Fundamentals K: Oxidation and Reduction

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Fundamentals K Oxidation and Reduction

• To date, we have look at 2 specific types of
  chemical reaction
• Acid-Base Reactions
• Precipitation Reactions
• There is one more reaction type we need to
  consider
• Oxidation-Reduction or Redox Reactions

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Oxidation-Reduction Reactions
Lets look at the reaction of magnesium metal
with molecular oxygen Mg (s) O2 --gt MgO
(s) But we know that magnesium oxide is an ionic
solid, so its chemical formula is actually Mg2
O2- The elemental magnesium became OXIDIZED or
LOST ELECTRONS after reactions with molecular
oxygen. The oxygen atoms GAINED ELECTRONS or
became REDUCED
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Oxidation-Reduction Reactions
Lets look at the reaction of magnesium metal
with molecular chlorine in a similar manner Mg
(s) Cl2 (g) --gt MgCl2 (s) But we know that
magnesium chloride is an ionic solid, so its
chemical formula is actually Mg2 2Cl- The
elemental magnesium became OXIDIZED or LOST
ELECTRONS after reactions with molecular
chlorine. The chlorine atoms GAINED ELECTRONS or
became REDUCED
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Oxidation-Reduction Reactions
LEO is GERman Loss of Electrons is
Oxidation Gain of Electrons is Reduction (Its
not politically correct, but it works)
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Oxidation-Reduction Reactions
Whenever a species is oxidized, another species
MUST be reduced Because of this phenomenon,
these types of reactions are frequently called
Redox reactions
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How do we know what is taking place in a Redox
Reaction?
Oxidation Numbers An oxidation number is a value
given to an atom that indicates its redox state
and allows us to track the atom during a chemical
reaction See Toolbox K.1 Assigning Oxidation
Numbers
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Oxidation Numbers Rules Summary

1. The oxidation number of an element in its
   elemental form is 0
2. The oxidation number of a monatomic ion is equal
   to its charge
3. Hydrogen has two possible oxidation numbers -1
   if bonded to a metal and 1 in nonmetal compounds
4. Oxygens oxidation number is almost always -2 in
   most compounds
5. Oxidation increases the oxidation number of an
   element
6. Reduction decreases the oxidation number of an
   element

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Oxidizing and Reducing Agents
The species that causes oxidation is an oxidizing
agent. IT IS REDUCED
The species that brings about a reduction is a
reducing agent. IT IS OXIDIZED
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Oxidizing and Reducing Agents
Cu2 is _______ to Cu. It is the _________
agent. Zn is _______ to Zn2. It is the
_________ agent. Zn (s) Cu2 (aq) ------gt Zn2
(aq) Cu (s)
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Balancing Redox Reactions
Remember hearing several times during the course
of the semester about how a reaction isnt
balanced until the charges are balanced? Heres
where that really becomes important. When
balancing the chemical equation for a redox
reaction involving ions, the total charge on each
side must be balanced.
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Examples

• i) Which species have been oxidized and which
  species have been reduced in the reaction
• 2Cu (aq) I2 (s) ---gt 2Cu2 (aq) 2I- (aq)
• To solve Calculate the oxidation s
• Find the oxidation numbers of sulfur, nitrogen
  and chlorine in
• a) SO32- b) NO2- c) HClO3

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Examples

• In an aqueous solution, Cerium (IV) ions oxidize
  iodide ions to solid diatomic iodine and are
  themselves reduced to Cerium (III) ions. Write
  the net ionic equation for the reaction.
• To solve
• What does the problem tell us?
• Balance the charges
• Check your work

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Examples

• A mixture of 5.00g of Cr(NO3)2 and 6.00 g of
  CuSO4 is dissolved in sufficient water to make
  25.00 mL of solution. In the reaction, copper
  metal is formed and each chromium ion loses one
  electron.
• Write the net ionic equation
• What is the number of electrons transferred in
  the balanced equation written with the smallest
  whole number coefficients?
• What are the molar concentrations of the two
  anions in the final solution?

