Download Unit 5 * Redox Reactions

Unit 5 – Redox Reactions
OXIDATION AND REDUCTION – A FUNDAMENTAL BASIS FOR
REACTIONS IN CHEMISTRY
Introduction
 We have learned much about chemical reactions, that they involve the
breaking and formation of chemical bonds.
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
 During the formation of these bonds, whether ionic or covalent, the
process in reality involves the movement of electrons.
Electron (e-) – a negatively charged particle
 In principle, this movement of electrons causes changes in the atom of
the element(s) involved.
Introduction (cont’d)
 Typically we understand this as the formation of ions, either:
 1. Positive (when losing electrons)
e.g. Fe3+, Na+, Mg2+
 2. Negative (when gaining electrons)
e.g. Cl-, S2-, F A common type of reaction where we easily observe an electron transfer
is the single-replacement reaction:
Cu(s) + AgCl(aq)  Ag(s) + CuCl(aq)
Introduction (cont’d)
 In order to really view the ions being formed is by writing the net
ionic equation for that reaction:
Cu(s) + Ag+(aq) + Cl-(aq)  Ag(s) + Cu+(aq) + Cl-(aq)
cancelling spectator ions, we get:
Cu(s) + Ag+(aq)  Ag(s) + Cu+(aq)
 It is easy to see that the ionic charge changes from one side to the
other.
 These charges that we are familiar with actually refer not just to the
fact that they are ions, but also refer to a change in what is called
the element’s oxidation number.
Oxidation Numbers
NO MATTER THE SITUATION, EVERY ELEMENT HAS ONE
Oxidation Numbers
 Oxidation numbers for elements are determined by using the following
rules:
Rule 1 – Free elements (including diatomic or polyatomic molecules like H2, O2, Fe,
or S8, among others) have an oxidation number of 0 (zero).
 Rule 2 – Monatomic ions have the same oxidation number as their charge (e.g. Na+
has an oxidation number of +1).
 Rule 3 – When molecular compounds are used, the more electronegative element
(remember that means it likes e- more) has the same oxidation number as if it were
an ion and is negative.
e.g. In CH , carbon is more electronegative, so it will have an oxidation number of – 4,
4
and hydrogen will be +1

Oxidation Numbers (cont’d)



Rule 4 – The oxidation number for oxygen is always – 2, except in peroxides (e.g.
H2O2) where it is – 1.
Rule 5 – Hydrogen has an oxidation number of +1, except in the form of hydrides
(when bound to a metal) where it is – 1.
Rule 6 – Metals of Groups 1 and 2 have an oxidation number same as their ionic
charge. Aluminum in Group 3 also follows that same rule.
 General Rules that help assign to oxidation numbers:
 1. The sum of the oxidation numbers for elements in a neutral molecule is zero (0).
 2. The sum of oxidation numbers of elements in a polyatomic ion is the same as the
charge for the entire ion.
For NaCl: Na (+1) + Cl (-1) = 0
For SO32-: S (+4) + 3 x O (-2) = -2
Assigning Oxidation Numbers
 Let’s try assigning oxidation numbers for the elements in molecules
and polyatomic ions.

Try these:
CO
nC = +2
nO = - 2
KCl
nK = +1
nCl = - 1
MgBr2
nMg = +2
nBr = - 1
nO = - 2
nH = +1
OHSince many of the elements do have more than one oxidation number, we have to rely
on the rules and the total for the group.
Assigning Oxidation Numbers (cont’d)
 Practice:
 Determine the oxidation numbers for the elements in the following:
1. CO2
2. H2O
3. NH3
4. SO425. BaCl2
6. K2CO3
nC = +4; nO = -2
nH = +1; nO = -2
nH = +1; nN = -3
nS = +6; nO = -2
nBa = +2; nCl = -1
nK = +1; nC = +4; nO = -2
Oxidation and Reduction
THE CONTEXT OF THE REACTION
Oxidation and Reduction
 Changes where an electron is gained or lost, have specific terms to
describe them:


