrogen peroxide undergoes reduction and oxidizes the compound. As we can see in the reaction the additional product is water.
We will now check the next reaction:
In the next reaction in the alkaline medium the compound undergoes reduction.
${[Fe{(CN)_6}]^{ - 3}}\, + \,{H_2}{O_2}\, \to \,{[Fe{(CN)_6}]^{4 - }}\, + \,{O_2}$
The oxygen atom in hydrogen peroxide is -1 whereas in the product side it is in elemental state so oxidation state is 0.
So as there is an increase in oxidation state it undergoes oxidation.
 The compound undergoes reduction as there is a decrease in oxidation number.
So the other compound in the reaction is oxygen.

The correct option is A ${H_2}O\,$and $\,({H_2}O\, + \,{O_2})\,\,$

Note:
Always remember that the oxidation state of a neutral compound is zero. To calculate oxidation state for an element of a compound we take the unknown oxidation state as x and then equate it with the total charge of the compound.