# 4.4: Redox reactions (2023)

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##### learning goals
• Identify oxidation-reduction reactions in solution.

The term oxidation was first used to describe reactions in which metals react with oxygen in the air to produce metal oxides. For example, when iron is exposed to air in the presence of water, the iron turns into rust, an oxide of iron. When exposed to air, metallic aluminum forms a continuous, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by donating electrons to the neutral oxygen atoms of an oxygen molecule. As a result, oxygen atoms acquire a negative charge and form oxide ions (O2-). As metals lost electrons to oxygen, they oxidized; Oxidation is the loss of electrons. On the other hand, the oxygen atoms were reduced when they gained electrons, so reduction is the gain of electrons. For every oxidation there must be an associated reduction. Therefore, these reactions are known as oxidation-reduction reactions, or "redox" reactions for short.

All oxidation mustALWAYSbe accompanied by a reduction and vice versa.

Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction widely used to extract metals from their ores. If e.g. For example, when solid copper(I) oxide is heated with hydrogen, its mass decreases because oxygen atoms are lost as a volatile product (water vapor) during the formation of pure copper. The reaction is as follows:

$\ce{Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g)} \label{4.4.1}$

Oxidation-reduction reactions are now defined as reactions that involve a change in the oxidation state of one or more elements in the reactants by a transfer of electrons, along the lines of "oxidation is loss, reduction is gain" or"Oil rig". Isoxidation stateof each atom in a compound is the charge that an atom would have if all of its bonding electrons were transferred to the atom with the greatest electron attraction. Atoms in their elemental form, such as O2Oh2get an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide

$\ce{ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s)} \label{4.4.2}$

Each neutral oxygen atom gains two electrons and becomes negatively charged, creating an oxide ion; thus the oxygen in the product has an oxidation state of -2 and has been reduced. Each neutral aluminum atom donates three electrons to produce an aluminum ion in the product with a +3 oxidation state, meaning the aluminum has been oxidized. In the formation of Al2Ö3, electrons are transferred as follows (the small overlapping number emphasizes the oxidation state of the elements):

$4 \overset{0}{\ce{Al}} + 3 \overset{0}{\ce{O2}} \rightarrow \ce{4 Al^{3+} + 6 O^{2-} } \label{4.4.3}$

Equation $$\ref{4.4.1}$$jEquation $$\ref{4.4.2}$$are examples of oxidation-reduction (redox) reactions. In redox reactions there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total number of electrons gained to maintain electrical neutrality. InEquation $$\ref{4.4.3}$$For example, the total number of electrons lost to aluminum equals the total number gained to oxygen:

\begin{align*} \text{missing electrons} &= \ce{4 Al} \, \text{atoms} \times {3 \, e^- \, \text{missing} \over \ce{ Al} \, \text{atom} } \\[4pt] &= 12 \, e^- \, \text{label{4.4.4a} \end{align*}

\begin{align*} \text{electrons gained} &= \ce{6 O} \, \text{atoms} \times {2 \, e^- \, \text{gained} \over \ce{ O} \, \text{atom}} \\[4pt] &= 12 \, e^- \, \text{ganado} \label{4.4.4b}\end{align*}

The same pattern is observed in all oxidation-reduction reactions: the number of electrons lost must equal the number of electrons gained. Another example of a redox reaction, the reaction of sodium metal with chlorine, is shown inFigure $$\PageIndex{1}$$.

In all oxidation-reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.

## Assignment of oxidation states

The assignment of oxidation states to elements in binary ionic compounds is straightforward: the oxidation states of elements are identical to the charges of monatomic ions. You have already learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge of monatomic ions formed by neutral elements. Examples of such compounds are sodium chloride (NaCl;Figure $$\PageIndex{1}$$), magnesium oxide (MgO) and calcium chloride (CaCl2). In covalent bonds, on the other hand, atoms share electrons. However, we can still assign oxidation states to the elements involved by treating them as if they were ionic (i.e. as if all bonding electrons were transferred to the more attractive element). Oxidation states in covalent compounds are somewhat arbitrary, but they are useful computational tools that will help you understand and predict many reactions.

