Chapter 2
The Reduction & Oxidation of Metals
Chapter A2
2.1 - Compounds and Chemical Change
Metallurgy
O Archeologist studies show that
ancient civilizations were not only
using metals, they learned to
manipulate it
O metallurgy – heating and hammering
copper into shapes
O by analyzing copper artifacts,
archeologists use chemical analysis to
track trade routes, locations of ancient
industry, etc.
Tarnishing
O Definition:
O the colour change, (and
loss of luster), associated
with metal over time
O silver tends to turn black
over time
O copper tends to turn green
Silver tarnish
O when a metal tarnishes, a
chemical reaction occurs
between the metal and
molecules in the air
O balance the following reaction
between pure silver, hydrogen
sulfide and oxygen in the air
4
2
1
2
2
__Ag
(s) + __H2S(g) + __O2(g)  __Ag2S(s) + __ H2O(l)
Silver tarnish
O Complete the Lewis dot diagram for
the reaction:
Balanced equations
O Note: the number of molecules drawn
corresponded to the coefficients of the balanced
equation.
O 4 Ag(s) + 2 H2S(g) + 1 O2(g)  2 Ag2S(s) + 2 H2O(l)
O when an equation is balanced, you can use the
coefficients to compare the ratios of one
compound to another
O Definition:
Mole Ratios
O the proportion of the coefficients in a balanced
chemical equation.
O This can be used to solve for the actual quantity of
the substance in moles using the formula:
n required
n given
=
coefficient required
coefficient given
Mole Ratios
O Since you will always need to solve for “n
the rearranged formula is:
n required = n given x
OR
nR = nG x
CoR
CoG
required”
coefficient required *
coefficient given
* Where “given” is the
value that is provided
for you in the question
and “required” is the
value that you need to
figure out.
Mole Ratios
O a ratio does not tell you the exact amounts, but
rather the proportions
O the ratio/ proportion is used to calculate the exact number of
moles of the required
O Ex. a mole ratio for hydrogen to oxygen (above) is 2:1; this means
TWO moles of hydrogen for every ONE mole of oxygen are used in
the reaction.
O If 100 moles of hydrogen was required how much oxygen would
you need?
O
50 moles…there is always HALF (2:1) as much oxygen as hydrogen
Steps to solving mole ratio
problems:
1.
write the balanced chemical equation for the
reaction
2 Fe+ 3 CuCl2 --> 2 FeCl3 + 3 Cu
2.
identify the substance that is “given” (value provided in
the question), and the substance that is “required” (the
value that you need to find) and write the numbers right
under the required substance
G
R
2 Fe+ 3 CuCl2 --> 2 FeCl3 + 3 Cu
nG = 0.75 mol
nR = ?
Steps to solving mole ratio
problems:
3. use the mole ratio formula: by filling in the
coefficients.
G
R
2 Fe+ 3 CuCl2 --> 2 FeCl3 + 3 Cu
nR = nG x CoR
CoG
nR = 0.75 mol x 3
2
Steps to solving mole ratio
problems:
4.
do it!! multiply the number of moles of “given” by
the mole ratio to find the number of moles of the
“required”
G
R
2 Fe+ 3 CuCl2 --> 2 FeCl3 + 3 Cu
nR = nG x CoR
CoG
nR = 0.75 mol x 3
2
nR = 1.125 = 1.1 mol of Cu
Practice problem #1:
Determine the amount of silver required to
make 0.876 mol of silver sulfide, based on the
reaction given above.
1. write the balanced chemical equation for the
reaction

4 Ag(s) + 2 H2S(g) + __ O2(g)  2 Ag2S(s) + 2 H2O(l)
Practice problem #1:
Determine the amount of silver required to
make 0.876 mol of silver sulfide, based on
the reaction given above.
2. Over the equation, write the “G’s” and “R’s”
to remind yourself which values/
coefficients go where. Fill them in!

R
G
4 Ag(s) + 2 H2S(g) + __ O2(g)  2 Ag2S(s) + 2 H2O(l)
nR = ?
nG = 0.876 mol
Practice problem #1:
Determine the amount of silver required to
make 0.876 mol of silver sulfide, based on the
reaction given above.
