oxidation- reduction reaction “redox reaction” part 1

Oxidation- Reduction Reaction“redox reaction”

Part 1

Objectives for the UnitObjectives for the Unit

• Identify elements that change oxidation state from one side of a chemical equation to the other.

• Identify reducing agents and oxidizing agents in a chemical equation.

• Use the half-reaction method to balance redox reactions.

OxidationOxidation

• Old definition: – combination of oxygen with other substances.

• Oxidation as defined today: – The loss of electrons from an atom or ion.

ee- - goes bye-bye

ReductionReduction

• Old definition:– “reducing" a substance into its components.

• Reduction as defined today: – The gaining of electrons by an atom or ion.

Reduction-Oxidation Reaction“Redox reaction”

• A reduction and oxidation reaction is commonly called redox reactionredox reaction for short

• Oxidation always accompanies reduction– One atom must lose electrons and another One atom must lose electrons and another

must pick up the electrons.must pick up the electrons.– The number of electrons lost (oxidation) must

be equal to the number of electrons gained (reduction)

Concept check:Concept check:(answer these questions on you paper using Cornell note format.

Answer the question first before advancing the slide)

1. Oxidation means gaining/ losing electronsAnswer: losing

2. Reduction means gaining/losing electronsAnswer: gaining

3. Can there be a oxidation without reduction? Explain.

Answer: no, one atom must give up electrons (oxidation) and another must be there to receive or gain (reduction) electrons

4. Compare a reducing agent with an oxidation agent.

Answer: reducing agents are atoms that provide the electrons (started out with electrons) and when they lose the electrons, then they are oxidized. Oxidation agents are atoms that

will accept the electrons, (started out with less electrons) and when they gain electrons, they will be reduced.

TipTip

LEOLEO The Lion Goes GER

Loose Electrons, Oxidize Gain Electrons, Reduce

How do you know thisis a "Redox" equation?

2 H2 H22 + O + O22 2 H 2 H22OOWhat is the charge of H2 here?

What is the charge of O2 here?

What is the charge of H2 and O here?

Think back, how does hydrogen bond with oxygen and why 2 hydrogen Think back, how does hydrogen bond with oxygen and why 2 hydrogen for every oxygen?for every oxygen?

0 0 +1 -2

This is a redox reaction because atoms are being reduced This is a redox reaction because atoms are being reduced (oxygen gained two electrons to go from a charge of 0 to -2) (oxygen gained two electrons to go from a charge of 0 to -2)

and oxidized (hydrogen loses one electron to go from a charge and oxidized (hydrogen loses one electron to go from a charge of 0 to +1)of 0 to +1)

How to identify a redox reactionHow to identify a redox reaction

Oxidation and Reduction must both occur in a Redox reaction.

If one particle gains electrons in a reaction, some other particle must lose them.

So?So?

Look for the change in oxidation number Look for the change in oxidation number (charges).(charges).

How to assign oxidation How to assign oxidation numbers.numbers.

• You have learned to read the oxidation numbers of many elements from the periodic table, ie: group 1 elements is +1, group 2 elements are +2, group 13 elements are +3, group 15 elements are -3, group 16 elements are -2 and group 17 elements are -1. While that information is important, the following rules are to be your guide when working with Redox equations.

Rules for assigning oxidation numbers: • The oxidation number of a free element = 0. • The oxidation number of a monatomic ion = the charge

on the ion. • The oxidation number of hydrogen = + 1 and rarely - 1. • The oxidation number of oxygen = - 2 and in peroxides

(H2O2)= - 1. • The sum of the oxidation numbers in a polyatomic ion

(ions made up of two or more elements) = charge on the ion.

• Elements in group 1, 2, and aluminum are always as indicated on the periodic table.

The oxidation number of elements not covered The oxidation number of elements not covered by the rules must be "calculated" using the by the rules must be "calculated" using the

known oxidation numbers in a compound.known oxidation numbers in a compound. • Example #1: K2CO3

To calculate C

By rule K is +1 and O is -2

The sum of all the oxidation numbers in this formula equal 0 because there is no charge on this polyatomic ion. Multiply the subscript by the oxidation number for each

element. K = (2) ( + 1 ) = + 2 O = (3) ( - 2 ) = - 6

therefore, C = (1) ( + 4 ) = + 4

• Example #2: HSO4-

To calculate S

by rule, H is + 1 by rule O is - 2 The sum of all the oxidation numbers in this formula equal -1 because this polyatomic ion

has a charge of -1. Multiply the subscript by the oxidation number for each element. H = (1) ( + 1 ) = + 1 O = (4) ( - 2 ) = - 8

therefore, S = (1) ( + 6 ) = + 6

Practice Problems: Use the rules above to determine the oxidation number of the element

indicated in each formula.1. Sb in Sb2O5

2. N in Al(NO3)3

3. P in Mg3(PO4)2

4. S in (NH4)2SO4

5. Cr in CrO4-2

6. Cl in ClO4-

7. B in NaBO3

8. Si in MgSiF6

9. I in IO3-

10. N in (NH4)2S 11. Mn in MnO4 -12. Br in BrO3 -

13. Cl in ClO –

14. Cr in Cr2O7 -2

15. Se in H2SeO3answers


Part 2

Reducing Agents and Oxidizing Reducing Agents and Oxidizing AgentsAgents

Reducing AgentsReducing Agents

• the reactant that gives up electrons.

