Soluble | |
---|---|
Insoluble | |
Chemical Reactions
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Balancing Equations
One should always work with a balanced chemical equation. Balancing a chemical equation is done by adding stoichiometric coefficients in front of the chemical species in a reaction. Stoichiometry deals with the relationship between the amounts of reactants and products in a chemical equation.
Not balanced
\[\mathrm{H_2} + \mathrm{O_2} \longrightarrow \mathrm{H_2O}\]
Two oxygen atoms appear on the left hand side of the equation but only one appears on the right.
Balanced
\[2~\mathrm{H_2} + \mathrm{O_2} \longrightarrow 2~\mathrm{H_2O}\]
\[\mathrm{H_2} + \dfrac{1}{2}~\mathrm{O_2} \longrightarrow \mathrm{H_2O}\]
Note: For the purpose of this class, balance using whole numbers (first example), and not fractions (second example), unless when instructed otherwise.
Practice
Balance the following chemical equation.
\[\mathrm{Fe(s)} + \mathrm{O_2(g)} \longrightarrow \mathrm{Fe_2O_3(s)}\]
Solution
\[4~\mathrm{Fe(s)} + 3~\mathrm{O_2(g)} \longrightarrow 2~\mathrm{Fe_2O_3(s)}\]
Practice
Balance the following chemical equation.
\[\mathrm{C_8H_{18}(l)} + \mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} + \mathrm{H_2O(g)}\]
Solution
\[2~\mathrm{C_8H_{18}(l)} + 25~\mathrm{O_2(g)} \longrightarrow 16~\mathrm{CO_2(g)} + 18~\mathrm{H_2O(g)}\]
Reversible Reactions
Many chemical reactions are written with a right-facing arrow separating the reactants and products. However, all reactions are reversible in theory, and occur in both the forward and reverse directions.
Consider the following reaction.
\[\mathrm{2~NO_2(g)} \longrightarrow \mathrm{2NO(g)} + \mathrm{O_2(g)}\]
We read this reaction as NO2(g) combines to form NO(g) and O2(g). The right-facing arrow may lead us to conclude that this reaction occurs in one direction only (toward the right). However, if carrying out this reaction in a closed vessel, one would realize that there would exist a mixture of both reactants and products at all times.
Not all the reactions will convert into products. In fact, products will combine to form more reactants (the reverse reaction) as seen in the following chemical equation.
\[\mathrm{2~NO(g)} + \mathrm{O_2(g)} \longrightarrow \mathrm{2~NO_2(g)}\]
Equilibrium
Equilibrium is a state, often a stable state, where a system does not change over time. Chemical equilibrium is a state where the amounts of reactants and products stop changing.
We can express a chemical reaction at equilibrium by using a double harpoon symbol instead of an arrow.
\[\mathrm{2~NO_2(g)} \rightleftharpoons \mathrm{2~NO(g)} + \mathrm{O_2(g)}\]
The equation above indicates that the stated reaction is at equilibrium.
If we were to run the reaction above starting with only NO2(g), we would initially observe
- The amount of NO2(g) decreasing
- The amount of NO(g) increasing
- The amount of O2 increasing
We can conclude that the rate of the forward reaction is greater than the rate of the reverse reaction.
As the reaction proceeds with time, if left undisturbed, it would reach equilibrium. At this state, we would observe
- The amount of NO2(g) remains constant
- The amount of NO(g) remains constant
- The amount of O2 remains constant
While one might assume that the reaction has stopped at equilibrium, this would be incorrect. At equilibrium, the reaction is still occurring. The rate of the forward reaction is equal to the rate of the reverse reaction.
Now we have our reaction at equilibrium and the amounts of reactants and products have stopped changing. We can peer in and measure the amounts of reactants and the amounts of products in the mixture. If there are more reactants than there are products, at equilibrium, the reaction is reactant favored. However, if there are more products than there are reactants, at equilibrium, the reaction is product favored.
