Learning Goals
• Today we are learning to:
1. Use terms and definitions related to
electrochemistry correctly
2. Explain redox reactions in terms of the loss
and gain of electrons and a change in
oxidation number
Unit 5: Electrochemistry
10.1: Oxidation and
Reduction
Oxidation & Reduction
Oxidation: loss of electrons
4Fe (aq)
3+
→ 4Fe
(aq)
+ 12e
Reduction: gain of electrons
3O2(aq) + 12 e- → 6O2- (g)
In redox equations, the number of atoms
and the CHARGE must balance!
Oxidation / Reduction
LEO goes GER !!
Loss of Electrons is Oxidation
Gain of Electrons is Reduction
OIL RIG
Oxidation Is Loss of electrons
Reduction Is Gain of electrons
Redox Reactions
2+
Zn
Zn(s) + I2(aq)
(aq)
+2I
(aq)
Zinc lost 2e and is oxidized:
Zn(s)
2+
Zn
+
2e
(aq)
Iodine gained 2e and is reduced:
I2(aq)
+ 2e
2I
(aq)
• Splitting a Full Equation into HALF
REACTIONS (as seen above) clearly
shows the oxidation/reduction
Oxidizing/Reducing Agents
A reactant that oxidizes another
reactant is called an oxidizing agent.
A reactant that reduces another
reactant is called a reducing agent.
Zinc reduced the iodine so it is called a
reducing agent.
Iodine oxidized the zinc, so it is called
an oxidizing agent.
Oxidizing/Reducing Agents
Zn(s) + I2(aq)
- reducing agent
- donates electrons
- undergoes oxidation
Zn2+(aq) + 2 I- (aq)
- oxidizing agent
- accepts electrons
- undergoes reduction
Oxidizing Agents
Gain of electrons to complete valence shells is typical
of non-metals.
Non-metals are usually OXIDIZING AGENTS.
Reducing Agents
Loss of electrons to form empty valence shells is
typical of metals.
metals behave as REDUCING AGENTS.
Oxidation number
(Oxidation State)
Oxidation number is the charge an atom would
have if it was ionically bonding.
Ex. In a water molecule the oxidation state of
hydrogen is +1 and oxygen is -2.
Rules for Oxidation States
The total oxidation state of the atoms in a
polyatomic ion are equal to the charge of the
polyatomic ion.
The total oxidation number of a neutral
molecule is zero.
Rules for Oxidation States
1. All elements in elemental state are zero
2. Metals of group 1 and 2 and polyatomic ions get their
desired charge.
3. Fluorine is -1 (no exceptions!)
4. Hydrogen is +1 (except when bonded to group 1 or 2)
5. Oxygen is -2 (some exceptions when bonded to F or H)
Rules for Oxidation States
6. The halogens (Cl, Br and I) are -1 (some exceptions
when bonded to F, H or O)
7. In general: In covalent compounds, the more
electronegative element is assigned its oxidation state
(the negative charge it usually has in its ionic
compounds)
Determining Oxidation Numbers
1. Use the rules to assign the oxidation numbers
2a. The sum of the oxidation number for a compound
is zero
2b. The sum of the oxidation numbers for a
polyatomic ion equals the charge on the ion
3. Any unknown oxidation number is determined
algebraically from the sum of the known
oxidation numbers and the net charge on the
entity.
Examples!!
Assign oxidation numbers to each element
CCl4
HBrO3
Cr2O72-
Assessment As Learning #15
• Answer the following questions to the best of your
ability and give yourself a mark out of 7 based on
your number of correct answers.
• This mark does not count towards your average, but
gives you feedback as to how well you are
understanding the concepts.
The oxidation number of:
Hydrogen in H2 is
Iron in Fe2+ is
Carbon in CH4 is
Sulfur in H2SO4 is
Phosphorus in PCl4+ is
Iodine in IO4- is
Carbon in CN- is
H.W. pg 606 #15-24
The oxidation number of:
Hydrogen in H2 is
Iron in Fe2+ is
0
+2
Carbon in CH4 is
-4
Sulfur in H2SO4 is
+6
Phosphorus in PCl4+ is
+5
Iodine in IO4- is
+7
Carbon in CN- is
+2
REDOX reactions with Oxidation
Numbers
Oxidation is an increase in oxidation number.
Reduction is a decrease in oxidation number.
Ex. 2MnO4 (aq) + 5 SO2(aq) + 2 H2O (l) →
2+
2+
2 Mn (aq) + 5 SO4 (aq) + 4 H (aq)
Determine which elements have been oxidized
and which reduced.
To do this, the oxidation numbers of all elements
must be calculated!
In the above reaction what is reduced and what is
oxidized?
Answer
Mn goes from +7 to +2
It has gained 5 e- so it is reduced.
Sulfur goes from +4 to +6
It lost 2 e- so it is oxidized.
