Transcript Spring.2009.week11.lesson2 - reich

Chemistry SM-1232
Week 11 Lesson 2
Dr. Jesse Reich
Assistant Professor of Chemistry
Massachusetts Maritime Academy
Spring 2008
Class Today
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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
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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
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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
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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
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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
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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
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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
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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
• Cr2O72-  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+ + Cr2O72-  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+ + Cr2O72-  2 Cr3+ + 7H2O
• Oxidation: 6I-  3I2 + 6e7. Add the two reactions together.
Balance the Redox
• Reduction: 6e- + 14H+ + Cr2O72-  2 Cr3+ + 7H2O
• Oxidation: 6I-  3I2 + 6e7. Add the two reactions together.
14H+ + Cr2O72- + 6I-  2 Cr3+ + 7 H2O + 3I2
When in doubt we add in
H2O and keep rebalancing
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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
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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
• MnO42- +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
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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.