Download Electrochemistry

ELECTROCHEMISTRY
SCH4U
Overview of Terminology
•
Electrochemistry – the study of the interchange of chemical
and electrical energy
•
Primarily concerned with the generation of an electrical current fro a
spontaneous chemical reaction and, the opposite process, the use of
current to produce a chemical reaction.
•
Oxidation–reduction (redox) reaction – involves a transfer
of electrons from the reducing agent to the oxidizing agent
Oxidation – loss of electrons
•
Reduction – gain of electrons
•
Reducing agent – electron donor
•
Oxidizing agent – electron acceptor
•
Redox Reactions
• How do we determine whether a given chemical reaction
is an oxidation-reduction reaction?
- by keeping track of oxidation numbers
(oxidation states) of all the elements
involved in the reaction.
• This procedure identifies whether any elements are
changing oxidation states.
Redox Reaction
• one or more elements change oxidation number
• all single displacement, and combustion,
• some synthesis and decomposition
• always have both oxidation and reduction
• split reaction into oxidation half-reaction and a reduction half-reaction
• aka electron transfer reactions
• half-reactions include electrons
• oxidizing agent is reactant molecule that causes
oxidation
• contains element reduced
• reducing agent is reactant molecule that causes
reduction
• contains the element oxidized
Oxidation & Reduction
• oxidation is the process that occurs when
• oxidation number of an element increases
• element loses electrons
• compound adds oxygen
• compound loses hydrogen
• half-reaction has electrons as products
• reduction is the process that occurs when
• oxidation number of an element decreases
• element gains electrons
• compound loses oxygen
• compound gains hydrogen
• half-reactions have electrons as reactants
Rules for Assigning Oxidation States
•
rules are in order of priority
free elements have an oxidation state = 0
1.
•
Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
monatomic ions have an oxidation state equal to their
charge
2.
•
Na = +1 and Cl = -1 in NaCl
(a) the sum of the oxidation states of all the atoms in
a compound is 0
3.
•
Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0
Rules for Assigning Oxidation States
(b) the sum of the oxidation states of all the
atoms in a polyatomic ion equals the charge
on the ion
3.
•
N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1
(a) Group I metals have an oxidation state of
+1 in all their compounds
4.
•
Na = +1 in NaCl
(b) Group II metals have an oxidation state of
+2 in all their compounds
4.
•
Mg = +2 in MgCl2
Rules for Assigning Oxidation States
in their compounds, nonmetals have oxidation states
according to the table below
5.
•
nonmetals higher on the table take priority
Nonmetal
Oxidation State
Example
F
-1
CF4
H **
+1
CH4
O
-2
CO2
Group 7A
-1
CCl4
Group 6A
-2
CS2
Group 5A
-3
NH3
Oxidation and Reduction
• oxidation occurs when an atom’s oxidation state increases
during a reaction
• reduction occurs when an atom’s oxidation state decreases
during a reaction
Identify all oxidation numbers below and label both reduction
and oxidation reactions.
CH4 + 2 O2 → CO2 + 2 H2O
Oxidation and Reduction
• oxidation occurs when an atom’s oxidation state increases
during a reaction
• reduction occurs when an atom’s oxidation state decreases
during a reaction
CH4 + 2 O2 → CO2 + 2 H2O
-4 +1
0
+4 –2
oxidation
reduction
Carbon is oxidized, therefore is a reducing agent.
Oxygen is reduced, therefore is an oxidizing agent.
