Chemistry SM-1232Week 11 Lesson 2

Dr. Jesse ReichAssistant Professor of Chemistry

Massachusetts Maritime AcademySpring 2008

Class Today• No class Friday• Tests are graded• Quizes are not yet graded. I’ll have them sometime this week.• A few wikis started!!! GET ON IT! At least start the page.• Chapter 16- just two days and a half: Monday, Wednesday,

part of Monday coming• Quiz on the Wiki on Friday• Start Chapter 18 on Monday, 18 on Wednesday, • Test on chapter 15 and 16 next Friday, so everyone can have

lab.

Simpler definition

• LeoGer• Aka Leo the Lion goes Ger• Lose electrons = oxidation• Gain Electrons = reduction

Redox reactions happen simultaneously

• As an oxidation occurs to lose electrons a reduction is happen to gain electrons.

• The substance that is oxidized is called the reducing agent because it causes the reduction of the other substance.

• The substance that is reduced is called the oxidizing agent because it causes the oxidation of the other substance.

Labels Applied

• 2H2 + O2 2H2O

• H2 is the reducing agent because it’s reducing oxygen.

• O2 is called the oxidizing agent because it’s oxidizing H2.

Rules!• 1. Pure elements have an oxydation state of 0• 2. Any charged ion has an oxydation state equal to its

charge• 3. If a compound is neutral the sum of all oxydation

states equals 0• 4. If a compound is charged the sum of all oxydation

states equals the charge• 5. Group 1,2,3 will always be +1,+2,+3• 6. Non metals get oxidation numbers on the next

slide.

More Rules: Oxidation numbers

• This is list is also in order of precedence! • Fluorine oxidation is always -1• Hydrogen is +1• Oxygen is -2• Group 7 is -1• Group 6 is -2• Group 5 is -3

Balancing Redox Equations

• Al(s) + Ag+(s) Al3(aq)+ + Ag(s)

• What are the oxidation numbers of each one of these?

• Let’s break this down into the oxidation and the reduction!

Oxidation

• Al Al+3

• What’s the oxidation state of Al? Al+3

• This isn’t balanced!• We’re missing the electrons!

• Al Al+3 + 3e-

• Now we are balanced!

Reduction

• Ag+ Ag(s)• What’s the oxidation state of Ag(s)? Ag+?• This isn’t balanced!• We’re missing the electrons!

• Ag + + e- Ag(s)

Let’s look at them together

• Al(s) Al+3 + 3e-

• 1 e- + Ag+ Ag(s)

• How many times must Ag+ be reduced in order for 1 Al(s) be oxidized?

3 Times!

• Al(S) Al+3 + 3 e-

• 3(1e- + Ag+ Ag(s))

• Al(s) Al+3 + 3e-

• 3e- + 3 Ag+ 3 Ag(s)• If we add them the electrons can cancel out.

• Al(s) + 3 Ag+ 3 Ag(s) + Al+3

You Try

• Al(s) + Cu+2 Al+3 + Cu(s)

• We’re looking to have charge and atom balanced!

When in doubt we add in H2O and keep rebalancing• I- + Cr2O7

-2 Cr3+ + I2(s)

• 1. Assign oxidation numbers• 2. Separate into half reactions• 3. Use H2O to balance out the Oxygens• 4. Add H+ to balance out the H in water• 5. Figure out the number of electrons that exchange

places• 6. Balance the Oxidation and the Reduction• 7. Add the two reactions together for the net

When in doubt we add in H2O and keep rebalancing• I- + Cr2O7

-2 Cr3+ + I2(s)

• 1. Assign oxidation numbers• I- = -1• Cr=+6• Cr+3=+3• I2=0

When in doubt we add in H2O and keep rebalancing• I- + Cr2O7

-2 Cr3+ + I2(s)

• 2. Separate into half reactions

Separate into Half RXNs

• 2. Separate into half reactions• 2I- I2

• Cr2O72- 2Cr3+

• 3. Use H2O to balance out the Oxygens

When in doubt we add in H2O and keep rebalancing• Cr2O7

2- 2Cr3+

• 3. Use H2O to balance out the Oxygens

• Cr2O72- 2 Cr3+ + 7H2O

• 4. Use H+ to balance out the Hydrogens

Balance out the Hs

• Cr2O72- 2 Cr3+ + 7H2O

• Becomes

• 14H+ + Cr2O72- 2 Cr3+ + 7H2O

• 5. Figure out the number of electrons that exchange places

# of electrons

• 14H+ + Cr2O72- 2 Cr3+ + 7H2O

• Cr+6 Cr+3 two times, so we need 6e- overall

• 6e- + 14H+ + Cr2O72- 2 Cr3+ + 7H2O

# of electrons for I-

• 2I- I2

• I- goes from -1 to I2 which is 0. So we lost two electrons.

