Redox Reaction.ppt

Oxidation and Reduction
Reactions

Oxidation (Read only)
Original definition:
When substances combined with oxygen.
Ex:
All combustion (burning) reactions
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
All “rusting” reactions
4Fe(s) + 3O2(g) 2Fe2O3(s)

Reduction (Read Only)
Original Definition:
Reaction where a substance “gave up” oxygen.
Called “reductions” because they produced
products that were “reduced” in mass because
gas escaped.
Ex:
2Fe2O3(l) + 3C(s) 4Fe(l) + 3CO2(g)

Deals with movement of ELECTRONS
during a chemical reaction.
(Oxygen doesn’t have to be present)
Oxidation/Reduction

Electron Transfer Reactions
Oxidation: LOSS of one or more electrons.
Reduction: GAIN of one or more electrons

Electron Transfer Reactions
Oxidation & reduction always occur together.
Electrons travel from what is oxidized towards
what is reduced.
One atom loses e-, the other gains e-

Redox Reactions:
ALWAYS involve changes in charge
A competition for electrons between atoms!

Conservation of “Charge”
Total electrons lost = Total electrons gained

Oxidizing/Reducing Agents
Oxidizing Agent:
substance reduced
 Gains electrons
Reducing Agent:
substance oxidized
 Loses electrons
The “Agent” is the “opposite”

Assigning Oxidation Numbers
Practice Problems
http://www.usca.edu/chemistry/genchem/ox
numb.htm
Animation of Oxidation and Reduction
http://www.ausetute.com.au/redox.html

Identify What is Changing in
Charge
What is oxidized and reduced?
What are the oxidizing and reducing agents?
Ex:
3Br2 + 2AlI3 2AlBr3 + 3I2

0 +3 -1 +3 -1 0
3Br2 + 2AlI3 2AlBr3 + 3I2
Br2 is reduced and is the oxidizing agent
I-1 is oxidized and is the reducing agent

What is oxidized and reduced?
What are the oxidizing and reducing agents?
Mg + CuSO4 MgSO4 + Cu
2K + Br2 2KBr
Cu + 2AgNO3 Cu(NO3)2 + 2Ag
NOTE:
Atoms in a polyatomic ion DO NOT change in charge!

0 +2 +2 0
Mg + CuSO4 MgSO4 + Cu
Mg oxidized (reducing agent)
Cu+2 reduced (oxidizing agent)
0 0 +1 -1
2K + Br2 2KBr
K oxidized (reducing agent)
Br2 reduced (oxidizing agent)
0 +1 +2 0
Cu + 2AgNO3 Cu(NO3)2 + 2Ag
Cu oxidized (reducing agent)
Ag+1 reduced (oxidizing agent)

Redox or Not Redox
(that is the question…)
Redox Reactions: must have atoms changing in
charge.
Not all reactions are redox.
Easy way to spot a redox reaction!!!
Look for elements entering and leaving
compounds.

Is it Redox?
Look for Changes in Charge!
Are elements entering and leaving compounds?
Synthesis:
Ex: 2H2 + O2 2H2O
Decomposition:
Ex: 2KClO3 2KCl + 3O2

Is it Redox?
Synthesis: YES
0 0 +1 -2
Ex: 2H2 + O2 2H2O
Decomposition: YES
+1 +5 -2 +1 -1 0
Ex: 2KClO3 2KCl + 3O2

Is it Redox?
Combustion:
CH4 + 2O2 CO2 + 2H20
Single Replacement:
Zn + CuCl2 ZnCl2 + Cu

Is it Redox?
Combustion: YES
-4 +1 0 +4 -2 +1 -2
CH4 + 2O2 CO2 + 2H20
Single Replacement: YES
0 +2 -1 +2 -1 0

Is it Redox?
Double Replacement:
AgNO3 + LiCl AgCl + LiNO3

Is it Redox?
Double Replacement: NO!!!!
Ions switch partners, but don’t change in charge
+1 +5 -2 +1 -1 +1 -1 +1 +5 -2
AgNO3 + LiCl AgCl + LiNO3
Remember charges of atoms inside polyatomic ions
do not change!

