Chapter 19, Problem 68A

(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 68A

$2MoC{l}_5+5S^{-2}\to 2Mo{S}_2+10C{l}^{-}+S$

Explanation of Solution

The given redox reaction is:

$MoC{l}_5+S^{-2}\to Mo{S}_2+C{l}^{-}+S$

Assigning the oxidation number-

$\stackrel{+5}{Mo}C{l}_5+\stackrel{-2}{S^{-2}}\to \stackrel{+4}{Mo}{S}_2+\stackrel{-1}{C{l}^{-}}+\stackrel{0}{S}$

The oxidation numbers of atoms shows that

$S$ is oxidized and $Mo$ is reduced.

Change in oxidation number of $S=+2$

Change in oxidation number of $Mo=-1$

Thus adjusting the coefficients:

$2MoC{l}_5+5S^{-2}\to 2Mo{S}_2+10C{l}^{-}+S$

(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 68A

$2TiC{l}_6{}^{2-}+Zn\to 2T{i}^{+3}+12C{l}^{-}+Z{n}^{+2}$

Explanation of Solution

The given redox reaction is:

$TiC{l}_6{}^{2-}+Zn\to T{i}^{+3}+C{l}^{-}+Z{n}^{+2}$

Assigning the oxidation number-

$\stackrel{+4}{Ti}{\stackrel{-1}{Cl}}_6{}^{2-}+\stackrel{0}{Zn}\to \stackrel{+3}{T{i}^{+3}}+\stackrel{-1}{C{l}^{-}}+\stackrel{+2}{Z{n}^{+2}}$

The oxidation numbers of atoms shows that

$Zn$ is oxidized and $Ti$ is reduced.

Change in oxidation number of $Zn=+2$

Change in oxidation number of $Ti=-1$

Thus adjusting the coefficients:

$2TiC{l}_6{}^{2-}+Zn\to 2T{i}^{+3}+12C{l}^{-}+Z{n}^{+2}$