Aqueous Solutions
Many reactions take place in solution. A solution is a homogeneous mixture composed of two or more substances. One substance is the solvent, the medium in which other substances, the solute, are dissolved. The solution is called an aqueous solution if the solvent is water.
Example: A glass of salt water (a saline solution) is an aqueous solution composed of water as the solvent and sugar as the solute.
Electrolytes
Substances can be classified as electrolytes, substances that produce ions with dissolved in water.
Sodium chloride, a soluble salt, readily dissociates when mixed with water.
\[\mathrm{NaCl(s)} \longrightarrow \mathrm{Na^+(aq)} + \mathrm{Cl^-(aq)}\]
In fact, NaCl effectively dissociates completely in water (i.e. every NaCl particle dissociates; product favored). Because NaCl effectively completely dissociates, it is classified as a strong electrolyte.
Weak electrolytes are substances that dissociate to some degree in water to produce ions but do not dissociate completely (reactant favored). One example is acetic acid, a weak acid. Only a small fraction of acetic acid molecules dissociate to produce ions in water.
\[\mathrm{CH_3COOH(aq)} + \mathrm{H_2O(aq)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{CH_3COO^-(aq)}\]
Nonelectrolytes are substances that do not produce ions when dissolved in water. Ethanol is an example of a nonelectrolyte.
\[\mathrm{CH_3CH_2OH(l)} \longrightarrow \mathrm{CH_3CH_2OH(aq)}\]
One can determine the electrolytic nature of substances by dissolving the substance in water and measuring the conductivity of the solution. Pure water is not a good conductor of electricity because pure water does not have an appreciable amount of ions present in solution. However, if many ions are present, the aqueous solution will be a better conductor of electricity.
Classifying Electrolytes
Strong Electrolytes
- Ionic compounds
- Strong acids
Weak Electrolytes
- Weak acids
- Weak bases
Nonelectrolytes
- Most molecular compounds
Solubility Rules
There exist many ionic compounds, some of which are soluble in water while others are not. We can draw general rules for determining if an ionic compound is soluble or not. Some solubility rules will only indicate if a compound is soluble or not. Others may additionally indicate if ionic compounds are “slightly” or “marginally” soluble.
See some solubility plots that demonstrate these rules.
Precipitation Reactions
Precipitation reactions are chemical reactions that take place in solution and produce a solid (i.e. precipitate) as a product. The solid that is created is a solid because it is insoluble in water. One can examine a chemical reaction and determine the phase of each substance by implementing the solubility rules (Table 1).
Consider the following reaction between silver nitrate and potassium chloride occurring in aqueous solution.
\[\mathrm{AgNO_3} + \mathrm{KCl} \longrightarrow \mathrm{AgCl} + \mathrm{KNO_3}\]
Using the solubility rules (Table 1), we can determine that
- AgNO3 is in the aqueous phase because compounds containing “nitrates are soluble without exception”.
- KCl is in the aqueous phase because compounds containing “alkali metals are soluble without exception”.
- AgCl is in the solid phase because ions containing “halides are soluble except when paired with Ag+”. AgCl is the precipitate of the reaction and the reaction is a precipitation reaction.
- KNO3 is in the aqueous phase because ions containing “alkali metals are soluble without exception” and correspondingly, “nitrates are soluble without exception”.
Therefore, the reaction can be written with the appropriate phase label on each substance.
\[\mathrm{AgNO_3(aq)} + \mathrm{KCl(aq)} \longrightarrow \mathrm{AgCl(s)} + \mathrm{KNO_3(aq)}\]
The above reaction can also be referred to as an exchange reaction because the cations swap out their anions with each other.
Practice
Determine the phase of each substance in the following reaction taking place in aqueous solution. Is the reaction a precipitation and/or exchange reaction? What is the precipitate (if applicable)?
\[\mathrm{Pb(NO_3)_2} + \mathrm{(NH_4)_2S} \longrightarrow \mathrm{PbS} + \mathrm{2~NH_4NO_3}\]
Solution
\[\mathrm{Pb(NO_3)_2(aq)} + \mathrm{(NH_4)_2S(aq)} \longrightarrow \mathrm{PbS(s)} + \mathrm{2~NH_4NO_3(aq)}\]
The reaction is a precipitation reaction and an exchange reaction. Lead(II) sulfide is the precipitate as it is not soluble in water.
