Chapter 4, Problem 89AP

Interpretation Introduction

Interpretation:

The given reactions are to be classified on the basis of their types.

Concept introduction:

A chemical reaction in which precipitates are formed on reacting two or more reactants is known as a precipitation reaction.

The reaction between a base and an acid is known as an acid–base reaction.

A reaction in which losing and gaining of electrons occur simultaneously is known as an oxidation–reduction reaction.

The reaction in which a metal displaces another metal in a salt solution is known as a displacement reaction.

The reaction in which a new compound is formed by oxidizing or reducing two or more its constituent elements is known as a combination reaction.

Answer to Problem 89AP

Solution:

a) Redox reaction

b) Precipitation reaction

c) Acid–base reaction

d) Combination reaction

e) Redox reaction

f) Redox reaction

g) Precipitation reaction

h) Redox reaction

i) Redox reaction

j) Redox reaction

Explanation of Solution

(a)

Given information: The given reaction is as follows:

${\text{Cl}}_2+2{\text{OH}}^{-}\to {\text{Cl}}^{-}+{\text{ClO}}^{-}+{\text{H}}_2\text{O}$

This is an oxidation–reduction reaction. The reaction can be divided into a reduction half and an oxidation half as follows:

$\begin{array}{l}{\text{Cl}}^{\text{0}}-e^{-}\to {\text{ClO}}^{-}\text{(oxidation)}\\ {\text{Cl}}^{\text{0}}+e^{-}\to {\text{Cl}}^{-}\text{(reduction)}\end{array}$

In the first equation, a neutral chlorine atom loses an electron to form ${\text{ClO}}^{-}$; this is the oxidation part of the reaction. In the second half of the equation, the neutral chlorine atom gains an electron to form a negatively charged chlorine atom ${\text{Cl}}^{-}$; this is the reduction part of the reaction.

Hence, as the reactants in the reaction show change in their oxidation states, the reaction is a redox reaction.

(b)

Given information: The given reaction is as follows:

${\text{Ca}}^{2+}+{\text{CO}}_3{}^{2-}\to {\text{CaCO}}_3$

This reaction is a precipitation reaction. The product is formed by the combination of an element, ${\text{Ca}}^{+2}$, and a compound ion, ${\text{CO}}_3^{-2}$. These ions combine to form an insoluble salt and hence, it forms precipitates on dissolving in water.

Thus, the reaction is precipitation reaction.

(c)

Given information: The given reaction is as follows:

${\text{NH}}_3+{\text{H}}^{+}\to {\text{NH}}^{4+}$

This reaction is an acid–base reaction because the reaction occurs between ${\text{NH}}_{\text{3}}$, which is a base, and ${\text{H}}^{+}$, an acid.

Hence, the reaction is an acid–base reaction.

(d)

Given information: The given reaction is as follows:

$2{\text{CCl}}_4+{\text{CrO}}_4{}^{2-}\to 2{\text{COCl}}_2+{\text{CrO}}_2{\text{Cl}}_2+2{\text{Cl}}^{-}$

The oxidation states of chromium, chlorine, and oxygen change in the reaction between carbon tetrachloride and ${\text{CrO}}_4^{-2}$. The combination of these two results in the formation of the product.

Hence, it is a combination reaction.

(e)

Given information: The given reaction is as follows:

$\text{Ca}+{\text{F}}_2\to {\text{CaF}}_2$

This reaction is an oxidation–reduction reaction. The reaction can be divided into a reduction half and an oxidation half as follows:

$\begin{array}{l}2{\text{F}}^{\text{0}}+2e^{-}\to {\text{2F}}^{-1}\text{(reduction)}\\ {\text{Ca}}^{\text{0}}-2e^{-}\to {\text{Ca}}^{2+}\text{(oxidation)}\end{array}$

In the present reaction, the fluorine molecule gains two electrons to get reduced and form a negatively charged ion. On the other hand, calcium loses two electrons to oxidize into a Ca²⁺ ion.

Hence, the reaction is a redox reaction.

(f)

Given information: The given reaction is as follows:

$2\text{Li}+{\text{H}}_2\to 2\text{LiH}$

This reaction is an oxidation–reduction reaction. The reaction can be divided into a reduction half and an oxidation half as follows:

$\begin{array}{l}{\text{H}}_2+2{\text{e}}^{-}\to 2{\text{H}}^{-1}\text{(reduction)}\\ 2\text{Li}-2{\text{e}}^{-}\to {\text{2Li}}^{+1}\text{(oxidation)}\end{array}$

The hydrogen molecule in this reaction gets reduced and lithium is oxidized by losing two electrons.

Thus, the reaction is a redox reaction.

(g)

Given information: The given reaction is as follows:

$\text{Ba}{\left({\text{NO}}_3\right)}_2+{\text{Na}}_2{\text{SO}}_4\to 2{\text{NaNO}}_3+{\text{BaSO}}_4$

This is a precipitation reaction. In this reaction, no change is observed in the oxidation states of the reacting elements but barium sulfate is obtained as a precipitate in the product.

Hence, it is a precipitation reaction.

(h)

Given information: The given reaction is as follows:

$\text{CuO}+{\text{H}}_{\text{2}}\to \text{Cu}+{\text{H}}_2\text{O}$

This reaction is an oxidation–reduction reaction. The reaction can be divided into a reduction half and an oxidation half as follows:

$\begin{array}{l}{\text{Cu}}^{2+}+2{\text{e}}^{-}\to {\text{Cu}}^0\left(\text{reduction}\right)\\ 2{\text{H}}^0-{\text{2e}}^{-}\to {\text{2H}}^{-1}\left(\text{oxidation}\right)\end{array}$

As the hydrogen molecule is oxidized to ${\text{H}}^{-1}$ and the copper is reduced to ${\text{Cu}}^0$, it classifies the reaction as oxidation-reduction reaction.

Hence, it is a redox reaction.

(i)

Given information: The given reaction is as follows:

$\text{Zn}+2\text{HCl}\to {\text{ZnCl}}_2+{\text{H}}_2$

This is an oxidation–reduction reaction. The reaction can be divided into a reduction half and an oxidation half as follows:

$$

$\begin{array}{l}2{\text{H}}^{+1}+2{\text{e}}^{-}\to 2{\text{H}}^0\left(\text{reduction}\right)\\ {\text{Zn}}^0-2{\text{e}}^{-1}\to {\text{Zn}}^{2+}\left(\text{oxidation}\right)\end{array}$

As the hydrogen molecule in this reaction gets reduced by gaining electrons and the zinc is oxidized by losing two electrons, the reaction is known as an oxidation-reduction reaction.

Hence, it is a redox reaction.

j)

Given information: The given reaction is as follows:

$2{\text{FeCl}}_2+{\text{Cl}}_2\to 2{\text{FeCl}}_3$

This is an oxidation–reduction reaction. The reaction can be divided into a reduction half and an oxidation half as follows:

$\begin{array}{l}{\text{Cl}}^0+e^{-1}\to {\text{Cl}}^{-1}\left(\text{reduction}\right)\\ {\text{Fe}}^{+2}-{\text{e}}^{-1}\to {\text{Fe}}^{+3}\left(\text{oxidation}\right)\end{array}$

The chlorine molecule in this reaction gets reduced by gaining an electron and the iron is oxidized by losing two electrons. Thus, the reaction is known as an oxidation-reduction reaction.

Therefore, it is a redox reaction.