Writing Half Reactions
Redox Reactions are composed of two parts
or half reactions.
Half Reactions Show:
Element being oxidized or reduced.
Change in charge
# of electrons being lost or gained

Writing Half Reactions
0 0 +1 -1
2Na + F2 2NaF
Oxidation: Na Na+1 + 1e-
or 2Na 2Na+1 + 2e-
Note: e- are “lost” (on the right of arrow)
Reduction: F + 1e- F-1
or F2 + 2e- 2F-1
Note: e- are “gained” (on the left of arrow)

Ox’s Have Tails!!
 Oxidation Half reactions always have
“tails” of electrons
Na Na+1 + 1e-

0 +2 -1 +2 -1 0
Ox: Zn Zn+2 + 2e-
Red: Cu+2 + 2e- Cu

Rules for Assigning Oxidation Numbers
1. All free, uncombined elements have an
oxidation number of zero. This includes
diatomic elements such as O2 or others like
P4 and S8.
2. Hydrogen, in all its compounds except
hydrides, has an oxidation number of +1
(positive one)
3. Oxygen, in all its compounds except
peroxides, has an oxidation number of -2
(negative two).

Practice Problems
 What is the oxidation number of . . .
 1) N in NO3¯
 2) C in CO3
2¯
 3) Cr in CrO4
2¯
 4) Cr in Cr2O7
2¯
 5) Fe in Fe2O3
 6) Pb in PbOH+
 7) V in VO2
+
 8) V in VO2+
 9) Mn in MnO4¯
 10) Mn in MnO4
2¯

 What you must be able to do is look at a
redox reaction and separate out the two half-
reactions in it. To do that, identify the atoms
which get reduced and get oxidized. Here are
the two half-reactions from the example:
Ag+ ---> Ag
Cu ---> Cu2+
 The silver is being reduced, its oxidation
number going from +1 to zero. The copper's
oxidation number went from zero to +2, so it
was oxidized in the reaction. In order to figure
out the half-reactions, you MUST be able to
calculate the oxidation number of an atom.

 When you look at the two half-reactions, you
will see they are already balanced for atoms
with one Ag on each side and one Cu on
each side. So, all we need to do is balance
the charge.
 To do this you add electrons to the more
positive side. You add enough to make the
total charge on each side become EQUAL.
 To the silver half-reaction, we add one
electron:
Ag+ + e¯ ---> Ag
 To the copper half-reaction, we add two
electrons:
Cu ---> Cu2+ + 2e¯

Half-reactions NEVER occur alone.
 notice that each half-reaction wound up
with a total charge of zero on each side.
This is not always the case. You need to
strive to get the total charge on each side
EQUAL, not zero.

Half-Reactions Practice
Problems
 Balance each half-reaction for atoms and
charge:
 1) Cl2 ---> Cl¯
 2) Sn ---> Sn2+
 3) Fe2+ ---> Fe3+
 4) I3¯ ---> I¯
 5) ICl2¯ ---> I¯
 6) Sn + NO3¯ ---> SnO2 + NO2
 7) HClO + Co ---> Cl2 + Co2+
 8) NO2 ---> NO3¯ + NO

Answers
 1) Cl2 + 2e¯ ---> 2Cl¯
 2) Sn ---> Sn2+ + 2e¯
 3) Fe2+ ---> Fe3+ + e¯
 4) I3¯ + 2e¯ ---> 3I¯
 5) ICl2¯ + 2e¯ ---> I¯ + 2Cl¯
 6) Sn ---> SnO2 and NO3¯ ---> NO2
 7) HClO ---> Cl2 and Co ---> Co2+
 8) NO2 ---> NO3¯ and NO2 ---> NO

Balancing Half-Reactions in Acid
Solution
 MnO4¯ ---> Mn2+ in an acid solution
 Before looking at the balancing technique,
the fact that it is in acid solution can be
signaled to you in several different ways:
1) It is explicitly said in the problem.
2) An acid (usually a strong acid) is
included as one of the reactants.
3) An H+ is written just above the
reaction arrow.

