Chapter 19, Problem 59A

(a)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Oxidation number method 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 balance the other elements.

Answer to Problem 59A

$C{l}_2+NaOH\to NaCl+HOCl$

Explanation of Solution

The given redox reaction is:

$C{l}_2+NaOH\to NaCl+HOCl$

Assigning the oxidation number-

$\stackrel{0}{C{l}_2}+NaOH\to Na\stackrel{-1}{Cl}+HO\stackrel{+1}{Cl}$

The oxidation numbers of atoms shows that

$Sb$ is oxidized and $Mn$ is reduced.

Change in oxidation number of $Cl=+2$

Change in oxidation number of $Cl=-2$

Thus adjusting the coefficients:

$C{l}_2+NaOH\to NaCl+HOCl$

(b)

Interpretation Introduction

Interpretation:

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

Concept introduction:

Oxidation number method 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 balance the other elements.

Answer to Problem 59A

$16HBr{O}_3\to 8B{r}_2+8H_{2}O+8O_2$

Explanation of Solution

The given redox reaction is:

$HBr{O}_3\to B{r}_2+H_{2}O+O_2$

Equation in ionic form:

$Br{O}_3{}^{-}\to B{r}_2+O_2$

Assigning the oxidation number-

$\stackrel{+5}{Br}{\stackrel{-2}{O}}_3{}^{-}\to \stackrel{0}{B{r}_2}+\stackrel{0}{O_2}$

The oxidation numbers of atoms shows that

$O$ is oxidized and $Br$ is reduced.

Change in oxidation number of $Operatom=+6$

Change in oxidation number of $Br=-5$

Half reactions are:

$Br{O}_3{}^{-}\to B{r}_2+O_2+6e^{-}$

$Br{O}_3{}^{-}+5e^{-}\to B{r}_2$

Now balancing the atoms and charges:

$2Br{O}_3{}^{-}+2H_{2}O\to B{r}_2+4O_2+6e^{-}+4H^{+}$

$2Br{O}_3{}^{-}+10e^{-}+12H^{+}\to B{r}_2+6H_{2}O$

Multiplying oxidation half reaction by 5 and reduction half by 3:

$16Br{O}_3{}^{-}+16H^{+}\to 8B{r}_2+8H_{2}O+8O_2$

Or

$16HBr{O}_3\to 8B{r}_2+8H_{2}O+8O_2$