+1 -2
Identify the Oxidizing and Reducing Agents in
Each of the Following
a) 3 H2S + 2 NO3– + 2 H+  3 S + 2 NO + 4 H2O
b)
MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O
Identify the Oxidizing and Reducing Agents in
Each of the Following
red ag
ox–ag
3 H2S + 2 NO3 + 2 H+ 3 S + 2 NO + 4 H2O
+1 -2
+5 -2
oxidation
+1
0
+2 -2
reduction
MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O
ox ag
+4 -2
red ag
+1 -1 +2 -1 0
oxidation
reduction
+1 -2
+1 -2
Common Oxidizing Agents (Gain
Electrons)
Oxidizing Agent
O2
Product when Reduced
O-2
H2 O2
H2O
F2, Cl2, Br2, I2
F-1, Cl-1, Br-1, I-1
ClO3-1 (BrO3-1, IO3-1)
H2SO4 (conc)
SO3-2
Cl-1, (Br-1, I-1)
SO2 or S or H2S
S2O3-2, or S or H2S
HNO3 (conc) or NO3-1
NO2, or NO, or N2O, or N2, or NH3
MnO4-1 (base)
MnO2
MnO4-1 (acid)
Mn+2
CrO4-2 (base)
Cr(OH)3
Cr2O7-2 (acid)
Cr+3
Common Reducing Agents (Lose Electrons)
Reducing Agent
H2
Product when Oxidized
H+1
H2O2
O2
I-1
I2
NH3, N2H4
S-2, H2S
N2
S
SO3-2
NO2-1
SO4-2
NO3-1
C (as coke or charcoal)
CO or CO2
Fe+2 (acid)
Fe+3
Cr+2
Cr+3
Sn+2
Sn+4
metals
metal ions
Balancing Redox Reactions
1)
assign oxidation numbers
a)
2)
write ox. & red. half-reactions, including
electrons
a)
3)
first balance elements other than H and O
add H2O where need O
add H+1 where need H
neutralize H+ with OH- in basic solutions
balance half-reactions by charge
a)
5)
6)
7)
ox. electrons on right, red. electrons on left of arrow
balance half-reactions by mass
a)
b)
c)
d)
4)
determine element oxidized and element reduced
balance charge by adjusting electrons
balance electrons between half-reactions
add half-reactions
check
Ex Balance the equation:
I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution
Assign
Oxidation
States
Separate
into halfreactions
Ox : I-(aq)  I2(aq)
red: MnO4-(aq)  MnO2(s)
Balance halfreactions by
mass
then O by
adding H2O
ox: 2 I-(aq)  I2(aq)
red: MnO4-(aq)  MnO2(s) + 2 H2O(l)
Balance
ox: 2 I-(aq)  I2(aq)
halfred: 4 H+(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l)
reactions
by mass
then H by
adding H+
Balance
ox: 2 I-(aq)  I2(aq)
halfred: 4 H+(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l)
reactions
by mass 4 H+(aq) + 4 OH-(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH-(aq)
in base,
4 H2O(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) + 4 OH-(aq)
neutralize
MnO4-(aq) + 2 H2O(l)  MnO2(s) + 4 OH-(aq)
+
the H
with OH-
19
I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution
Balance
Halfreactions
by
charge
Balance
electrons
between
halfreactions
ox: 2 I-(aq)  I2(aq) + 2 ered: MnO4-(aq) + 2 H2O(l) + 3 e-  MnO2(s) + 4 OH-(aq)
ox: 2 I-(aq)  I2(aq) + 2 e- } x3
red: MnO4-(aq) + 2 H2O(l) + 3 e-  MnO2(s) + 4 OH-(aq) }x2
ox: 6 I-(aq)  3 I2(aq) + 6 ered: 2 MnO4-(aq) + 4 H2O(l) + 6 e-  2 MnO2(s) + 8 OH-(aq)
Balance the equation:
I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution
Add the ox: 6 I-(aq)  3 I2(aq) + 6 eHalfred: 2 MnO4-(aq) + 4 H2O(l) + 6 e-  2 MnO2(s) + 8 OH-(aq)
reactions tot: 6 I-(aq)+ 2 MnO4-(aq) + 4 H2O(l)  3 I2(aq)+ 2 MnO2(s) + 8 OH-(aq)
Check
Reactant
Count
Element
Product
Count
6
I
6
2
Mn
2
12
O
12
8
H
8
2
charge
2
Homework
• Grade 12 Chemistry
pg. 467 # 2-4
pg. 468 # 5-8
pg. 478 # 9, 12
pg. 480 # 13-15
pg. 484 # 17-20
pg. 486 # 21-24
pg. 490 # 25-28
• Ap Chemistry :
Q’s 11- 22
Galvanic Cell/ Voltaic Cell
•
•
•
Device in which chemical energy is changed to electrical
energy.