Oxygen and hydrogen do not change their
oxidation numbers!
Disproportionation Redox reactions
Occurs when the same species is
both oxidized and reduced.
Ex.
+
2 Cu
(aq)
2+
→ Cu
(aq) + Cu (s)
Balancing Redox Equations
Using Oxidation Numbers
Example
Balance the following redox equation using the oxidation
number method.
HNO3(aq) + H3AsO3(aq) --> NO(g) + H3AsO4(aq) + H2O(l)
1. Check if it is a redox reaction
The N atoms change from +5 to +2, so they are reduced. The
As atoms, which change from +3 to +5, are oxidized.
Balancing Redox Equations Using
Oxidation Numbers
2. 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.
As +3 to +5
Net Change = +2
N +5 to +2
Net Change = -3
Balancing Redox Equations Using
Oxidation Numbers
3. 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. (Charge needs to balance)
We need 3 As atoms for every 2 N atoms.
Balancing Redox Equations Using
Oxidation Numbers
4. Use the ratio identified in Step 3, and add coefficients to the
formulas which contain the elements whose oxidation number
is changing.
2HNO3(aq) + 3H3AsO3(aq) --> 2NO(g) + 3H3AsO4(aq) + H2O(l)
5. Check to see if other atoms are balanced. If they are not,
balance them by inspection.
2HNO3(aq) + 3H3AsO3(aq) -> 2NO(g) + 3H3AsO4(aq) + H2O(l)
Example
Balance the following redox equation using the oxidation
number method.
Cu(s) + HNO3(aq) --> Cu(NO3)2(aq) + NO(g) + H2O(l)
1. Check if it is a redox reaction
The copper atoms are changing their oxidation number from 0
to +2, and some of the nitrogen atoms are changing from +5 to
+2.
Balancing Redox Equations Using
Oxidation Numbers
2. 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.
Cu 0 to +2
N +5 to +2
Net Change = +2
Net Change = -3
Balancing Redox Equations Using
Oxidation Numbers
3. 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.
We need 3 Cu atoms for every 2 nitrogen atoms
Balancing Redox Equations Using
Oxidation Numbers
4. Use the ratio identified in Step 3, and add coefficients to the
formulas which contain the elements whose oxidation number
is changing.
3Cu(s) + 2HNO3(aq) --> 3Cu(NO3)2(aq) + 2NO(g) + H2O(l)
5. Balance the rest of the equation by inspection.
3Cu(s) + 8HNO3(aq) --> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
Balancing Redox Equations
Using Oxidation Numbers
In Acid or Base
•
If you are told to balance a redox reaction in acid or base the
following additional steps are necessary:
6. If the reaction takes place in acid or base, balance the H’s and
O’s last.
7. In acid, balance the oxygen atoms by adding H2O to the side
that is short oxygen. Balance the hydrogen's by adding H+, to
the side that is short hydrogen.
Balancing Redox Equations
Using Oxidation Numbers
In Acid or Base
8. In base complete step 7, then add the same number of OH- (to
each side of the equation) as there are H+. If there are H+ and
OH- on the same side, combine them to make water.
9. Cancel out things that are the same on the left and right side
of the equation.
Balance the following redox reactions in acidic solution using
the oxidation number method.
a. Al(s) + MnO2(s) --> Al2O3(s) + Mn(s)
b. SO2(g) + HNO2(aq) --> H2SO4(aq) + NO(g)
Balance the following redox reactions in acidic solution using
the oxidation number method.
a. 4Al + 3MnO2 → 2Al2O3 + 3Mn
b. SO2 + 2HNO2 → H2SO4 + 2NO
Balance the following redox reactions in acidic solution using
the oxidation number method.
•
c. HNO3 (aq) + H2S (aq) → NO(g) + S(s) + H2O(l)
•
d. Al(s) + H2SO4 (aq) → Al2(SO4)3 (aq) + H2(g)
Balance the following redox reactions in acidic solution using
the oxidation number method.
•
c. 2HNO3 + 3H2S → 2NO + 3S + 4H2O
•
d. 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Balance the following redox reactions in basic solution using
the oxidation number method.
•
e. Cl2 (aq) → Cl- (aq) + ClO3- (aq)
•
f. I- (aq) + Cr2O72- (aq) → I2 (s) + Cr3+ (aq)
Balance the following redox reactions in basic solution using
the oxidation number method.
•
e. 3Cl2 + 6OH- → 5Cl- + ClO3- + 3H2O
•
f. 6I- + Cr2O72- + 7H2O → 3I2 + 2Cr3+ + 14OH-
Full solution to e.
Balance the following redox equation using the oxidation number
method.
e. Cl2 (aq) → Cl- (aq) + ClO3- (aq)
1. Check if it is a redox reaction
Some Cl atoms are changing their oxidation number from 0 to -1,
and some Cl atoms are changing their oxidation number from 0
to +5.
