Chapter 8, Problem 47A

Interpretation Introduction

Interpretation :

Oxidation-reduction reactions must be written for each of the given metals and non-metals.

Concept Introduction :

In an oxidation-reduction reaction, there is change in oxidation numbers of the involved species takes place.

Answer to Problem 47A

Oxidation-reduction reactions are given below.

  $\begin{array}{l}\text{Ba(s)+}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g)}\to \text{BaO(s)}\text{.}\\ \text{Ba(s)+S(g)}\to \text{BaS(s)}\text{.}\\ {\text{Ba(s)+Cl}}_{\text{2}}\text{(g)}\to {\text{BaCl}}_{\text{2}}\text{(s)}\text{.}\\ {\text{3Ba(s)+N}}_{\text{2}}\text{(g)}\to {\text{Ba}}_{\text{3}}{\text{N}}_{\text{2}}\text{(s)}\text{.}\\ {\text{Ba(s)+Br}}_{\text{2}}\text{(g)}\to {\text{BaBr}}_{\text{2}}\text{(s)}\text{.}\end{array}$

  $\begin{array}{l}{\text{K(s)+O}}_{\text{2}}\text{(g)}\to {\text{KO}}_{\text{2}}\text{(s)}\text{.}\\ \text{K(s)+2S(g)}\to {\text{KS}}_{\text{2}}\text{(s)}\text{.}\\ \text{K(s)+}\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\to \text{KCl(s)}\text{.}\\ \text{3K(s)+}\frac{1}{2}{\text{N}}_{\text{2}}\text{(g)}\to {\text{K}}_{\text{3}}\text{N(s)}\text{.}\\ \text{K(s)+}\frac{1}{2}{\text{Br}}_{\text{2}}\text{(g)}\to \text{KBr(s)}\text{.}\end{array}$

  $\begin{array}{l}\text{Mg(s)+}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g)}\to \text{MgO(s)}\text{.}\\ \text{Mg(s)+S(g)}\to \text{MgS(s)}\text{.}\\ {\text{Mg(s)+Cl}}_{\text{2}}\text{(g)}\to {\text{MgCl}}_{\text{2}}\text{(s)}\text{.}\\ {\text{3Mg(s)+N}}_{\text{2}}\text{(g)}\to {\text{Mg}}_{\text{3}}{\text{N}}_{\text{2}}\text{(s)}\text{.}\\ {\text{Mg(s)+Br}}_{\text{2}}\text{(g)}\to {\text{MgBr}}_{\text{2}}\text{(s)}\text{.}\end{array}$

  $\begin{array}{l}{\text{Rb(s)+O}}_{\text{2}}\text{(g)}\to {\text{RbO}}_{\text{2}}\text{(s)}\text{.}\\ \text{Rb(s)+2S(g)}\to {\text{RbS}}_{\text{2}}\text{(s)}\text{.}\\ \text{Rb(s)+}\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\to \text{RbCl(s)}\text{.}\\ \text{3Rb(s)+}\frac{1}{2}{\text{N}}_{\text{2}}\text{(g)}\to {\text{Rb}}_{\text{3}}\text{N(s)}\text{.}\\ \text{Rb(s)+}\frac{1}{2}{\text{Br}}_{\text{2}}\text{(g)}\to \text{RbBr(s)}\text{.}\end{array}$

  $\begin{array}{l}\text{Ca(s)+}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g)}\to \text{CaO(s)}\text{.}\\ \text{Ca(s)+S(g)}\to \text{CaS(s)}\text{.}\\ {\text{Ca(s)+Cl}}_{\text{2}}\text{(g)}\to {\text{CaCl}}_{\text{2}}\text{(s)}\text{.}\\ {\text{3Ca(s)+N}}_{\text{2}}\text{(g)}\to {\text{Ca}}_{\text{3}}{\text{N}}_{\text{2}}\text{(s)}\text{.}\\ {\text{Ca(s)+Br}}_{\text{2}}\text{(g)}\to {\text{CaBr}}_{\text{2}}\text{(s)}\text{.}\end{array}$

  $\begin{array}{l}\text{2Li(s)+}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{Li}}_{\text{2}}\text{O(s)}\text{.}\\ \text{2Li(s)+S(g)}\to {\text{Li}}_{\text{2}}\text{S(s)}\text{.}\\ \text{Li(s)+}\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\to \text{LiCl(s)}\text{.}\\ \text{3Li(s)+}\frac{1}{2}{\text{N}}_{\text{2}}\text{(g)}\to {\text{Li}}_{\text{3}}\text{N(s)}\text{.}\\ \text{Li(s)+}\frac{1}{2}{\text{Br}}_{\text{2}}\text{(g)}\to \text{RbBr(s)}\text{.}\end{array}$

Explanation of Solution

Valency of Ba, Mg, Ca, S and O is 2.

Valency of K, Li, Rb, Cl and Br is 1.

Valency of N is 3.

Using these valencies all the oxidation-reduction reactions are written.

Here, Ba, Mg, Ca, Li, Rb, K are oxidized to Ba²⁺, Mg²⁺, Ca²⁺, Li⁺, Rb+ and K⁺.

S, O, Cl, N and Br are reduced to S^2-, O²- or O₂^-, Cl^-, N³- and Br^-.

Conclusion

Metals are oxidized and non-metals are reduced.