Chapter 20.3, Problem 31LC

Interpretation Introduction

Interpretation : The equations for the production of sodium, bromine, and iodine are to be reviewed. The oxidizing agent and the reducing agent in each reaction are to be identified.

Concept Introduction : A chemical process in which electrons are moved between two reactants is referred to as a redox reaction. The redox reactions include combination reactions, combustion reactions, decomposition reactions, and single-replacement reactions.

Answer to Problem 31LC

In the production of sodium, the oxidizing agent and the reducing agents are sodium and chlorine respectively.

In the production of bromine, the oxidizing agent and the reducing agents are chlorine and bromine respectively.

In the production of iodine, the oxidizing agent and the reducing agents are iodine and sulfur respectively.

Explanation of Solution

An element is oxidized and acts as a reducing agent when its oxidation number rises.

The oxidizing agent is the element whose oxidation number decreases.

Consider the chemical equation used for the production of sodium.

  $2\text{NaCl}\left(l\right)\to 2\text{Na}\left(l\right)+{\text{Cl}}_{\text{2}}\left(g\right)$

The above equation is written with its charge.

  $2\stackrel{\text{+1}}{\text{Na}}\stackrel{-\text{1}}{\text{Cl}}\left(l\right)\to 2\stackrel{0}{\text{Na}}\left(l\right)+\stackrel{0}{{\text{Cl}}_2}\left(g\right)$

There is a change in the oxidation number of sodium and chlorine.

The oxidation number of sodium is changed from $+1$ to $0$ . So, sodium is reduced and acts as an oxidizing agent in the reaction.

The oxidation number of chlorine is changed from $-1$ to $0$ . So, chlorine is oxidized and acts as the reducing agent in the reaction.

Consider the chemical equation used for the production of bromine.

  $2\text{NaBr}\left(aq\right)+{\text{Cl}}_2\left(g\right)\to 2\text{NaCl}\left(aq\right)+{\text{Br}}_{\text{2}}\left(l\right)$

The above equation is written with its charge.

  $2\stackrel{\text{+1}}{\text{Na}}\stackrel{-\text{1}}{\text{Br}}\left(aq\right)+\stackrel{0}{{\text{Cl}}_2}\left(g\right)\to 2\stackrel{0}{\text{Na}}\stackrel{-1}{\text{Cl}}\left(aq\right)+\stackrel{0}{{\text{Br}}_2}\left(l\right)$

There is a change in the oxidation number of bromine and chlorine.

The oxidation number of bromine is changed from $-1$ to $0$ . So, sodium is oxidized and acts as the reducing agent in the reaction.

The oxidation number of chlorine is changed from $0$ to $-1$ . So, chlorine is reduced and acts as the oxidizing agent in the reaction.

Consider the chemical equation used for the production of iodine.

  $2{\text{NaIO}}_3\left(aq\right)+5{\text{NaHSO}}_4\left(aq\right)\to {\text{I}}_2\left(g\right)+2{\text{Na}}_2{\text{SO}}_4\left(aq\right)+3{\text{NaHSO}}_4\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The above equation is written with its charge.

  $2\stackrel{+1}{\text{Na}}\stackrel{+5}{\text{I}}\stackrel{-2}{{\text{O}}_3}\left(aq\right)+5\stackrel{+1}{\text{Na}}\stackrel{+1}{\text{H}}\stackrel{+4}{\text{S}}\stackrel{-2}{{\text{O}}_4}\left(aq\right)\to \stackrel{0}{{\text{I}}_2}\left(g\right)+2\stackrel{+1}{{\text{Na}}_2}\stackrel{+6}{\text{S}}\stackrel{-2}{{\text{O}}_4}\left(aq\right)+3\stackrel{+1}{\text{Na}}\stackrel{+1}{\text{H}}\stackrel{+6}{\text{S}}\stackrel{-2}{{\text{O}}_4}\left(aq\right)+\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}\left(l\right)$

There is a change in the oxidation number of sulfur and iodine.

The oxidation number of iodine is changed from $+5$ to $0$ . So, iodine is reduced and acts as the oxidizing agent in the reaction.

The oxidation number of sulfur is changed from $+4$ to $+6$ . So, sulfur is oxidized and acts as the reducing agent in the reaction.