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Chapter 12 Electrochemistry

• Expressing Redox Reactions in terms of Half
  Reactions
• Complicated redox reactions require us to think
  of the reactions in terms of the 2 half reactions
  occurring oxidation and reduction
• A Half Reaction is the oxidation or reduction
  part of the reaction considered alone
• Complete reaction Zn (s) 2Ag (aq) --gt Zn2
  (aq) 2Ag (s)
• Oxidation half reaction Zn (s) --gt Zn2 (aq)
  2e-
• Reduction half reaction Ag (aq) 1e- --gt Ag
  (s)

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Balancing Redox Reactions
Because of differences in charge between
reactants and products, it is not always possible
to balance redox reactions using simple
stoichiometric coefficients. For example,
Au3 (aq) I- (aq) --gt Au (s) I2 (s) At
first glance, it seems that this equation can be
balanced by placing a 2 in front of the iodide.
Au3 (aq) 2I- (aq) --gt Au (s) I2
(s) The stoichiometry looks right, but is the
charge?
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Balancing Redox Reactions
We know that this is a redox reaction, so lets
examine the half reactions 2I- (aq) --gt I2
(s) 1e- Au3 (aq) 3e- --gt Au (s) The
half reactions give us an indicator of what is
going on here. Iodide is losing a single
electron, but in order to reduce a single gold
ion to elemental gold, we need 3 electrons. We
would need 6 total iodide ions in order to reduce
a single gold ion, right? 2Au3 (aq) 6I-
(aq) --gt 2Au (s) 3I2 (s)
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Balancing Redox Reactions The Rules

• We will deal with 3 types of redox reactions.
• Straight Reactions These reactions can be
  balanced by simply evaluating the oxidation
  numbers of the reactants and products
• Redox reactions that occur in acidic conditions
  Just as the name implies. Use the rules on the
  next slide.
• Redox reactions that occur in basic conditions
  Hmm. What do you think this means. Use the
  rules for acidic reactions WITH ONE ADDITIONAL
  STEP AT THE END

