Chapter U5.112, Problem 4E

(a)

Interpretation Introduction

Interpretation:

Balanced equations for the reaction of copper and tin with oxygen must be given along with the charge for cation and anion in the products. The species which gets oxidized must also be written.

Concept Introduction:

Copper and tin are oxidized by oxygen to their corresponding oxides.

Answer to Problem 4E

Balanced equations are given below.

  $\begin{array}{l}\text{Cu(s)+}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g)}\to \text{CuO(s)}\text{.}\\ {\text{Sn(s)+O}}_{\text{2}}\text{(g)}\to {\text{SnO}}_2\text{(s)}\text{.}\end{array}$

Charges of cations are +2 and +4 for Cu and Sn respectively.

Charge of oxide anion is -2.

Cu and tin are oxidized to CuO and SnO₂ respectively.

Explanation of Solution

  $\begin{array}{l}\text{Cu(s)+}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g)}\to \text{CuO(s)}\text{.}\\ {\text{Sn(s)+O}}_{\text{2}}\text{(g)}\to {\text{SnO}}_2\text{(s)}\text{.}\end{array}$

Cu and Sn are oxidized to corresponding oxides and oxygen is reduced.

Oxidation number of Cu and Sn is increased from 0 to +2 and +4 respectively.

Cu and Sn have valency 2 and 4 respectively. Accordingly these have reacted with different number of oxygen atoms. The charges are determined in cation and anion as per the valency of metals.

b)

Interpretation Introduction

Interpretation:

Balanced equations for the reaction of silver and aluminium with oxygen must be given along with the charge for cation and anion in the products. The species which gets oxidized must also be written.

Concept Introduction:

Silver and aluminium are oxidized by oxygen to their corresponding oxides.

Answer to Problem 4E

Balanced equations are given below.

  $\begin{array}{l}\text{2Ag(s)+}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{Ag}}_{\text{2}}\text{O(s)}\text{.}\\ \text{2Al(s)+}\frac{3}{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{Al}}_{\text{2}}{\text{O}}_3\text{(s)}\text{.}\end{array}$

Charges of cations are +1 and +3 for Ag and Al respectively.

Charge of oxide anion is -2.

Ag and Al are oxidized to Ag₂O and Al₂O₃ respectively.

Explanation of Solution

  $\begin{array}{l}\text{2Ag(s)+}\frac{1}{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{Ag}}_{\text{2}}\text{O(s)}\text{.}\\ \text{2Al(s)+}\frac{3}{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{Al}}_{\text{2}}{\text{O}}_3\text{(s)}\text{.}\end{array}$

Ag and Al are oxidized to corresponding oxides and oxygen is reduced.

Oxidation number of Ag and Al is increased from 0 to +1 and +3 respectively.

Ag and Al have valency 1 and 3 respectively. Accordingly these have reacted with different number of oxygen atoms. The charges are determined in cation and anion as per the valency of metals.

c)

Interpretation Introduction

Interpretation:

Balanced equations for the reaction of Cu and Pb with oxygen must be given along with the charge for cation and anion in the products. The species which gets oxidized must also be written.

Concept Introduction:

Copper and lead are oxidized by oxygen to their corresponding oxides.

Answer to Problem 4E

Balanced equations are given below:

  $\begin{array}{l}\text{Cu(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to \text{CuO(s)}\text{.}\\ \text{Pb(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to \text{PbO(s)}\text{.}\end{array}$

Charges of cations are +2.

Charge of oxide anion is -2.

Cu and Pb are oxidized to CuO and PbO respectively.

Explanation of Solution

  $\begin{array}{l}\text{Cu(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to \text{CuO(s)}\text{.}\\ \text{Pb(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)}\to \text{PbO(s)}\text{.}\end{array}$

Cu and Pb are oxidized to corresponding oxides and oxygen is reduced.

Oxidation number of Cu and Pb is increased from 0 to +2.

Cu and Pb have valency 2 and 4 respectively. However lead is stable in +2 state instead of +4 state due to inert pair effect for which 6s² electron of lead don’t participate in reaction. Accordingly these have reacted with same number of oxygen atoms. The charges are determined in cation and anion as per the valency of metals.