There are three other chemical
species available in an acidic
solution:
 H2O
 H+
 e¯
 water is present because the reaction is taking
place in solution
 the hydrogen ion is available because it is in
acid solution
 electrons are available because that's what is
transferred in redox reactions
 All three will be used in getting the final
answer.

1. Balance the atom being
reduced/oxidized. In our example, there
is already one Mn on each side of the
arrow, so this step is already done.
MnO4¯ ---> Mn2+
 Balance the oxygens. Do this by adding
water molecules (as many as are needed)
to the side needing oxygen. In our case,
the left side has 4 oxygens, while the right
side has none, so:
MnO4¯ ---> Mn2+ + 4H2O

 Balance the hydrogens. Do this by
adding hydrogen ions (as many as are
needed) to the side needing hydrogen. In
our example, we need 8 (notice the water
molecule's formula, then consider 4 x 2 =
8).
8H+ + MnO4¯ ---> Mn2+ + 4H2O
 Balance the total charge. This will be
done using electrons. It is ALWAYS the
last step.
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

Here is a second half-reaction,
also in acid solution:
Cr2O7
2¯ ---> Cr3+
1. Balance the atom being reduced/oxidized.
Cr2O7
2¯ ---> 2Cr3+
2. Balance the oxygens.
Cr2O7
2¯ ---> 2Cr3+ + 7H2O
3. Balance the hydrogens.
14H+ + Cr2O7
2¯ ---> 2Cr3+ + 7H2O
4. Balance the total charge.
6e¯ + 14H+ + Cr2O7
2¯ ---> 2Cr3+ + 7H2O

Another example (in acid solution)
SO2 ---> SO4
2¯
1. SO2 ---> SO4
2¯ (the sulfur is already
balanced)
2. 2H2O + SO2 ---> SO4
2¯ (now there are 4
oxygens on each side)
3. 2H2O + SO2 ---> SO4
2¯ + 4H+ (2 x 2 from
the water makes 4 hydrogens)
4. 2H2O + SO2 ---> SO4
2¯ + 4H+ + 2e¯ (zero
charge on the left; +4 from the hydrogens
and -2 from the sulfate, so 2 electrons
gives the -2 charge required to make zero

Practice Problems
 1) Re ---> ReO2
 2) Cl2 ---> HClO
 3) NO3¯ ---> HNO2
 4) H2GeO3 ---> Ge
 5) H2SeO3 ---> SeO4
2¯
 6) Au ---> Au(OH)3 (this one is a bit odd!)
 7) H3AsO4 ---> AsH3
 8) H2MoO4 ---> Mo
 9) NO ---> NO3¯
 10) H2O2 ---> H2O

Answers:
1. 2H2O + Re ---> ReO2 + 4H+ + 4e¯
2. 2H2O + Cl2 ---> 2HClO + 2H+ + 2e¯
3. 2e¯ + 3H+ + NO3¯ ---> HNO2 + H2O
4. 4e¯ + 4H+ + H2GeO3 ---> Ge + 3H2O
5. H2O + H2SeO3 ---> SeO4
2¯ + 4H+ + 2e¯
6. 3H2O + Au ---> Au(OH)3 + 3H+ + 3e¯
7. 8e¯ + 8H+ + H3AsO4 ---> AsH3 + 4H2O
8. 6e¯ + 6H+ + H2MoO4 ---> Mo + 4H2O
9. 2H2O + NO ---> NO3¯ + 4H+ + 3e¯
10. 2e¯ + 2H+ + H2O2 ---> 2 H2O

Balancing Half-Reactions in
Basic Solution
 Before looking at the balancing technique,
the fact that it is in basic solution can be
signaled to you in several different ways:
 It is explicitly said in the problem.
 A base (usually a strong base) is included
as one of the reactants.
 An OH¯ is written just above the reaction
arrow.

There are three other chemical
species available in a basic solution
besides the ones shown above.
They are:
 H2O
 OH¯
 e¯
 water is present because the reaction is
taking place in solution
 the hydroxide ion is available because it is in
basic solution
 electrons are available because that's what is
transferred in redox reactions.
 All three will be used in getting the final
answer.