Ionic Equations
We can express chemical reactions involving ionic compounds in three ways.
Molecular Equation
The molecular equation is a chemical equation that expresses each substance in the reaction as though they all exist as molecules.
\[\mathrm{AgNO_3(aq)} + \mathrm{KCl(aq)} \longrightarrow \mathrm{AgCl(s)} + \mathrm{KNO_3(aq)}\]
Complete Ionic Equation
The complete ionic equation is a chemical equation that expresses each soluble ionic compound as if they exist in their dissociated state as free ions.
\[\mathrm{Ag^+(aq)} + \mathrm{NO_3^-(aq)} + \mathrm{K^+(aq)} + \mathrm{Cl^-(aq)} \longrightarrow \mathrm{AgCl(s)} + \mathrm{K^+(aq)} + \mathrm{NO_3^-(aq)}\]
Net Ionic Equation
The net ionic equation is a chemical equation that only shows the particles that are directly involved in the chemical reaction. It is determined by starting with the complete ionic equation and omitting particles that appear as both a reactant and product. These omitted particles are called spectator ions.
\[\begin{align*} \mathrm{Ag^+(aq)} + \cancel{\mathrm{NO_3^-(aq)}} + \cancel{\mathrm{K^+(aq)}} + \mathrm{Cl^-(aq)} &\longrightarrow \mathrm{AgCl(s)} + \cancel{\mathrm{K^+}(aq)} + \cancel{\mathrm{NO_3^-(aq)}}\\[2ex] \mathrm{Ag^+(aq)} + \mathrm{Cl^-(aq)} &\longrightarrow \mathrm{AgCl(s)} \end{align*}\]
The final chemical equation (above) is the net ionic reaction. The spectator ions are K+ and NO3–.
Note: Weak electrolytes are not written as their individual ionic components in an ionic equation. It is written in its molecular form.
Acids and Bases
Below is a table of common strong and weak acids and bases. Note that many acids exist naturally as a gas (e.g. HCl(g) is hydrogen chloride). Sometimes, if the substance is dissolved in a solvent (such as water), and meets the definition of an acid, only then is it called an acid (e.g. HCl(aq) is hydrochloric acid). Exceptions include organic compounds such as carboxylic acids (e.g. acetic acid) which are still referred to as their acid name even in the absence of water. Anhydrous acetic acid is also called “glacial acetic acid”.
Strong Acid | |
---|---|
Weak Acid | |
Strong Base | |
Weak Base | |
Acid Base Definitions
Let us define a general notation for acids and bases
- Acid: HA, HB+
- Base: B, A–
There are different definitions for acids and bases.

An Arrhenius acid is a substance that dissociates in water to produce H+ (or H3O+) ions.
\[\begin{aligned} \require{color} {\color{green}\mathrm{HCl}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + {\color{red}\mathrm{Cl^-}}\mathrm{(aq)} \end{aligned}\]
or generally
\[{\color{green}\mathrm{HA}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + {\color{red}\mathrm{A^-}}\mathrm{(aq)}\]
An Arrhenius base is is a substance that dissociates in water to produce OH– ions.
\[{\color{red}\mathrm{NH_3}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{OH^-(aq)} + {\color{green}\mathrm{NH_4^+}}\mathrm{(aq)}\]
or generally
\[{\color{red}\mathrm{B}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{OH^-(aq)} + {\color{green}\mathrm{HB^+}}\mathrm{(aq)}\]
A Brønsted-Lowry acid is a substance that donates a proton (i.e. is a proton donor).