Full solution to e.
2. 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.
Cl 0 to -1
Cl 0 to +5
Net Change = -1
Net Change = +5
Full solution to e.
3. 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.
We need 1 Cu+5 ion for every 5 Cu-1 ions.
Full solution to e.
4. Use the ratio identified in Step 3, and add coefficients to the
formulas which contain the elements whose oxidation number
is changing. (notice the 5 to 1 ratio on the right side of the
equation)
e. Cl2 (aq) → 5Cl- (aq) + 1ClO3- (aq)
5. Now balance the Cl atoms on the left by inspection.
e. 3Cl2 (aq) → 5Cl- (aq) + 1ClO3- (aq)
Full solution to e.
6. Now balance the O atoms by adding water.
e. 3H2O + 3Cl2 (aq) → 5Cl- (aq) + ClO3- (aq)
7. Now balance the H atoms by adding H+.
e. 3H2O + 3Cl2 (aq) → 5Cl- (aq) + ClO3- (aq) + 6H+
8. Now add 6 OH- to the left and right side of the equation.
e. 6OH- + 3H2O + 3Cl2 (aq) → 5Cl- (aq) + ClO3- (aq) + 6H+ +
6OH-
Full solution to e.
9. Combine the 6OH- with 6H+ (on the right side) to make 6H2O.
e. 6OH- + 3H2O + 3Cl2 (aq) → 5Cl- (aq) + ClO3- (aq) + 6H2O
10. Cancel 3H2O from the left and right side to get the final
answer!
e. 6OH- + 3Cl2 (aq) → 5Cl- (aq) + ClO3- (aq) + 3H2O
Success Criteria for Learning
Goals
• I know I was successful because:
Learning Goals
• Today we are learning to:
1. Balance redox reactions using the half
reaction method
Balancing Half Reactions
Using Half Reaction Method
Example
Magnesium metal displaces aluminum from an
aqueous solution of aluminum nitrate, Al(NO3)3.
Mg(s) + Al(NO3)3(aq)→ Al(s) + Mg(NO3)2
Write the balanced redox equation for this
reaction.
Step 1: Write the 2 half-reactions.
Mg → Mg2+ + 2eAl3+ + 3e- → Al
Step 2: Balance the atoms in each half-reaction
equation. (done)
Step 3: Multiply each half-reaction equation by simple
whole numbers to balance the electrons lost and gained
3Mg → 3Mg2+ + 6e2Al3+ + 6e- → 2Al
Step 4: Add the two half-reactions equations canceling
the electrons and anything else that is EXACTLY the
same on both sides of the equation. Put the spectator
ions back in
3Mg + 2Al3+ + 6e- → 2Al + 3Mg2+ + 6e3Mg(s) + 2Al(NO3)3(aq)→ 2Al(s) + 3Mg(NO3)2
Balancing a Redox Reaction in
Acidic Solution (Using Half Rxns)
Write the balanced redox equation for this reaction under
acidic conditions.
Cr2O72-(aq) + HNO2(aq) --> Cr3+(aq) + NO3-(aq)
Step 1: Separate the chemical equation into the start of two
half-reaction equations
Cr2O72- --> Cr3+
HNO2 --> NO3Step 2: Balance atoms, other than O and H.
Cr2O72- --> 2Cr3+
HNO2 --> NO3-
Step 3: Balance O by adding H2O
Cr2O72- --> 2Cr3+ + 7H2O
HNO2 + H2O --> NO3-
Step 4: Balance H by adding H+.
Cr2O72- + 14H+ --> 2Cr3+ + 7H2O
HNO2 + H2O --> NO3- + 3H+
•
Step 5: Balance the charge on each side by adding eand cancel anything that is the same on both sides
6e- + Cr2O72- + 14H+ --> 2Cr3+ + 7H2O
HNO2 + H2O --> NO3- + 3H+ + 2e-
Step 6: Multiply each half-reaction equation by simple whole
numbers to balance the electrons lost and gained
6e- + Cr2O72- + 14H+ --> 2Cr3+ + 7H2O
3HNO2 + 3H2O --> 3NO3- + 9H+ + 6e-
Step 7: Add the half reactions
6e- + Cr2O72- + 3HNO2 + 3H2O + 14H+ --> 6e- + 2Cr3+ + 3NO3- +
7H2O + 9H+
Step 8: Cancel things that are the same
Cr2O72- + 3HNO2 + 5H+ --> 2Cr3+ + 3NO3- + 4H2O
Balancing a Redox Reaction in
Acidic Solution (Using Half Rxns)
•
ClO4- + NO2 → Cl- + NO3-
Step 1: Separate the chemical equation into the start of two
half-reaction equations
•
NO2 → NO3-
•
ClO4- → Cl-
Step 2: Balance atoms, other than O and H.