3 & 4. use the mole ratio formula, fill in the
coefficients & values, and do the math!

R
G
4 Ag(s) + 2 H2S(g) + __ O2(g)  2 Ag2S(s) + 2 H2O(l)
nR = nG x CoR
CoG
nR = 0.876 mol x 4
2
nR = 1.75 mol of Ag
Practice problem #2:
O How many moles of oxygen are required to burn 5.0mol
of propane (C3H8(g))?
O Remember the four steps
Practice problem #2:
O How many moles of oxygen are required to burn 5.0mol
of propane (C3H8(g))?
O 1. Write out/ balance the equation
G
____ C3H8(g)
R
5 O2(g)  ____
3 CO2(g) + ____
4 H2O(g)
+ ____
2. Identify “given” &
“required”
Practice problem #2:
O How many moles of oxygen are required to burn 5.0mol
of propane (C3H8(g))?
O 3 & 4. use the mole ratio formula, fill in the
coefficients & values, and do the math!
G
R
5 O2(g)  ____
3 CO2(g) + ____
4 H2O(g)
____ C3H8(g) + ____
n
= n G x Co R
Co G
n R = 5.0 mol x 5
1
nR = 25 mol
R
O Practice problems (page 63)
O Q’s 4-6
O 2.1 Summary (page 68)
O Q’s 2 (b), 3 & 5 pg 68
6 CO2(g) + 6 H2O(g) → C6H12O6(aq) + 6 O2(g)
6 CO2(g) + 6 H2O(g) → C6H12O6(aq) + 6 O2(g)
n
= n G x Co R
Co G
n R = 9.00 mol x 6
1
R
nR = 54.0 mol of H2O(g)
6 CO2(g) + 6 H2O(g) → C6H12O6(aq) + 6 O2(g)
n
= n G x Co R
Co G
n R = 9.00 mol x 6
1
R
nR = 54.0 mol of O2(g)
6 CO2(g) + 6 H2O(g) → C6H12O6(aq) + 6 O2(g)
 54.0 moles of CO2(g).
 CO2(g), H2O(g) and O2(g) all have a
coefficient of 6; glucose has a
coefficient of 1.
 Since the mole are 6 : 1, each substance
requires six times as many moles as
glucose to balance the equation.
O Complete Pre-Lab Analysis for
Investigation: Mole Ratios in
Chemical Reactions.
Ca(OH)2
Solid precipitate
NaCl
solution
O MUST BE COMPLETED in order to
participate in lab.
Chapter A2
2.2 – The Gain & Lose of Electrons
Making metal useful:
O Metals are an important resource, but pure metals are
rarely found in nature.
O the term ore refers to a rock containing enough useful
metal to be mined
Formation of an ionic compound
O a metal is made of cations surrounded by free
floating electrons
O some of these electrons will be lost to other atoms in
the environment
O result: the number of free-floating electrons is no longer equal to
the number of positive charges
O metals that lose the electrons become positively-charged ions
Calcium atom
20 p+
20 eNet charge = 0
Calcium ion
20 p+
18 eNet charge = 2+
Formation of an ionic compound
O the atoms that picked up the extra electrons
become negatively-charged ions
O these are called anions
O result: the number of free-floating electrons
is no longer equal to the number of positive
charges
Phosphorus atom
15 p+
15 eNet charge = 0
Phosphorus ion
15 p+
18 eNet charge = 3 –
Formation of an ionic compound
O positively-charged
cations are attracted
to the negativelycharged anions.
O the complete
transfer of valence
electrons results in
an ionic bond.
Oxidation: The Loss of Electrons
O when an atom loses an electron it is oxidized
O oxidation is a chemical process which a substance
loses electrons
O two types of oxidation reactions
O metal atom (neutral)  metal ion + electrons
O e.g. Fe(s)  Fe2+(aq) + 2e-
O non-metal ion (charged)  non-metal atom +
electrons
O e.g. 2Cl-(aq)  Cl2(g) + 2e-
* Notice how the electrons are always
on the PRODUCT side of the reaction
Reduction: The Gain of Electrons
O when an atom gains an electron it is reduced
O two types of reduction reactions
O metal ion (charged) + electrons  metal atom
O e.g. Fe2+(aq) + 2e-  Fe(s)
O non-metal atom (neutral)  non-metal ion + electrons
O e.g. Cl2(g) + 2 e-  2Cl-(aq)
* Notice how the electron are always
on the REACTANT side of the reaction
L.E.O the lion says G.E.R
REDOX Reactions
O Following the Law of Conservation of Matter;
matter CAN NOT be created or destroyed
O The electrons that are gained by one element must
come from the element that loses them.