Uses a spontaneous redox reaction to produce a current that
can be used to do work.
The key is to prevent the reactants in a redox reaction from
coming into direct contact with each other. Instead electrons
flow from one reactant to the other through an external
circuit. The flow of electrons through the external circuit is
electric current.
How Galvanic Cells Work
A Galvanic Cell
Sample Galvanic Cell
Molecular View of Electrodes
(a) A Cu2+ ion comes in contact with the surface of the Zn strip and gains two
electrons from a Zn atom; the Cu+ ion is reduced, and the Zn atom is oxidized.
(b) The resulting Zn2+ ion enters the solution, and the Cu atom remains deposited
on the strip.
Galvanic Cell Terminology
•
•
•
•
•
Conductors that carry electrons into and out of the
cell are named electrodes.
Electrolytes are substances that conduct electricity in
water.
Oxidation occurs at the anode (-) .
Reduction occurs at the cathode (+) .
Electrons flow through out external circuit form the
negatively charged electrode to the positively charged
electrode.
Galvanic Cell Terminology
•
Salt bridge or porous disk (semi permeable membrane)
– devices that allow ions to flow to neutralize the charge
imbalance without extensive mixing of the solutions.
•
•
Salt bridge – contains a strong electrolyte solution held in a Jello–like matrix
Porous Disk – contains tiny passages that allow hindered flow of ions.
Inert Electrodes
• Some redox reactions involve substances that cannot act
as electrodes, such as gases or dissolved electrolytes.
These inert electrodes are made from a material that is
neither a reactant for a product of the cell reaction.
Platinum and graphite are common examples.
Galvanic Cell Notation
Zn | Zn 2+ || Cu2+| Cu
ANODE
PHASE BOUNDARY
BEWEEN ELECTRODE
AND SOLUTION
CATHODE
POROUS
BARRIER OR
SALT BRIDGE
BETWEEN HALF
CELLS
Electromotive Force (EMF)
• Water flows
spontaneously one way
in a waterfall.
• Comparably, electrons
flow spontaneously one
way in a redox reaction,
from high to low
potential energy.
Unit of electrical potential is the volt (V).
 1 joule of work per coulomb of charge
transferred. (1 V = 1 J/C).
Electromotive Force (EMF)
• This potential difference between two electrodes is
called the electromotive force (causing electron
motion), or emf.
• The emf of a cell, Ecell, is also called the cell
potential.
• Ecell is measured in volts, and is often referred to as
cell voltage.
• For any cell reaction that is spontaneous, (voltaic cell)
the potential is positive.
• A cell potential of 0 V, has no electric potential meaning no electrons will
flow.
Standard Reduction Cell Potentials
• Under standard conditions, (i.e. 25°C, 1 M , 1 atm) , the
emf is called the standard emf, or standard cell
potential, E °cell
• EMF depends on the particular cathode and anode halfcells involved.
• The standard reduction potential, E°red can be assigned
for individual half –reactions
• The potential is compared to the standard hydrogen
electrodede, sometimes referred to as the normal
hydrogen electrode (NHE), which has an E°red = 0
Standard Reduction Potential
The standard hydrogen electrode (SHE) is used as a
reference electrode.