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Balancing Redox Reactions by Oxidation Number
Step 1 Try to balance the atoms in the
equation by inspection, that is, by the standard
technique for balancing non-redox equations.
(Many equations for redox reactions can be easily
balanced by inspection.) If you successfully
balance the atoms, go to Step 2. If you are
unable to balance the atoms, go to Step 3.
Step 2 Check to be sure that the net charge is
the same on both sides of the equation. If it is,
you can assume that the equation is correctly
balanced. If the charge is not balanced, go to
Step 3. Step 3 If you have trouble
balancing the atoms and the charge by inspection,
determine the oxidation numbers for the atoms in
the formula, and use them to decide whether the
reaction is a redox reaction. If it is not redox,
return to Step 1 and try again. If it is redox,
go to Step 4. Step 4 Determine the net
increase in oxidation number for the element that
is oxidized and the net decrease in oxidation
number for the element that is reduced. Step
5 Determine a ratio of oxidized to reduced
atoms that would yield a net increase in
oxidation number equal to the net decrease in
oxidation number (a ratio that makes the number
of electrons lost equal to the number of
electrons gained). Step 6 Add coefficients
to the formulas so as to obtain the correct ratio
of the atoms whose oxidation numbers are
changing. (These coefficients are usually placed
in front of the formulas on the reactant side of
the arrow.) Step 7 Balance the rest of the
equation by inspection.
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Balancing Redox Reactions that occur in Acidic
conditions
Step 1 Write the skeletons of the oxidation
and reduction half-reactions. (The skeleton
reactions contain the formulas of the compounds
oxidized and reduced, but the atoms and electrons
have not yet been balanced.) Step 2
Balance all elements other than H and O.
Step 3 Balance the oxygen atoms by adding H2O
molecules where needed. Step 4 Balance
the hydrogen atoms by adding H ions where
needed. Step 5 Balance the charge by
adding electrons, e-. Step 6 If the
number of electrons lost in the oxidation
half-reaction is not equal to the number of
electrons gained in the reduction half-reaction,
multiply one or both of the half- reactions by a
number that will make the number of electrons
gained equal to the number of electrons lost.
Step 7 Add the 2 half-reactions as if they
were mathematical equations. The electrons will
always cancel. If the same formulas are found on
opposite sides of the half-reactions, you can
cancel them. If the same formulas are found on
the same side of both half-reactions, combine
them. Step 8 Check to make sure that the
atoms and the charges balance.
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Balancing Redox Reactions that occur in Basic
conditions
Steps 1-7 Begin by balancing the equation as if
it were in acid solution. If you have H ions in
your equation at the end of these steps, proceed
to Step 8. Otherwise, skip to Step 11.
Step 8 Add enough OH- ions to each side to
cancel the H ions. (Be sure to add the OH- ions
to both sides to keep the charge and atoms
balanced.) Step 9 Combine the H ions and
OH- ions that are on the same side of the
equation to form water. Step 10 Cancel or
combine the H2O molecules. Step 11 Check
to make sure that the atoms and the charge
balance. If they do balance, you are done. If
they do not balance, re-check your work in Steps
1-10.
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Balancing Redox Reactions Examples
Balance the following redox equation HNO3 (aq)
H3AsO3 (aq) --gt NO (g) H3AsO4 (aq) H2O (l)
Step 1 Can we balance it eyeballometrically?
Maybe, but it will take a while. The hydrogens
are problematic. Step 2 Are we sure it is a
redox reaction? Check the oxidation
numbers. Step 3 Find a shared number between
the two elements such that the gained electron
number equals the number of lost electrons. Step
4 Place stoichiometric coefficient in front of
the appropriate chemical formula. Step 5
Balance the rest of the equation by inspection.
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Balancing Redox Reactions Examples
Balance the following redox equation Cu (s)
HNO3 (aq) --gt Cu(NO3)2 (aq) NO (g) H2O (l)
Step 1 Can we balance it eyeballometrically?
Nicht. Step 2 Are we sure it is a redox
reaction? Check the oxidation numbers. Step 3
Find a shared number between the two elements
such that the gained electron number equals the
number of lost electrons. Step 4 For chemical
formulas of compounds involved in redox couples
(NOT HNO3), put the coefficients of the common
electron count from step 3. Step 5 Balance the
rest of the equation by inspection.
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Balancing Redox Reactions Examples
Balance the following redox equation NO2 (g)
H2 (g) --gt NH3 (g) 2H2O (l)
Step 1 Can we balance it eyeballometrically?
Si. Easy peasy. Step 2 Done!
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Balancing Redox Reactions Acid Conditions
Balance the following redox equation Cr2O72-
(aq) HNO2 (aq) --gt Cr3 (aq) NO3- (aq)
(acidic) Step 1 We need to identify the 2 half
reactions involved in the redox
reaction Oxidation numbers Cr 6 --gt 3 Gain
of 3 e-, Reduction N 3 --gt 5 Loss of 2
e-, Oxidation Step 2 Write the skeletal half
reactions and i) Balance all atoms other than
H and O ii) Balance O by adding H2O to one
side iii) Balance H by adding H to the other
side iv) Balance charge by adding electrons to
the side that needs it
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Balancing Redox Reactions Acid Conditions
Balance the following redox equation Cr2O72-
(aq) HNO2 (aq) --gt Cr3 (aq) NO3- (aq)
(acidic) Oxidation Reaction Cr2O72- (aq)
----gt Cr3 (aq) Cr2O72- (aq) ----gt 2Cr3
7H2O Balance Cr and O Cr2O72- (aq) 14H
----gt 2Cr3 7H2O Balance H Cr2O72- (aq) 14H
6e- ----gt 2Cr3 7H2O Balance
charge Reduction Reaction HNO2 (aq) ---gt NO3-
(aq) HNO2 (aq) H2O ---gt NO3- (aq) HNO2 (aq)
H2O ---gt NO3- (aq) 3H HNO2 (aq) H2O ---gt
NO3- (aq) 3H 2e-
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Balancing Redox Reactions Acid Conditions
Balance the following redox equation Cr2O72-
(aq) HNO2 (aq) --gt Cr3 (aq) NO3- (aq)
(acidic) Step 3 Multiply the half reaction
with the lowest number of electrons by the
stoichiometric coefficient to make the number of
electrons in each half reaction the
same. Cr2O72- (aq) 14H 6e- ----gt 2Cr3
7H2O HNO2 (aq) H2O ---gt NO3- (aq) 3H
2e- Multiply by 3
Cr2O72- (aq) 14H 6e- 3HNO2 (aq) 3H2O
---gt 2Cr3 7H2O 3NO3- (aq) 9H 6e-
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Balancing Redox Reactions Acid Conditions
Cr2O72- (aq) 14H 6e- 3HNO2 (aq) 3H2O
---gt 2Cr3 7H2O 3NO3- (aq) 9H
6e- Step 4 Cancel the compounds found on both
side of the arrow Cr2O72- (aq) 5H 3HNO2
(aq)--gt 2Cr3 4H2O 3NO3- (aq)