O In a single replacement reaction, an element
reacts with a compound to produce a new element
and a new compound
REDOX Reactions
O That means there are three elements involved
altogether:
O One element will be oxidized (and give up/lose electrons)
O One element will be reduced (and gain the excess electrons)
O One element will be the “spectator” (it will remain the same, as
an ion, on both sides of the reactions
Example:
Single Replacement Reactions
O in order to get a valuable metal our of
an ionic compound, a less valuable
metal can be used to take its place in
the compound.
O most silver is found as the
compound silver sulfate. The pure
silver can be collected by reacting
the silver sulfate with zinc metal.
O Identify the substance being:
O oxidized (losing electrons)
O reduced (gaining electrons)
O spectating (staying the same)
Example:
Single Replacement Reactions
O Most silver is found as the compound
silver sulfate. The pure silver can be
collected by reacting the silver sulfate
with zinc metal.
O step 1: write out the reaction
Ag2SO4(s) + Zn(s)  ZnSO4(s) + 2 Ag(s)
Example:
Single Replacement Reactions
O
step 2: identify each atom, and classify them as
either an element or an ion on BOTH sides of
the equation
O Ag2SO4(s) + Zn(s)  ZnSO4(s) + 2 Ag(s)
2Ag+(aq)  2Ag(s)
Zn(s)  Zn2+(aq)
SO42-(aq)  SO42-(aq)
Example:
Single Replacement Reactions
O
step 3: determine whether each reactant had to gain or
lose electrons to become the product
O Ag2SO4(s) + Zn(s)  ZnSO4(s) + 2 Ag(s)
2Ag+(aq)+ 2e-
 2Ag(s)
the silver had to gain electrons = reduced
Zn(s)  Zn2+(aq) + 2ethe zinc loses electrons = oxidized
SO42-(aq)  SO42-(aq)
the sulfate is unaffected = spectator ion
O Practice problems (page 72)
O Q 15
O Practice problems (page 74)
O Q 18
O Practice problems (page 75)
O Q 20& 21
O 2.2 Summary (page 75)
O Q’s 2 & 4
Chapter A2
2.3 – The Reactivity of Metals
Stability vs. Reactivity
O Some metals are very stable –
they can be found in pure form,
and don’t often form compounds.
O e.g. gold
O Metals that are reactive, tend to
corrode easily.
O e.g. iron
O corrosion is the oxidation of a metal
O e.g. tarnishing and rusting
Simulation
O Four different metals
are immersed into four
different ionic
solutions, and the
results recorded.
O If nothing occurs,
then “no reaction” will
be noted.
O If a change occurs,
then “reaction” will be
noted.
O The results should be
as follows:
Simulation - results
Solutions
Metals
Mg2+(aq)
magnesium
copper
zinc
silver
Cu2+(aq)
Zn2+(aq)
Ag+(aq)
no reaction reaction
reaction
reaction
no reaction no reaction no reaction reaction
no reaction reaction no reaction reaction
no reaction no reaction no reaction no reaction
O rank the ions in order from most to least reactive
O Ag+  Cu2+  Zn2+  Mg2+
O rank the metal atoms in order from most to least reactive
O Mg  Zn  Cu  Ag
O what connection do you see between the reactivity of an ion and
its stability as an atom?
O The MORE STABLE a metal ATOM, the MORE REACTIVE the ION is
Activity Series
O A list that organizes
metal atoms from
most stable to least
stable
O Simultaneously
organizes metal ions
from most reactive to
least reactive
O see textbook page 80 or
page 4 of your data
booklet
Activity Series
O when read left to right
you get a reduction halfreaction
O Example:
Cd2+(aq) + 2e-  Cd(s)
* Electron on REACTANT side = gain
G.E.R = Reduction
Activity Series
O If read right to left you
get an oxidation halfreaction
O flip the arrow around in your
head
Example:
Cd(s)  Cd2+(aq) + 2e
* Electron on PRODUCT side = lose
L.E.O = oxidation
Activity Series
O an activity series allows you to:
compare the relative reactivity of metal ions
1.
