Oxidation-Reduction

Electron Transfer Reactions

Preliminary Exercise

As a review, in the following equations for familiar reactions, underline the reducing agent once and the oxidizing agent twice. At the right, indicate the change in oxidation number per atom for each element concerned. If there is no change, write in “no valence change.”

Reactions	Element Oxidized	Oxidation number increase per atom	Element Reduced	Oxidation number decrease per atom
2Al(s) + 3Cl2(g) à 2Al^3+ + 6Cl^-
Ag^+ + Cl^- à AgCl(s)
Mg(s) + 2H^+à Mg^2+ + H2(g)
Cu^2+ + H2S(g) à CuS(s) + 2H^+
Ba(s) + 2H2O à Ba^2+ + 2OH^- +H2(g)
3Na2O2(s) + Cr2O3(s) + H2O à 6Na^+ + 2CrO4^2- + 2OH^-
H2O2(aq) à H2O + ½ O2(g)
CO2(g) + H2O à H2CO3(aq)
CO2(g) + C(s) à 2CO(g)
HCl(g) + NH3(g) à NH4Cl(s)
2ZnS(s) + 3O2(g) à 2ZnO(s) + 2SO2(g)
4H^+ + 2Cl^- + MnO2(s) à Mn^2+ + Cl2(g) + 2H2O
10H^+ + SO4^2- + 8I^- à 4I2 + H2S(g) + 4H2O

Table –1

Element	Oxidation state	Formulas	Comments
SULFUR COMPOUNDS	+6 +4 +2 0 -1 -2	SO3, H2SO4, SO4^2- SO2, H2SO3, SO3^2- S2O3^2- S S2^2- H2S, S^2-	Concentrated acid is a strong oxidizing agent. Active either as oxidizing or reducing agent. Thiosulfate ion decomposes to S and H2SO3 in acid solution. Polysulfide ion decomposes to S and H2S in acid. Strong reducing agent usually oxidized to S.
OXYGEN COMPOUNDS PEROXIDES	0 -1 -2	Acidic Basic O2 H2O2 HO2^- H2O OH^-	Active as oxidizing or reducing agent.
CHLORINE COMPOUNDS	+7 +5 +4 +3 +1 0 -1	Cl2O7, HClO4,ClO4^- HClO3, ClO3^- ClO2 HClO2, ClO2^- Cl2O, HClO, ClO^- Cl2 Cl^-	Cl2O7 is unstable. HClO4 is a strong oxidizer, reduced to Cl^-. Strong oxidizing agent Reduced to Cl^-. Unstable explosive. Good oxidizing agent. Reduced to Cl^-. Good oxidizing agent. Reduced to Cl^-. Good oxidizing agent. Reduced to Cl^-.
NITROGEN COMPOUNDS	+5 +4 +3 +2 +1 0 -3	N2O3, HNO3, NO3^- NO2, N2O4 N2O3, HNO2, NO2 NO N2O N2 NH3, NH4^+	Strong oxidizing agent, usually reduced to NO2 and NO; largely to NO2 in concentrated acid and to NO in dilute acid. May go to NH3 with strong reducer. A heavy brown gas. N2O3, and HNO2 are unstable. Nitirtes are fairly stable. Active as oxidizing or reducing agent. Oxidized by air to NO2 Supports combustion quite vigorously. Good reducing agents.
CHROMIUM COMPOUNDS	+6 +3 +2 0	Acidic Basic Cr2O7^2- CrO4^2- Cr^3+ Cr(OH)4^- Cr^2+ Cr	Strong oxidizing agents. Amphoteric. An uncommon ion, because it is such a strong reducer that it reduces water to hydrogen gas. The metal.
MANGANESE COMPOUNDS	+7 +6 +4 +3 +2 0	MnO4^- MnO4^2- MnO2, MnO(OH)2 Mn^3+, Mn(OH)3 Mn^2+, Mn(OH)2 Mn	Permanganate ion. Strong oxidizing agent, reduced to Mn^2+ in acid solution and MnO2 in neutral or basic sol’n. Manganate ion. Stable only in base. Easily reduced to manganese dioxide. Brown as ppt from solution. Mn^3+ is unstable, gives Mn^2+ and MnO2. Mn(OH)2 is oxidized in air to Mn(OH)3 The metal.

