CHEM 191

MODULE 5

CHEMICAL REACTIONS: PRECIPITATION and OXIDATION - REDUCTION

Learning Objectives.

By the end of this module you should be able to:

• Classify chemical reactions as decomposition, combination, displacement or exchange

• Recognise spectator ions in a chemical reaction and write ionic equations

• Use solubility rules to identify the products of precipitation reactions

• Describe the dissolving process and account for solubility in terms of solute and solvent interactions

• Recognise oxidation-reduction reactions by identifying electron transfer between reactants

• Write balanced half-equations and use these to write balanced equations for redox reactions

• Identify oxidants and reductants in redox reactions

• Determine and use oxidation numbers to identify redox processes

Reference: ESA Chapters 5, 21, 4

CLASSIFYING CHEMICAL REACTIONS

Chemical reactions can be classified in two ways; according to the way the reactant particles behave in relation to each other or according to changes that occur to the reacting particles.

If we focus on the number and nature of the reactants and products and on the type of transformation needed to convert the reactant(s) into the product(s), reactions can be classified as:

•     Combination (Synthesis) - elements or less complex compounds come together to form a single more complex compound

•    Decomposition - a compound breaks apart into either elements or less complex compounds

•    Displacement - a single element replaces another one in a compound

•    Exchange – ions from one compound switch places with ions from another compound to form two new compounds

Examples of different types of reactions

Decomposition

•      This is the simplest type of reaction.

•      A single compound is broken down into several different compounds.

Example 1:  When green crystals of copper(II) carbonate are heated, a gas (which can be identified as carbon dioxide) is released and black copper(II) oxide is left as a residue.

i.e.             CuCO3(s)  →  CuO(s)  +  CO2(g)

Since one reactant has visibly become two products, decomposition must have occurred.

Combination

•      This is the opposite of decomposition

•      It involves the joining of two or more elements or compounds to form a single compound.

Example: Hydrogen and chlorine react to produce hydrogen chloride H2(g)  +  Cl2(g)  →  2HCl(g)

In a combination reaction, there will be several reactants but only one product.

Displacement Reactions

•      In a displacement reaction, one element is displaced from a compound by another element in its elemental form.

•      The equation describing the reaction must have an element and a compound as reactants and a different element and compound as products.

Example: If zinc metal is placed in a solution of copper(II) sulfate, the blue solution is decolourised. Colourless zinc sulfate is formed and copper metal forms a brown deposit.

Zn(s)  +  CuSO4(aq)    →    Cu(s)  +  ZnSO4  (aq)

Exchange Reactions

•         An exchange reaction is one in which atoms or groups of atoms are interchanged between two compounds.

•         For ionic compounds, an exchange reaction can be identified when two reactants give two products by a change ofpartners between the anions and cations.

Example:     When solutions of barium chloride and sodium sulfate are mixed, a precipitate of barium sulfate

is formed and sodium chloride is left in solution.

BaCl2(aq)  +  Na2SO4(aq)  →  BaSO4(s)  +  2NaCl(aq)   Written as an ionic equation:  Ba2+(aq)  +  SO42-(aq) →  BaSO4(s)

Activity 5.1 Types of reactions

Questions

1.    List the number of reactants and the number of products found in the combination reaction (as shown in the table) and link this to the definition of a combination reaction.

2.    List the number of reactants and the number of products found in the decomposition reaction (as shown in the table) and link this to the definition of a combination reaction

3.    Explain how you would compare a displacement reaction to an exchange reaction.

Exercise 5.1

Identify the type of reaction shown in each of the following chemical equations. Justify your answers.