• 2I- I2 + 2e

• 6. Balance the Oxidation and the Reduction

Balance the Redox• Reduction: 6e- + 14H+ + Cr2O7

2- 2 Cr3+ + 7H2O

• Oxidation: 2I- I2 + 2e

• How many times does the Oxidation have to happen so that the number of electrons are equal?

• 3 TIMES!

Balance the Redox• Reduction: 6e- + 14H+ + Cr2O7

2- 2 Cr3+ + 7H2O

• Oxidation: 6I- 3I2 + 6e-

7. Add the two reactions together.

Balance the Redox• Reduction: 6e- + 14H+ + Cr2O7

2- 2 Cr3+ + 7H2O

• Oxidation: 6I- 3I2 + 6e-

7. Add the two reactions together.

14H+ + Cr2O72- + 6I- 2 Cr3+ + 7 H2O + 3I2

When in doubt we add in H2O and keep rebalancing• CN- + MnO4

-2 CNO- + MNO2

• 1. Assign oxidation numbers• 2. Separate into half reactions• 3. Use H2O to balance out the Oxygens

• 4. Add H+ to balance out the H in water• 5. Figure out the number of electrons that exchange

places• 6. Balance the Oxidation and the Reduction• 7. Add the two reactions together for the net

When in doubt we add in H2O and keep rebalancing• CN- + MnO4

-2 CNO- + MNO2

• 1. Assign oxidation numbers

When in doubt we add in H2O and keep rebalancing• CN- + MnO4

-2 CNO- + MNO2

• 1. Assign oxidation numbers• CN-, N=-3 total = -1, C=2• MnO42-, O=-2, 4O=-8, total =-2, Mn-8=-2, Mn=+6

• 2. Separate into half reactions

When in doubt we add in H2O and keep rebalancing• CN- + MnO4

-2 CNO- + MnO2

• 2. Separate into half reactions

• CN- CNO-

• MnO42- MnO2

• 3. Add water to make up the difference in unbalanced Os

When in doubt we add in H2O and keep rebalancing• CN- CNO-• MnO42- MnO2• 3. Add water to make up the difference in unbalanced Os• CN- + H2O CNO-• MnO42- MnO2 + 2H2O

• 4. Add in the Hydrogen to make up the difference in unbalanced Hs

When in doubt we add in H2O and keep rebalancing• 4. Add in the Hydrogen to make up the difference in

unbalanced Hs• CN- + H2O CNO- + 2H+

• 4H+ + MnO42- MnO2 + 2H2O

• 4a. Add in OH- to neutralize H+ to make water molecules

When in doubt we add in H2O and keep rebalancing• 4a. Add in OH- to neutralize H+ to make water molecules• CN- + H2O + 2OH- CNO- + 2H+ + 2OH-

• 4H+ + MnO42- +4OH- MnO2 + 2H2O + 4OH-

• 4b. Cancel the waters on each side.• CN- + 2OH- CNO- + H2O• MnO4

2- +2H2O MnO2 + 4OH-

• 5. Check the oxidation state of carbon on the left and right as well as Manganese on the left and right. Figure out the electron flow.

When in doubt we add in H2O and keep rebalancing• 5. Figure out the electron flow• CN- + 2OH- CNO- + H2O + 2e-

• 3e- + MnO42- +2H2O MnO2 + 4OH-

• 6. Balance the oxidation and reduction. What’s the lease common denominator between 2 and 3.

When in doubt we add in H2O and keep rebalancing• 6. Balance the oxidation and reduction. What’s the lease

common denominator between 2 and 3. • CN- + 2OH- CNO- + H2O + 2e-

• 3e- + MnO42- +2H2O MnO2 + 4OH-

• 6! So, we need six electrons on both sides. We’ll multiply the top by 3 and the bottom by 2.

• 3CN + 6OH- 3CNO- + 3H2O + 6e-

• 6e- + 2 MnO42- + 4H2O 2 MnO2 + 8 OH-

• 7. We have to add these two half reactions together

When in doubt we add in H2O and keep rebalancing• 7. We have to add these two half reactions together• 3CN + 6OH- 3CNO- + 3H2O + 6e-

• 6e- + 2 MNO42- + 4H2O 2 MnO2 + 8 OH-

• 3CN + 2 MnO42- + H2O 3CNO- + 2MnO2 + 2OH-

Activity Series

• Lab is all about the activity series. Some of you won’t have lab until next week.

• You’ll learn everything you need about this in Lab, so I’m skipping this section in class.

Read all of chapter 16

• Work on your wikis• Work on your homework• Homework due Monday• Take Home Quiz on Friday Too• Read 16.6-16.8 for Monday• Review Monday- Start Chapter 18• Test on Friday of next week.