PbO2 ---> PbO
It is to be balanced in basic
solution.
 Step One to Four: Balance the half-
reaction AS IF it were in acid solution.
1. Balance the atom being
reduced/oxidized.
2. Balance the oxygens (using H2O).
3. Balance the hydrogens (using H+).
4. Balance the charge. When you do that to
the above half-reaction, you get:
2e¯ + 2H+ + PbO2 ---> PbO + H2O

 Step Five: Convert all H+ to H2O.
 Do this by adding OH¯ ions to both sides.
 The side with the H+ will determine how
many hydroxide to add. In our case, the
left side has 2 hydrogen ions, while the
right side has none, so:
2e¯ + 2H2O + PbO2 ---> PbO + H2O +
2OH¯

 Notice that, when the two hydroxide ions
on the left were added, they immediately
reacted with the hydrogen ion present.
The reaction is:
H+ + OH¯ ---> H2O

 Step Six: Remove any duplicate
molecules or ions.
 In our example, there are two water
molecules on the left and one on the right.
 This means one water molecule may be
removed from each side, giving:
2e¯ + H2O + PbO2 ---> PbO + 2OH¯
 The half-reaction is now correctly
balanced.

Here is a second half-reaction,
also in basic solution:
MnO4¯ ---> MnO2
 Step One: Balance the half-reaction AS IF
it were in acid solution.
 3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
 Step Two: Convert all H+ to H2O.
 3e¯ + 4H2O + MnO4¯ ---> MnO2 + 2H2O +
4OH¯
 Step Three: Remove any duplicate
molecules or ions.
 3e¯ + 2H2O + MnO4¯ ---> MnO2 + 4OH¯

Balancing Simple Redox Rxns
Must be:
Balanced for Mass
ATOMS balance
Balanced for Charge
Total e- Lost = Total e- Gained

Balancing Harder
Redox Reactions
(Honors)

Oxidation Number Method
(Balancing in Acid Solution)
• Find ox #’s and use brackets to connect elements
changing in charge.
• Balance atoms changing in charge
• Find total e- involved in each change
• If necessary balance e- by multiplication
• Balance all other atoms except H and O
• Balance oxygen by adding H2O to side deficient
• Balance hydrogen by adding H+1 to side deficient
• Check for balance with respect to atoms and charge.

Half Reaction Method
(Ion/Electron Method)
(In acid solution)
 Separate equation into two “basic” half reactions
 Balance all atoms except H and O
 Balance oxygen by adding H2O
 Balance hydrogen by adding H+1
 Balance charge by adding electrons to more
positive side
 If necessary balance e- by multiplication
 Add together half reactions and simplify
 Check for balance of atoms and charge

Applications of Redox Reactions
Corrosion of Metals
the metal gets oxidized
forming metal oxides on
the surface
Prevention: Use paint, oil,
plating or attach to negative
terminal of a battery.
Gold doesn’t
rust…Why?

Photograph Development involves oxidation and
reduction of silver atoms and ions

Bleach acts on
stains by oxidizing
them, getting reduced
in the process
Explosives form
neutral gases like N2
from compounds!

Reactivity of Metals
Reference Table J
Metals Higher on Table J
are more ‘active”
It is easier for more “active”
metals to be oxidized
or lose electrons.

Copper replaces
silver!
Cu0(s) + AgNO3(aq) Ag0(s) + CuNO3(aq)
Ag0(s) + CuNO3(aq) wouldn’t happen!!!

Reactivity of Nonmetals
Reference Table J
Nonmetals higher on Table J
are more “active”
It is easier for more “active”
nonmetals to “gain” electrons
and be reduced.

Electrochemical Cells (Batteries)
Chemical reaction that produces electricity.
Called “voltaic cells” as they produce voltage
This happens SPONTANEOUSLY.

Moving Electrons = Electricity
Electrons given off by oxidized substance
travel towards substance being reduced.
Traveling electrons move
through “external circuit”
where they do work.