Oxidation – occurs when an atom loses one or more electrons
e.g. Na  Na+ + eReduction – occurs when an atom gains one or more electrons
e.g. Cl + 2 e-  2 Cl2
 We can use a pneumonic device to help us remember these concepts:
 LEO says GER
Loss of Electrons - Oxidation
Gain of Electrons - Reduction
Identifying Oxidation and Reduction
 For the following changes in oxidation numbers, indicate whether the
change is an oxidation, or a reduction:






1. Cl2  2 Cl2. K  K+
3. Fe3+  Fe2+
4. Al  Al3+
5. Cu+  Cu2+
6. CH4  CO2
reduction
oxidation
reduction
oxidation
oxidation
oxidation
Oxidation-Reduction Reactions
 In chemical reactions where changes in oxidation numbers occur, we
call them oxidation-reduction reactions, or Redox reactions for short.
 In such reactions, we will observe that there is a balance of sorts
happening.
 If an element in a chemical is oxidized, another element in a different
chemical will be reduced. We cannot have one occur without the other.
oxidized
Cu(s) + AgCl(aq)  CuCl(aq) + Ag(s)
reduced
Oxidation and Reduction (cont’d)
 In a Redox reaction, we identify the chemicals with elements that
undergo oxidation or reduction.
 For such chemicals, we call one chemical the oxidizing agent, while
another is called the reducing agent. Both of these must be reactants.

Oxidizing Agent – the reactant chemical that contains the element that is reduced
(i.e. gains electrons), so it causes the oxidation of another element.

Reducing Agent – the reactant chemical that contains the element that is oxidized
(i.e. loses electrons), so it causes the reduction of another element.
Oxidation and Reduction (cont’d)
 Let’s try to identify the oxidizing and reducing agent in the following
reactions:
0
Oxidizing agent
+1
2 Na(s) + Cl2(g)  2 NaCl(s)
Reducing agent
+4
0
+2
-1
3 SnCl4(s) + 2 Fe(s)  3 SnCl2(s) + 2 FeCl3(s)
Oxidizing agent
0
Reducing agent
+3
Oxidizing and Reducing Agents:
 Practice determining oxidation numbers and identifying both the
oxidizing and reducing agents in the following reactions:
Oxidizing Agent
Reducing Agent
H2 + O2  H2O
O2
H2
FeO + O2  Fe2O3
O2
FeO
FeCl3
Zn
O2
CH4
Zn + FeCl3  Fe + ZnCl2
CH4 + O2  CO2 + H2O
Redox Reactions
BALANCING A REDOX REACTION
Balancing Redox Reactions
 Up to this point, we have been used to balancing chemical equations
fairly easily.
 Balancing Redox equations are a bit more complicated. For example,
consider the following reaction:
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
 There are a number of methods that we can use to balance redox
equations. The first one we will learn is called the Oxidation-Number
Method.
Oxidation Number Method
EQUAL POSITIVE AND NEGATIVE
# OXIDATION = # REDUCTION
Balancing by Oxidation-Numbers
 In order to properly balance a redox reaction, we must ensure that the
total number of e- that are gained by an element is equal to the total
number of e- lost by another.
 For most basic redox reactions, we will do that by using the
conventional balancing method, which, depending on the coefficients,
may prove to be time-consuming.
 Another method we can use, can make this job quicker, balancing by
oxidation-numbers.
Oxidation-Number Method
 Step 1 – Assign oxidation numbers to all elements in the chemical




equation.
Step 2 – Identify which atoms are oxidized and which are reduced.
Step 3 – Determine the size of the change in the oxidation numbers
for each affected element.
Step 4 – By using some factor, adjust the change so that they are
the same but opposite sign (a lowest common multiple). We do this
by changing/adding a coefficient to those chemicals.
Step 5 – If necessary, we then use the regular method to balance
the remaining elements in the equation.
Oxidation Number Method (cont’d)
 It is also worth mentioning that charges need to be balanced in
these equations (same on both sides).
reduced
 Example:
0
(+5)
+1
-2
+2
(+5)
-2
+4 -2
+1 -2
Cu + HNO3  Cu(NO3)2 + NO2 + H2O
oxidized



