Chapter 20.3, Problem 24SP

(a)

Interpretation Introduction

Interpretation: The redox equation using the oxidation-number-change method is to be balanced.

Concept Introduction: A chemical process in which electrons are moved between two reactants is referred to as a redox reaction. The redox reactions include combination reactions, combustion reactions, decomposition reactions, and single-replacement reactions.

Answer to Problem 24SP

The redox equation using the oxidation-number-change method is given as follows:

  ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}\left(s\right)+8{\text{HNO}}_{\text{3}}\left(aq\right) \to 2\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\left(aq\right)+2\text{NO}\left(g\right)+3\text{S}\left(s\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

In the oxidation-number-change method, balancing the redox equation is done by comparing the increase and decrease in oxidation numbers.

The given redox equation is as follows:

  ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}\left(s\right)+{\text{HNO}}_{\text{3}}\left(aq\right) \to \text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\left(aq\right)+\text{NO}\left(g\right)+\text{S}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The redox equation is now written with its charge as follows:

  $\stackrel{+3}{{\text{Bi}}_{\text{2}}}\stackrel{-2}{{\text{S}}_{\text{3}}}\left(s\right)+\stackrel{+1}{\text{H}}\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_{\text{3}}}\left(aq\right) \to \stackrel{+3}{\text{Bi}}{\left(\stackrel{+5}{\text{N}}\stackrel{-2}{{\text{O}}_{\text{3}}}\right)}_{\text{3}}\left(aq\right)+\stackrel{+2}{\text{N}}\stackrel{-2}{\text{O}}\left(g\right)+\stackrel{0}{\text{S}}\left(s\right)+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}\left(l\right)$

There is change in oxidation number of nitrogen from $+5$ to $+2$ .

There is change in oxidation number of sulfur from $-2$ to $0$ .

Balance the lost and gained electrons.

  $\begin{array}{c}\text{N:}{\text{+3e}}^{-}{\text{×2=6e}}^{-}\\ \text{S:}-{\text{2e}}^{-}\text{×3=}-{\text{6e}}^{-}\end{array}$

The redox equation is given as follows:

  ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}\left(s\right)+2{\text{HNO}}_{\text{3}}\left(aq\right) \to \text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\left(aq\right)+2\text{NO}\left(g\right)+3\text{S}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Now adjust the coefficients to get the balanced redox equation.

  ${\text{Bi}}_{\text{2}}{\text{S}}_{\text{3}}\left(s\right)+8{\text{HNO}}_{\text{3}}\left(aq\right) \to 2\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\left(aq\right)+2\text{NO}\left(g\right)+3\text{S}\left(s\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(b)

Interpretation Introduction

Interpretation: The redox equation using the oxidation-number-change method is to be balanced.

Concept Introduction: A chemical process in which electrons are moved between two reactants is referred to as a redox reaction. The redox reactions include combination reactions, combustion reactions, decomposition reactions, and single-replacement reactions.

Answer to Problem 24SP

The redox equation using the oxidation-number-change method is given as follows:

  ${\text{SbCl}}_{\text{5}}\left(aq\right)+2\text{KI}\left(aq\right) \to {\text{SbCl}}_{\text{3}}\left(aq\right)+2\text{KCl}\left(aq\right)+{\text{I}}_{\text{2}}\left(s\right)$

Explanation of Solution

In the oxidation-number-change method, balancing the redox equation is done by comparing the increase and decrease in oxidation numbers.

The given redox equation is as follows:

  ${\text{SbCl}}_{\text{5}}\left(aq\right)+\text{KI}\left(aq\right) \to {\text{SbCl}}_{\text{3}}\left(aq\right)+\text{KCl}\left(aq\right)+{\text{I}}_{\text{2}}\left(s\right)$

The redox equation is now written with its charge as follows:

  $\stackrel{+5}{\text{Sb}}\stackrel{-1}{{\text{Cl}}_{\text{5}}}\left(aq\right)+\stackrel{+1}{\text{K}}\stackrel{-1}{\text{I}}\left(aq\right) \to \stackrel{+3}{\text{Sb}}\stackrel{-1}{{\text{Cl}}_{\text{3}}}\left(aq\right)+\stackrel{+1}{\text{K}}\stackrel{-1}{\text{Cl}}\left(aq\right)+\stackrel{0}{{\text{I}}_{\text{2}}}\left(s\right)$

There is change in oxidation number of antimony from $+5$ to $+3$ .

There is change in oxidation number of iodine from $-1$ to $0$ .

Balance the lost and gained electrons.

  $\begin{array}{c}\text{Sb:}{\text{+2e}}^{-}{\text{×1=2e}}^{-}\\ \text{I:}-1{\text{e}}^{-}\text{×2=}-2{\text{e}}^{-}\end{array}$

The redox equation is given as follows:

  ${\text{SbCl}}_{\text{5}}\left(aq\right)+2\text{KI}\left(aq\right) \to {\text{SbCl}}_{\text{3}}\left(aq\right)+\text{KCl}\left(aq\right)+{\text{I}}_{\text{2}}\left(s\right)$

Now adjust the coefficients to get the balanced redox equation.

  ${\text{SbCl}}_{\text{5}}\left(aq\right)+2\text{KI}\left(aq\right) \to {\text{SbCl}}_{\text{3}}\left(aq\right)+2\text{KCl}\left(aq\right)+{\text{I}}_{\text{2}}\left(s\right)$