O
Use LEFT SIDE; whichever is listed closer to
the top= more reactive
compare the relative reactivity of metal atoms
2.
O
use RIGHT SIDE, whichever is listed closer to
the bottom= more reactive
determine if a reaction will occur
spontaneously
3.
O
a spontaneous reaction is one that will occur
by itself without the addition of energy
Activity Series
O to determine if a
reaction is
spontaneous, locate the
two half-reactions on
the table
O if the reduction reaction
(LR) is located above
the oxidation reaction
(RL) it will occur
spontaneously
O a non-spontaneous
reaction will have the
reduction reaction
located below the
oxidation reaction
Activity Series-Example
O A piece of solid
magnesium is placed in a
solution containing silver
ions.
O Step 1: identify reactants
O solid magnesium & silver
ions
O Step 2: identify half-
reactions
O reduction reaction (LR)
= silver ions
O oxidation reaction (RL)
= solid magnesium
Activity Series-Example
O A piece of solid magnesium is
placed in a solution containing
silver ions.
O Step 3: determine
spontaneity…
O If the reduction reaction (LR)
is above the oxidation reaction
(RL) = spontaneous
O reduction reaction (LR)
below the oxidation reaction
(RL) = NON-spontaneous
O This reaction IS spontaneous b/c
the silver reduction rxn is above the
magnesium oxidation reaction
Practice Problem #1
a.
O a piece of zinc is
element/ion that gains
electrons).
placed in a solution
containing Cu2+ ions
a.
write the oxidation
and reduction half
reactions.
b.
determine if the
reaction will occur
spontaneously
c.
explain how you
knew it was
spontaneous or not.
reduction half reaction (the
b.
c.
Cu2+(aq) + 2e-  Cu(s)
oxidation half reaction (the
element/ion that loses
electrons).
Zns)  Zn2+(aq) + 2eit is spontaneous
b/c the copper reduction
reaction is ABOVE the
zinc oxidation reaction in the
activity series
Practice Problem #2
a.
O a piece of gold jewelry
element/ion that gains
electrons).
is placed in a solution
containing H+ ions
a.
write the oxidation
and reduction half
reactions.
b.
determine if the
reaction will occur
spontaneously
c.
explain how you
knew it was
spontaneous or not.
reduction half reaction (the
b.
c.
2H+(aq) + 2e-  H2(g)
oxidation half reaction (the
element/ion that loses
electrons).
Au(s)  Au3+(aq) + 3eit is NON spontaneous
b/c the hydrogen reduction
reaction is BELOW the
gold oxidation reaction in the
activity series
Oxidizing and Reducing agents
O
In keeping with the Law of Conservation, one substance
cannot gain electrons (be reduced) without another
substance providing the electrons (being oxidized).
O
A reducing agent must stimulate the loss of electrons, in
order for them to be available for the substance being
reduced.
Oxidizing and Reducing Agents
O the reducing agent is
the entity that CAUSES
another substance to
be reduced
O it is itself oxidized
O the oxidizing agent is
the entity that CAUSES
another substance to
be oxidized
O it is itself reduced
Oxidizing and Reducing Agents
O Consider the following analogy:
O A travel agent does not travel
themselves, but rather their
actions allow someone else to
travel.
O in a similar way, the oxidizing
agent is the one that enables
another substance to be
oxidized,
O and the reducing agent enables
the other substance to be
reduced.
O Practice Problems (page 79)
O Q 24 & 25
O Practice Problems (page 82)
O Q 28 & 29
O Practice Problems (page 83)
O Q 30(a,c,e,g,i) & 31
O 2.3 Summary (page 85)
O Q’s 3 & 5
O Complete the
A 2.1-2.3 Worksheet:
Redox & Activity Series
O You must submit your work
by the end of class
Homework
Check-up