Oxidation and Reduction Study Guide

Non-Redox:

Exchange Reactions

Previously, we studied Exchange reactions where ions in solutions exchange partners to form new compounds. Some were precipitation reactions between two aqueous salts to form at least one insoluble product.

K₂CrO₄ (aq) + PbCl₂ (aq) à PbCrO₄ (s) + 2KCl (aq)

In Module 5 Part I, you had to use a table of solubility to determine if precipitation occurred between the two reacting compounds:

Another Exchange Reaction is a process where a gas molecule was formed as a product:

Some Common Gas Forming Exchange Reactions:

The Last Exchange Reaction is a special reaction between an acid and a base to form a salt and the covalent molecule, water (Neutralization). Sometimes this reaction is thought to be a proton transfer reaction.

Neutralize Stomach Acid with Malox:

Al(OH)₃ (s) + 3HCl (aq) à AlCl₃ (aq) + 3HOH (l)

Net Ionic:

Al(OH)₃ (s) + 3H ¹⁺ (aq) à Al ³⁺ (aq) + 3HOH (l)

Neutralize Stomach Acid With Milk of Magnesia:

Mg(OH)₂ (s) + 2 HCl (aq) à MgCl₂ (aq) + 2HOH (l)

Net Ionic:

Mg(OH)₂ (s) + 2H ¹⁺ (aq) à Mg ²⁺ (aq) + 2HOH (l)

Oxidation and Reduction (Redox)

Oxidation-Reduction Reactions occur when covalent bonds are broken and new covalent bonds are formed. In the process, one of the reactants gains electrons and one of the reactants loses electrons:

Look at the possibilities of element X breaking apart from element Y:

Most covalent bond break unevenly, if X & Y have different electronegativities so that one atom gains electrons and the other loses.

When covalent bonds are broken and reformed, this may be a simple atom of an element changing to an ion or vice versa, a polyatomic ion rearranging to another ion or molecule, or a molecule rearranging to another molecule or an ion.

For example: Molecule to Molecule:

Oxalic Acid Molecule changes to 2 Carbon Dioxide Molecules:

H₂C₂O₄ à 2 CO₂ + 2 H ¹⁺ + 2 e ^1-

Oxidation has taken place! Covalent Bonds are broken and reformed with 2 electrons lost by the oxalic acid molecule when making the two carbon dioxide molecules and two hydrogen ions as products.

Example: Polyatomic Ion to a Metallic Cation:

Reduction has taken place! Covalent Bonds are broken to form a metallic ion with 5 electrons gained by the manganese central atom of the permanganate ion to become the Manganese II (Manganous) ion.

Example: Metallic Atom to a Metallic Cation:

In oxidation reduction reactions, called redox reactions for short, the reactant which loses electrons in the bond breaking and making process is called oxidation. We say that substance is oxidized. However oxidation can not happen without reduction, so we say the substance undergoing oxidation is the agent of reduction, or the substance causing reduction. The reducing agent is sometimes called the reducer. The reducer is oxidized. You have to find a way to separate these definitions between nouns and verbs as it can be confusing.

For example: Metallic ion to a Metallic Atom:

Which substance above is oxizided? Which substance is reduced?

The way your instructor figures out the vocabulary:

Reduce means get smaller. On the number line +2 is larger than 0. In changing, the Copper II ion at +2 charge changes to a Copper atom which has no charge, or the charge gets smaller. Therefore this is the Reduction. Copper II ion is reduced, but Copper II is the oxidizer or oxidizing agent.

If reduce means get smaller, than oxidize means the opposite, get larger. The hydrogen atoms in hydrogen gas have zero charge, but in the compound H₂S they have changed to 1+ charged ions. As the hydrogen atoms change, they have gained in charge. To gain in charge, they must lose electrons. Therefore hydrogen is oxizided, or the hydrogen change is the oxidation step. The hydrogen atoms are the reducer or the reducing agent.

The Iron II ion increases from +2 to +3 charge in the reaction. Therefore, it undergoes oxidation to the Iron III ion. Iron II is oxidized. Iron II is the reducer.

The permanganate ion, MnO₄ ^1-, is reduced. It is the oxidizer.

There are several methods to balance oxidation reduction equations. You can not balance the equations by inspection as you may get a chemical balance, total numbers of each atoms equal on both sides of the equation, but the equation must also be charged balanced. That is the total electrons gained in the reduction MUST equal the total electrons lost in the oxidation.