(done)
Step 3: Balance O by adding H2O
•
NO2 + H2O → NO3-
•
ClO4- → Cl- + 4H2O
Step 4: Balance H by adding H+.
•
NO2 + H2O → NO3- + 2H+
•
ClO4- + 8H+ → Cl- + 4H2O
Step 5: Balance the charge on each side by adding e•
NO2 + H2O → NO3- + 2H+ + e-
•
ClO4- + 8H+ + 8e- → Cl- + 4H2O
Step 6: multiply top rxn by 8, to balance the electrons
•
8NO2 + 8H20 → 8NO3- + 16H+ + 8e-
•
ClO4- + 8H+ + 8e- → Cl- + 4H2O
Step 7: Add the half reactions
•
8NO2 + 8H2O + ClO4- + 8H+ + 8e- → 8NO3- + 16H+ + 8e- + Cl- + 4H2O
Step 8: Cancel things that are the same
•
8NO2 + 4H2O + ClO4- → 8NO3- + 8H+ + Cl-
H.W. pg 615 #39-41 (use half reaction method)
Assessment As Learning #17
• Answer the following questions to the best of your
ability and give yourself a mark out of 10 based on
your number of correct answers.
• This mark does not count towards your average, but
gives you feedback as to how well you are
understanding the concepts.
Balance the following redox reactions in acidic solution using
the half reaction method.
•
a. MnO4- (aq) + Br- (aq) --> MnO2 (s) + BrO3- (aq)
•
b. I2 (s) + OCl- (aq) --> IO3- (aq) + Cl- (aq)
•
c. Cr2O72- (aq) + C2O42-(aq) --> Cr3+(aq) + CO2 (g)
•
d. Mn (s) + HNO3 (aq) --> Mn2+(aq) + NO2 (g)
Balance the following redox reactions in acidic solution using
the half reaction method.
•
a. 2MnO4- + Br- + 2H+ → 2MnO2 + BrO3- + H2O
•
b. I2 + 5OCl- + H2O → 2IO3- + 5Cl- + 2H+
•
c. Cr2O72- + 3C2O42- + 14H+ → 2Cr3+ + 6CO2 + 7H2O
•
d. Mn + 2HNO3 + 2H+ → Mn2+ + 2NO2 + 2H2O
Balancing a Redox Reaction in
Basic Solution (Using Half Rxns)
•
Cr(OH)3(s) + ClO3-( aq) --> CrO42-(aq) + Cl-(aq)
Step 1: Separate the chemical equation into the start of two
half-reaction equations
Cr(OH)3 --> CrO42ClO3- --> ClStep 2: Balance atoms, other than O and H.
(done)
Step 3: Balance O by adding H2O
•
Cr(OH)3 + H2O --> CrO42-
•
ClO3- --> Cl- + 3H2O
Step 4: Balance H by adding H+.
•
Cr(OH)3 + H2O --> CrO42- + 5H+
•
ClO3- + 6H+ --> Cl- + 3H2O
Step 5: Balance the charge on each side by adding e- and
cancel anything that is the same on both sides
•
Cr(OH)3 + H2O --> CrO42- + 5H+ + 3e-
•
ClO3- + 6H+ + 6e- --> Cl- + 3H2O
Step 6: Multiply by a whole number to get the electrons to
balance. (multiply eqn 1 by 2)
•
2Cr(OH)3 + 2H2O --> 2CrO42- + 10H+ + 6e-
•
ClO3- + 6H+ + 6e- --> Cl- + 3H2O
Step 7 and 8: Add the half reactions and cancel the things
that are the same.
2Cr(OH)3(s) + ClO3-(aq) --> 2CrO42-(aq)
4H+(aq)
+ Cl-(aq)
+ H2O(l) +
Step 9: Because there are 4 H+ on the right side of our
equation above, we add 4 OH- to each side of the
equation.
2Cr(OH)3 + ClO3- + 4OH- --> 2CrO42- + Cl- + H2O + 4H+ + 4OH-
Step 10: Combine the 4 H+ ions and the 4 OH- ions on the
right of the equation to form 4 H2O. Then cancel or
combine the H2O molecules.
•
2Cr(OH)3 + ClO3- + 4OH- --> 2CrO42- + Cl- + H2O + 4H2O
•
2Cr(OH)3(s) + ClO3- + 4OH-(aq) --> 2CrO42-(aq) + Cl-(aq) + 5H2O(l)
H.W. pg 615 #42-44 (use half reaction method)
Balance the following redox reactions in a basic solution using
the half reaction method.
•
a. CrO42-(aq) + S2-(aq) --> Cr(OH)3(s) + S(s)
•
b. MnO4-(aq) + I-(aq) --> MnO2(s) + IO3-(aq)
•
c. H2O2(aq) + ClO4-(aq) --> O2(g) + ClO2-(aq)
•
d. S2-(aq) + I2(s) --> SO42-(aq) + I-(aq)
Balance the following redox reactions in a basic solution using
the half reaction method.