Below are a set of rules for assigning oxidation states to atoms in chemical compounds.

##### Rules for assigning oxidation states
1. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.
2. The oxidation state of a monatomic ion is equal to its charge, e.g. Na+= +1, Kl= −1.
3. The oxidation state of fluorine in chemical compounds is always -1. Other halogens also typically have −1 oxidation states, except when combined with oxygen or other halogens.
4. Hydrogen is assigned the +1 oxidation state in its compounds with nonmetals and -1 in its compounds with metals.
5. Oxygen is usually assigned an oxidation state of -2 in compounds, with two exceptions: In compounds containing oxygen-fluorine or oxygen-oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.
6. The sum of the oxidation states of all atoms in a molecule or neutral ion must equal the charge on the molecule or ion.

Occasionally non-integer (fractional) oxidation states are encountered. They are usually due to the presence of two or more atoms of the same element with different oxidation states.

In any chemical reaction, the net charge must be conserved; That is, in a chemical reaction, the total number of electrons is constant, as is the total number of atoms. Consequently, Rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. For example, in NaCl, Na has an oxidation state of +1 and Cl is -1. The net charge is zero as it should be for any compound.

Rule 3 is necessary because fluorine attracts electrons more than any other element, for reasons you will find inChapter 6. Therefore, fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for hydrogen bonds with nonmetals (such as chlorine) versus hydrogen bonds with metals (such as sodium). For example, NaH contains the Hion while HCl forms H+y KlIons when dissolved in water. Rule 5 is necessary because fluorine attracts electrons more than oxygen; this rule also prevents violations of rule 2. Therefore, the oxidation state of oxygen in OF is +22but −1⁄2 in KO2. Note that an oxidation state from -½ to O is in KO2it is perfectly acceptable.

The reduction of copper(I) oxide shown in Equation $$\ref{4.4.5}$$ shows how these rules apply. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which is true for H2and ass. From rule 4, hydrogen in H2O has an oxidation state of +1 and by rule 5 oxygen in Cu2O and H2O has an oxidation state of -2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu2Or it must have a +1 charge: 2(+1) + (-2) = 0. Hence the oxidation states are as follows:

$\overset {\color{ref}{+1}}{\ce{Cu_2}} \overset {\color{ref}-2}{\ce{O}} (s) + \overset {\color{ ref}0}{\ce{H_2}} (g) \rightarrow 2 \overset {\color{ref}0}{\ce{Cu}} (s) + \overset {\color{ref}+1}{ \ce{H}}_2 \overset {\color{ref}-2}{\ce{O}} (g) \label{4.4.5}$

The assignment of the oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). It is therefore a redox reaction. Again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge:

$\text{missing electrons} = 2 \, H \, \text{atoms} \times {1 \, e^- \, \text{missing} \over H \, \text{atoms} } = 2 \ , e^- \, \text{Error} \label{4.4.6a}$

$\text{gained electrons} = 2 \, Cu \, \text{atoms} \times {1 \, e^- \, \text{gains} \over Cu \, \text{atoms}} = 2 \ , and^- \, \text{won} \label{4.4.6b}$

Remember that oxidation states are useful for visualizing electron transfer in oxidation-reduction reactions, but an atom's oxidation state and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way to assign electrons to atoms and are useful for predicting the types of reactions that substances will undergo.

##### Example $$\PageIndex{1}$$: oxidation states

Assign oxidation states to all atoms in each compound.

1. Sulfur hexafluoride (SF6)
2. Methanol (CH3OH)
3. Ammoniumsulfat [(NH4)2SO4]
4. Magnetit (Fe3Ö4)
5. Ethanoic acid (acetic acid) (CH3CO2H)

given: Molecular or empirical formula

Strategy:

Start with atoms whose oxidation states can be determined unambiguously from the rules presented (e.g. fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of the other atoms present using Rule 1.