The most popular method of balancing redox is the Electron Transfer Method. This is the process usually taught in high school chemistry, and if you have done this before, it is the method you have probably used.

Some times the Electron Transfer Process is called the Oxidation Number Method. You assign oxidation numbers to atoms in a covalent bonded group to identify the atom gaining electrons and the atom losing electrons.

You then balance the electrons transferred first by applying coefficients to the oxidizer and the reducer and their corresponding products to make the Total electron Gain equal to the Total electron Loss in the first part of the balancing. Then you balance the remaining atoms, ions, and/or molecules by inspection.

For example: iodide ion is oxidized by Dichromate ion:

6KI (aq) + K₂Cr₂O₇ (aq) + 14HCl (aq) à 2CrCl₃ (aq) +3I₂ (aq) + 8KCl (aq) + 7H₂O

Electron Transfer Method:

This process requires you to assign an oxidation number to the chromium in the dichromate ion or to the chromium in the ionic compound, Potassium dichromate. To do this you need some guidelines. Arrgh! A new set of rules:

From the Kotz & Treichel textbook (p 171):

From another textbook:

1. A metal or a nonmetal in the free state has an oxidation number of 0.

2. A monoatomic ion has an oxidation number equal to it ionic charge.

3. A hydrogen atom is usually assigned an oxidation number of +1,

except hydrides where you assign -1.

4. An oxygen atom is usually assigned the oxidation number of -2, except in

peroxides where you assign -1.

5. For a molecular compound, the more electronegative element is assigned an

oxidation number equal to the charge of its anion

6. For an ionic compound, the sum of the oxidation numbers of each atom totals 0.

7. For a polyatomic ion, the sum of the oxidation numbers for each atom total to the

ionic charge of the polyatomic ion.

Do you see why the chromium in K₂Cr₂O₇ is +3?

Electron Transfer Method:

Kotz & Treichel Reference: Chapter 5 Section 5.7

Ion Electron Method:

Kotz & Treichel Reference: Chapter 20 Section 20.1 p 830-835

All redox equations must be atom for atom balance on both sides of the equations but also the electron gain must equal the electron loss of the reactants. In the Electron Transfer Method electron gain and loss is balanced first, then the remaining atoms, ions, and/or molecules are balanced. This requires an additional task of understanding oxidation numbers as explained above.

In the Ion-Electron Method, sometimes called the Half Equation Method, the atoms, ions, and/or molecules are balanced first, then the electron gain and loss is balanced last. Each atom, ion, and/or molecule is treated as its whole group with its charge including zero for atoms and molecules. No Oxidation numbers are needed. The following is the 10 step method:

Step 1: Rewrite the equation ionically( p167):

(a) Soluble salts are written as ions

[salts with (aq) phase]

(b) Strong Acids are written as ions

[HClO₄, H₂SO₄, HNO₃, HCl, HBr, HI are strong]

(c) Leave insoluble salts in formula unit form

[salts with (s) phase]

(d) Weak acids are written in molecular form

[those acids that are not strong]

(e) Covalent molecules such as water, diatomic gases, organic

compounds, binary molecular and all oxides are left in

molecular form.

Step 1 will be done by your instructor for the test items. The problems will be given in Net Ionic Form.

Molecular Equation:

KI (aq) + K₂Cr₂O₇ (aq) + HCl (aq) à CrCl₃ (aq) +I₂ (aq) + KCl (aq) + H₂O (l)

Net Ionic Equation:

I ^1-(aq) + Cr₂O₇ ^2-(aq) + H ¹⁺(aq) à Cr ³⁺ (aq) + I₂(aq) + H₂O(l)

Step 2: Acid, Basic, or Neutral media?