\[{\color{green}\mathrm{HCl}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + {\color{red}\mathrm{Cl^-}}\mathrm{(aq)}\]
Here, HCl, an acid, donates its proton to water. A general scheme can be written as
\[{\color{green}\mathrm{HA}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + {\color{red}\mathrm{A^-}}\mathrm{(aq)}\]
A Brønsted-Lowry base is a substance that accepts a proton (i.e. is a proton acceptor).
\[{\color{red}\mathrm{NH_3}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{OH^-(aq)} + {\color{green}\mathrm{NH_4^+}}\mathrm{(aq)}\]
Here, NH3, an base, accepts a proton from water. A general scheme can be written as
\[{\color{red}\mathrm{B}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{OH^-(aq)} + {\color{green}\mathrm{HB^+}}\mathrm{(aq)}\] or
\[{\color{red}\mathrm{A^-}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{OH^-(aq)} + {\color{green}\mathrm{HA}}\mathrm{(aq)}\]
A Lewis acid is a substance that accepts a pair of electrons (i.e. a lone-pair acceptor).
\[{\color{green}\mathrm{HCl}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + {\color{red}\mathrm{Cl^-}}\mathrm{(aq)}\]
Here, HCl dissociates into H+ and Cl–. The oxygen on water donates a lone pair of electrons to the free proton and forms a bond to give the hydronium ion, H3O+.
Lewis Acid Example
Water behaves as a Lewis base by donating an electron pair to HCl. HCl behaves as a Lewis acid by accepting the lone electron pair from water.
A Lewis base is a substance that donates a lone pair of electrons (i.e. lone-pair donor).
\[{\color{red}\mathrm{NH_3}}\mathrm{(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{OH^–(aq)} + {\color{green}\mathrm{NH_4^+(aq)}}\]
Here, the lone pair on the nitrogen in ammonia, NH3, is donated to a proton on water forming a bond to give NH4+.
Lewis Base Example
Water behaves as a Lewis acid by accepting the lone electron pair from ammonia. Ammonia behaves as a Lewis base by donating an electron pair to water.
In this example, ammonia forms a Lewis adduct with boron triflouride. No proton is transferred in this addition reaction. The N–B covalent bond is a dative bond, one that forms from an electron pair from one species.
Strong vs Weak
- A strong acid is an acid that effectively dissociates completely. The equilibrium is very product favored.
- A weak acid is an acid that dissociates but not to a large extent. Its equilibrium is very reactant favored.
- A strong base is an base that effectively dissociates completely. The equilibrium is very product favored.
- A weak base is an base that dissociates but not to a large extent. Its equilibrium is very reactant favored.
- Amphiprotic substances can behave as both an acid or a base (such as water) depending on the chemical environment it is in.
Acid-Base Neutralization Reaction
An acid-base neutralization reaction is a reaction between an acid and a base that produces a salt and oftentimes water.
Strong Acid + Strong Base
In the reaction below, a strong acid reacts with a strong base to produce water (hence neutralization) and sodium chloride salt, a neutral salt.
Molecular Equation
\[\mathrm{HCl(aq)} + \mathrm{NaOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{NaCl(aq)}\]
Complete Ionic Equation
\[\mathrm{H^+(aq)} + \mathrm{Cl^-(aq)} + \mathrm{Na^+(aq)} + \mathrm{OH^-(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{Na^+(aq)} + \mathrm{Cl^-(aq)}\]
Net Ionic Equation
\[\mathrm{H^+(aq)} + \mathrm{OH^-(aq)} \longrightarrow \mathrm{H_2O(l)}\]
Weak Acid + Strong Base
In the reaction below, a weak acid reacts with a strong base to produce water and sodium fluoride, a basic salt.
Molecular Equation
\[\mathrm{HF(aq)} + \mathrm{NaOH(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{NaF(aq)}\]
Complete Ionic Equation
\[\mathrm{HF(aq)} + \mathrm{Na^+(aq)} + \mathrm{OH^-(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{Na^+(aq)} + \mathrm{F^-(aq)}\]
Net Ionic Equation
\[\mathrm{HF(aq)} + \mathrm{OH^-(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{F^-(aq)}\]
Strong Acid + Weak Base
In the reaction below, a strong acid reacts with a weak base to produce an ammonium chloride salt, an acidic salt. This reaction is an example of an acid-base reaction that does not explicitly “produce water”.