•
a. 2CrO42- + 3S2- + 8H2O → 2Cr(OH)3 + 3S + 10OH-
•
b. 2MnO4- + I- + H2O → 2MnO2 + IO3- + 2OH-
•
c. 2H2O2 + ClO4- → 2O2 + ClO2- + 2H2O
•
d. S2- + 4I2 + 8OH- → SO42- + 8I- + 4H2O
Success Criteria for Learning
Goals
• I know I was successful because:
Learning Goals
• Today we are learning to:
1. Draw fully labelled galvanic cells and explain
how they work
2. Identify the components of a galvanic cell
and the function of each component
11.1: Technology of Cells and
Batteries
Cells and Batteries
• Definitions:
•
Electric current is the flow of charge in a circuit.
•
Electrochemistry is the study of the processes involved
in converting chemical energy to electrical energy.
•
Cell Potential is a measure of the driving force behind
an electrochemical reaction (in volts). The potential of
an electrochemical cell measures how far an oxidationreduction reaction is from equilibrium! What if cell
potential = 0?
Definitions
•
Electric Cell: a device that continuously converts
chemical energy into electrical energy.
•
Battery: group of two or more electric cells connected in
series.
Basic Cell Design and Properties
• Each cell is composed of two electrodes and one
electrolyte.
• Electrode: a solid electrical conductor.
• Electrolyte: an aqueous electrical conductor.
Galvanic Cell
(aka Voltaic Cell)
Device used to convert chemical
energy into electrical energy.
Uses a spontaneous redox reaction as a source of
electrical energy.
Consist of two half-cells separated by a porous
boundary with solid electrodes connected by an
external circuit. Basis of commercial batteries.
*Remember*
RED CAT & AN OX
• REDuction occurs at the CAThode (notice
that both reduction and cathode start with
a consonant)
• OXidation occurs at the ANode (notice
that both oxidation and anode start with a
vowel)
Galvanic Cells
Copper is gaining electrons and is the CATHODE
Zinc is losing electrons and is the ANODE
Electrons travel in the external circuit from the
anode to the cathode.
Salt Bridge
H.W. pg 641 #1-5
• The salt bridge is often a piece of filter paper
dipped in a solution of a strong electrolyte, such a
potassium nitrate.
• anions move towards the anode and cations move
toward the cathode as the cell operates.
• The salt bridge maintains electrical neutrality.
Inert Electrode
• When a half reaction does not involve a solid
metallic surface (ex. for gaseous or aqueous half
reactions), an inert electrode must be used.
• It provides a solid conductive surface that will
allow the given half reaction to occur, but itself
does not participate in the reaction.
• Usually, platinum is used
since it is unreactive.
How Does It Work?
A metal in contact with an aqueous solution of its
ions will establish a redox equilibrium and acquire
an electrical charge.
M (s) → Mn+(aq) + n eThe greater the difference in reactivity of the
metals, the greater the cell potential.
A Galvanic cell always has a positive cell potential
which means the reaction is spontaneous!
Anode & Cathode
ANODE: negative electrode
The strongest reducing agent undergoes
oxidation to produce the electrons.
The more reactive metal will ALWAYS be the
NEGATIVE electrode.
CATHODE: positive electrode
The strongest oxidizing agent undergoes
reduction to consume the electrons.
Cell Notation
Development of a Redox Table
When a species gives a particular product as a
result of gaining/losing electrons, the pair is called
a REDOX COUPLE.
Ex. Zn2+ / Zn
If one species is a powerful oxidizing agent, then
the other will be a very weak reducing agent
(similar to conjugate acid-base pairs).
Standard Reduction Potentials Table
• A ranking of relative reactivity’s.
• Large positive numbers strong oxidizing agents.
• Large negative numbers weak oxidizing agents.
• Spontaneity Rule: can use a redox table to predict
whether or not a reaction is spontaneous.
• If the OA is higher than the RA , spontaneous rxn.
• If the OA is lower than the RA, nonspontaneous rxn.
•
The half reaction which is higher in the table will be the
reduction, occurring at the cathode.
•
The half reaction which is lower in the table will be the
oxidation, occurring at the anode.
Standard Cells and Cell Potentials
•
Standard Cell: A galvanic cell in which each half-cell
contains all entities shown in the half-reaction equations
at SATP conditions, with a concentration of 1.0 M for all
aqueous solutions.
•
Standard cell potential (∆Eo ) is the energy difference
between the cathode and the anode.
•
Standard reduction potential (Er o ) represents the ability
of a standard half-cell to attract electrons, thus
undergoing a reduction.
•
o
∆E
=
cell
o
E
cathode
o
E
anode
Calculating a Standard Cell Potential
•
Calculate the standard cell potential for the galvanic
cell in which the following reaction occurs.
