Question

Given $XN{a_2}HAs{O_3} + YNaBr{O_3} + ZHCl \to NaBr + {H_3}As{O_4} + NaCl$
The values of X, Y, Z in the given redox reaction are respectively
(A) $2,1,2$
(B) $2,1,3$
(C) $3,1,6$
(D) $3,1,4$

Answer 1

A balanced chemical equation contains an equal number of atoms on both sides i.e. reactants and products side following the law of conservation of mass. The atoms other than oxygen and hydrogen need to be balanced first then after hydrogen and oxygen.

Complete step by step answer:
We need to balance the given redox reaction for the given reaction
$XN{a_2}HAs{O_3} + YNaBr{O_3} + ZHCl \to NaBr + {H_3}As{O_4} + NaCl$. First, we will have to check whether the given reaction is balanced or not.
We will use the half-reaction method for balancing the reaction. In this process, we split the equation into two halves, then balanced both parts of the reaction separately. Finely we add them together to get the balanced chemical equation.
We will follow the given steps to get the balanced chemical equation. First, we will have to oxidize and reduce half of the equation.
Now we divide the equation into a separate half-reaction that is oxidation half and reduction half. We write the oxidation half and reduction half and we get the following reaction.
The oxidation half is as follows and the reduction half are as follows
$AsO_3^{3 - } \to AsO_4^{3 - }$. Here arsenic is oxidised from $+ 3$ to $+ 5$
$BrO_3^ - \to B{r^ - }$ . Here bromine is reduced from $+ 5$ to $- 1$
Now we will balance the atoms of the reaction except hydrogen and oxygen. As we can see that arsenic and bromine are already balanced so we will move to the next step.
As the reaction takes place in an acidic medium so we will have to add water molecules for balancing the oxygen atoms of the equation and ${H^ + }$ ions for balancing the hydrogen atoms in the equation.
Writing the required equations, we get
$AsO_3^{3 - } + {H_2}O \to AsO_4^{3 - } + 2{H^ + }$
$6{H^ + } + BrO_3^ - \to B{r^ - } + 3{H_2}O$
Now we will balance the charges in the given equation. We will multiply and make the number of electrons the same. The equations are as follows
$[AsO_3^{3 - } + {H_2}O + 2{e^ - } \to AsO_4^{3 - } + 2{H^ + }] \times 3$
$[6{H^ + } + BrO_3^ - \to B{r^ - } + 3{H_2}O + 6{e^ - }] \times 1$
Writing the equation after multiplication and equating it we get,
$3AsO_3^{3 - } + 3{H_2}O + 6{e^ - } \to 3AsO_4^{3 - } + 6{H^ + }$
$6{H^ + } + BrO_3^ - \to B{r^ - } + 3{H_2}O + 6{e^ - }$
Cancelling the equal terms on both sides of the equation, we get
$3AsO_3^{3 - } + BrO_3^ - \to AsO_4^{3 - } + B{r^ - }$
Writing the given equation in normal form, we get
$3N{a_2}HAs{O_3} + NaBr{O_3} + HCl \to NaBr + 3{H_3}As{O_4} + 6NaCl$
As we can see nine hydrogens are present on the right side so we need to add six to the coefficient of HCl to balance the equation so the final equation is
$3N{a_2}HAs{O_3} + NaBr{O_3} + 6HCl \to NaBr + 3{H_3}As{O_4} + 6NaCl$.Hence the values of x, y, z are $3,1,6$ respectively .

Hence the correct answer is option C.

Note: Redox reactions can also be balanced by the oxidation number method. It includes calculating the oxidation number based on the number of the atoms and multiplying it to keep both equations the same. We then add hydrogen ion or hydroxide ion to further balance the oxygen and hydrogen atom in the reaction.