Transcript Physical Properties - Winthrop University

Fundamentals K: Oxidation and
Reduction
To date, we have look at 2 specific types of chemical
reaction:
1. Acid-Base Reactions
2. Precipitation Reactions
There is one more reaction type we need to consider:
Oxidation-Reduction or Redox Reactions
Oxidation-Reduction Reactions
Let’s look at the reaction of magnesium metal with
molecular oxygen:
Mg (s) + O2 --> 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
Oxidation-Reduction Reactions
Let’s look at the reaction of magnesium metal with
molecular chlorine in a similar manner:
Mg (s) + Cl2 (g) --> 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
Oxidation-Reduction Reactions
LEO is GERman
Loss of Electrons is Oxidation
Gain of Electrons is Reduction
(Its not politically correct, but it works)
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
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
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. Oxygen’s 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
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
Oxidizing and Reducing Agents
Cu2+ is _______ to Cu. It is the _________ agent.
Zn is _______ to Zn2+. It is the _________ agent.
Zn (s) + Cu2+ (aq) ------> Zn2+ (aq) + Cu (s)
Balancing Redox Reactions
Remember hearing several times during the course
of the semester about how a reaction isn’t balanced
until the charges are balanced? Here’s 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.
Examples
i) Which species have been oxidized and which
species have been reduced in the reaction:
2Cu+ (aq) + I2 (s) ---> 2Cu2+ (aq) + 2I- (aq)
To solve: Calculate the oxidation #’s
ii) Find the oxidation numbers of sulfur, nitrogen
and chlorine in:
a) SO32-
b) NO2-
c) HClO3
Examples
iii) 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:
a) What does the problem tell us?
b) Balance the charges
c) Check your work
Examples
iv) 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.
a) Write the net ionic equation
b) What is the number of electrons transferred in
the balanced equation written with the smallest
whole number coefficients?
c) What are the molar concentrations of the two
anions in the final solution?
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) --> Zn2+ (aq) + 2Ag (s)
Oxidation half reaction: Zn (s) --> Zn2+ (aq) + 2eReduction half reaction: Ag+ (aq) + 1e- --> Ag (s)
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) --> 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) --> Au (s) + I2 (s)
The stoichiometry looks right, but is the charge?
Balancing Redox Reactions
We know that this is a redox reaction, so let’s examine the
half reactions:
2I- (aq) --> I2 (s) + 1eAu3+ (aq) + 3e- --> 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) --> 2Au (s) + 3I2 (s)
Balancing Redox Reactions: The Rules
We will deal with 3 types of redox reactions.
1. Straight Reactions: These reactions can be balanced by
simply evaluating the oxidation numbers of the reactants
and products
2. Redox reactions that occur in acidic conditions: Just as
the name implies. Use the rules on the next slide.
3. 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
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.
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.
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.
Balancing Redox Reactions: Examples
Balance the following redox equation:
HNO3 (aq) + H3AsO3 (aq) --> 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.
Balancing Redox Reactions: Examples
Balance the following redox equation:
Cu (s) + HNO3 (aq) --> 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.
Balancing Redox Reactions: Examples
Balance the following redox equation:
NO2 (g) + H2 (g) --> NH3 (g) + 2H2O (l)
Step 1: Can we balance it eyeballometrically? Si. Easy peasy.
Step 2: Done!