• The reducing agent contains the element that is oxidized (looses electrons).

• If a substance gives up electrons easily, it is said to be a strong reducing agent.

Oxidizing and Reducing Agents

• Reducing agent: atoms that provide electrons to reduce the other atom

Oxidizing agentsOxidizing agents

• the reactant that gains electrons.

• The oxidizing agent contains the element that is reduced (gains electrons).

• If a substance gains electrons easily, it is said to be a strong oxidizing agent.

Oxidizing and Reducing Agents

• Oxidizing agent: atoms that will gain electrons thus oxidizing the other atom.

• Example:

Fe2O3 (s)+ 3CO (g) 2Fe(s) + 3CO2 (g)

• Notice that the oxidation number of C goes from +2 on the left to +4 on the right.

• The reducing agent is CO, because it contains C, which loses e -. • Notice that the oxidation number of Fe goes from +3 on the left to 0

on the right.

• The oxidizing agent is Fe2O3, because it contains the Fe, which

gains e -.

+3 -2 +2 -2 0 +4 -2

Charting Reducing Agents and Oxidizing Agents

• Practice Problems:In any Redox equation, at least one particle will gain electrons and at least one particle will lose electrons. This is indicated by a change in the particle's oxidation number from one side of the equation to the other. For each reaction below, draw arrows and show electron numbers as in the example here. The top arrow indicates the element that gains electrons, reduction, and the bottom arrow indicates the element that looses electrons, oxidation. An arrow shows what one atom of each of these elements gains or looses.

This is the first thing that must be done in balancing a Redox reaction. Learn to do it well.

Charting Reducing Agents and Oxidizing Charting Reducing Agents and Oxidizing Agents Practice ProblemsAgents Practice Problems

1. Mg + O2 MgO 2. Cl2 + I- Cl-+ I2 3. MnO4 - + C2O4 -2 Mn+2 + CO2 4. Cr + NO2 - CrO2 - + N2O2 -2 5. BrO3 - + MnO2 Br - + MnO4 - 6. Fe+2 + MnO4 - Mn+2 + Fe+3 7. Cr + Sn+4 Cr+3 + Sn+2 8. NO3 - + S NO2 + H2SO4 9. IO4

- + I- I2 10. NO2 + ClO - NO3 - + Cl -

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Part 3

Balancing Redox Equations by Balancing Redox Equations by the Half-reaction Methodthe Half-reaction Method

Balancing Redox Equations by the Half-reaction MethodBalancing Redox Equations by the Half-reaction Method1. Decide what is reduced (oxidizing agent) and what is oxidized (reducing agent).

Do this by drawing arrows as in the practice problems.

2. Write the reduction half-reaction. – The top arrow in step #1 indicates the reduction half-reaction. Show the electrons gained

on the reactant side. – Balance with respect to atoms / ions.

• To balance oxygen, add H2O to the side with the least amount of oxygen. • THEN: add H + to the other side to balance hydrogen.

Remember that the arrow in step #1 indicates the number of electrons gained by one

atom.

3. Write the oxidation half-reaction. – The bottom arrow in step #1 indicates the oxidation half-reaction. – Show the electrons lost on the product side. – Balance with respect to atoms / ions.

• To balance oxygen, add H2O to the side with the least amount of oxygen. • THEN: add H + to the other side to balance hydrogen.

Remember that the arrow in step #1 indicates the number of electrons lost by one atom.

Balancing Redox Equations by the Half-reaction Method ContinueBalancing Redox Equations by the Half-reaction Method Continue4. The number of electrons gained must equal the number of electrons lost.

– Find the least common multiple of the electrons gained and lost. – In each half-reaction, multiply the electron coefficient by a number to reach the common

multiple. – Multiply all of the coefficients in the half-reaction by this same number.

5. Add the two half-reactions. – Write one equation with all the reactants from the half-reactions on the left and all the

products on the right. – The order in which you write the particles in the combined equation does not matter.

6. Simplify the equation. – A) Cancel things that are found on both sides of the equation as you did in net ionic

equations. – B) Rewrite the final balanced equation.

Check to see that electrons, elements, and total charge are balanced.

– There should be no electrons in the equation at this time. – The number of each element should be the same on both sides. – It doesn't matter what the charge is as long as it is the same on both sides.

If any of these are not balanced, the equation is incorrect. The only thing to do is go back to step #1 and begin looking for your mistake.

Practice Problems:Identify the oxidizing agent and reducing agent in each equation: Answers

1. H2SO4 + 8HI H2S + 4I2 + 4H2O2. Au2S3 + 3H2 2Au + 3H2S 3. Zn + 2HCl H2 + ZnCl2

• To make working with redox equations easier, we will omit all physical state symbols. However, remember that they should be there. An unbalanced redox equation looks like this:

MnO4- + H2SO3 + H + Mn+2 + HSO4

- + H2O

• Study how this equation is balanced using the half-reaction method on the next slide. It is important that you understand what happens in each step.