Step 5 Check work. Done!
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Balancing Redox Reactions Acid Conditions
MnO4- (aq) Br- (aq) ---gt MnO2 (aq) BrO3-
(aq) Step 1 Half reactions Mn 7 --gt 4 Gain
3e-, Reduction Br -1 --gt 5 Lose 6e-,
Oxidation Step 2 Balance half
reactions Oxidation Br- (aq) --gt BrO3- 3H2O
Br- (aq) --gt BrO3- 3H2O Br- (aq) --gt BrO3-
6H 3H2O Br- (aq) --gt BrO3- 6H 6e-
Reduction MnO4- --gt MnO2 MnO4- --gt MnO2
2H2O MnO4- 4H --gt MnO2 2H2O MnO4- 4H
3e- --gt MnO2 2H2O

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Balancing Redox Reactions Acid Conditions
MnO4- (aq) Br- (aq) ---gt MnO2 (aq) BrO3-
(aq) Step 3 We have 6 electrons transferred in
the oxidation reaction and 3 electrons in the
reduction reactions. Multiply the reduction
reaction by 2 and combine with the oxidation
reaction. 2MnO4- Br- 3H2O 8H 6e- --gt
BrO3- 6H 6e- 2MnO2 4H2O Step 4 Cancel
molecules found on both sides of the reaction
arrow. 2MnO4- Br- 2H --gt BrO3- 2MnO2
H2O Step 5 Done!

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Balancing Redox Reactions Basic Conditions
Cr(OH)3 (s) ClO3- (aq) ---gt CrO42- (aq) Cl-
(aq) Step 1 Same as acidic conditions.
Identify the 2 half reactions Cr 3 --gt 6 Lose
3 electrons, Oxidation Cl 5 --gt -1 Gain 6
electrons, Reduction Step 2 Balance the half
reactions Oxidation reaction Cr(OH)3 (s) ---gt
CrO42- Cr(OH)3 (s) H2O ---gt CrO42- Cr(OH)3 (s)
H2O ---gt CrO42- 5H Cr(OH)3 (s) H2O ---gt
CrO42- 5H 3e-

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Balancing Redox Reactions Basic Conditions
Cr(OH)3 (s) ClO3- (aq) ---gt CrO42- (aq) Cl-
(aq) Step 2 (contd) Balance the half
reactions Reduction reaction ClO3- (aq) ---gt
Cl- ClO3- (aq) ---gt Cl- 3H2O ClO3- (aq) 6H
---gt Cl- 3H2O ClO3- (aq) 6H 6e- ---gt Cl-
3H2O Step 3 Multiply the oxidation reaction by
2 to get the same of electrons and combine the
equations 2Cr(OH)3 (s) 2H2O ClO3- (aq) 6H
6e- --gt Cl- 3H2O 2CrO42- 10H 6e-

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Balancing Redox Reactions Basic Conditions
2Cr(OH)3 (s) 2H2O ClO3- (aq) 6H 6e-
--gt Cl- 3H2O 2CrO42- 10H 6e-
Step 4 Remove things on both sides of the
reaction arrow 2Cr(OH)3 (s) ClO3- (aq) --gt
Cl- H2O 2CrO42- 4H Step 5 Now for the
tricky part. Count the number of protons and add
the same number of hydroxide ions to BOTH
sides. 2Cr(OH)3 (s) ClO3- (aq) 4 OH- --gt
Cl- H2O 2CrO42- 4H 4 OH- OR 2Cr(OH)3
(s) ClO3- (aq) 4 OH- --gt Cl- 5H2O
2CrO42-