Molecular Equation
\[\mathrm{HCl(aq)} + \mathrm{NH_3(aq)} \longrightarrow \mathrm{NH_4Cl(aq)}\]
Complete Ionic Equation
\[\mathrm{H^+(aq)} + \mathrm{Cl^-(aq)} + \mathrm{NH_3(aq)} \longrightarrow \mathrm{NH_4^+(aq)} + \mathrm{Cl^-(aq)}\]
Net Ionic Equation
\[\mathrm{H^+(aq)} + \mathrm{NH_3(aq)} \longrightarrow \mathrm{NH_4^+(aq)}\]
Weak Acid + Weak Base
In the reaction below, a weak acid reacts with a weak base to produce ammonium fluoride, an acidic salt.
Molecular Equation
\[\mathrm{HF(aq)} + \mathrm{NH_3(aq)} \longrightarrow \mathrm{NH_4F(aq)}\]
Complete Ionic Equation
\[\mathrm{HF(aq)} + \mathrm{NH_3(aq)} \longrightarrow \mathrm{NH_4^+(aq)} + \mathrm{F^-(aq)}\]
Net Ionic Equation
\[\mathrm{HF(aq)} + \mathrm{NH_3(aq)} \longrightarrow \mathrm{NH_4^+(aq)} + \mathrm{F^-(aq)}\]
Oxidation-Reduction Reactions
An oxidation-reduction reaction (redox reaction) is a reaction that involves the transfer of electrons. We refer to the loss of electrons as oxidation and the gain of electrons as reduction. The species losing electrons is the reducing agent/reducant whereas the species gaining electrons is the oxidizing agent/oxidant. There is no oxidation process without a corresponding reduction process and vice-versa.
We can readily determine the oxidant and reductant in a redox reaction by first assigning oxidation states to each atom in the reaction. If the oxidation state increases in a species as it proceeds from reactants to products, the species underwent oxidation (lost electron(s)). If the oxidation state decreases in a species as it proceeds from reactants to products, the species underwent reduction (gained electron(s)).
\[\Huge{\mathrm{OIL~RIG}}\] Oxidation is losing electrons. Reduction is gaining electrons.
An oxidation state or oxidation number is a hypothetical charge that an atom has if all of its bonds to other atoms were fully ionic. It gives insight into the degree of oxidation an atom has in a compound.
For example, each hydrogen atom in dihydrogen (H2) has an oxidation state of zero. The electrons shared between each hydrogen are equally shared. If the single bond in H2 were broken, each hydrogen would retain its one electron (homolytic cleavage).
However, in chloromethane (CH3Cl), breaking the carbon-chlorine bond could result in a methyl cation and chloride anion. Both electrons involved in the single bond are captured by the chlorine atom (heterolytic cleavage). One could rationalize this outcome by considering the electronegative nature of halides and their tendency to become negatively charged.
As represented here, chlorine “gained” electrons whereas the CH3 group “lost” electrons.
Oxidation number rules
- Oxidation number = 0 for any atom in its pure, elemental state.
- Zn(s): Zn = 0
- H2(g): H = 0
- Oxidation number = charge on a monatomic ion
- Na+: Na = +1
- Cl–: Cl = –1
- Sum of all oxidation numbers for the atoms in a polyatomic ion = charge on the polyatomic ion
- OH–: O + H = –2 + 1 = –1
- Sum of all oxidation numbers = charge on the species
- H2O: O + H + H = –2 + 1 + 1 = 0
- NaCl: Na + Cl = +1 + –1 = 0
Additionally, the rules below are prioritized in the following order:
Element | Oxidation Number | Exceptions |
---|---|---|
Fluorine | –1 | none |
Group 1 or 2 metal | +1 or +2, respectively | none |
Hydrogen | +1 | If paired with Group 1A or 2A metal to form a metal hydride such as LiH and CaH2 |
Oxygen | –2 | If paired with any element listed above that would violate rules 1-4 such as H2O2 or KO2 |
Group 17 (excluding Fluorine) | –1 | If paired with any element listed above that would violate rules 1-4 such as ClF, BrO4–, and IO3– |
Note: IUPAC recommends the use of Roman numerals to denote oxidation states. For example, the oxidation states of each atom in FeCl2 should be represented as
- iron(II) chloride
- FeIICl–I
In the first bullet point, a Roman numeral is not assigned to “chloride” as its oxidation state is equal to its charge as a monatonic ion.
