Chapter 8, Problem 8H.1E

(a)

Interpretation Introduction

Interpretation:

The chemical equation for burning of lithium in oxygen has to be given.

Answer to Problem 8H.1E

The chemical equation for burning of lithium in oxygen is,

  ${\text{4Li}}_{\left(\text{s}\right)}{\text{+O}}_{\text{2(g)}}\stackrel{\text{Δ}}{\to }{\text{2Li}}_{\text{2}}{\text{O}}_{\text{(s)}}$

Explanation of Solution

Lithium burns in the presence of oxygen to form lithium oxide $\left({\text{Li}}_{\text{2}}\text{O}\right)\text{.}$  The chemical equation is given as,

  ${\text{Li}}_{\left(\text{s}\right)}{\text{+O}}_{\text{2(g)}}\stackrel{\text{Δ}}{\to }{\text{Li}}_{\text{2}}{\text{O}}_{\text{(s)}}$

Four moles of lithium are required in the reactant side and two moles of lithium oxide are required in the product side to make the equation a balanced one.  The balanced chemical equation for burning of lithium in oxygen is,

  ${\text{4Li}}_{\left(\text{s}\right)}{\text{+O}}_{\text{2(g)}}\stackrel{\text{Δ}}{\to }{\text{2Li}}_{\text{2}}{\text{O}}_{\text{(s)}}$

(b)

Interpretation Introduction

Interpretation:

The chemical equation for sodium metal with water has to be given.

Answer to Problem 8H.1E

The chemical equation for sodium metal with water is,

  ${\text{2Na}}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{2NaOH}}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2(g)}}$

Explanation of Solution

Sodium reacts with water to form sodium hydroxide with the evolution of hydrogen gas.  The chemical equation is given as,

  ${\text{Na}}_{\left(\text{s}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{NaOH}}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2(g)}}$

Two moles of sodium, two moles of water are required in the reactant side and two moles of sodium hydroxide are required in the product side to make the equation a balanced one.  The balanced chemical equation for sodium metal with water is,

  ${\text{2Na}}_{\left(\text{s}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{2NaOH}}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2(g)}}$

(c)

Interpretation Introduction

Interpretation:

The chemical equation for fluorine gas with water has to be given.

Answer to Problem 8H.1E

The chemical equation for fluorine gas with water is,

  ${\text{2F}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{4HF}}_{\left(\text{aq}\right)}{\text{+O}}_{\text{2(g)}}$

Explanation of Solution

Fluorine reacts with water to form hydrogen fluoride $\left(\text{HF}\right)$ with evolution of oxygen gas.  The chemical equation is given as,

  ${\text{2F}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{4HF}}_{\left(\text{aq}\right)}{\text{+O}}_{\text{2(g)}}$

Two moles of fluorine, two moles of water are required in the reactant side and four moles of hydrogen fluoride are required in the product side to make the equation a balanced one.  The balanced chemical equation for fluorine gas with water is,

  ${\text{2F}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{4HF}}_{\left(\text{aq}\right)}{\text{+O}}_{\text{2(g)}}$

(d)

Interpretation Introduction

Interpretation:

The chemical equation for oxidation of water at the anode of an electrolytic cell has to be given.

Answer to Problem 8H.1E

The chemical equation for oxidation of water at the anode of an electrolytic cell is,

  ${\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{O}}_{\text{2(g)}}{\text{+4H}}^{\text{+}}{\text{+4e}}^{\text{-}}$

Explanation of Solution

The oxidation of water in anode of electrolytic cell results in the formation of gaseous oxygen and hydrogen ions.  The chemical equation for oxidation of water at the anode of an electrolytic cell is,

  ${\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{O}}_{\text{2(g)}}{\text{+4H}}^{\text{+}}{\text{+4e}}^{\text{-}}$