(a)        2KClO3(s)  →     2KCl(s) + 3O2(g)

(b)        AgNO3(aq) + KCl(aq) →  AgCl(s) + KNO3(aq)

(c)        2H2(g) + O2(g)  →   2 H2O(g)

(d)        2AgNO3(aq) + Cu(s)  →  Cu(NO3)2(aq) + 2Ag(s)

(e)        CaCO3(s)   → CaO(s)   + CO2(g)

(f)         2NaI(aq) + Cl2(g)  →   2NaCl(aq)   + I2(s)

Ionic Equations and Spectator Ions

When ionic compounds are dissolved in water, the anions and cations dissociate and so essentially exist independently. For example, when solid sodium chloride, NaCl(s), is dissolved in water we can write:

This implies that the ions have separated and dispersed through the solution so they will be surrounded by water molecules. A solution of sodium chloride is usually written as NaCl(aq) which is misleading as it does not show the separated ions. However, recognising the dissociation of ions can simplify the way equations for some reactions are written.

For example: For the reaction of sodium chloride and silver nitrate solutions

NaCl(aq)  +  AgNO3(aq)  →  AgCl(s)  +  NaNO3(aq)

We can consider the dissociation of the salts in solution and write the equation as:

Na+(aq)  +   Cl-(aq)  +  Ag+(aq)  +  NO3-(aq)  →  AgCl(s)  +  Na+(aq)   +  NO3-(aq)

Since nothing has happened to the Na+ and NO3- ions during the reaction they are known as spectator ions so they can be eliminated from the equation as follows:

Ag+(aq)  +  Cl-(aq)   →  AgCl(s)

This is known as a net ionic equation, or more often, an ionic equation. (Note that the formation of solid

AgCl implies a new compound as the ions have “changed partners” from those in the original solutions and the formation of the solid removes the ions from the solution).

Activity 5.2 – ionic equations

This activity is designed to help you understand ionic equations

Questions

1.    Consider the reactions modelled in the beakers drawn in the box

(a)   (i)    Which beakers contain solid, insoluble substances? Give reasons for your answers.

(ii)   Which beakers contain solutions of ionic substances? Give reasons for your answers.

(iii) Which of the reactions produces a gas? How do you know this?

(b)   Indicate which of the equations (i) to (iii) corresponds to the reactions 1, 2 or 3 above.

(i).     Zn (s) + 2HCl (aq) → ZnCl2  (aq) + H2 (g)

(ii).    Zn (s) + Cu(NO3)2  (aq) → Zn(NO3)2  (aq) + Cu (s)

(iii)   Zn(NO3)2  (aq) + Na2CO3  (aq) → ZnCO3  (s) + 2NaNO3  (aq)

(c)   In each of the reactions 1 to 3, there are ions present in the solutions that do not participate in the chemical reaction. In other words, they exist in the same form. both before and after the reaction.  These substances are called spectator ions. Use the equations (i) to (iii) to identify the spectator  ions for each reaction.

2.    In the table below, the equation for one of the previous reactions is written in three different ways.

Molecular Equation     Zn(s) + Cu(NO3)2(aq) → Zn(NO3)2(aq) + Cu(s)

Ionic Equation            Zn(s) + Cu2+ (aq) + 2NO3 −(aq) → Zn2+ (aq) + 2NO3 −(aq) + Cu(s)

Net Ionic Equation       Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s)

(a)   (i)    Which equation represents the ionic substances as bonded together in the solution?

(ii)  Which equation represents the ionic substances as separate ions in solution?

(iii) Which equations is a better representation of how the species take part in the reaction?

(iv)  Compare the net ionic equation to the other two equations.

a.  What chemical species is missing in the net ionic equation?

b.  Explain why it is valid to remove this species from the equation.

(b)        Write ionic and net ionic equations for the remaining reactions in 1: Zn(s) + 2HCl (aq) → ZnCl2  (aq) + H2  (g)

Zn(NO3)2  (aq) + Na2CO3  (aq) → ZnCO3  (s) + 2NaNO3 (aq)

Exercise 5.2

(a)   Write the following equations as balanced net ionic equations.

(i)         BaCl2(aq) + Na2SO4(aq) ⎯→   BaSO4(s) + 2NaCl(aq)

(ii)        Zn(s) + H2SO4(aq) ⎯→  zinc sulfate (aq) + hydrogen (g)

(iii)       HNO3(aq) + Ba(OH)2(s) ⎯→ Ba(NO3)2(aq) + H2O(ℓ)

(b)   Balance the following ionic equations (the spectator ions have been eliminated).  Some of these maybe quite difficult but they would be worse if the spectator ions had been included.