-
2I (aq) + Br2(l)
-
I2(s) + 2Br (aq)
1. Write the oxidation and reduction half reactions.
Br2 + 2e-  2Br2I-  I2 + 2e2. Locate reduction potentials in the table (SLIDE 79).
Br2 + 2e-  2Br-
Eo = 1.09 V
2I-  I2 + 2e-
Eo = 0.54 V
Calculating a Standard Cell Potential
-
-
2I (aq) + Br2(l)
I2(s) + 2Br (aq)
3. Subtract the reduction potential of the cathode
and the reduction potential of the anode.
o
∆E
cell
=
o
∆E
cell
=
o
E
cathode
-
o
E
1.09 – 0.54 = 0.55 V
anode
•
2Au(NO3)3 (aq) + 3Co (s) → 3Co(NO3)2 (aq) + 2Au (s)
•
Calculate the standard cell potential for the galvanic
cell in which the above reaction occurs.
•
Given: Co2+ (aq) + 2e-  Co (s) Eo = -0.28V
•
Given: Au3+ (aq) + 3e-  Au (s) Eo = 1.50V
Predicting Spontaneity
• If the cell potential is positive, it is a spontaneous
cell (galvanic).
• If the cell potential is negative, it is a nonspontaneous cell (electrolytic).
Predicting Spontaneity
• Predict whether the following reaction is
spontaneous or non-spontaneous under standard
conditions.
2+
Cd(s) + Cu (aq)
2+
Cd (aq) + Cu(s)
• The 2 half reactions are:
•
2+
2e + Cu (aq)
Reduction (at cathode)
• Oxidation (at anode)
-
Cd(s)
o
Cu(s) E = 0.342 V
2+
Cd (aq) + 2e
o
o
o
E cell = E cathode - E anode
= 0.342 V - (-0.403) V
= 0.745 V
o
E = -0.403 V
• The above reaction had a positive cell potential,
so the reaction is spontaneous under standard
conditions.
• Predict whether the following reaction is
spontaneous -
I2(s) + 2Cl (aq)
2I (aq) + Cl2(g)
• Answer = -0.822 V
• The standard cell potential is negative, so the
reaction is non-spontaneous
Assessment As Learning #18
• Answer the following questions to the best of your
ability and give yourself a mark out of 3 based on
your number of correct answers.
• This mark does not count towards your average, but
gives you feedback as to how well you are
understanding the concepts.
Example
Calculate the standard cell potential for the
galvanic cell in which the following reaction
occurs.
3Cu(NO3)2 (aq) + 2Al (s) → 2Al(NO3)3 (aq) + 3Cu (s)
∆Eocell =
Eocathode
∆Eocell = 0.342 – (-1.66)
∆Eocell = 2.0 V
-
Eoanode
Example
Calculate the standard cell potential for the
electrochemical cell in which the following
reaction occurs.
•
2Ag(s) + Pb2+(aq) → Pb(s) + 2Ag+(aq)
•
2Ag(s) + Pb2+(aq) → Pb(s) + 2Ag+(aq)
∆Eocell =
Eocathode
∆Eocell = – 0.1262 – 0.800
∆Eocell =
–0.926 V
-
Eoanode
Example
Calculate the standard cell potential for the
electrochemical cell in which the following
reaction occurs.
•
2Ag(s) + Pb2+(aq) → Pb(s) + 2Ag+(aq)
•
Is this a spontaneous cell?
•
2Ag(s) + Pb2+(aq) → Pb(s) + 2Ag+(aq)
∆Eocell =
Eocathode
-
Eoanode
∆Eocell = – 0.1262 – 0.800
∆Eocell =
–0.926 V
• This is not a spontaneous cell, since the ∆Eocell is
negative!
Electrolytic Cells
• Electrolytic Cell: is a device
that converts electrical energy
to chemical energy
• Electrolysis – is the process that takes place in an
electrolytic cell
• Same set up as a galvanic cell, but they require an
external source of electricity (external voltage)
Electrolytic Cells (differences)
•
Opposite reaction occurs in electrolytic cell
•
Cell potential is negative
•
Electrical energy is converted to chemical energy
•
Anode is now (+) and cathode is (-)
Electrolytic Cells (similarities)
•
There are two electrodes. Reduction still occurs at
cathode and oxidation still occurs at anode
•
Electrons still flow from anode to cathode
•
There is still a salt bridge in which the negative ions flow
towards anode and positive ions flow towards cathode
Summary
Summary
Electrolysis of molten salts
• NaCl can be separated into sodium metal and
chlorine gas in an electrolytic cell.
• The external source of electricity forces electrons
onto one electrode. This electrode becomes
negatively charged and attracts the Na+ ions
where they gain an electron and become sodium
metal
• The –ve chloride ions move towards the positive
electrode where they lose electrons and become
chlorine gas.
