Chapter 19.2, Problem 19PP

Interpretation Introduction

Interpretation:

The given redox reaction is to be balanced using oxidation number method.

Concept introduction:

Oxidation number method to balance ionic equation in acidic solution involve:-

• Assign the oxidation number to each element.
• Check which element is oxidized and which element is reduced.
• Check the change in oxidation number for the elements which are oxidized and reduced.
• Balance the change in oxidation number by adjusting the coefficients in the equation.
• Then add enough hydrogen ions and water molecules to the equation to balance hydrogen atoms on both sides.

Answer to Problem 19PP

$3H_{2}S+2N{O}_3{}^{-1}+2H^{+}\to 3S+2NO+4H_{2}O$

Explanation of Solution

The given redox reaction is:

$H_{2}S+N{O}_3{}^{-1}\to S+NO$

Assigning the oxidation number-

${\stackrel{+1}{H}}_2\stackrel{-2}{S}+\stackrel{+5}{N}{\stackrel{-2}{O}}_3{}^{-1}\to \stackrel{0}{S}+\stackrel{+2}{N}\stackrel{-2}{O}$

The oxidation numbers of atoms shows that

$S$ is oxidized and $N$ is reduced.

Change in oxidation number of $S=+2$

Change in oxidation number of $N=-3$

Thus adjusting the coefficients:

$3H_{2}S+2N{O}_3{}^{-1}\to 3S+2NO$

Now adding hydrogen ions on left side of equation to balance negative charge

$3H_{2}S+2N{O}_3{}^{-1}+2H^{+}\to 3S+2NO$

Adding water to right side:

$3H_{2}S+2N{O}_3{}^{-1}+2H^{+}\to 3S+2NO+4H_{2}O$

Conclusion

Thus the balanced redox equation is:

$3H_{2}S+2N{O}_3{}^{-1}+2H^{+}\to 3S+2NO+4H_{2}O$