(i)   Ca2+(aq)  +  PO43-(aq)  →  Ca3(PO4)2(s)    (ii)  Cu2+(aq)  + NH3(aq)  →  Cu(NH3)42+(aq)

(iii) MnO4-(aq)  +  H+(aq)  +  Fe2+(aq)  →  Mn2+(aq)  +  H2O(ℓ)  +  Fe3+(aq)

(iv)  S2O32-(aq)  +  I2(aq)  →  S4O62-(aq)  + I-(aq)

(v)   Cr2O72-(aq)  +  Fe2+(aq)  +  H+(aq)  →  Cr3+(aq)  +  Fe3+(aq)  +  H2O(ℓ)

DISSOLVING

Dissolving a solute in a solvent to make a solution is a physical process. A solution is a mixture and the properties of the components are retained. For example, when salt is added to water the solution tastes salty    in the same way as do the salt crystals. Since many chemical reactions occur in aqueous solutions, it is useful to understand the interactions between the particles in a solution.

•     For aqueous solutions, the activity of the water solvent is related to the nature of the water molecules.

•     The water molecule, H2O, is bent with the H-O-H angle being about 105o. The O-H bond is polar  because the oxygen atom is more electronegative than the hydrogen atom and attracts the electron density in the bond towards itself.

•     As a result, the water molecule has a negative “end” at the oxygen atom, and positive “ends” at the

hydrogen atoms.  (In the diagram the δ symbol means ‘slightly’ – showing that the ends are only

slightly positive and slightly negative). The shape of the water molecule is important in determining its solvent properties.

•      Substances dissolve in one another because the solution they form together is more stable than if the two substances exist separately.

•      This implies that the attractive forces between the solvent molecules and the solute particles (which can be molecules or ions) are greater than the attractive forces within the solute structure and the attractive   forces between the solvent molecules.

•      Ionic solids normally have strong attractive forces between their oppositely charged ions so for water to dissolve an ionic substance the water molecules must have some way of overcoming these attractive forces within the solid.

•      The polar nature of the water molecule enables it to reduce the attraction between the charged ions

because the negatively charged oxygen atoms in the water molecule interact with the positive ions in the ionic solid and the positively charged hydrogen atoms in the water molecule interact with the negative ions.

Fig 5.1 Dissolving sodium chloride in water. Note the orientation of the water molecules around the ions

•   Solvents that are composed of polar molecules are able to dissolve both ionic and polar substances but those with non-polar molecules cannot do so because they cannot overcome the attractive forces

between the ions. In contrast non-polar solvents are more able to dissolve non-polar solutes.

•  Non-polar solutes generally have weak intermolecular interactions in the solid or liquid and they interact weakly with the solvent. If their interactions with solvent molecules is similar to or stronger than their

interactions with each other in the solid state, then the solute molecules maybe soluble.

•   Thus a general rule emerges - like dissolves like.  That is, polar solvents are more likely to dissolve ionic and polar solutes, and non-polar solvents are more likely to dissolve non-polar solutes.

Focussing Questions 1:

1.    How would you know that a chemical reaction has taken place?

2.    What do the terms “solute” and “solvent” refer to?

3.    Explain, in terms of energy, why some substances dissolve in water.

4.    What needs to happen for an ionic solid to dissolve in water?

5.    To what does the rule “like dissolves like” refer?

6.    Why are two layers observed when oil and water are mixed?

PRECIPITATION REACTIONS

Recall that when ionic solids dissolve in water they completely dissociate (see fig.5. 1) – that is the ions remain separated from each other in solution.

NaCl(aq)  →   Na+(aq)  +  Cl-(aq)

So an aqueous solution of sodium chloride, NaCl(aq), contains the ions Na+  and Cl– surrounded by water molecules.

Not all ionic solids (salts) are soluble in water, but even insoluble salts will dissolve to a very small extent.