Oxidation States for Common Reagents

Much useful information about the behavior of oxidizing and reducing agents under various conditions can be summarized in the form of charts. Such charts for some common elements are presented in Table-1, and are repeated again in later experiments where these elements are studied in greater detail. The charts, with their comments on the behavior of the various compounds, will help you predict the probable changes in oxidation state in a particular reaction. Note that the oxidation state is given just before each formula in the chart. Let us comment briefly on the charts for sulfur and oxygen compounds.

It should be obvious from the chart of sulfur compounds that since H₂S represents the lowest possible oxidation state it can act only as a reducing agent in a redox process. In such a process it can be oxidized to free sulfur or to some higher state, such as a sulfate, its extent of oxidation, depending on the conditions and the strength of the oxidizing agent. Sulfuric acid (H₂SO₄), representing the highest oxidation state, can act only as an oxidizing agent, its reduction products being any lower states of sulfur. However, since sulfurous acid (H₂SO₃) represents an intermediate state of sulfur, it can act either as a reducing agent with substances that can take on electrons, or as an oxidizing agent with substances that can lose electrons.

Hydrogen peroxide and the peroxides are important oxidizing agents, both commercially and in the laboratory. Note, from the chart, that when hydrogen peroxide acts as an oxidizing agent it is reduced to water or, in basic solution, to hydroxide ion. It is oxidized to free oxygen only when it acts as a reducing agent, in the presence of stronger oxidizing agents. The instability of hydrogen peroxide, especially in the presence of certain catalysts, is due to the ability of one molecule to oxidize another molecule of the same substance (auto-oxidation-reduction). The half-reactions corresponding to these statements are

2H⁺ + H₂O₂ + 2e^- à 2H₂O (reduction)

H₂O₂ à 2H⁺ + O₂ + 2e^- (oxidation)

2 H₂O₂ à 2 H₂O + O₂ (auto-oxidation-reduction)

Listed below are some important categories of common oxidizing and reducing agents.

Some Common Oxidizing Agents

1. The halogens and oxygen-reduced to their negative ions, such as the following:

F₂ reduced to F^-

Cl₂ reduced to Cl^-

O₂ reduced to O^2-

2. Ions in which the metal ion has a stable lower oxidation state such as the following:

Ce^{4+ reduced} to Ce²⁺

Mn³⁺ reduced to Mn²⁺

Co³⁺ reduced to Co²⁺

3. Oxygen containing complex ions, where the central atom is in a high oxidation state, for example:

MnO^- reduced to Mn²⁺

ClO₃^- reduced to Cl^-

Cr₂O₇^2- reduced to Cr³⁺

NO₃^- reduced to NO

Some Common Reducing Agents

1. The metals oxidized to their positive ions – such as these

Sn oxidized to Sn²⁺

Zn oxidized to Zn²⁺

2. Ions in which the metal has another higher oxidation state, such as the following:

Sn²⁺ oxidized to Sn⁴⁺

Fe²⁺ oxidized to Fe³⁺

Hg₂²⁺ oxidized to Hg²⁺

3. Carbon and organic compounds – may be oxidized to other organic compounds or to CO₂ and H₂O; for example:

C (coke, much used in industry) – oxidized to CO or to CO₂

CH₃CH₂OH (alcohol) – may be oxidized to CH₃COOH + 4H⁺ + 4e^-

HCHO (formaldehyde) – oxidized to HCOOH (formic acid); thus,

HCHO + H₂O à HCOOH + 2H⁺ + 2e^-

The Balancing of Oxidation-Reduction Equations

In balancing any oxidation-reduction reaction, you must first know all of the reactants and products. If you do not, you will not be able to balance it correctly. Once the reactants and products are known, balance the reaction by keeping in mind that, in an oxidation-reduction reaction which is essentially an electron –transfer reaction, relative amounts of reactants must be taken in such a way that all the electrons supplied by the oxidation process are used in the reduction process. There are several methods for doing this, each differing in mechanics of operation, but all based on the same principle.

