Download 2 2e Mg + S Mg S

Chemistry 102
Chapter 18
OXIDATION-REDUCTION REACTIONS

Some of the most important reaction in chemistry are oxidation-reduction (redox) reactions.
In these reactions, electrons transfer from one reactant to the other. The rusting of iron,
bleaching of hair and the production of electricity in batteries involve redox reactions.

Synthesis, decomposition, single replacement and combustion reactions are all examples of
redox reactions. In these reactions, one substance loses electrons (oxidized) while another
substance gain electrons (reduced). Therefore, oxidation is defined as loss of electrons,
while reduction is defined as gain of electrons.

For example in reaction of magnesium metal and sulfur to form magnesium sulfide:
– 2e (oxidation)
0
+2
Mg + S
MgS
0
-2
+ 2e (reduction)
Mg :
- is oxidized (loses electrons)
- causes the reduction of S
- called Reducing Agent
S:
- is reduced (gains electrons)
- causes the oxidation of Mg
- called Oxidizing Agent
2e–
Mg0 +

S0
Mg 2+S2
Oxidation-Reduction reactions can be represented by half-reactions:
Oxidation Half-Reaction
Mg
 Mg
2+
+ 2e
Reduction Half-Reaction
–
S
1
+
2e –
 S2
Chemistry 102
Chapter 18
OXIDATION NUMBERS OR STATES

Identifying the oxidation-reduction nature of reactions between metals and non-metals is
straight forward because of ion formation. However, redox reactions that occur between
two non-metals is more difficult to characterize since no ions are formed.

Chemists have devised a scheme to track electrons before and after a reaction in order to
simply this process. In this scheme, a number (oxidation state or number) is assigned to
each element assuming that the shared electrons between two atoms belong to the one
with the most attraction for these electrons. The oxidation number of an atom can be
thought of as the “charge” the atom would have if all shared electrons were assigned to it.

The following rules are used to assign oxidation numbers to atoms in elements and
compounds. (Note: these rules are hierarchical. If any two rules conflict, follow the rule that
is higher on the list)
1. The oxidation number of an atom in a free element is 0.
2. The oxidation number of a monatomic ion is equal to its charge.
3. The sum of the oxidation number of all atoms in:

A neutral molecule or formula is equal to 0.

An ion is equal to the charge of the ion.
4. In their compounds, metals have a positive oxidation number.

Group 1A metals always have an oxidation number of +1.

Group 2A metals always have an oxidation number of +2.
5. In their compounds non-metals are assigned oxidation numbers
according to the table at right. Entries at the top of the table take
precedence over entries at the bottom of the table.
2
Chemistry 102
Chapter 18
OXIDIDATION-REDUCTION REACTIONS

Oxidation numbers (or states) can be used to identify redox reactions and determine which
substance is oxidized and which is reduced.

To do so, assign oxidation numbers to all elements in the reactants and products, then track
which elements change oxidation numbers from reactants to products.

Elements that increase their oxidation numbers lose electrons, and are therefore oxidized.
Elements that decrease their oxidation numbers gain electrons, and are therefore reduced.
OXIDATION (LOSS OF ELECTRONS)
-6
-5
-4
-3
-2
-1
0
+1
+2
+3
+4
+5
+6
REDUCTION (GAIN OF ELECTRONS)
Examples:
1. Using oxidation numbers, identify the element is oxidized and the element that is reduced
and the number of electrons transferred, in each of the following redox reactions:
a) Sn (s) + 4 HNO3 (aq) → SnO2 (s) + 4 NO2 (g) + 2 H2O (g)
b) Hg(NO3)2 (aq) + 2 KBr (aq) → Hg2Br2 (s) + 2 KNO3 (aq)
3
+7
Chemistry 102
Chapter 18
HALF–REACTIONS