2 H+ (aq, 1 M) + 2 e-  H2 (g, 1 atm)
E red = 0V
Using a Standard Hydrogen Electrode
Calculating Cell Potential, E°cell
E°cell = E°red (cathode) – E°red (anode)
Example:
Zn2+(aq) + 2 e-  Zn(s) = -0.76 V
2 H+ (aq) +2 e-  H2(g) = 0.00 V
E°cell = E°red (cathode) – E°red (anode)
E°cell = 0.00 - (-0.76V)
E°cell = 0.76
Standard cell potential of a
voltaic cell.
The cell potential measures the
difference in the standard
reduction potentials of the
cathode and anode reactions:
E°cell = E°red
(anode)
(cathode)
- E°red
For all spontaneous reactions at standard
conditions, E °red > 0
Electrolytic Cells
• Although water flows downhill spontaneously, you can also
pump water uphill. This process requires energy.
• Today you will learn about a type of cell that uses energy to
move electrons from low to higher potential energy.
• This type of cell is called an electrolytic cell. The process that
takes place in an electrolytic cell is electrolysis. The overall
reaction in an electrolytic cell is non-spontaneous, and
requires energy to occur.
• Unlike galvanic cells, electrolytic cells require an external
source of electricity, called external voltage. Also, some
electrolytic cells are not separated, and take place in the same
container.
Electrolysis of Molten Salts
• This electrolytic cell
decomposes sodium
chloride into its elements. To
melt NaCl, the temperature
must be roughly 800° C.
• As in an aqueous solution of
sodium chloride, the ions in
molten sodium chloride have
some freedom of movement,
making molten NaCl the
electrolyte of this cell.
Electrolytic Cells Continued
• The external source of electricity forces electrons onto one
electrode. As a result, this electrode becomes negative relative
to the other electrode.
• The Na+ ions move toward the (-) electrode, where they gain
electrons are are reduced to the element sodium (as a liquid at
this temperature). Happens at cathode.
{ Na+(l) + e-  Na(l) } x 2
Eo = -2.71 V
• The Cl- ions move towards the positive electrode where they
lose an electron and are oxidized to the element, chlorine as a
gas, Cl2. Happens at anode.
2 Cl-(l)  Cl2(g) + 2 e-
Eo = 1.36 V
Electrolytic Cells Continued
• Because of the external voltage of the electrolytic cell, the
electrodes do not have the same polarities in electrolytic
and galvanic cells.
• In a galvanic cell, cathode  ( + )
anode  ( - )
• In an electrolytic cell, the anode (oxidation – electrons are
lost) is positive (connected to the positive terminal of a
external source) and the cathode (reduction- electrons
are gained) is negative.
Industrial Electrolytic Cell
• The large cell used
for the electrolysis of
sodium chloride in
industry is known as a
Downs Cell.
• To decrease heating
costs, CaCl2 is added
to lower the melting
point of sodium
chloride to
approximately 600° C.
• The reaction
produces sodium and
calcium by reduction
at the cathode, and
chlorine by oxidation
at the anode.
Electrolysis of Water
CATHODE
ANODE
• When electrolyzing aqueous
solution, there are two compounds
present: water and the dissolved
electrolyte. Water maybe
electrolyzed as well, producing
oxygen and hydrogen gas.
• Oxidation at the Anode
2 H2O(l)  O2 (g) + 4 H+ (aq) + 4 e• Reduction at the Cathode
2 H2O(l) + 2 e-  H2(g) + 2 OH- (aq)
Water is present at both electrodes- it can be oxidized and reduced simultaneously.
Electrolysis of Water
x2
The standard reduction potentials are as follows:
E* cell = E*red (cat) – E* red (an )
= (-0.828 V) – (1.229) = - 2.057 V ( NON SPONT REACTION )
Electrolysis of Water under NonStandard Conditions
• The standard reduction potentials used to calculate E* cell, for the
decomposition of water , apply only to reactants and products in their
standard states.
• In pure water at 25 * C, the hydrogen ions and hydroxide ions each
have concentrations of 1 x 10-7 mol/L, which is not the standard state
value of 1 mol/L.