The Half-Reaction Method

Separate half-reactions, or electron reactions, are written for the oxidation and for the reduction processes. In developing these, we can first determine the number of electrons required from the change in oxidation number, then insert H⁺ (or OH^- if the solution is basic) to balance the charges, and finally add H₂O to balance the atoms.

The reverse process is sometimes used. First, balance the atoms in half-reaction by inserting H⁺ and H₂O as needed, then insert as many electrons as needed to balance the charges. Study the following example.

Example 1. Let us consider the oxidation of sodium sulfite (Na₂SO₃) by potassium dichromate (K₂Cr₂O₇) in an acid solution. Referring to the charts of oxidation states for sulfur compounds and chromium compounds (Table –1), we find that sulfite ion will be oxidized to sulfate ion, and that the dichromate ion will be reduced to the chromic ion (Cr³⁺), as will be evidenced by the green color of the solution. We may write first a partial equation including only sulfite ion and its oxidation product:

Step (I) SO₃^2- à SO₄^2-+

We need another oxygen atom on the left; this is supplied by water, and we write the hydrogen as 2H⁺ on the right. (Note that we keep oxidation states of oxygen

-2 and hydrogen +1 on both sides of the equation, since they are not the substances oxidized and reduced.) Our partial equation then becomes

Step (2) H₂O + SO₃^2- à SO₄^2- + 2H⁺

We still need to balance the charges so they are the same on both sides of the equation. To do so we add two electrons on the right,

Step (3) H₂O + SO₃^2- à SO₄^2- + 2H⁺ + 2e^-

Thereby completing the oxidation half-reaction, and demonstrating the fact that in an oxidation process, electrons are lost.

The reduction of dichromate ion to chromic ion may be expressed similarly. The several steps are as follows.

Step (1) Cr₂O₇^2- à 2Cr³⁺

Then balance the hydrogen and oxygen by inserting 14H⁺ to react with the 7 oxygen atoms to form 7H₂O,

Step (2) Cr₂O₇^2- + 14H⁺ à 2Cr³⁺ + 7H₂O

Finally, we balance the charges. As above written, we have 12 positive on the left and 6 positive charges on the right, or a net charge of 6+ on the left. We therefore need 6 electrons on the left to complete the reduction half-reaction:

Step (3) Cr₂O₇^2- + 14H⁺ + 6e^- à 2Cr³⁺ + 7H₂O

This emphasizes the fact that in a reduction process electrons are gained. Note the requirement of 6 electrons corresponds to the charge in oxidation state of chromium from +6 to +3, so two chromium atoms decrease by a total of 6 charges. Finally, we may combine the two processes in such a way that we balance electrons gained against electrons lost. To do this we multiply the oxidation half-reaction by 3 and add algebraically to the reduction half-reaction:

3(H₂O + SO₃^2- à SO₄^2- + 2H⁺ + 2e^-)

Cr₂O₇^2- + 14H⁺ + 6e^- à 2Cr³⁺ + 7H₂O

Cr₂O₇^2- + 3SO₃^2- + 8H⁺ à 2Cr³⁺ + 3SO₃^2- + 4H₂O

Always check your results to make certain that both atoms and charges are balanced in the equation.

Application of Principle

1. Show the change in oxidation number (give number of electrons gained or lost per atom: e.g. 3e^- gained ) in the following:

Reaction number of e^- lost or gained

NO₂^- à NO₃^-

SO₂ à S₂^2-

MnO₄^- à MnO₂

KClO₂ à KCl

NH₃ à NH₄⁺

HCOOH à HCHO

2. Give the formula of a product (derived from the first substance in the statement) that may be formed in the following reactions. (Note: in the example any lower oxidation state compound is possible; but not any higher one. Some are more probable others.)