Note if the solution is Acid, Basic, or Neutral media:

Acid: H ¹⁺ (or H₃O ¹⁺) ions are present

I ^{1- +} Cr₂O₇ ^2- + H ¹⁺ à Cr ³⁺ + I₂ + H₂O

MnO₄ ^1- + C₂O₄ ^{2- +} H ¹⁺ à Mn ²⁺ + CO₂ + H₂O

Basic: OH ^1- ions are present

MnO₄ ^1- + C₂O₄ ^2- + OH ^1- + H₂O à MnO₂ + CO₃ ^2-

Cr ³⁺ + ClO₃ ^1- + OH ^1- à CrO₄ ^{2- +} Cl ^1- + H₂O

Neutral: Neither H ¹⁺ or OH ^1- ions are present

Cu(NH₃)₄Cl₂ + KCN + HOH à NH₃ + NH₄Cl + K₂Cu(CN)₃ + KCNO + KCl

H ^{1+ ions}, OH ^1- ions and H₂O molecules will be used as helpers to balance half equations (unless O₂, O₃, or H₂O₂ are present)

Step 3: Write Half Equations:

Neglecting H ^{1+ ions}, OH ^1- ions and H₂O molecules , select the atoms, ions, and/or molecules undergoing a redox change. Write each as a reactant in a ½ equation:

I ^{1- +} Cr₂O₇ ^2- + H ¹⁺ à Cr ³⁺ + I₂ + H₂O

½ equation: Cr₂O₇ ^2- à Cr ³⁺

½ equation: I ^1- à I₂

Step 4: Balance the oxidizer and reducer

Neglecting hydrogen and oxygen atoms, balance the atoms undergoing change in the ½ reaction:

½ equation: Cr₂O₇ ^2- à 2 Cr ³⁺

½ equation: 2 I ^1- à I₂

Step 5: Balance Oxygen by:

Acid Media:

Add water molecules to the opposite side of the ½ equation that has excess unbalanced oxygen [O]:

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If Unbalanced oxygen on Left:

If the two unbalanced oxygen atoms are on the left side of the equation, then add two water molecules to the right side of the equation:

½ equation: 2 [O] à 2 H₂O

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For example:

½ equation: MnO₄ ^1- à Mn ²⁺

has four unbalanced oxygen atoms on the left side of the ½ equation in the permanganate ion. Therefore add four water molecules to the right side of the ½ equation to balance the oxygen:

½ equation: MnO₄ ^1- à Mn ²⁺ + 4 H₂O

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Or Unbalanced oxygen on the Right:

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If the three unbalanced oxygen atoms are on the right side of the equation, then add three water molecules to the left side of the equation:

½ equation: 3 H₂O à 3 [O]

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For example:

½ equation: Cr ³⁺ à CrO₄ ^-2

has four unbalanced oxygen atoms on the right side of the ½ equation in the chromate ion. Therefore add four water molecules to the left side of the ½ equation to balance the oxygen:

½ equation 4 H₂O + Cr ³⁺ à CrO₄ ^-2

Basic Media

Add hydroxide ions to the opposite side of the ½ equation that has excess unbalanced oxygen [O]

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If Unbalanced oxygen on the Left:

If the three unbalanced oxygen atoms are on the left side of the equation, then add three hydroxide ions to the right side of the equation:

½ equation: 3 [O] à 3 OH ^1-

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For example:

½ equation: MnO₄ ^1- à Mn ²⁺

has four unbalanced oxygen atoms on the left side of the ½ equation in the permanganate ion. Therefore add four hydroxide ions to the right side of the ½ equation to balance the oxygen:

½ equation: MnO₄ ^1- à Mn ²⁺ + 4 OH ^1-

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Or Unbalanced oxygen on the Right:

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If the two unbalanced oxygen atoms are on the right side of the equation, then add two hydroxide ions to the left side of the equation:

½ equation: 2 OH ^1- à 2 [O]

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For example:

½ equation: Cr ³⁺ à CrO₄ ^-2

has four unbalanced oxygen atoms on the right side of the ½ equation in the chromate ion. Therefore add four hydroxide ions to the left side of the ½ equation to balance the oxygen:

½ equation 4 OH ^1- + Cr ³⁺ à CrO₄ ^-2

Balance Oxygen in Neutral Media:

If the media is not acidic or basis, and there are no H⁺ or OH^- ions present in the equation, then you have to create your own help equations from what is present. Use the ions/molecules in the total equations that are not part of the ½ equations to create help equations to balanced unbalanced oxygen atoms.