Solution:

for. From rule 3 we know that fluorine in its compounds always has the oxidation state -1. The six fluorine atoms in sulfur hexafluoride give it an overall negative charge of -6. Since Rule 1 requires that in a neutral molecule (here SF6), the oxidation state of sulfur must be +6:

[(6 Atome F)(−1)] + [(1 Atom S) (+6)] = 0

B. According to rules 4 and 5, hydrogen and oxygen have oxidation states +1 and -2, respectively. Since methanol has no net charge, carbon must have an oxidation state of -2:

[(4 H-Atom)(+1)] + [(1 O-Atom)(−2)] + [(1 C-Atom)(−2)] = 0

C. Observe que (NH4)2THEN4is an ionic compound composed of a polyatomic cation (NH4+) and a polyatomic anion (SO42) (verTable 2.4). We separately assign oxidation states to the atoms in each polyatomic ion. to NH4+, hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of -3:

(Video) 4.4 Oxidation Reduction Reactions

[(4 H atoms)(+1)] + [(1 N atom)(-3)] = +1, without NH charge4+rein

For SO42-, oxygen has an oxidation state of -2 (rule 5), so sulfur must have an oxidation state of +6:

[(4 atoms O) (−2)] + [(1 atom S)(+6)] = −2, the charge of the sulfate ion

D. Oxygen has an oxidation state of -2 (Rule 5), giving a total charge of -8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:

[(4 Atome O)(−2)]+[(3 Atome Fe)$$\left (+{8 \over 3} \right )$$]= 0

Partial oxidation states are allowed because oxidation states are a somewhat arbitrary way of tracking electrons. indeed faith3Ö4can be considered as having two Fe3+ions and an Fe2+ion per formula unit, giving a net positive charge of +8 per formula unit. Believe3Ö4It is a magnetic iron mineral commonly referred to as magnetite. In ancient times, magnetite was known as a magnet because it could be used to make primitive compasses that pointed to the Pole Star (the North Star), known as "pads."

My. First, we assign oxidation states to the CH components3CO2H like any other connection. Hydrogen and oxygen have oxidation states of +1 and -2 (rules 4 and 5, respectively), resulting in a total charge of hydrogen and oxygen of

[(4 H-Atome)(+1)] + [(2 O-Atome)(−2)] = 0

Therefore, the oxidation state of carbon must also be zero (Rule 6). However, this is an average oxidation state for the two carbon atoms present. Since each carbon atom has a different set of atoms attached, it is likely that they have different oxidation states. To determine the oxidation states of individual carbon atoms, we use the same rules as before, but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (-CH3) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has a +1 oxidation state, and we just said that the carbon-carbon bond can be ignored when calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of -3. Likewise, the carbon atom of the carboxylic acid group (-CO2H) is bonded to a carbon atom and two oxygen atoms. Again, we ignore the attached carbon atom and assign the -2 and +1 oxidation states to the oxygen and hydrogen atoms, respectively, resulting in a net charge of

[(2 Atom O)(−2)] + [(1 Atom H)(+1)] = −3

To get an electrically neutral carboxylic acid group, the charge on that carbon must be +3. The oxidation states of the individual atoms in acetic acid are therefore

$\underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \sin número$

So the sum of the oxidation states of the two carbon atoms is actually zero.

##### Task $$\PageIndex{1}$$: oxidation states

Assign oxidation states to all atoms in each compound.