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MnO4- + H2SO3 + H+ Mn+2 + HSO4

- + H2O

+7 -2 +1 +4 -2 +1 +2 +1 +7 -2 +1 -2

Oxidizing agentOxidizing agent Reducing agentReducing agent

+7 to +3 gain 5 e-

Reducing half reaction: Step 2

MnO4- + 5e- Mn+2

+4 to +7 lose 3 e-

Oxidation half reaction: Step 3

H2SO3 HSO4- + 3e-

Reducing half reaction: Step 5a

3MnO4- + 5H2SO3

- +15 e- 3Mn+2 + 5HSO4- + 15e-


3 (MnO4- + 5e- Mn+2) = 3MnO4

- + 15e- 3Mn+2

Oxidation half reaction: Step 4

5 (H2SO3 HSO4- + 3e-) = 5H2SO3 5HSO4

- + 15e-


3MnO4- + 5H2SO3

- +15 e- 3Mn+2 + 5HSO4- + 15e-

Reducing half reaction: Step 5b

3MnO4- + 5H2SO3

- 3Mn+2 + 5HSO4-

3MnO4- + 5H2SO3 + H+ 3Mn+2 + 5HSO4

- + H2O

Notice, some of the above is balance but the H and O are not. Balance the equation the way that you know, by counting the number and kind of atoms.

3MnO4- + 5H2SO3 + 9H+ 3Mn+2 + 5HSO4

- + 7H2O

Practice Problems: Balance these Redox equations using the half-reaction method.

The balanced equation and the detailed half-reaction solution are provided for each equation. Use these helps only after doing your best to balance the equation without help.

It is to your advantage to work all these problems to gain experience with different situations that arise when working with redox equations.

1. HNO3 + H3PO3 NO + H3PO4 + H2O 2. Cr2O7

-2 + H + + I - Cr+3 + I2 + H2O 3. As2O3 + H + + NO3

- + H2O H3AsO4 + NO 4. CuS + NO3

- Cu+2 + NO2 + S 5. H2SeO3 + Br - Se + Br2 6. Fe+2 + Cr2O7

-2 Fe+3 + Cr+3

7. HS - + IO3- I - + S

8. CrO4-2 + I - Cr+3 + I2

9. IO4- + I - I2

10. MnO4- + H2O2 Mn+2 + O2

11. H3AsO4 + Zn AsH3 + Zn+2

Click below to see answer

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oxidation number answers: 1. Sb + 5

2. N + 5

3. P + 5

4. S + 6

5. Cr + 6 6. Cl + 7

7. B + 5

8. Si + 4

9. I + 5

10. N – 3

11. Mn + 7

12. Br + 5

13. Cl + 1

14. Cr + 6

15. Se + 4

Back

Balanced Redox Equations Answers:

1. 2HNO3 + 3H3PO3 2NO + 3H3PO4 + H2O2. 14H+ + Cr2O7

-2 + 6I - 2Cr+3 + 3I2 + 7H2O3. 3As2O3 + 4H+ + 4NO3

- + 7H2O 6H3AsO4 + 4NO4. CuS + 2NO3

- + 4H+ Cu+2 + 2NO2 + S + 2H2O5. H2SeO3 + 4Br - + 4H+ Se + 2Br2 + 3H2O6. 6Fe+2 + Cr2O7

-2 + 14H+ 6Fe+3 + 2Cr+3 + 7H2O7. 3HS - + IO3

- + 3H+ I - + 3S + 3H2O8. 16H+ + 2CrO4

-2 + 6I - 2Cr+3 + 3I2 + 8H2O9. 8H+ + IO4

- + 7I - 4I2 + 4H2O 10. 6H+ + 2MnO4

- + 5H2O2 2Mn+2 + 5O2 + 8H2O11. 8H+ + H3AsO4 + 4Zn AsH3 + 4Zn+2 + 4H2O

Click below to go back to problems.

Charting Charting Reducing AgentsReducing Agents and and Oxidizing Oxidizing AgentsAgents Practice Problems Practice Problems

1. Mg + O2 MgO

2. Cl2 + I- Cl-+ I2

3. MnO4 - + C2O4 -2 Mn+2 + CO2

0 0 +2 -2

0 to -2, lose 2 e-

0 to -2, gain 2 e-

0 -1 -1 0

0 to -1, gain 1 e-

-1 to 0, lose 1 e-

+7 -2 +3 -2 +2 +4 -2

+7 to +2, gain 5 e-

+3 to +4, lose 1 e-






- + I- I2 10. NO2 + ClO - NO3 - + Cl -








Part 3 Redox Agents Answers

1) ox: H2SO4 red: HI

2) ox: Au2S3 red: H2

3) ox: HCl red: Zn

oxidation- reduction reaction “redox reaction” part 1

Documents

oxidation state

oxidation old definition

chemical equation

halfreaction method

redox reactions

oxidizing agents

combination of oxygen

loss of electrons