We will adopt the use of Arabic numerals to denote oxidation states as is commonly done.
- Fe+2Cl–1
Oxidation states are always written as the sign followed by the numeral (positive signs are not included when using Roman numerals but negative signs are included) whereas charges are written by the numeral followed by the sign.
- Charge: Fe2+
- Oxidation state: Fe+2
If an atom has an oxidation state of zero, a superscript “0” is used since there is no Roman numeral for the number zero.
Apply these rules to the following reaction and assign oxidation numbers to each atom.
Molecular Equation
\[\begin{alignat*}{3} & \mathrm{Zn(s)} ~+~ && ~\mathrm{CuCl_2(aq)} ~ \longrightarrow ~ && ~~\mathrm{ZnCl_2(aq)} ~+~ && \mathrm{Cu(s)} \\ & ~(0) && (+2)(-1) && (+2)(-1) && ~(0) \end{alignat*}\]
Next, analyze the change in oxidation states and determine what underwent oxidation (the reducing agent) and what underwent reduction (the oxidizing agent).
- Zinc’s oxidation state increased from 0 → +2; zinc lost electrons and underwent oxidation (reducing agent)
- The copper in copper(II) chloride experiences a decrease in oxidation state from +2 → 0; copper gained electrons and underwent reduction (oxidizing agent)
You can reach the same conclusion by analyzing a redox reaction in its complete ionic or net ionic form.
Complete Ionic Equation
\[\mathrm{Zn(s)} + \mathrm{Cu^{2+}(aq)} + 2~\mathrm{Cl^-(aq)} \longrightarrow \mathrm{Zn^{2+}(aq)} + \mathrm{2~Cl^-(aq)} + \mathrm{Cu(s)}\]
Net Ionic Equation
\[\mathrm{Zn(s)} + \mathrm{Cu^{2+}(aq)} \longrightarrow \mathrm{Zn^{2+}(aq)} + \mathrm{Cu(s)}\]
Where does “Oxidation” come from?
Oxidation-reduction reactions were well known well before the electrons were well understood. These reactions were observed through the reaction of a metal with oxygen. For example, a well known redox reaction is the conversion of iron to rust (iron(III) oxide).
\[4~\mathrm{Fe(s)} + \mathrm{3~O_2(g)} \longrightarrow 2~\mathrm{Fe_2O_3(s)}\]
One could carry out a simple experiment where a piece of pure iron metal with a known mass would be allowed to rust. Upon the conclusion of the experiment, the mass of the rusted iron metal would be weighed. The mass difference provided the mass of oxygen that was involved in the reaction.
In the reaction above, iron experiences an increase in oxidation state (0 → +2) and lost electrons. Oxygen “took” electrons from iron and oxygen was referred to as the oxidizing agent. The reference stands today but has been extended to anything that “takes” or “gains” electrons.
Anything that “gains oxygen” “loses electrons”. Anything that “loses oxygen” “gains electrons”.