1. Bariumfluorid (BaF2)
2. Formaldehyde (CH2Ö)
3. Potassium dichromate (K2kr2Ö7)
4. Cesium oxide (CsO2)
5. Ethanol (CH3CH2OH)

Ba, +2; F, −1

C, 0; H, +1; O, −2

K, +1; Cr, +6; O, −2

responder

CS, +1; O, −½

C, –3; H, +1; C, -1; H, +1; O, −2; h, +1

(Video) 4.4 Redox Reactions

## Types of redox reactions

Many types of chemical reactions are classified as redox reactions, and it would be impossible to remember them all. However, there are some important types of redox reactions that you are likely to encounter and that you should be familiar with. These include:

• synthesisReactions: The formation of any compound directly from the elements is a redox reaction, for example the formation of water from hydrogen and oxygen: $\ce{ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g) } \ no number \ ] • decompositionReactions: On the other hand, the decomposition of a compound into its elements is also a redox reaction, as in the electrolysis of water: \[\ce{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)} \ number \ ] • combustionReactions: Many chemicals combust (burn) with oxygen. In particular, organic chemicals such as hydrocarbons burn in the presence of oxygen to form carbon dioxide and water as products: \[\ce{ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g ) } \ no number$

The following sections describe another important class of redox reactions: simple metal displacement reactions in solution.

## Redox reactions of solid metals in aqueous solution

• A very common class of oxidation-reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as rust on a car (Figure $$\PageIndex{2}$$). Rust forms from a complex oxidation-reduction reaction with dilute acidic solutions containing ClIons (effectively diluted HCl), metallic iron and oxygen. When an object rusts, metallic iron reacts with HCl(aq) to form ferrous chloride and hydrogen gas:

$\ce{ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)} \label{4.4.81}$

In further steps, $$\ce{FeCl2}$$ is oxidized to a red-brown precipitate of $$\ce{Fe(OH)3}$$.

Many metals are dissolved by reactions of this type, which have the general form

$\text{metal} + \text{acid} \rightarrow \text{salt} + \text{hydrogen} \label{4.4.82}$

Some of these reactions have important consequences. For example, it has been suggested that one factor contributing to the fall of the Roman Empire was the widespread use of lead in pots and pipes that carried water. As we've seen, rainwater is slightly acidic, and foods like fruit, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:

$\ce{Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) } \label{4.4.83}$

Consequently, it has been speculated that both the water and food consumed by the Romans contained toxic levels of lead, leading to widespread lead poisoning and eventual insanity. Maybe that explains why the Roman Emperor Caligula named his favorite horse Consul!

• ## Simple displacement reactions

Certain metals are oxidized by aqueous acids while others are oxidized by aqueous solutions of various metal salts. Both types of reactions are referred to as simple displacement reactions, in which the ion in solution is displaced by oxidation of the metal. Two examples of simple displacement reactions are the reduction of iron salts by zinc (equation $$\ref{4.4.84}$$) and the reduction of silver salts by copper (equation $$\ref{4.4.85}$$ and figure $$\PageIndex{3}$$):

$\ce{Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)} \label{4.4.84}$

$\ce{ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)} \label{4.4.85}$

The reaction of equation $$\ref{4.4.84}$$ is often used to prevent (or at least delay) the corrosion of iron or steel objects such as nails and sheet metal. "Electroplating" involves covering iron or steel with a thin layer of zinc that is protected from oxidation while the zinc remains in the object.

• ## The set of activities

Chemists have observed what happens when samples of different metals are contacted with solutions of other metals and have ranked the metals according to the relative ease or difficulty with which they can be oxidized in a single displacement reaction. For example, metallic zinc reacts with iron salts and metallic copper with silver salts. Experimentally it was found that zinc reacts with both copper and silver salts and produces $$\ce{Zn2+}$$. Zinc therefore tends to oxidize more than iron, copper or silver. While zinc does not react with magnesium salts to form metallic magnesium, metallic magnesium reacts with zinc salts to form metallic zinc:

$\ce{Zn(s) + Mg^{2+}(aq) \xcancel{\rightarrow} Zn^{2+}(aq) + Mg(s)} \label{4.4.10}$

$\ce{Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s)} \label{4.4.11}$

(Video) AP chem 4.4 - redox reactions

Magnesium tends to oxidize more than zinc.