For example:

if there are cyanide and cyanate ions present, then unbalanced oxygen could be balanced as follows:

3 [O] + 3 CN ^1- à 3 CNO ^1-

or

3 CNO ^1- à 3 [O] + 3 CN ^1-

Acid Media Homework:

Zn + NO₃ ^1- + H ¹⁺ à Zn ²⁺ + NH₄ ¹⁺ + H₂O

MnO₄ ^1- + C₂O₄ ^{2- +} H ¹⁺ à Mn ²⁺ + CO₂ + H₂O

Cr₂O₇ ^2- + C₂H₅OH + H ¹⁺ à Cr ³⁺ + HC₂H₃O ₂ + H₂O

SO₄ ^2- + CH₂O + H ¹⁺ à H₂S + CO₂ + H₂O

FeS + NO₃ ^1- + H ¹⁺ à NO + Fe ²⁺ + SO₄ ^2- + H₂O

Cr₂O₇ ^2- + Cl ^1- + H ¹⁺ à Cr ³⁺ + Cl₂ + H₂O

H₂O₂ + MnO₄ ^1- + H ¹⁺ à Mn ²⁺ + O₂ + H₂O

Step 6: Balance Hydrogen by

Acid Media:

Add hydrogen ions to the opposite side of the ½ equation that has excess unbalanced hydrogen [H]

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If Unbalanced hydrogen on the Left:

If the three unbalanced hydrogen atoms are on the left side of the equation, then add three hydrogen ions to the right side of the equation:

½ equation: 3 [H] à 3 H ¹⁺

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For example:

½ equation 4 H₂O + Cr ³⁺ à CrO₄ ^-2

has eight unbalanced hydrogen atoms on the left side of the ½ equation in the four water molecules. Therefore add eight hydrogen ions to the right side of the

½ equation to balance the hydrogen:

½ equation 4 H₂O + Cr ³⁺ à CrO₄ ^-2 + 8 H ¹⁺

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Or Unbalanced hydrogen on the Right:

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If the two unbalanced hydrogen atoms are on the right side of the equation, then add two hydrogen ions to the left side of the equation:

½ equation: 2 H ¹⁺ à 2 [H]

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For example:

½ equation: MnO₄ ^1- à Mn ²⁺ + 4 H₂O

has eight unbalanced hydrogen atoms on the right side of the ½ equation in the four water molecules. Therefore add eight hydrogen ions to the left side of the ½ equation to balance the hydrogen:

½ equation: 8 H ¹⁺ + MnO₄ ^1- à Mn ²⁺ + 4 H₂O

Basic Media

Add a hydroxide ion to the same side of the ½ equation that has excess unbalanced hydrogen, then add a water molecule to the opposite of the equation (The extra hydrogen is combined with a hydroxide ion on one side to make a water molecule on the other side of the ½ equation)

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If Unbalanced hydrogen on the Left:

If the three unbalanced hydrogen atoms are on the left side of the equation, then add three hydroxide ions to the left side and three water molecules to the right side of the equation:

½ equation: 3 [H] + 3 OH ^1- à 3 H₂O

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For example:

½ equation 4 OH ^1- + Cr ³⁺ à CrO₄ ^-

has four unbalanced hydrogen atoms on the right side of the ½ equation in the four hydroxide ions. Therefore, add four more hydroxide ion to the right side and four waters to the left side of the ½ equation to balance the hydrogen:

½ equation 4 OH ^1- + 4 OH ^1- + Cr ³⁺ à CrO₄ ^- + 4 H₂O

or combine hydroxides:

½ equation: 8 OH ^1- + Cr ³⁺ à CrO₄ ^- + 4 H₂O

Or Unbalanced hydrogen on the Right:

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If the two unbalanced hydrogen atoms are on the right side of the equation, then add two hydroxide ions to the right side to combine with the unbalanced hydrogen and then add water molecules to the left side of the equation to balance the hydroegn:

½ equation: 2 H₂O à 2 [H] + 2 OH ^1-

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For example:

½ equation: MnO₄ ^1- à Mn ²⁺ + 4 OH ^1-

has four unbalanced hydrogen atoms on the left side of the ½ equation.

Therefore add four hydroxide ions to the left side of the ½ equation and four water to the right to balance the hydrogen:

½ equation: 4 H₂O + MnO₄ ^1- à Mn ²⁺ + 4 OH ^1- + 4 OH ^1-

or combine the hydroxide ions:

½ equation: 4 H₂O + MnO₄ ^1- à Mn ²⁺ + 8 OH ^1-

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Balance Hydrogen in Neutral Media:

If the media is not acidic or basis, and there are no H⁺ or OH^- ions in the reaction, then you have to create your own help equations from what is present. Use the ions/molecules in the total equations that are not part of the ½ equations to create help equations to balanced oxygen or hydrogen.