Electrolytic Cells
• As in a galvanic cell, reduction occurs at the
cathode, and oxidation occurs at the anode. The
half reactions are as follows
+
• Reduction (at cathode) Na + e
• Oxidation (at anode)
-
2Cl (aq)
-
Na(l)
Cl2(g) + 2e
• Answer = -4.069 V
The standard cell potential is negative, so the
reaction is non-spontaneous
• H.W. pg 664 #19-22
-
Electrolysis of Water
• Water can be electrolyzed to give oxygen gas and
hydrogen gas.
-
• Reduction (at cathode) 2H2O(l) + 2e
• Oxidation (at anode)
2H2O(l)
-
H2(g) + 2OH
+
-
O2(g) + 4H + 4e
Success Criteria for Learning
Goals
•
I know I was successful because:
Learning Goals
• Today we are learning to:
1. Explain how the hydrogen half cell is used
as a standard reference to determine
voltages of other half cells
Electrolysis of Water
• Cell potentials are as follows: (DO NOT use the
cell potentials on pg 748 they are for a 1 mol/L
solution of water. Pure water is 55 mol/L)
-
-
2H2O(l) + 2e
2H2O(l)
• ∆Ecell =
H2(g) + 2OH
+
-
E= -0.414V
O2(g) + 4H + 4e
E= 0.815 V
Ecathode
Eanode
• ∆Ecell = -0.414 – 0.815
• ∆Ecell = -1.229
-
Electrolysis of Water
• The negative cell potential shows that the
reaction is not spontaneous.
• Electrolytic cells are used for
non-spontaneous redox reactions, so all
electrolytic cells have negative cell potentials.
Predicting the Products of
Electrolysis
for an aqueous solution
• You must examine the half reactions and
their cell potentials
• Find the overall reaction that requires the
lowest external voltage. (ie. The one with a
negative cell potential, closest to zero)
Example
• Predict the products of electrolysis of 1 mol/L
LiBr(aq)
1. List the 4 relevant half-reactions and their
reductions potentials from the table (pg 748)
Br2(l) + 2e
-
2Br
+
O2(g) + 4H + 4e
2H2O(l) + 2e
+
Li (aq) + e
-
-
-
-
o
E = 1.066 V
2H2O(l) Eo = 0.815 V
-
H2(g) + 2OH (aq) Eo = -0.414 V
o
Li(s) E = -3.040 V
Example
• There are 2 possible oxidation half-reactions at
the anode (appear higher in the table on 748
opposite to the galvanic cell)
2Br
-
Br2(l) + 2e
+
O2(g) + 4H + 4e
2H2O(l)
o
E = 1.066
-
o
E = 0.815
• There are 2 possible reduction half-reactions at
the cathode (appear lower in the table on 748)
-
2H2O(l) + 2e
+
Li (aq) + e
H2(g) + 2OH (aq)
Li(s)
o
E = -0.414
o
E = -3.040
Example
2. Predict the products by finding the cell
reaction that requires the lowest external
voltage.
+
2 Li (aq) + 2Br
2Li(s) + Br2(l)
o
o
o
E cell = E cathode - E anode
= -3.040 V - 1.066 V
= -4.106 V
Cont.
2H2O(l)
O2(g) + 2H2(g)
Ecell = Ecathode - Eanode
= -0.414 V - 0.815V
= -1.229V
+
4Li (aq) + 2H2O(l)
o
o
o
E cell = E cathode - E anode
= -3.040 V - 0.815V
= -3.855V
+
4Li(s) + O2 + 4H
Cont.
-
2Br (aq) + 2H2O(l)
-
Br2(l) + H2 + 2OH
o
o
o
E cell = E cathode - E anode
= -0.414 V - 1.066 V
= -1.480 V
• The electrolysis of water requires the lowest
external voltage. Therefore, the predicted
products of this electrolysis are hydrogen and
oxygen.
Predicting Redox Reactions in
Solution
1. List all entities present and classify each as an OA, RA or both.
Do not label spectator ions.
2. Choose the strongest oxidizing agent as indicated in appendix
pg 748 and write the equation for its reduction. (most negative
reduction potential)
3. Choose the strongest reducing agent as indicated in the table,
and write the equation for its oxidation.
4. Balance the number of electrons lost and gained in the halfreaction equations by multiplying one or both equations by a
number. Then add the two balanced half-reaction equations to
obtain a net ionic equation.
5. Using the spontaneity rule, predict whether the net ionic equation
represents a spontaneous or nonspontaneous redox reaction
Hints for Listing and Labelling
Entitites
• Aqueous solutions contain H2 O molecules
• Acidic solutions contain H+ ions
• Basic solutions contain OH- ions
• Some oxidizing and reducing agents are combinations,
for example, MnO4- and H+ (i.e. Must be acidic
conditions for permanganate ion to be an OA)
• H2 O, Fe2+ , Cu+ , Sn2+ and Cr2+ may act as either
oxidizing or reducing agents. Label both possibilities on
your list
Example
• Suppose a solution of potassium permanganate is
slowly poured into an acidified iron (II) sulfate
solution. Does a redox reaction occur and if it
does, what is the reaction equation?