Step 1 – Assign oxidation numbers.
Step 2 – Determine which is oxidized, which is reduced.
Step 3 – Determine the size of the changes for oxidation and reduction. If they
are not equal, multiply one, or both so that they are. These numbers will be the
coefficients in front of the compounds affected.
Oxidation Number Method: (cont’d)
In the oxidation: Cu  Cu2+ has a change of +2
In the reduction: N5+ (in HNO3) N4+ (in NO2) has a change of -1
So we much place a coefficient of 2 in front of both N compounds, HNO3
and NO2.
2 HNO3  Cu(NO3)2 + 2 NO2 + 2 H2O
Cu + 4

Step 4 – Now we must balance the rest by the regular method of
balancing.

We see that the N and O atoms are not balanced. So we will need to fix that.
 We now have this equation balanced.
Oxidation-Number Method v. 2.0
WHEN THE SOLUTION IS ACIDIC OR BASIC
Reactions That Are Acidic or Basic
 Very often, we have to have reactions occurring under either acidic or
basic conditions. Certain reactions will work better.
 What this allows is that we can now freely add H2O, H+, and/or OHwhen we need them (since there are lots of these around).
 So, when can we do this?
 Once we have completed a balance of the elements (other than oxygen
and hydrogen) by using either the oxidation number method, or the
conventional method, we may find there are not enough O atoms (or
any at all):
Cr2O72- + 2 Cl-  2 Cr3+ + Cl2
(acidic)
Acidic or Basic Reactions: (cont’d)




When we balance by the oxidation number method, we end up with the following
result:
Cr2O72- + 6 Cl-  2 Cr3+ + 3 Cl2 (acidic)
We find that the O atoms cannot be balanced. Here we now add H2O to balance the O
atoms (water is freely available in aqueous solution).
Cr2O72- + 6 Cl-  2 Cr3+ + 3 Cl2 + 7 H2O (acidic)
We now have added H atoms from the water. Since the solution is acidic, we have lots
of H+ floating around, so we can balance the H atoms using these.
Cr2O72- + 6 Cl- + 14 H+  2 Cr3+ + 3 Cl2 + 7 H2O (acidic)
We will now see that the charge is also balanced on both sides.
What if the Reaction is in Basic Conditions?
 If a chemical reaction occurs in basic conditions, we actually only
have to go one step further than if it is in acidic conditions.
 Once we have it to the addition of H+, it is already balanced (all
atoms and charges), but it will be in acidic conditions (H+ will be
present somewhere).
 How do we change it to basic conditions?



Step 6 – Add H2O to balance O atoms.
Step 7 – Add H+ to balance the H atoms from H2O.
Step 8 – Add OH- to remove H+ (turns into water), but you must add OH- to
both sides (to maintain balance).
Basic Conditions: (cont’d)
7H
Cr2O72- + 6 Cl- + 14
14
H+2O+ 14 OH- 
2 Cr3+ + 3 Cl2 + 7 H2O + 14 OH(basic)

Step 9 – Cancel H2O on both sides until one side runs out of water.
 Amazingly, the equation will now be balanced (in both charge and in
the elements/atoms)
Redox Balancing of Net Ionic Equations
AN EASIER WAY OUT…SIMPLIFYING THE CHEMICAL
EQUATION FIRST
Review: Writing Net Ionic Equations
 Writing a net ionic equation involves splitting up any chemical that are
aqueous (e.g. NaCl(aq)) into their component ions (remember you cannot
split up a polyatomic ion).
e.g.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  Na+(aq) + Cl-(aq) + H2O(l)
 All other types stay in their written form (i.e. solid, liquid, or gas).
Review: Writing Net Ionic Equations (cont’d)
 Once you have finished, you can now cancel ions from both sides that
are the same. These cancelled ions are called “spectator” ions.
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  Na+(aq) + Cl-(aq) + H2O(l)
So your net ionic equation for
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
is:
H+(aq) + OH-(aq)  H2O(l)
Redox Balancing of Net Ionic Equations
 Another way of redox balancing uses the equation having only the