Paired reactions of this type form the basis of the activity series (Figure $$\PageIndex{4}$$), which lists metals and hydrogen in the order of their relative tendency to oxidize. The metals at the top of the list that are most prone to donating electrons are the alkali metals (Group 1), the alkaline earth metals (Group 2), and Al (Group 13). On the other hand, the metals at the bottom of the row that are less prone to oxidation are the precious or decomposable metals: platinum, gold, silver, and copper and mercury, which are in the lower-right metals row on the periodic table. . In general, you should know which types of metals are active metals, which have a higher tendency to oxidize. (at the top of the list) and are the inert metals, which are less susceptible to oxidation. (at the end of the row).

Keep this in mind when using the activity series to predict the outcome of a reactionEach element reduces the connections of the elements below it in the row. Since magnesium is above zinc in figure $$\PageIndex{4}$$, metallic magnesium reduces zinc salts, but not vice versa. Precious metals are also at the lower end of the spectrum of activity, so virtually every other metal reduces precious metal salts to pure precious metals. Hydrogen is included in the series, and a metal's tendency to react with an acid is indicated by its position relative to hydrogen in the activity series.Only metals higher in the activity series than hydrogen dissolve in acids to make themH2.Because precious metals are below hydrogen, they do not dissolve in dilute acid and therefore do not corrode easily. The $$\PageIndex{2}$$ example shows how familiarity with the activity series allows you to predict the products of many individual shift responses.

##### Example $$\PageIndex{2}$$: Activity

Use the activities to predict what will happen in each situation. If a reaction occurs, write the net ionic equation.

1. A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
2. A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
3. Some sulfuric acid was accidentally spilled on the terminals from a car battery.

Given:reagents

On request of:general reaction and net ion equation

Strategy:

1. Locate the reactants in the series of activities in Figure $$\PageIndex{4}$$ and use their relative positions to predict whether a reaction will take place. When a reaction occurs, identify which metal is being oxidized and which is being reduced.
2. Write the net ionic equation for the redox reaction.

Solution:

1. AAluminum is an active metal, ranking above silver in the activity series, so we expect a reaction to occur. Depending on their relative positions, the aluminum will oxidize and dissolve, and the silver ions will be reduced to metallic silver.BThe net ion equation is as follows:

$\ce{ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)} \nonumber$

Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and do not participate in the reaction.

2. AMercury is below lead in the activity series, so no reaction occurs.
3. ALead is higher in the activity series than hydrogen, so the lead ends oxidize and the acid is reduced to H2.BRecall from our discussion of solubilities that Pb2+and then42form insoluble lead(II) sulfate. In this case, the sulfate ionsNOspectators, and the reaction is as follows:

$\ce{Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) } \nonumber$

Lead(II) sulfate is the white solid that forms on corroded battery terminals. Corroded battery poles.The white solid is lead(II) sulfate, formed by the reaction of solid lead with a solution of sulfuric acid.

##### Exercise $$\PageIndex{2}$$

Use the activities to predict what will happen in each situation. If a reaction occurs, write the net ionic equation.

1. A strip of metallic chromium is placed in an aqueous solution of aluminum chloride.
2. A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
3. A piece of aluminum foil is placed in a jar with vinegar (the active ingredient is acetic acid).

$$No reaction$$

$$3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)$$

(Video) 4.4 Oxidation-Reduction Reactions

## Videos

1. 4.4a Redox Reactions and Oxidation Numbers
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3. 4.4 Oxidation Reduction Reactions and Assigning Oxidation Numbers Part 1
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4. 4.4 Oxidation Reduction Reactions and Assigning Oxidation Numbers Part 2
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5. 4.4b Redox reactions- Metal Activity and Spontaneity
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6. AP Chem - Section 4.4 - Redox Reactions
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