Question

In the balanced reaction,
\[xBr{O_3}^ - + yC{r^{3 + }} + z{H_2}O \to pB{r_2} + qCr{O_4}^{2 - } + r{H^ + }\]
What are x,y,z respectively?
(A) 6, 10, 11
(B) 6, 10, 20
(C) 6, 8, 22
(D) 6, 10, 22

Answer 1

Hint: In order to balance this redox reaction, first find the oxidation number of the atoms in all the species. Then try to multiply the number of moles of the species in a way that the number of such atoms are equal in both left and right hand side of the equation.

Complete step by step solution:
We will use the oxidation number method to balance the given reaction.
For $Br{O_3}^ - $ ,
Overall charge = Oxidation number of Br + 3( Oxidation number of O)
-1 = Oxidation number of Br + 3(-2)
Oxidation number of Br = 6-1 = +5
For $B{r_2}$,
- We know that if any molecule is made up of the same elements, then its oxidation number is taken as zero.
For $Cr{O_4}^{2 - }$,
Overall charge = Oxidation number of Cr + 4(Oxidation number of O)
-2 = Oxidation number of Cr + 4(-2)
Oxidation number of Cr = 8 – 2 = +6
- Now, to balance the equation, we need to multiply the number of moles of the species by some number according to the change in oxidation number.
- We can see that the oxidation number of Br atom decreases from +5 to 0. Oxidation number of $Cr$ atoms increases from +3 to +6. So, to make the reaction balanced, we need to multiply $B{r_2}$ by 6 and $Br{O_3}^ - $ by 6. Then we need to multiply $C{r^{3 + }}$ and $Cr{O_4}^{2 - }$ with 10.
- Then we also need to balance the hydrogen and oxygen atoms in the reaction as well. Here, ${H^ + }$ ions are given in the right hand side as water acts as an acid here. So, if we multiply ${H^ + }$ with 44 and ${H_2}O$with 22, then we will get the balanced redox reaction.
So, we can re-write the balanced reaction as
\[6Br{O_3}^ - + 10C{r^{3 + }} + 22{H_2}O \to 3B{r_2} + 10Cr{O_4}^{2 - }44{H^ + }\]
Now, we can say that the values of x,y and z are 6, 10 and 22 respectively.

Thus, the correct answer of the question is (D).

Note: Note that to balance this kind of redox reaction, mainly two methods are available. One is the oxidation number method and the other is the half reaction method. In half reaction method, we write the oxidation and reduction half reaction first and then balance the reaction using those reactions.