oxidation and reduction processes being shown .
This is a simpler method, ignoring the species (basically the
spectator ions) that don’t undergo oxidation or reduction.
In these types of reactions, we will often also see the terms, in acid
solution, or in basic solution.
These statements allow us to add H+ and H2O (acidic), or H+, H2O,
and OH- (basic) to help aid in balancing.
The reasons for adding these will become obvious shortly when we
investigate an example.
Balancing Net Ionic Redox Equations: (cont’d)
 Example:
+7 -2
reduced
-1
-8
-1
0
ClO4-(aq) + Br-(aq)  Cl-(aq) + Br2(g)
(acidic)
oxidized +1




Step 1 – Assign oxidation numbers to all elements.
Step 2 – Identify which is oxidized and which is reduced.
Step 3 – Determine size of change, adjust by adding an
appropriate coefficient(s) to the respective chemicals.
ClO4-(aq) + 8 Br-(aq)  Cl-(aq) + 4 Br2(g)
Step 4 – Add enough H2O to balance the O atoms, then add
H+ to balance the added H atoms from the H2O.
ClO4-(aq) + 8 Br-(aq) + 8 H+(aq)  Cl-(aq) + 4 Br2(g) + 4 H2O(l)
Balancing Net Ionic Redox Equations: (cont’d)

Step 5 – Verify that the numbers of atoms for each element are equal on both sides,
and the charge on both sides is the same.
ClO4-(aq) + 8 Br-(aq) + 8 H+(aq)  Cl-(aq) + 4 Br2(g) + 4 H2O(l)
Atoms are balanced (same number of Cl, O, Br, and H)
Charge is balanced (net -1 on both sides)
 Let’s now look at balancing when we have basic conditions.
Balancing Net Ionic Redox Equations: (cont’d)
 Example:
reduced
-2 +1 -2 +1
+7 -2
+4 -2
-1
+6
-2
CH3OH(l) + MnO4-(aq)  CO32-(aq) + MnO42-(aq)
oxidized



(basic)
+6
Step 1 – Assign oxidation numbers for every element.
Step 2 – Determine which is oxidized and which is reduced.
Step 3 – Determine the magnitude of the change, add coefficients to make equal.
CH3OH(l) + 6 MnO4-(aq)  CO32-(aq) + 6 MnO42-(aq)
Balancing Net Ionic Redox Equations: (cont’d)

Step 4 – Now add an appropriate number of H2O molecules to balance the O atoms.
Now 8 H atoms here
No H atoms here
2 H2O(l) + CH3OH(l) + 6 MnO4-(aq)  CO32-(aq) + 6 MnO42-(aq) + 8 H+(aq)
25 O atoms here
27 O atoms here
Step 5 – Add the appropriate number of H atoms as H+ to balance these.
 Step 6 – Since we are in basic conditions, we must add OH- (on both sides to
maintain balance) to cancel the H+ ions. When both are present, they cancel to form
H2O.
+
8
OH
CH3OH(l) + 6 MnO4-(aq) + 2 H2O(l) + 8 OH-(aq) CO32-(aq) + 6 MnO42-(aq) + 8 H+2O
(l)
(aq)
(aq)

Balancing Net Ionic Redox Equations: (cont’d)
Step 7 – Now we can cancel H2O (if necessary).
CH3OH(l) + 6 MnO4-(aq) + 2 H2O(l) + 8 OH-(aq)  CO32-(aq) + 6 MnO42-(aq) + 68 H2O(l)
Check that atoms and charge are balanced.
Final Balanced Equation:

CH3OH(l) + 6 MnO4-(aq) +8 OH-(aq)  CO32-(aq) + 6 MnO42-(aq) + 6 H2O(l)
 So now we have two tools to balance redox equations.
 We will next look at a method that is good at balancing both the
atoms and charge at the same time, balancing by what are called,
Half-reactions.
Half-Reactions
HE SAID, SHE SAID…
THE OXIDATION SIDE
THE REDUCTION SIDE
Redox Half-Reactions
 You may recall that we can show a reaction as having an oxidation, and
a reduction (hence redox).
 We can write an oxidation simply as:
Na(s)  Na+(aq) + e-
 We can then write a reduction as:
Cl2(g) + 2 e-  2 Cl-(aq)
 We can more easily see the oxidation and the reduction.
 So when we have a redox reaction, we can balance more easily if we
concentrate on only the elements that are being oxidized and being
reduced.
Balancing by Half-Reactions
 How do we balance by half reactions?
 The process goes like this:
 Step 1 – Write the Net Ionic Equation for the redox reaction.
 Step 2 – Write the oxidation and the reduction half reactions for the Net Ionic
Equation.
 Step 3 – Balance the atoms and charges in each half reaction.
 Step 4 – Balance the charges between the half reactions by finding a common
multiple (same size, opposite charge).
 Step 5 – Add the balanced half-reactions, then return the spectator ions.
 Note to consider acidic or basic conditions where they occur.
Balancing by Half-Reactions: (cont’d)
 Example:
Fe2+(aq) + Cr2O72-(aq)  Fe3+(aq) + Cr3+(aq)
(acidic)
 The reaction above occurs under acidic conditions. Balance this equation by using the
half-reaction method.
 Step 1 – It is already in an ionic equation form (no spectators)
 Step 2 – Write the two half reactions: (remember acidic)
Oxidation: Fe2+(aq)  Fe3+(aq)
(we balance the oxidation
numbers by adding electrons
Fe2+(aq)  Fe3+(aq) + e
to the appropriate side)
Reduction: Cr2O72-(aq)  Cr3+(aq)
14 H+ + Cr2O72-(aq) + 6 e-  2 Cr3+(aq) + 7 H2O
(under acidic conditions, we can balance the O atoms by adding
H2O and add H+ to balance the H)
Balancing by Half-Reactions: (cont’d)


Step 4 – Find a common multiple, 6 will work here for the Fe.
6 Fe2+(aq)  6 Fe3+(aq) + 6 eStep 5 a – Add the half-reactions together.
6 Fe2+(aq)  6 Fe3+(aq) + 6 e14 H+(aq) + Cr2O72-(aq) + 6 e-  2 Cr3+(aq) + 7 H2O(l)
_________________________________________________
14 H+(aq) + Cr2O72-(aq) + 6 Fe2+(aq)  6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l)

Step 5 b – Since we didn’t have any spectator ions to add back from earlier, we
are done at this point. Check your balance in atoms and charges.
Balancing by Half-Reactions: (cont’d)
 So in general, we:
 1. Get the reaction in Net Ionic form (if not already). This removes the spectator
ions, which we will add back and balance at the end.
 2. Determine the oxidation number changes, then split the reaction into the two
half reactions (oxidation and reduction).
 3. We balance the oxidation number/charge by the addition of electrons (e-) for
each half reaction.
 4. Add a multiplier to each coefficient to match # of e- in both half reactions.
 5. Add the half reactions together. Add back the spectator ions and balance these
if necessary.
Redox Half-Unit Overview:
 1. Oxidation Numbers for the Elements
 Following the rules to find them
 2. Oxidation & Reduction
 LEO – loss of electrons is Oxidation
 GER – gain of electrons is Reduction
 3. Oxidizing & Reducing Agents
 4. Redox Reactions
 Reactions with oxidation and reduction going on.
 5. Balancing Redox Reactions
 Oxidation Number Method (need to know oxidation #)
 Half Reaction Method (what is a half reaction?)
 Acidic and Basic Conditions – what do you do?
Electrochemistry
THE PRACTICAL APPLICATION OF REDOX CHEMISTRY
Introduction and Voltaic Cells
 Electrochemistry is the practical application of redox chemistry.
 Since electrons are transferred in redox reactions, we should see that we
can generate charge, or the movement of charge (recall that electricity
typically involves the movement of charge, or electrons).
 We are already familiar with the common form of practical
electrochemistry, batteries.
 Let us first start with understanding the basic principles of
electrochemistry.
Voltaic Cells
 We have learned that electrons are transferred in what we call redox
reactions.
 We have found that we can use this property to generate this flow of
electrons to create electrical energy.
 We recall this by remembering half-reactions:


Fe  Fe2+ + 2 eZn2+ + 2 e-  Zn
 We can combine these two reactions:
 Fe(s) + Zn2+(aq)  Fe2+(aq) + Zn(s)
Electrochemical Cells
 In order to capture or measure this flow of electrons, we can’t have the
two solutions together, instead we need to make an electrochemical cell:
Parts of an Electrochemical Cell
 1. Anode – the electrode where the oxidation reaction occurs
 2. Cathode – the electrode where the reduction reaction occurs
 3. Salt Bridge – a pathway that allows the transfer of ions from one cell
to the other. This usually involves either a wet paper strip or a tube of
agar
 4. Voltmeter – measures the potential generated from the combination
of the two cells
Cell Potential
WHAT BATTERY WOULD GIVE THE MOST BANG FOR THE
BUCK?
HOW DOES ONE DETERMINE CELL POTENTIAL?
Cell Potential
 You will be familiar with the battery example mentioned earlier.
This is a useful example of an electrochemical cell.
 We may know that the common types generally have an electrical
potential of 1.50 V (D, C, AA, AAA sizes).
 However, if we need a battery with a specific voltage, how can we
figure out which pair of cells to use?
Standard Reduction Potentials
 In order to determine what the electrical potential for a pair of half-
reaction “cells” is, we need electrochemical data for the half-reactions
we would like to try.
 These have been determined under standard conditions (here, 298 K,
1.0M, and 1.0 atm), so are termed Standard Reduction Potentials
(Eo).
 You have a chart that lists and compares the different reduction values
for different half-cells measured in volts (V).
Using Standard Reduction Potentials
 Here is a version of that table:
 Note that the half-reactions that are stronger
oxidizers (accept electrons) are at the top of the
chart.
 The stronger reducers (donate electrons) are at
the bottom of the chart.
 All of the half-reactions are reversible (sign
changes when reversed)
F2(g) + 2 e-  2 F-(aq)
Eo = +2.87 V
2 F-(aq)  F2(g) + 2 eEo = -2.87 V
Calculating Electrical Potential
 The basic principle behind this calculation is determining the
potential difference between the two half-reactions.
 Potential Difference – this is the measure of the energy of the
electrons that are used by one chemical to reduce another chemical.
This is measured in volts (V).
 This electrical energy is also termed emf (electromotive force).
 In order for a cell potential to be favored or spontaneous, it must be
positive. If it is negative (reaction as written), we must reverse it.
Calculating Electrical Potential (cont’d)
 Deciding which half-reaction is the anode and which reaction is the
cathode is relatively simple:


Anode – is where the electrons are formed (so if we look at the table, it will be the
one lower on the chart, the stronger reducer) so this half-reaction is favored (more
negative V)
Cathode – where the electrons are accepted, so it will be the half-reaction that is
higher on the chart (more positive V).
 To calculate the Cell Potential, we can then sum these two half-
reactions to determine the net reaction and calculate the potential.
Cell Electrical Potential (cont’d)
 Example:
 What is the net final reaction and the cell potential using the following halfreactions?
Hg22+ + 2 e-  2 Hg(l)
Cd2+(aq) + 2 e-  Cd(s)
 Solution: Find the Eo values from the Standard Reduction Table:
Hg22+ + 2 e-  2 Hg(l)
Eo = +0.85 V
Cd2+(aq) + 2 e-  Cd(s)
Eo = -0.40 V
So the cathode will be the first reaction, the anode will be the second (so must be
reversed to form electrons).
Cell Electrical Potential (cont’d)
 So we can write the Net Reaction:
Eo
Hg22+ + 2 e-  2 Hg(l)
+ 0.85 V
Cd(s)  Cd2+(aq) + 2 e+ 0.40 V
______________________________________
So it is favored
+ 1.25 V
Hg22+ + Cd(s)  Cd2+(aq) + 2 Hg(l)
 The Cell Potential Difference can also determined by the following
formula:
𝑜
𝐸𝑐𝑒𝑙𝑙
𝑜
𝑜
𝑜
𝐸𝑐𝑒𝑙𝑙
= 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒
− 𝐸𝑎𝑛𝑜𝑑𝑒
= (+0.85 V) – (-0.40 V) = + 1.25 V
Cell Electrical Potential (cont’d)
 Example:
 What is the net final reaction and the cell potential using the following halfreactions?
Fe2+(aq) + 2 e-  Fe(s)
Au3+(aq) + 3 e-  Au(s)

Solution: Find the Eo values from the Standard Reduction Table:
Fe2+(aq) + 2 e-  Fe(s)
Eo = -0.44 V
Au3+(aq) + 3 e-  Au(s)
Eo = +1.50 V
So the cathode will be the second reaction, the anode will be the first (so that one must
be reversed in order to form electrons).
Cell Electrical Potential (cont’d)
 So we can write the Net Reaction:
6 e- is the common multiple
Eo
(Fe(s)  Fe2+(aq) + 2 e- ) x 3
+0.44 V
(Au3+(aq) + 3 e-  Au(s) ) x 2
+1.50 V
______________________________________
3 Fe(s) + 2 Au3+(aq)  3 Fe2+(aq) + 2 Au(s) + 1.94 V
 Using the Cell Potential formula again:
𝑜
𝐸𝑐𝑒𝑙𝑙
𝑜
𝑜
𝑜
𝐸𝑐𝑒𝑙𝑙
= 𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒
− 𝐸𝑎𝑛𝑜𝑑𝑒
= (+1.50 V) – (-0.44 V) = + 1.94 V
The potential values do
NOT change, as the energy
of the electrons actually
don’t change, not matter
how many we need.
Cell Notation
SIMPLIFIED WAY TO DESCRIBE AN ELECTROCHEMICAL CELL
Cell Notation
 A standard way of writing electrochemical cells is by using what is
called cell notation.
 Cell Notation is written in the form of the anode to cathode direction:
Represent phase boundaries
Anode metal | anode ion (1 M) || cathode ion (1 M) | cathode metal
 If we use our previous example:
Represents salt bridge
Cd(s) | Cd2+ (1 M) || Hg22+ (1 M) | Hg(l)
Cell Notation (cont’d)
 There are the odd occasions where H2 gas is involved in the reaction.




The notation changes a little bit:
First of all, we have the ionic form of hydrogen, H+ (so it is an acid and
also at 1 M).
Next, H2 gas must be at 1 atm.
Note the
addition of
The electrode that is used for H2 is Pt metal.
the Pt here
Example:

Cell Notation for a Zn/H electrode:
Zn(s) | Zn2+ (1 M) || H+ (1 M) | H2(g) (1 atm) | Pt(s)
Hydrogen Electrode in Electrochemical Cells
 Here are two diagrams showing how a hydrogen electrode actually
works:
Practical Cells
APPLYING THEORY TO PRACTICE: BATTERIES
Batteries
 Now that we have a basis for electrochemistry, let’s use it in the form of
the common household battery.
 Batteries basically have multiple “cells” in series that allow it to
generate voltages that are higher than is possible for the individual
cells:
Primary and Secondary Batteries
 There are two kinds of batteries, based on whether we can
regenerate the battery (i.e. reuse, or recharge them):