Solubility rules can be used to determine whether a salt will dissolve or not. (In this context insoluble means that very little will dissolve and soluble means that a reasonable amount will dissolve). The rules are based on observations of salts of common cations and anions.

Solubility Rules

1. All salts of the Group 1 metals and of the ammonium ion, NH4 + are soluble.

2. All nitrates, chlorates and ethanoates are soluble.

3. All chlorides, bromides and iodides are soluble except those of Ag+ , Hg2 2+, Cu+ and Pb2+ .

4. All sulfates are soluble except those of Ca2+, Ba2+ and Pb2+ .

5. All hydroxides, carbonates and sulfides are insoluble (except when Rule 1 applies).

Mixing two ionic solution sometimes results in the formation of a solid known as a precipitate. The solubility rules can be used to help identify a precipitate.

For example:  A mixture of silver nitrate solution, AgNO3(aq),,and sodium chloride solution, NaCl(aq), results in a precipitate.

•     The ions present on mixing will be Ag+, Na+, NO3-  and Cl- .

•     The two new combinations of ions in the mixture are AgCl and NaNO3.

•     Using the solubility rules, we can see that the only combination that will give an insoluble salt is AgCl.

•     Rule 1 says that all salts of group 1 metals are soluble so NaNO3  is a soluble salt. Rule 3 lists the chlorides of Ag+  as being insoluble.

The equation for the reaction:   NaCl(aq)  +  AgNO3(aq)  →  AgCl(s)  +  NaNO3(aq)

This can be written as an ionic equation: Ag+(aq)  +  Cl-(aq)  →  AgCl(s)

Activity 5.3 Precipitation Reactions

Answer the questions below using the equations in the box and the solubility rules given above.

EQUATIONS 1

A. AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

B. 2HNO3(aq) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2H2O(ℓ)

C. Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl(aq)

D. Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq)

E. FeBr2(aq) + K2SO4(aq) → FeSO4(aq) + 2KBr(aq)

Questions:

1.    Use the equations to predict which of the reactions A to E would result in a precipitate. Give a reason for your answer.

2.    Precipitation reactions are sometimes called “double replacement reactions” because the ionic solids dissolved in water are said to ‘switch partners ’. Use equation A to illustrate the idea of double displacement.

3.    In equation A, only one of the products of the reaction will form a precipitate. Identify the solubility rules that apply to the reaction products.

4.    Predict the precipitate that forms in the following reactions and write a balanced equation for the reaction.

(a)       FeCl3(aq)  +  NaOH(aq)    →

(b)       KI(aq)  +  Pb(NO3)2       →

Exercise 5.3

(a)      (i)    Solid iodine is composed of the molecules I2. Explain why iodine is more likely to be soluble in

the non-polar solvent, carbon tetrachloride, CCl4,  than in water.

(ii)  Explain why it is possible for ionic solids to dissolve in water in spite of the strong forces of attraction between the ions.

(b)     Determine whether a precipitate will form. when the aqueous solutions of the following salts are mixed. Write a balanced equation for each reaction where a precipitate is observed.

(i)      Calcium chloride and potassium sulfate

(ii)     Lead nitrate and calcium chloride

(iii)    Ammonium sulfate and sodium nitrate

(iv)    Copper(II) sulfate and potassium hydroxide

(v)     Sodium carbonate and barium nitrate

(c)     Write a balanced net ionic equation for the following reactions.

(i)         Calcium chloride and potassium sulfate

(ii)        Lead nitrate and calcium chloride

(iii)       Copper(II) sulfate and potassium hydroxide

(iv)       Sodium carbonate and barium nitrate

OXIDATION AND REDUCTION REACTIONS

Antioxidants

The term antioxidant will be a familiar term for you from the advertising of such things as ‘superfoods’ and anti-wrinkle creams. Antioxidants in biological systems are substances which may prevent some types of cell damage. Oxidation reactions in cells produce free radicals (molecules with an unpaired electron) that can damage or kill cells. Antioxidants stop oxidation reactions from occurring. Some commonly recognised antioxidants in biological systems include the Vitamins A, C and E.