Balancing Redox Reactions: Acid Conditions
Balance the following redox equation:
Cr2O72- (aq) + HNO2 (aq) --> Cr3+ (aq) + NO3- (aq) (acidic)
Step 1: We need to identify the 2 half reactions involved in
the redox reaction
Oxidation numbers: Cr: +6 --> +3 Gain of 3 e-, Reduction
N: +3 --> +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
Balancing Redox Reactions: Acid Conditions
Balance the following redox equation:
Cr2O72- (aq) + HNO2 (aq) --> Cr3+ (aq) + NO3- (aq) (acidic)
Oxidation Reaction:
Cr2O72- (aq) ----> Cr3+ (aq)
Cr2O72- (aq) ----> 2Cr3+ + 7H2O
Balance Cr and O
Cr2O72- (aq) + 14H+ ----> 2Cr3+ + 7H2O Balance H
Cr2O72- (aq) + 14H+ + 6e- ----> 2Cr3+ + 7H2O Balance charge
Reduction Reaction:
HNO2 (aq) ---> NO3- (aq)
HNO2 (aq) + H2O ---> NO3- (aq)
HNO2 (aq) + H2O ---> NO3- (aq) + 3H+
HNO2 (aq) + H2O ---> NO3- (aq) + 3H+ + 2e-
Balancing Redox Reactions: Acid Conditions
Balance the following redox equation:
Cr2O72- (aq) + HNO2 (aq) --> 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- ----> 2Cr3+ + 7H2O
HNO2 (aq) + H2O ---> NO3- (aq) + 3H+ + 2e-
Multiply by 3
Cr2O72- (aq) + 14H+ + 6e- + 3HNO2 (aq) + 3H2O --->
2Cr3+ + 7H2O + 3NO3- (aq) + 9H+ + 6e-
Balancing Redox Reactions: Acid Conditions
Cr2O72- (aq) + 14H+ + 6e- + 3HNO2 (aq) + 3H2O --->
2Cr3+ + 7H2O + 3NO3- (aq) + 9H+ + 6e-
Step 4: Cancel the compounds found on both side of the
arrow
Cr2O72- (aq) + 5H+ + 3HNO2 (aq)--> 2Cr3+ + 4H2O + 3NO3- (aq)
Step 5: Check work. Done!
Balancing Redox Reactions: Acid Conditions
MnO4- (aq) + Br- (aq) ---> MnO2 (aq) + BrO3- (aq)
Step 1: Half reactions
Mn: +7 --> +4
Gain 3e-, Reduction
Br: -1 --> +5
Lose 6e-, Oxidation
Step 2: Balance half reactions
Oxidation:
Br- (aq) --> BrO33H2O + Br- (aq) --> BrO33H2O + Br- (aq) --> BrO3- + 6H+
3H2O + Br- (aq) --> BrO3- + 6H+ + 6eReduction:
MnO4- --> MnO2
MnO4- --> MnO2 + 2H2O
MnO4- + 4H+ --> MnO2 + 2H2O
MnO4- + 4H+ + 3e- --> MnO2 + 2H2O
Balancing Redox Reactions: Acid Conditions
MnO4- (aq) + Br- (aq) ---> 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- --> BrO3- + 6H+ + 6e- + 2MnO2 + 4H2O
Step 4: Cancel molecules found on both sides of the reaction arrow.
2MnO4- + Br- + 2H+ --> BrO3- + 2MnO2 + H2O
Step 5: Done!
Balancing Redox Reactions: Basic Conditions
Cr(OH)3 (s) + ClO3- (aq) ---> CrO42- (aq) + Cl- (aq)
Step 1: Same as acidic conditions. Identify the 2 half reactions
Cr: +3 --> +6
Lose 3 electrons, Oxidation
Cl: +5 --> -1
Gain 6 electrons, Reduction
Step 2: Balance the half reactions
Oxidation reaction:
Cr(OH)3 (s) ---> CrO42Cr(OH)3 (s) + H2O ---> CrO42-
Cr(OH)3 (s) + H2O ---> CrO42- + 5H+
Cr(OH)3 (s) + H2O ---> CrO42- + 5H+ + 3e-
Balancing Redox Reactions: Basic Conditions
Cr(OH)3 (s) + ClO3- (aq) ---> CrO42- (aq) + Cl- (aq)
Step 2 (cont’d): Balance the half reactions
Reduction reaction:
ClO3- (aq) ---> ClClO3- (aq) ---> Cl- + 3H2O
ClO3- (aq) + 6H+ ---> Cl- + 3H2O
ClO3- (aq) + 6H+ + 6e- ---> 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- -->
Cl- + 3H2O + 2CrO42- + 10H+ + 6e-
Balancing Redox Reactions: Basic Conditions
2Cr(OH)3 (s) + 2H2O + ClO3- (aq) + 6H+ + 6e- -->
Cl- + 3H2O + 2CrO42- + 10H+ + 6eStep 4: Remove things on both sides of the reaction arrow
2Cr(OH)3 (s) + ClO3- (aq) --> 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- --> Cl- + H2O + 2CrO42- + 4H+ + 4 OHOR
2Cr(OH)3 (s) + ClO3- (aq) + 4 OH- --> Cl- + 5H2O + 2CrO42-