Chapter U5.15, Problem 3E

(a)

Interpretation Introduction

Interpretation:

Balanced equations for the reaction of calcium, chromium and lithium with oxygen must be given and for each case the charge of cation and anion in the products must also be given.

Concept Introduction:

Calcium, chromium and lithium are oxidized by water to their corresponding oxides.

Answer to Problem 3E

Balanced equations are given below:

  $\begin{array}{l}\text{Ca(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)=CaO(s)}\text{.}\\ {\text{2Cr(s)+3O}}_{\text{2}}{\text{(g)=Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{.}\\ \text{2Li(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}{\text{(g)=Li}}_{\text{2}}\text{O(s)}\text{.}\end{array}$

Charges of cations are +2, +3 and +1 for Ca, Cr and Li respectively.

Charge of oxide anion is -2.

Explanation of Solution

All the metals are oxidized to corresponding oxides and oxygen is reduced.

Ca, Cr and Lithium have valency 2, 3 and 1 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:

Elements must be selected which are oxidized.

Concept Introduction:

Elements for which the oxidation numbers are increased are those which get oxidized.

Answer to Problem 3E

Calcium (Ca), chromium (Cr) and lithium (Li) are oxidized.

Balanced equations are given below.

  $\begin{array}{l}\text{Ca(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)=CaO(s)}\text{.}\\ {\text{2Cr(s)+3O}}_{\text{2}}{\text{(g)=Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{.}\\ \text{2Li(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}{\text{(g)=Li}}_{\text{2}}\text{O(s)}\text{.}\end{array}$

Explanation of Solution

Balanced equations are given below.

  $\begin{array}{l}\text{Ca(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\text{(g)=CaO(s)}\text{.}\\ {\text{2Cr(s)+3O}}_{\text{2}}{\text{(g)=Cr}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}\text{.}\\ \text{2Li(s)+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}{\text{(g)=Li}}_{\text{2}}\text{O(s)}\text{.}\end{array}$

Oxidation numbers of Ca, Cr and Li are increased from 0 to +2, +3 and +1 respectively in the corresponding oxides.