• Below are the reduction potential values for non-standard conditions
in pure water are given below. The superscript (nought) is now
omitted from the E symbol, because the values are no longer
standard.
• Using these new half- cell potentials, E
for the
decomposition of pure water at 25* C, is -1.229 V.
cell
• Therefore, the calculated value of the external voltage needed
is 1.229 V.
• In practice, the external voltage needed for electrolytic cells is
always greater than the calculated value, especially for
reactions involving gases. Therefore the actual voltage need to
electrolyze pure water is GREATER than 1.229 V. The excess
voltage required above the value is called overvoltage.
Overvoltage depends on the gases involved and on the
materials in the electrodes.
• When electrolyzing pure water poor conductor) , an electrolyte
that does not interfere in the reaction is added to the water to
increase the conductivity.
Electrolysis in Aqueous Solutions
• By added in a external electrical supply,
with a greater voltage than the voltage of
the Daniell Cell, you can push electrons
in the opposite direction and therefore,
reverse the chemical reaction.
Galvanic vs. Electrolytic Cells
Predicting the Products of Electrolysis
for an Aqueous Solution
• The electrolysis of an aqueous solution, may involve the
electrolysis of water. How can we predict the actual
products for this type of reaction?
• To predict the products of an electrolysis involving an
aqueous solution, you must examine all possible halfreactions and their reduction potentials. Then, you must
find the overall reaction that requires the lowest external
voltage.
( i.e., which reduction occurs FIRST, and which oxidation
occurs FIRST? )
• That is, you must find the overall cell reaction with a
negative cell potential that is closest to zero.
Practice Problem : Electrolysis of an
Aqueous Solution
• Predict the products of the electrolysis of 1 mol/L LiBr
(aq).
• Step 1: List the four relevant half-reactions and their
reduction potentials.
Hint : Use non-standard values for water.
Oxidation
Reduction
Practice Problem : Electrolysis of an
Aqueous Solution
• Predict the products of the electrolysis of 1 mol/L LiBr
• Step 1: List the four relevant half-reactions and their
reduction potentials.
Hint : Use non-standard values for water.
(aq).
Practice
Problem :
Electrolysis of
an Aqueous
Solution
• Combine pairs of
half-reactions to
produce four possible
overall reactions.
• Reaction 2 requires
the lowest external
voltage.
Practice Problem Continued
Homework
• Grade 12 :
Read 11.3 (only pgs. 524-535. Stop
at Rechargeable Batteries)
Q’s pg. 525 # 9-12,
pg. 531 # 13-16
pg. 534 # 17-20
pg. 537 # 1-3
• AP : Read 20.9 pg 876- 878
q’s 85, 87
• Both Classes:
Worksheet with Problems
Electrochemistry Workbook
Package!
(Electrolytic Cells on page 7)
Meanwhile …. Darcy is busy
ordering some
lunch  !
Faraday’s Law
• “ 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”
Measurements in Electricity
• Electric Current: flow of electrons through an external
circuit, measurement in amperes, A.
• Electric Charge: product of current flowing through a
circuit and the time for which it flows, measured in
coulombs, C.
Charge (C) = current (A) x time (s)
• Charge on 1 electron = 1.602 x 10-19 C
• Charge on 1 mol of electrons = 9.647 x 104 C/mol
(often rounded to 96 500 C/mol)
Calculating the Mass of an Electrolysis
Product
• Calculate the mass of Aluminum produced by the
electrolysis of molten aluminum chloride, in a current of
500 mA passes for 1.50 h.
Calculating the Mass of an Electrolysis
Product SOLUTION
Calculating the Mass of an Electrolysis
Product SOLUTION
Relating to Faraday’s Law
• “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”
• Relating to the practice problem, suppose that the
quantity were doubled by using the same current for twice
the time, 3 hours.
• Grade 12 : Read Section 11.4 ( Faraday’s Law )
q’s pg. 541 # 21-24, pg.
Chapter 11 Review
• AP Class :