Oxidation and Reduction Study Guide

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

two electrons lost by the Oxalic Acid molecule

when making the two Carbon Dioxide molecules

and two hydrogen ions as products.

Permanganate Ion changes to Manganese II Cation:

5 e ^1- + 8 H¹⁺ + MnO₄^1- à Mn^{2+ +} 4 H₂O

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.

Ion Electron Method:

All REDOX equations must have 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 states as explained in McMurry Chapter 4, Kotz Chapter 5, Bishop Chapter 9, and Corwin Chapter 17.

In the Ion-Electron Method, sometimes called the Half Equation Method or Help 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!.

References:

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

4^th Corwin Reference: Chapter 17 Section 17.4 p 475-478

5^th McMurry Reference: Chapter 4 Sections 4.6, 4.7, 4.9, 4.10

The following is the 10 step method:

Step 1: Rewrite the equation ionically( p167):

(a) Soluble salts are written as ions including Strong Bases

[salts with (aq) phase][NaOH, KOH, Ca(OH)₂, Ba(OH)₂]

(b) Strong Acids are written as ions

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

(c) Leave insoluble salts (and bases) as formula units (compounds)

[salts with (s) phase]

(d) Weak acids and bases 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. In REDOX challenge only the first 8 are ionic

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:

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

Step 6: Balance Hydrogen by

Acid Media:

Add hydrogen ions to the opposite side of the ½ equation

that has excess unbalanced hydrogen [H] atoms.

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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:

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

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

or

+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

Now the ½ equation is chemically and electronically balanced.

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 molecules which occur on both sides of the equation:

Only the 6 electrons are common to both sides,calcel the six electrons. Therefore the balanced net ionic reaction is:

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:

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

Balance the Potassium Chloride (spectators) by inspection:

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

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

Mn ²⁺ + S₂O₈ ^2- + H₂O à MnO₂ + SO₄ ^2- + H ¹⁺

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

CdS + I₂ + H ¹⁺ à Cd ²⁺ + HI + S

Zn + NO₃ ^1- + H ¹⁺ à Zn ²⁺ + N₂O + H₂O

Zn + VO₂ ¹⁺ + H ¹⁺ à Zn ²⁺ + VO²⁺ + H₂O

V(OH)₄ ¹⁺ + Fe ²⁺ + H ¹⁺ à VO²⁺ + Fe³⁺ + H₂O

---------Basic Media------Taylor’s Method-----

Steps #1 through #4, #7- #10, are identical, but Steps 5 & 6 are the only difference between Acid and Basic Media. It is shown below:

(See Step #9½ for the current method to Balance oxygen-via acid media)

However, many books today (McMurry) instruct you to balance the BASIC media, exactly like ACID media including steps #5 & #6 above (not below), except another step is added between #9 & #10 (or Step #9½) to change the H ¹⁺ ions to OH ^1-

Basic Media Step #5: Balance Oxygen:

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:

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]

-------------------------------------------------------------------------------

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

Basic Media Step #6: Balance Hydrogen

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)

------------------------------------------------------------------------------

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:

-------------------------------------------------------------------------------

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- (from Step#5)

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-

-------------------------------------------------------------------------------

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)

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Basic Media Step # 9½: Change Final H¹⁺ to OH^1-

Suppose from Step #9 above, the solution is basic media. The result from Step#9 Acid Media is:

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

Now we need to neutralize the H ¹⁺ ions with OH^1- Ions by adding hydroxides to both sides (see pages 131-132 McMurry):

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

Now combine the [ 14 OH^1- + 14 H ¹⁺ ] to 14 H₂O molecules on the left:

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

This creates 7 H₂O molecules common to both sides.

Next subtract - 7 H₂O from both sides for final basic media answer:

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

This is the net balanced ionic REDOX EQUATION in basic media.

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Neutral Media:

Suppose there are no hydrogen or hydroxide ions in the equation, then this is called Neutral Media (by the definition of Bronsted`-Lowry Acids-Bases). In Neutral media Steps #5 and #6 need for you to be creative by using the concepts of Lewis Acids and Bases (Electron Pair Acceptors and Electron Pair Donars):

5. 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 the 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-

6. 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