1. Primary Battery – this kind of battery is the classical use up and throw away
style. Once the battery’s charge (energy) is depleted, it has to be discarded.
Basically the redox reaction involved cannot be reversed.
2. Secondary Battery – this kind of battery is becoming more common. These
kind of batteries involve redox reactions that can be reversed, so we can
recharge the battery.
 Under these kinds of batteries, there are several types:
Types of Batteries
 There several types of batteries. They are differentiated by the
nature of the “electrodes” and the medium which allows the
movement of the electrons (called the “electrolyte”):
 1. Dry Cell



– is the common form of a battery where the electrolyte is a moist paste.
Examples of electrodes used in this type include:
a. Zinc-Carbon – the case is Zn (anode) while the C rod electrode (cathode) is
found at the center. The electrolyte paste is NH4Cl.
b. Zinc-Steel – the case is steel (Fe and MnO2 mix), while the center rod is Zn.
The electrolyte is KOH
Types of Batteries: (cont’d)
 2. Lead-Acid Storage Battery

– in this form, the electrolyte is a liquid. The common form would be a car battery
where the acid is H2SO4 (sulfuric acid).
 3. Lithium Battery

– these are newer batteries that are smaller and have a higher capacity (more voltage).
Types of Batteries: (cont’d)
 4. Fuel Cell
 – this type of battery is known as the hydrogen fuel cell. Being used as power sources
for vehicles.
Corrosion
RUSTING AS AN APPLICATION OF REDOX
The Nasty Side of Redox - Corrosion
 Moving away from a practical use of redox, we have the concept of
corrosion.
 We best understand corrosion by using the example of “rusting”:
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
 We can easily see that it is, in fact, a redox reaction.
 The Half-Reactions involved are and require 2 steps:



Fe2+ + 2 e-  Fe
O2 + 4 H+ + 4 e-  2 H2O
Fe3+ + e-  Fe2+
Eo = -0.44 V
Eo = +0.68 V
Eo = +0.77 V
How Can We Protect Our Fe?
 Steel is a very common material in our society. It is made up of Fe and C
(most commonly used, from 0.2% to 2.1% C).
 Still, Fe is susceptible to being oxidized very easily.
 A process was developed to protect this material from the environment
(rain-catalyzed oxidation) by the addition of a thin layer of Zn metal to
the surface exposed. This process is called “galvanization”.

Zn does oxidize as well, however, instead of the Zn oxide (ZnO) coming off the steel
surface, it actually binds tightly to it.
Other Forms of Processed Steel

Galvanized steel looks like this:
 There are other common forms of steel, for example
Stainless steel.


This form of steel is an alloy of Fe and Cr (between 10.5% to
11.0%)
This form is resistant to oxidation (but it will actually oxidize
under certain conditions).
Electrolysis
ANOTHER APPLICATION OF REDOX
Electrolysis
 Electrolysis – is the process of adding electrons (i.e. using electricity) to
a material in order to convert it into its component elements.
 We can use this process to perform this reaction:
2 H2O  2 H2 + O2
 A diagram is shown here:



At the anode, H2 gas forms,
At the cathode, O2 gas forms
This is called the Hoffman apparatus
Electrochemistry Half-Unit Review
 1. Half-Reactions (review), Half-Cells
 2. Electrochemical Cells, Voltaic Cells
 Electrodes: Anode, Cathode
 Salt Bridge
 3. Electrical Potential
 Potential Difference - Voltage
 4. Cell Potential
 Standard Reduction Potentials (Standard Conditions)
 Spontaneous or Favored Reactions
 Cell Potential Calculation
 Cell Notation
Electrochemistry Review (cont’d)
 5. Corrosion
 Fundamental basis of corrosion
 Galvanized Steel
 6. Electrolysis
 Process of electrolysis
 Application