Download Oxidation- Reduction Reaction “redox reaction”

Oxidation- Reduction Reaction
“redox reaction”
Part 1
Objectives 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.
Oxidation
• Old definition:
– combination of oxygen with other substances.
• Oxidation as defined today:
– The loss of electrons from an atom or ion.
e- goes bye-bye
Reduction
• 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 reaction for short
• Oxidation always accompanies reduction
– One atom must lose electrons and another
must pick up the electrons.
– The number of electrons lost (oxidation) must
be equal to the number of electrons gained
(reduction)
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 electrons
Answer: losing
2.
Reduction means gaining/losing electrons
Answer: 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.
Tip
LEO The Lion Goes GER
Loose Electrons, Oxidize
Gain Electrons, Reduce
How do you know this
is a "Redox" equation?
0
0
+1
-2
2 H2 + O2  2 H2O
What 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
for every oxygen?
This is a redox reaction because atoms are being reduced
(oxygen gained two electrons to go from a charge of 0 to -2) and
oxidized (hydrogen loses one electron to go from a charge of 0
to +1)
How 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?
Look for the change in oxidation number
(charges).
How to assign oxidation
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
by the rules must be "calculated" using the
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: HSO4To 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.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Sb in Sb2O5
N in Al(NO3)3
P in Mg3(PO4)2
S in (NH4)2SO4
Cr in CrO4-2
Cl in ClO4B in NaBO3
Si in MgSiF6
I in IO3N in (NH4)2S
Mn in MnO4 Br in BrO3 Cl in ClO –
Cr in Cr2O7 -2
Se in H2SeO3
answers
Oxidation- Reduction Reaction
“redox reaction”
Part 2
Reducing Agents and Oxidizing
Agents
Reducing 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 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:
+3
-2
+2 -2
0
+4 -2
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 -.
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
Agents Practice Problems
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O4 -2  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -
Oxidation- Reduction Reaction
“redox reaction”
Part 3
Balancing Redox Equations by
the Half-reaction Method
Balancing Redox Equations by the Half-reaction Method
1. 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 Continue
4. 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 + 4H2O
2. 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.
+7 to +3 gain 5 e+7
-2
+1 +4 -2
+1
+2
+1 +7 -2
+1 -2
MnO4- + H2SO3 + H+  Mn+2 + HSO4- + H2O
Oxidizing agent Reducing agent
+4 to +7 lose 3 eReducing half reaction: Step 2
Reducing half reaction: Step 5
MnO4- + 5e-  Mn+2
3MnO4- + 5H2SO3- +15 e-  3Mn+2 + 5HSO4- + 15e-
Oxidation half reaction: Step 3
Reducing half reaction: Step 5a
H2SO3  HSO4- + 3e-
3MnO4- + 5H2SO3- +15 e-  3Mn+2 + 5HSO4- + 15e-
Reducing half reaction: Step 4
Reducing half reaction: Step 5b
3 (MnO4- + 5e-  Mn+2) = 3MnO4- + 15e-  3Mn+2
3MnO4- + 5H2SO3-  3Mn+2 + 5HSO4-
Oxidation half reaction: Step 4
5 (H2SO3  HSO4- + 3e-) = 5H2SO3  5HSO4- + 15e-
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
oxidation number answers:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Sb + 5
N +5
P +5
S +6
Cr + 6
Cl + 7
B +5
Si + 4
I +5
N –3
Mn + 7
Br + 5
Cl + 1
Cr + 6
Se + 4
Back
Balanced Redox Equations
Answers:
1. 2HNO3 + 3H3PO3  2NO + 3H3PO4 + H2O
2. 14H+ + Cr2O7-2 + 6I -  2Cr+3 + 3I2 + 7H2O
3. 3As2O3 + 4H+ + 4NO3- + 7H2O  6H3AsO4 + 4NO
4. CuS + 2NO3- + 4H+  Cu+2 + 2NO2 + S + 2H2O
5. H2SeO3 + 4Br - + 4H+  Se + 2Br2 + 3H2O
6. 6Fe+2 + Cr2O7-2 + 14H+  6Fe+3 + 2Cr+3 + 7H2O
7. 3HS - + IO3- + 3H+  I - + 3S + 3H2O
8. 16H+ + 2CrO4-2 + 6I -  2Cr+3 + 3I2 + 8H2O
9. 8H+ + IO4- + 7I -  4I2 + 4H2O
10. 6H+ + 2MnO4- + 5H2O2  2Mn+2 + 5O2 + 8H2O
11. 8H+ + H3AsO4 + 4Zn  AsH3 + 4Zn+2 + 4H2O
Click below
to go back
to
problems.
Charting Reducing Agents and Oxidizing
Agents Practice Problems
0 to -2, lose 2 e0
0
+2
-2
1. Mg + O2  MgO
0 to -2, gain 2 e-1 to 0, lose 1 e-1
-1
0
0
2. Cl2 + I-  Cl-+ I2
0 to -1, gain 1 e+7 to +2, gain 5 e+7
-2
+3
-2
+2
+4 -2
3. MnO4 - + C2O4 -2  Mn+2 + CO2
+3 to +4, lose 1 e-
Charting Reducing Agents and Oxidizing
Agents Practice Problems
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O4 -2  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -
Charting Reducing Agents and Oxidizing
Agents Practice Problems
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O4 -2  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -
Charting Reducing Agents and Oxidizing
Agents Practice Problems
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O4 -2  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -
Charting Reducing Agents and Oxidizing
Agents Practice Problems
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Mg + O2  MgO
Cl2 + I-  Cl-+ I2
MnO4 - + C2O4 -2  Mn+2 + CO2
Cr + NO2 -  CrO2 - + N2O2 -2
BrO3 - + MnO2  Br - + MnO4 Fe+2 + MnO4 -  Mn+2 + Fe+3
Cr + Sn+4  Cr+3 + Sn+2
NO3 - + S  NO2 + H2SO4
IO4- + I-  I2
NO2 + ClO - NO3 - + Cl -
Part 3 Redox Agents Answers
1) ox: H2SO4 red: HI
2) ox: Au2S3 red: H2
3) ox: HCl red: Zn