Question

Consider a redox reaction :

$xN{a}_{2}HAs{O}_3+yNaBr{O}_3+zHCl\to NaBr+H_{3}As{O}_4+NaCl$

where $x,y,z$ are the stoichiometric coefficients of $N{a}_{2}HAs{O}_3,NaBr{O}_3,HCl$ respectively.
The value of $x+y+z$ in the above balanced redox reaction is :

• A. $3$
• B. $9$
• C. $10$
• D. $12$

Answer 1

The correct option is C $10$

The equation before balancing is:
$N{a}_{2}HAs{O}_3+NaBr{O}_3+HCl\to NaBr+H_{3}As{O}_4+NaCl$

Oxidation state $As$ in $N{a}_{2}HAs{O}_3=+3$
Oxidation state $As$ in $H_{3}As{O}_4=+5$
Oxidation state $Br$ in $NaBr{O}_3=+5$
Oxidation state $Br$ in $NaBr=-1$

$N{a}_{2}HAs{O}_3$ is undergoing oxidation and is the reducing agent
$NaBr{O}_3$ is undergoing reduction and is the oxidising agent..


$n_f$ of $N{a}_{2}HAs{O}_3=2$
$n_f$ of $NaBr{O}_3=6$

here 2 is common in both n-factors.

$\Rightarrow$Ratio of n-factors=1:3

Cross multiplying these with ratio of $n_f$'s of each other.
we get,
$3N{a}_{2}HAs{O}_3+NaBr{O}_3+HCl\to NaBr+H_{3}As{O}_4+NaCl$

Balancing $As$,
$3N{a}_{2}HAs{O}_3+NaBr{O}_3+HCl\to NaBr+3H_{3}As{O}_4+NaCl$

As of now oxygen already got balanced,

Balancing $Na$,
$3N{a}_{2}HAs{O}_3+NaBr{O}_3+6HCl\to NaBr+3H_{3}As{O}_4+6NaCl$

After this when we observe every element is balanced. so no need to follow further steps.
We can also see that charge on both sides is 0. which also indicates that the equation is balanced now.


$3N{a}_{2}HAs{O}_3+NaBr{O}_3+6HCl\to NaBr+3H_{3}As{O}_4+6NaCl$

This is the final balanced equation.

Comparing to the equation given in question, we get,
The value of $x,y,z$ to be $3,1,6$ respectively.
Value of $x+y+z=3+1+6=10$