How do the Electrons Move?
Batteries often contain 2 metals.
Start with Table J
Electrons travel from the more “Active metal”
toward the less active metal.
Metal above = oxidized
Ion on Metal below = reduced

Electrons flow
“Down Table J”
From metal
above to ion of
metal below
e-

Parts of a Simple Battery
(Voltaic Cell)
Made of Two “Half Cells” containing:
2 Metal Electrodes
2 Solutions of Ions
External Wire
Salt Bridge

Electrons need to flow in a “circuit” that
is connected.
External Wire:
allows electrons to flow
between metal electrodes
Salt Bridge: allows ions
to flow between solutions

Zn/Zn+2//Cu+2/Cu
What is Ox/Red?
See Table J
Metal above is oxidized
Zn
Ion of metal below reduced
Cu+2

Which way do electrons
flow in the external wire?
See Table J
Electrons flow “Down” the
table from what is oxidized
towards what is reduced.
from Zn to Cu
e-

Which electrode is negative?
Which electrode is positive?
Electrons flow from negative to
positive electrode.
Negative electrode: Zn
Positive electrode: Cu
e-

Which electrode is the
anode and cathode?
Anode: metal electrode
where oxidation occurs
Zn
Cathode: metal electrode
where reduction occurs
Cu

Remember
AN OX
RED CAT
Anode is where oxidation happens
Cathode is where reduction happens

What are the Half Reactions?
What is the Net Equation?
Ox:
Zn0 Zn+2 + 2e-
Red:
Cu+2 + 2e- Cu0
Net: (add ½ reactions)
Zn0 + Cu+2 Zn+2 + Cu0
Make sure final net equation is
balanced for electrons and atoms!
e-

Which electrode gains/loses weight?
Look at half reactions!!
Which one forms solid metal?
Which one forms dissolved ions?
Ox:
Zn0 Zn+2 + 2e-
Red:
Cu+2 + 2e- Cu0
Zinc electrode loses mass
Copper electrode gains mass

Which way to do the ions in the salt
bridge “migrate” or move?
Remember:
“The negative ions complete the circuit”
(The ions actually end up moving towards
the solution of opposite charge that forms.)

 http://www.mhhe.com/physsci/chemistry/e
ssentialchemistry/flash/galvan5.swf

Dead Battery
Voltage = 0
Means the reaction in the battery has
reached EQUILIBRIUM.

You try it…
Mg/Mg+2//Al+3/Al
• Draw and label Battery
• What is oxidized/reduced?
• What are the half reactions and net(balanced)?
• What is the neg/pos electrode?
• What is the anode/cathode?
• Which way do e- flow in wire?
• Which way do -/+ ions flow in salt bridge?
• Which electrode gains/loses mass?

Finding Voltage of a Battery
(Honors)
Use Voltage Table
Find your half reactions and record voltage
Note:
All ½ reactions shown are reductions.
For oxidation, reverse the sign of the voltage

Nerntz Equation (Honors)
Find voltage of a battery when the conc. of dissolved
ions is not 1 Molar (as on “standard voltage” table)
Ecell = E0 – 0.0592 log [product ion]x
n [reactant ion]y
n = total # of moles electrons being transferred
The concentration of dissolved ions can affect voltage.
Greater concentration of reactant ions (see net)
increases the overall voltage.

Electrolytic Cells &
Electrolysis Reactions
Uses electricity to “split” or “lyse” a
compound into it’s neutral elements
An outside electrical source provides electrons
to force a non-spontaneous redox reaction to
occur.

Electrolysis Set Up
Single Cell filled with electrolyte with +/- ions
Attach battery to two electrodes.
Electrodes are made of an inert substance
(like platinum or graphite) that conducts.
Electrodes don’t chemically change like in a battery,
they just provide current

Pulls electrons off one electrode
Making it POSITIVE
Adds electrons onto one electrode
Making it NEGATIVE
Role of the Battery

Which Way do the Ions Move?
To electrode of
opposite charge

At neg. electrode
electrons are gained by ion
(reduction at CATHODE)
At positive electrode
electrons are lost by ion
(oxidation at ANODE)
What is Oxidized/Reduced?