Example: H2SO3 is treated with a reducing agent.	S, S2^2- or H2S
HClO2 is treated with a reducing agent.
H2SO3 is treated with an oxidizing agent.
SnCl4 is treated with zinc dust.
Cr2O7^2- is treated with SnCl2.
KMnO4 is treated with FeSO4
MnO2 is treated with concentrated HCl

3. Write the half-reaction equation for the oxidation of:

NO₂^- to NO₃^- (acidic)

H₂S to SO₄^2- (acidic)

NH₄⁺ to NO₃^-(acidic)

H₂O₂ to O_2(g) (basic)

Cr(OH)₄^- to CrO₄^2- (basic)

ClO^- to ClO₄^-(acidic)

HCOOH to CO₂ (acidic)

4. Write the half-reaction equation for the reduction of:

SO₃^2- to H₂S (acidic)

MnO₄^2- to MnO₂ (basic)

HO₂^- to OH^- (basic)

HCOOH to CH₃OH (acidic)

ClO₃^- to Cl^- (acidic)

Cr₂O₇^2- to Cr³⁺ (acidic)

CH₃NO₂ to CH₃NH₂ (acdic)

5. Given the reactants and products, write balanced net ionic equations for the following reactions. Supply H₂O, H⁺, or OH^- as needed.

1. Iron fillings are added to an aqueous FeCl₃ solution.

Fe + Fe³⁺ à Fe²⁺

2. Bismuth metal is dissolved in hot, concentrated HNO₃, and a brown gas is given off.

Bi + NO₃^- à Bi³⁺ + NO_2(g)

3. A mixture of Na₂S, NaClO, and NaOH solutions is warmed, giving a suspended precipitate.

S^2- + ClO^- à S + Cl^-

4. SO₂ gas is bubbled into K₂Cr₂O₇ solution (acidic)

SO₂ + Cr₂O₇^2- à Cr³⁺ + SO₄^2-

6. Predict the products and write balanced net ionic equations for the following reactions:

1. SnCl₂ is added to KMnO₄ solution (acidic)

2. Zinc dust is treated with dilute HNO₃, forming NH₄⁺

3. KMnO₄ oxidizes oxalate in CaC₂O₄ to CO₂ in acidic solution.

Redox Reactions – The Activity Series

Redox reactions are defined as reactions in which electrons move from one element to another. The element that gains one or more electrons is said to be reduced and the element that loses one or more electrons is said to be oxidized. Single replacement reactions are one type of redox reactions.

Some examples of redox reactions are

Cu + 2AgNO₃ à 2Ag + Cu(NO₃)₂

In this reaction copper metal is oxidized to the copper(II) ion and silver ion is reduced to silver metal.

AgNO₃ + Fe(NO₃)₂ à Ag + Fe (NO₃)₃

In this reaction, silver is reduced and the ferrous ion is oxidized to the ferric ion.

2KI + F₂ à I₂ + 2KF

In this reaction the iodide ion is oxidized to elemental iodine and fluorine is reduced to the fluoride ion.

In this experiment the goal is to perform some redox reactions and determine the relative ease of oxidation of each of these substances. You will build an activity table based on your results. You will determine the relative ease of oxidation pair-wise for a number of substances. For example, if a strip of copper metal is placed into a solution of mercury(II) chloride, the solution will become blue over time and a small pool of liquid mercury will collect at the bottom of the container. The reaction is as follows:

Cu + HgCl₂ à Hg + CuCl₂

The copper is more easily oxidized than the mercury because we see the copper replaces mercury as the more oxidized substance. If we were try to reverse the reaction, that is, to pour liquid mercury in a solution of copper(II) chloride, there would be no reaction. Again, we would conclude that copper is more easily oxidized because mercury is not able to replace copper as the most easily oxidized.

A strip of copper metal is dropped into a solution of lead(II) chloride. There is no visible reaction. Which is more active, copper or lead? Combining the results of this experiment with the results of the first experiment (copper and mercury), Can you determine whether lead is more or less active than mercury?

Go back up to the sample reactions given at the beginning of the lab. Determine which specie of each pairs of substances is more active.

Copper or silver?

Silver or iron(II)?

Fluoride or iodide ions?

Experimental Procedure

You will mix the following sets of substances and determine the more easily oxidized specie in each case. From the relative oxidizability of each set, you will develop a comprehensive activity series.