Practice
Write out the oxidation numbers for each atom in the following reaction
\[\mathrm{Zn(s)} + \mathrm{Cl_2(g)} \longrightarrow \mathrm{ZnCl_2(s)}\]
and determine the following
- Species oxidized
- Species reduced
- Oxidizing agent
- Reducing agent
Solution
\[\begin{alignat*}{3} & \mathrm{Zn(s)} ~+~ && \mathrm{Cl_2(g)} ~ \longrightarrow ~ && \mathrm{ZnCl_2(s)} \\ & ~(0) && ~(0) && (2)(-1) \end{alignat*}\]
- Species oxidized: Zn(s)
- Species reduced: Cl2(g)
- Oxidizing agent: Cl2(g)
- Reducing agent: Zn(s)
Below is a table of some oxidizing and reducing agents and their corresponding common reaction products.
Oxidizing Agent | |
---|---|
Reducing Agent | |
Consider the following redox reaction between copper and nitric acid.
\[\mathrm{Cu(s)} + \mathrm{2~NO_3^-(aq)} + \mathrm{4~H_3O^+(aq)} \longrightarrow \mathrm{Cu^{2+}(aq)} + \mathrm{2~NO_2(g)} + \mathrm{6~H_2O(l)}\]
Nitric acid is a strong acid and completely ionizes into hydronium and nitrate. Nitrate becomes nitrogen dioxide and nitrogen is reduced (oxidation state: 5 → 4) whereas copper is oxidized to copper(II). Hydrogen and oxygen experience no change in oxidation state.
Balancing Simple Redox Reactions
When balancing redox reactions, one must mass balance and charge balance the reaction.
Consider the following net ionic equation:
\[\mathrm{Cu^+(aq)} + \mathrm{Fe(s)} \longrightarrow \mathrm{Fe^{3+}(aq)} + \mathrm{Cu(s)}\]
As written, the reaction is mass balanced (one copper and iron on the left; one copper and iron on the right). However, notice that the total charge of the reactants equals “1” whereas the total charge of the products is “3+”. These charges must equal each other. To do this, split the redox reaction into two half-reactions where one is an oxidation half-reaction and the other is a reduction half-reaction. To do this, you must identify what is undergoing oxidation and what is undergoing reduction by assigning and evaluating the oxidation states of each atom.
The half-reactions are as follows:
\[\begin{alignat*}{2} &\text{oxidation half-reaction:} \quad \mathrm{Fe(s)} &&\longrightarrow \mathrm{Fe^{3+}(s)} \\[1.5ex] &\text{reduction half-reaction:} \quad \mathrm{Cu^+(aq)} &&\longrightarrow \mathrm{Cu(s)} \end{alignat*}\]
Next, charge balance each individual half reaction by adding an appropriate number of electrons. For the oxidation half-reaction, iron is losing electrons; therefore, electrons are introduced as a product of reaction. We add three electrons (i.e. 3 units of negative charge) to the right side of the oxidation half-reaction. One electron is added to the left side of the reduction half-reaction.
\[\begin{alignat*}{3} &\text{oxidation half-reaction:} \quad &&\mathrm{Fe(s)} &&\longrightarrow \mathrm{Fe^{3+}(s)} + 3~\mathrm{e^-} \\[1.5ex] &\text{reduction half-reaction:} \quad \mathrm{Cu^+(aq)} ~ + ~&&\mathrm{e^-} &&\longrightarrow \mathrm{Cu(s)} \end{alignat*}\]
Now the charges for each individual half-reaction are balanced (0 charge on both sides of the reaction arrows). Next, balance the number of electrons between each half-reaction. Notice that the oxidation half-reaction has 3 electrons while the reduction half-reaction only has 1 electron. We can multiply the reduction half-reaction by the number 3 to net us 3 electrons resulting in 3 electrons in both half-reactions.
\[\begin{alignat*}{3} &\text{oxidation half-reaction:} \quad &&\mathrm{Fe(s)} &&\longrightarrow \mathrm{Fe^{3+}(s)} + 3~\mathrm{e^-} \\[1.5ex] &\text{reduction half-reaction:} \quad 3~\mathrm{Cu^+(aq)} ~ + ~ &&\mathrm{3e^-} &&\longrightarrow 3~\mathrm{Cu(s)} \end{alignat*}\]
Now that the electrons are balanced, combine both half-reactions back into a single reaction. Notice how the electrons will cancel out.
\[3~\mathrm{Cu^+(aq)} + \mathrm{Fe(s)} \longrightarrow 3~\mathrm{Cu(s)} + \mathrm{Fe^{3+}(s)}\]
The redox reaction is now mass and charge balanced.