Half Reactions & Net Equation
Rxn at Anode: (Ox)
Cl- Cl + 1e-
Or more correctly DIATOMIC!!!!!!
2Cl- Cl2 + 2e-
Rxn at Cathode: (Red)
Na+ + 1e- Na
(Multiply by 2 to balance electrons)
NET:
2Na+ + 2Cl- 2Na + Cl2

Determining Voltage Needed
Use the Voltage Table to determine the total
voltage needed to run the Electrolytic cell.
Total voltage should be a NEGATIVE number

Electrolysis of Molten NaCl (l)

Electrolysis of PbCl2(l)
What is oxidized?
What is reduced?
What are the ox/red
half reactions?
What is the net
equation?
Negative
Electrode
Positive
Electrode

Electrolysis of PbCl2(l)
Oxidized: Cl- Reduced: Pb+2
Half Reactions
Ox: Cl-1 Cl + 1e-
2Cl-1 Cl2 + 2e-
Red: Pb+2 + 2e- Pb
Net: Pb+2 + 2Cl-1 Pb + Cl2

Electrolysis of Water
At Positive Electrode:
Ox: O-2 O + 2e-
but there is a diatomic!
2O-2 O2 + 4e-
At Negative Electrode
Red: H+1 + 1e- H
but there is a diatomic!
2H+1 + 2e- H2
Net: 2H2O 2H2 + O2

 Lemon Battery Demo
 http://youtu.be/AY9qcDCFeVI
 Electrolysis of Copper Sulfate
 http://youtu.be/xBz9HJ32Ouw
 Electrolysis of Water/ Silver Nitrate and Cu reaction
 http://youtu.be/Bcfp8VtcrSA
 Electrolysis of Water (Animation)
 http://youtu.be/2t13S-KpGeE
 Electrolysis of Water (Simple)
 http://youtu.be/HQ9Fhd7P_HA

Electroplating
Electrolysis reaction
used to coat a
substance with a thin
layer of metal.
Often coating is a less
reactive metal that is
not easily oxidized or
corroded.

Electroplating
 Negative Electrode
 Is the OBJECT TO BE
PLATED
 so the positive metal ions
would go towards it and be
REDUCED.
 It is the CATHODE
Red: Ag+ + 1e- Ag0

Electroplating
 Positive Electrode
 Made of plating metal
 It dissolves into solution as
metal strip gets OXIDIZED.
 It is the ANODE
 This replenishes the ions for
plating.
Ox: Ag0 Ag+ + 1e-

Electroplating Problems
(Honors)
Coulomb = measure of electrical charge
1 mole e- = 96,500 coulombs
# coulombs = # amps x seconds

(Honors)
Reduction:
Happens on object to be plated
Look at Reduction half reaction
Look at mole relationships
between electrons and metal atoms.
Ex: Ag+ + 1e- Ag0

(Honors)
 You can now answer questions regarding
the amount of a substance in moles or
grams that can be electroplated over a
certain amount of time.

(Honors)
If 10 amps are run through a CuSO4 solution for
5 minutes, calculate the grams of Cu that will plate
onto the spoon.
We Know:
1 mole e- = 96,500 coulombs
# coulombs = # amps x seconds
Red: Cu+2 + 2e- Cu0
1 mole Cu = 63.5 grams

So….Let’s start here
# coulombs = 10 amps x 300 seconds
= 3000 coulombs
3000 coul. x 1 mole e- x 1 mole Cu x 63.5g Cu = .987 grams
96,500 coul 2 mole e- 1 mole Cu
Mole ratio from
Reduction half reaction

You Try One
How long will it take to deposit 20 grams of
silver from a solution of AgCl onto a copper
tray if a current of 5 amps is used?
Answer = 3, 574 sec
or 59.5 minutes or about 1 hour

You Try One
How many amps are needed to deposit .504g.
of Iron in 40 minutes by passing a current
through a solution of Iron II Sulfate?
Answer: .72 amps

Redox Reaction.ppt

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