Experimental Design

In this experiment you will react various types of metal with salt solutions to see if the metals are more or less active than the salt cations. As you perform the experiments, keep in mind the following:

1. Be sure to sand all of the metal pieces to remove any oxidation that may prevent the reaction from proceeding.

2. Watch the reactions for at least 15 minutes before deciding that no reaction took place. Some of the reactions are slow.

3. Describe the reaction that takes place so that you could describe to me exactly what it looked like. Describe texture, color, location, etc. Be detailed in these descriptions, they are your observations and therefore your data.

4. Note that some of the metals are reacted with acids. Here, the cation is H⁺, which will be reduced to H₂ gas if a reaction occurs.

Reactions

1. Place a strip of copper metal into a 3 mL sample of 0.1 M ZnSO₄.

2. Place a strip of copper metal into a 3 mL sample of 0.1 M FeSO_4.

3. Place a strip of zinc metal into a 3 mL sample of 0.1 M CuSO₄.

4. Place a strip of zinc metal into a 3 mL sample of 0.1 M FeSO_4.

5. Place a piece of steel wool, (iron) into 3 mL sample of 0.1 M CuSO_4.

6. Place a piece of steel wool, (iron) into 3 mL sample of 0.1 M ZnSO_4.

7. Place a piece of copper metal into 3 mL sample of 1.0 M H₂SO₄ (or what he stock room has available for this experiment)_.

8. Place a strip of zinc metal into 3 mL sample of 1.0 M H₂SO₄ (or what he stock room has available for this experiment)_.

9. Place a strip of steel wool (iron) into 3 mL sample of 1.0 M H₂SO₄ (or what he stock room has available for this experiment)_.

Add a few drops of I₂ (in methanol) to a mixture of 3mL of D.I. water and 1 mL hexane, shake well (this may be already made up for you by the stock room), and record the color of the hexane layer. Do the same with the Br₂ (in water). Now you know how to detect elemental iodine and bromine. This will help you identify the presence of these elements. When you are finished working with the halogen solutions, dispose of them in the appropriate container in the hood.

10. Mix 3 mL of 0.1 M KBr with 1 mL of hexane (this may be already prepared for you by the stockroom). Add few drops of I_2.

11. Mix 3 mL of 0.1 M KI with 1 mL of hexane (this may be already prepared for you by the stockroom). Add few drops of Br_2.

In the next experiments you will examine the redox chemistry of iron(III) ions and copper metal with the halogen/halides. Use the information gained in these experiments to get a complete activity series.

12. Mix 1 mL of 0.1 M FeCl₃ with2 mL of 0.1 M KBr.

13. Mix 1 mL of 0.1 M FeCl₃ with2 mL of 0.1 M KI.

14. Add some copper turnings to 10 mL of Br₂ water. Shake well for several minutes. Allow any precipitate that forms to settle and pour off some of the liquid phase into each of two new test tubes. To the first tube add few drops of 0.1 M AgNO_3,

Does a AgBr precipitate form? To the second tube add 1 mL of 6 M NH₃, does a blue Cu(NH₃)₄²⁺ precipitate form?

15. Add some copper turnings to 15 mL of 0.05 M I₂ methanol. Shake well for several minutes. Allow any precipitate that forms to settle and pour off some of the liquid phase into each of two new test tubes. To the first tube, add a few drops of 0.1 M AgNO₃, does a AgI precipitate form? ( Note that you may see a precipitate of CuI here, do the test for Cu²⁺ to be sure that the expected reaction occurred.) To the second tube, add 1 mL of 6 M NH₃. Does a blue Cu(NH₃)₄²⁺ precipitate form?

16. From the data you have gathered in this experiment, develop an activity series.

( As was done for Pb, Cu and Hg in the introduction.)

Data Table

Reactants		Observation	Reaction	E.O.	E.R.	More active.
Cu	Zn^2+
Cu	Fe^2+
Zn	Cu^2+
Zn	Fe^2+
Fe	Cu^2+
Fe	Zn^2+
Cu	H^+
Zn	H^+
Fe	H^+
I2	Br^-
Br2	I^-
Fe^3+	Br^-
Fe^3+	I^-
Cu	I2
Cu	Br2