For example:

If there are ammonia and ammonium ions present, then unbalanced hydrogen could be balanced as follows:

3 [H] + 3 NH₃ à 3 NH₄ ¹⁺

or

3 NH₄ ¹⁺ à 3 [H] + 3 NH₃

Basic Media Homework:

Cr ³⁺ + ClO₃ ^1- + OH ^1- à CrO₄ ^{2- +} Cl ^1- + H₂O

Mn ^2- + Br₂ + OH ^1- à MnO₂ + Br ^1- + H₂O

Fe(OH)₂ + O₂ + H₂O à Fe ³⁺ + OH ^1-

Zn + NO₃ ^1- + OH ^1- à ZnO₂ ^{2- +} NH₃ + H₂O

AsO₂ ^1- + ClO ^1- à AsO₃ ^1- + Cl ^1- (basic solution)

MnO₄ ^1- + C₂O₄ ^2- + OH ^1- + H₂O à MnO₂ + CO₃ ^2-

N₂H₄ + O₂ à N₂ + H₂O₂ (basic solution)

Step 7: Charge Balance the ½ Equations:

Charge balance the ½ equation by adding electrons to the side of the equation so that the total electronic charge on the left side of the equation is equal to the total electronic charge on the right side of the equation:

For the iodide to the iodine molecule, the charge balance is simple:

½ equation: 2 I ^1- à I₂

2 x (-1) = 1 x 0

or

-2 = 0

Think of the number line, which is the larger: -2 or 0?

0 is the larger! Therefore add two electrons to right side of the equation to bring the charge down from 0 to -2:

½ equation: 2 I ^1- à I₂ + 2 e ^1-

Now the ½ equation is chemically and electronically balanced.

Since the electrons were added to the right, then electrons are lost by the iodide to become iodine molecules. This step is the oxidation. Iodide is oxidized. The iodide is the reducer or the reducing agent.

Look at the other ½ equation, we need a little more work here:

½ equation: 14 H ¹⁺ + Cr₂O₇ ^2- à 2 Cr ³⁺ + 7 H₂O

14 x (+1) + 1 x (-2) = 2 x (+3) + 7 x (0)

or

+14 + (-2) = +6 + 0

+12 = +6

+12 is larger, so add six electrons to the left:

½ equation: 6 e ^1- + 14 H ¹⁺ + Cr₂O₇ ^2- à 2 Cr ³⁺ + 7 H₂O

The dichromate ion gains electrons because the electrons are added to the left side of the equation. Dichromate is the reduction step or dichromate is reduced. It is the oxidizer or the agent of oxidation.

Step 8: Make electron gain = electron loss

We now have two chemically and electronically balanced half equations. We must make the electron gain = the electron loss. The oxidation gives up two electrons each time it occurs, while the reduction requires six electrons to make it occur just once.

½ equation: 6 e ^1- + 14 H ¹⁺ + Cr₂O₇ ^2- à 2 Cr ³⁺ + 7 H₂O

½ equation: 2 I ^1- à I₂ + 2 e ^1-

Therefore it takes three oxidations to equal one reduction. Multiple the entire oxidation step by three and leave the reduction alone:

½ equation: 6 I ^1- à 3 I₂ + 6 e ^1-

Step 9: Combine the ½ equations

Add the two half equations:

6 e ^1- + 14 H ¹⁺ + Cr₂O₇ ^2- + 6 I ^1- à 2 Cr ³⁺ + 3 I₂ + 7 H₂O + 6 e^1-

Cancel the electrons and any other atoms, ions, and/or molecule which occur on both sides of the equation:

14 H ¹⁺ + Cr₂O₇ ^2- + 6 I ^1- à 2 Cr ³⁺ + 3 I₂ + 7 H₂O

Step 10: Balance the Molecular Equation:

Balance the total molecular equation by substituting the net balanced ionic equation into the molecular equation and balance the spectator ions by inspection:

6KI (aq) + K₂Cr₂O₇ (aq) + 14HCl (aq) à 2CrCl₃ (aq) +3I₂ (aq) + 8KCl (aq) + 7H₂O