Chapter 19, Problem 63A

Interpretation Introduction

Interpretation:

The redox reaction in the sapphire, oxidizing and reducing agents are to be written

Concept introduction:

Redox reaction is the reaction in which both oxidation and reduction takes place.

Oxidation is loss of electrons and reduction is gain of electrons.

Answer to Problem 63A

The reaction is:

$2F{e}^{2+}+T{i}^{+4}\to 2F{e}^{3+}+T{i}^{+2}$

Oxidizing agent: $T{i}^{+4}$

Reducing agent: $F{e}^{+2}$

Explanation of Solution

$F{e}^{2+}+T{i}^{+4}\to F{e}^{3+}+T{i}^{+2}$

Assigning oxidation state:

$\stackrel{+2}{F{e}^{2+}}+\stackrel{+4}{T{i}^{+4}}\to \stackrel{+3}{F{e}^{3+}}+\stackrel{+2}{T{i}^{+2}}$

Oxidation number of iron is increasing and titanium is decreasing. Therefore,

Oxidizing agent: $T{i}^{+4}$ Reducing agent: $F{e}^{+2}$

Conclusion

Oxidation number of iron is increasing and titanium is decreasing. Therefore,

Oxidizing agent: $T{i}^{+4}$ Reducing agent: $F{e}^{+2}$