Oxidation - reduction reactions, commonly known as redox reactions, are classified according to the changes that take place in the reacting particles. They can be described at particle level as electron transfer reactions – one species loses electrons which are gained by another species.

Examples of redox reactions include the rusting of iron (iron metal (Fe) is oxidised to form. rust, a compound containing Fe3+ ions), the burning of coal or wood (carbon,or carbon containing compounds are converted to CO2) or the action of a household bleach. In biological systems, respiration and photosynthesis are processes that involve oxidation-reduction reactions.

Definitions

If, in a chemical reaction, a species (atom, ion or molecule) loses electrons, it is said to be oxidised. This term originally meant ‘reacted with oxygen’ as atoms of all elements (except fluorine) lose electrons when they react with the very electronegative oxygen atom. The electrons cannot be lost completely, they have to go somewhere, and the species that gains the electrons is said to be reduced. Oxidation of one species cannot happen without the corresponding reduction of another species.

To help you remember you can use one of the following mnemonics:

• Oxidation Is Loss of electrons Reduction Is Gain of electrons    (OIL RIG)

• Loss of Electrons is Oxidation Gain of Electrons is Reduction    (LEO GER)

In the reaction, the species that is oxidised (loses electrons) causes another species to be reduced (by  providing it with electrons). Hence the oxidised species is known as a reducing agent or reductant.

Similarly, the reduced species removes electrons from another species causing it to be oxidised. Hence the reduced species is known as an oxidising agent or oxidant.

Representing Oxidation-Reduction Reactions

Consider the reaction of magnesium with oxygen.  2Mg(s)   +   O2(g)    →    2MgO(s)

The product, magnesium oxide, is an ionic compound made up of Mg2+  and O2-  ions. We can consider each reactant separately:

Mg atoms lose electrons to become Mg2+  ions:     Mg    →    Mg2+     +   2e–

O atoms gain electrons to become O2-  ions:          O2      +   4e-       →    2O2–

During the reaction, electrons are transferred from Mg atoms to O atoms.

When considered separately these steps are known as half reactions represented by half equations.

The half reaction equations explicitly show the number of electrons involved in the process. When the same number of electrons are involved in each half reaction then the sum of the two half-reactions (above) gives   the overall reaction.

2Mg   +   O2     +   4e-       →    2Mg2+     +   2O2-     +   4e-

Cancelling the electrons on both sides of the equation and combining the ions as MgO(s) gives the equation,

2Mg(s)   +   O2(g)    →   2MgO(s)

In this reaction, magnesium is oxidised (loses electrons) and is a reducing agent or reductant because it causes the oxygen to be reduced. Oxygen is reduced (gains electrons), and is an oxidising agent or oxidant because it causes the magnesium to be oxidised.

Example: When zinc metal is added to hydrochloric acid, bubbles of hydrogen gas are produced and the zinc metal eventually all disappears.

Zn(s)   +   2HCl(aq)    →   ZnCl2(aq)   +   H2(g)

In this reaction the zinc is oxidised (its atoms lose electrons) to Zn2+(aq) and the H+(aq) of the acid is reduced (the ions accept electrons) and it becomes molecular hydrogen. Zinc is thereductant, and H+  is the oxidant,

To identify the redox processes (as shown in the half equations) it is helpful to first, consider the species present in the reactants and products: Since we have ZnCl2(aq) and HCl(aq) we can write:

Zn(s)   +   2H+(aq)   +   2Cl-(aq)    →    Zn2+(aq)   +   2Cl-(aq)   +   H2(g)

The chloride ion, remains unchanged - it is a spectator ion and can therefore be omitted from the equation:

Zn(s)   +   2H+(aq)    →   Zn2+(aq)   +   H2(g)

So the half reactions can be identified as: Oxidation:       Zn  →  Zn2+    +   2e- Reduction:       2H+   +   2e-  →   H2

Adding these half equations and cancelling out the electrons gives:

Zn(s)   +   2H+(aq)    →   Zn2+(aq)   +   H2(g)