Step 1: Separate the chemical equation into the start
of two half-reaction equations
Step 2: Balance atoms, other than O and H
•
Step 3: Balance O by adding H2O
•
Step 4: Balance H by adding H+.
•
Step 5: Balance the charge on each side by adding eand cancel anything that is the same on both sides
Step 6: multiply by the lowest common multiple
Step 7: Add the half reactions
Step 8: Cancel things that are the same
Faraday’s Law
• Electric Current: is the flow of electrons through an
external circuit
• Quantity of Electricity, also known as the electric
charge: is the product of the current flowing
through a circuit and the time for which it flows
• Q = I(t)
Electric Current (I)
• I=Q=Nxe
t
t
Quantity
current
amount of charge
time interval
number of
elementary
charges
elementary charge
Symbol
I
Q
t
N
SI Unit
A (ampere)
C (coulomb)
s (second)
no unit
e = 1.6 x 10-19
C (coulomb)
Faraday’s Law
• For stoichiometric calculations you need to know
the electric charge on a mole of electrons.
• To do this you multiply the charge on an electron
and multiple that by the number of atoms in a
mole.
• Charge on 1 mole = 1.6 x 10-19(6.022 x 1023)
= 9.647 x 104 C/mol
Example
• Calculate the mass of aluminum produced by the
electrolysis of molten aluminum chloride, if a
current of 500 mA passes for 1.5 h.
• You know the charge on 1 mole of electrons is
96500 C/mol
Example
1. Use the current and time to find the quantity of electricity used
I = Q/t
Q = I(t)
Q = 0.5 A(5400s)
Q = 2700C
2. From the quantity of electricity, find the amount of electrons that
passed through the circuit
Example
Amount of electrons = 2700C x 1 mol e- = 0.028 mol e96500 C
3. The half-reaction for the reduction of aluminum ions to
aluminum is Al3+ + 3e-  Al
Amount of Al = 0.028 mol e- x 1mol Al = 0.00933 mol Al
3 mol e-
Example
4. Convert the amount of aluminum to mass
mass of Al = 0.00933 mol Al x 27 g Al = 0.252 g Al
1 mol Al
H.W. pg 541 #22-24
Faraday’s Law
• Faraday’s Law: States that the amount of a
substance produced or consumed in an
electrolysis reaction is directly proportional to the
quantity of electricity that flows through the circuit
• We used this law to solve the previous sample
problem
Another Example
• Calculate the mass if zinc plated onto the cathode
of an electrolytic cell by a current of 750 mA in
3.25 h.
• Answer: 2.97 g of Zn
Review
• Rank the following in order from most effective to
least effective oxidizing agents: Zn2+(aq), Co3+(aq),
Br2(l), H+(aq)
• Answer: Co3+(aq), Br2(l), H+(aq), Zn2+(aq)
Review
• Write the half-reactions and calculate the standard
cell potential for each reaction. Identify each
reaction as spontaneous or non-spontaneous
• Answer: a) 0.315 V, spontaneous
b) –0.918 V non-spontaneous
a) Zn(s) + Fe
2+
2+
(aq)
b) Cr(s) + AlCl3(aq)
Zn
(aq) + Fe(s)
CrCl3(aq) + Als)
Review
• Calculate the mass of magnesium that can be
plated onto the cathode by the electrolysis of
molten magnesium chloride, using a current of
3.65 A for 55.0 min.
• Answer: 1.52 g
Review
•
The two half-cells in a galvanic cell consist of on iron
electrode in a 1 mol/L iron(II) sulfate solution, and a
silver electrode in a 1 mol/L silver nitrate solution.
Assuming the cell is acting as a galvanic cell. State the
cell potential, the oxidation half reaction, the reduction
half-reaction, and the overall cell reaction
o
E cell = 1.247 V
2+
Fe (aq)
oxidation: Fe(s)
+
-
Ag(s)
overall: 2Ag + Fe(s)
2+
reduction: Ag + e
+
Fe
+ 2e
-
(aq) + 2Ag(s)
Review
•
The ions Fe2+, Ag+ and Cu2+ are present in the half cell that
contains the cathode of an electrolytic cell. The concentration of
each of these ions is 1 mol/L. If the external voltage is very
slowly increased from zero, in what order will the three metals
Fe, Ag and Cu be plated onto the cathode? Explain why.
•
Answer: Ag+, Cu2+, Fe2+
Review
• Predict the products of the electrolysis of a 1 mol/L
aqueous solution of copper(I) bromide.
• Answer: copper and oxygen