Practice
Mass/charge balance the following redox reaction.
\[\mathrm{Pb^{2+}(aq)} + \mathrm{Ag(s)} \longrightarrow \mathrm{Pb(s)} + \mathrm{Ag^+(aq)}\]
Solution
Silver is oxidized. Lead is reduced.
\[\mathrm{Pb^{2+}(aq)} + 2~\mathrm{Ag(s)} \longrightarrow \mathrm{Pb(s)} + 2~\mathrm{Ag^+(aq)}\]
Combustion
Combustion reactions are a type of oxidation-reduction reaction that occurs between a fuel and an oxidant that is typically O2(g). These reactions are high-temperature, exothermic redox reactions.
Carbon containing compounds such as hydrocarbons are a common fuel in combustion reactions. If the reacting compound contains carbon, carbon dioxide is a likely product of reaction as well as water.
\[\mathrm{CH_4(g)} + 2~\mathrm{O_2(g)} \longrightarrow \mathrm{CO_2(g)} + 2~\mathrm{H_2O(g)}\]
Here is a chemical reaction for the combustion of glucose, a carbon-containing compound.
\[\mathrm{C_6H_{12}O_6(s)} + 6~\mathrm{O_2(g)} \longrightarrow 6~\mathrm{CO_2(g)} + 6~\mathrm{H_2O(g)}\]
Sometimes combustion reactions do not involve carbon and, therefore, do not result in the formation of carbon dioxide.
\[4~\mathrm{NH_3(g)} + 5~\mathrm{O_2(g)} \longrightarrow 4~\mathrm{NO(g)} + 6~\mathrm{H_2O(g)}\]
\[2~\mathrm{H_2(g)} + \mathrm{O_2(g)} \longrightarrow 2~\mathrm{H_2O(g)}\]
Sometimes the reaction does not involve O2(g) as the oxidant. For example, hydrogen gas burns explosively with fluorine gas, a stronger oxidant than oxygen gas.
\[\mathrm{H_2} + \mathrm{F_2} \longrightarrow 2~\mathrm{HF}\]
Other oxidants can be chlorine trifluoride (ClF3)
\[ \mathrm{Fe} + 2~\mathrm{ClF_3} \longrightarrow \mathrm{FeF_3} + \mathrm{Cl_2}\]
and nitrous oxide (N2O).
\[\mathrm{C} + 2~\mathrm{N_2O} \longrightarrow \mathrm{CO_2} + \mathrm{2~N_2}\]
Other Reaction Classifications
Synthesis
A synthesis reaction combines multiple reactants into a single product. The example reaction below is also a redox reaction.
\[\mathrm{Zn(s)} + \mathrm{Cl_2(g)} \longrightarrow \mathrm{ZnCl_2(s)}\]
Decomposition
A decomposition reaction decomposes a reactant into multiple products. The example reaction below is also a redox reaction.
\[\mathrm{H_2O_2(aq)} \longrightarrow \mathrm{H_2O(l)} + \mathrm{O_2(g)}\]
Single displacement
In a single displacement reaction, an ion from one compound is “displaced” to another compound. The example reaction below is also a redox reaction.
\[\mathrm{CuCl_2(aq)} + \mathrm{Zn(s)} \longrightarrow \mathrm{Cu(s)} + \mathrm{ZnCl_2(aq)}\]
Double displacement/Exchange
In a double displacement (or exchange) reaction, an two ionic compounds swap one of their ionic components (e.g. is “displaced”) with each other.
\[\mathrm{CH_3COOH(aq)} + \mathrm{NaF(aq)} \longrightarrow \mathrm{NaCH_3COO(s)} + \mathrm{HF(aq)}\]