Redox Reactions are discussed (sometimes balanced) by writing two Half-Reactions:
OXIDATION HALF-REACTION
 involves loss of electrons
 increase in Oxidation Number
Cu(s)
REDUCTION HALF-REACTION
 involves gain of electrons
 decrease in Oxidation Number
Cu2+(aq) + 2 e–
2Ag+(aq) + 2e–
2Ag0(s)
NOTE:
Number of electrons
lost in
Oxidation reaction
=
Number of electrons
gained in
Reduction reaction
CONCLUSION:
Oxidation–Reduction (Redox) Reactions
 are reactions in which the Oxidation Numbers of at least two elements change
 involve transfer of electrons:
 from :
 to:
the element that is oxidized (called Reducing Agent)
the element that is reduced (called Oxidizing Agent)
Examples:
1. Determine the oxidized and reduced species and write half-reactions for the reaction of
zinc and sulfur to form zinc sulfide:
Zn (s) +
S (s)
4
ZnS (s)
Chemistry 102
Chapter 18
Examples:
2. Determine the oxidized and reduced species and write half-reactions for the reaction of
aluminum and iodine to form zinc aluminum iodine:
Al (s) + I2 (s)
AlI3 (s)
3. Determine the oxidized and reduced species and write half-reactions for the reaction
shown below:
Mg (s) + HCl (aq)
MgCl2 (aq) + H2 (g)
5
Chemistry 102
Chapter 18
BALANCING REDOX REACTIONS IN ACIDIC & BASIC SOLUTIONS

When balancing redox reactions, the following must be observed:
 Atoms must balance
 Nr. of electrons lost = Nr. of electrons gained
 Net charges of ions on both sides of the equation must balance
1. Acidic Solution (H+ ions and H2O molecules may be added as needed)
Balance the following redox reaction that takes place in acidic solution:
Zn(s) + NO3(aq)  Zn2+(aq) + NH4+(aq)
(acidic solution)
Step 1: Assign oxidation numbers and determine the oxidized and reduced species and number
of electrons involved.
0
Zn
+5
+2
–3

2+
+ NO3  Zn
+ NH4+
Step 2: Set up half-reactions.
 Oxidation loses electrons (write as product)
 Reduction gains electrons (write as reactant)
Oxidation half-reaction
Zn  Zn2+ + 2 e–
Reduction half-reaction
NO3 + 8 e–  NH4+
–
Step 3: Balance atoms; H2O molecules and H+ ions may be added as needed
Oxidation half-reaction
Zn  Zn2+ + 2 e–
(balanced; no action needed)
Reduction half-reaction
NO3– + 8 e–  NH4+
Balance O atoms by adding H2O molecules:
NO3 + 8 e–  NH4+ + 3 H2O
Balance H atoms by adding H+ ions:
NO3 + 10 H+ + 8 e– NH4+ + 3 H2O

Check to see that atoms and charges are balanced for each half-reaction,
6
Chemistry 102
Chapter 18
Step 4: Equalize loss and gain of electrons.
Oxidation half-reaction
Reduction half-reaction
[ Zn  Zn2+ + 2 e– ] x 4
[ NO3 + 10 H+ + 8 e– NH4+ + 3 H2O ] x 1
4 Zn  4 Zn2+ + 8 e–
NO3 + 10 H+ + 8 e– NH4+ + 3 H2O
Step 5: Add up half-reactions and cancel electrons.
4 Zn + NO3 + 10 H+ + 8e

4 Zn2+ + 8e + NH4+ + 3 H2O
Check to see that atoms and charges are balanced for the overall reaction
Examples:
Balance the following redox reaction in acidic solution:
1.
Br2 (l) + SO2 (g)  Br– (aq) + SO42– (aq)
2.
H2O2 (aq) + MnO4– (aq)  Mn2+ (aq) + O2 (g)
7
Chemistry 102
Chapter 18
2. Basic Solution (OH ions and H2O molecules may be added as needed)
Basic Idea:
O and H will be balanced as though the solution were acidic
In a later step, the acidic solution will be converted to a basic solution
Balance the following redox reaction that takes place in basic solution:
MnO4(aq)
+
SO32 (aq)
MnO2(s)
+
SO42(aq)
Step 1: Assign oxidation numbers and determine the oxidized and reduced species and number
of electrons involved.
+7
MnO4(aq)
+
+4
SO32 (aq)
+4
MnO2(s)
+6
+ SO42(aq)
Step 2: Set up half-reactions.
 Oxidation loses electrons (write as product)
 Reduction gains electrons (write as reactant)
Oxidation half-reaction
SO32–  SO42– + 2 e–
Reduction half-reaction
MnO4– + 3 e–  MnO2
Step 3: Balance atoms; H2O molecules and H+ ions may be added as needed
Oxidation half-reaction
Reduction half-reaction
SO32–  SO42– + 2 e–
MnO4– + 3 e–  MnO2
Balance O atoms by adding H2O molecules: Balance O atoms by adding H2O molecules:
SO32– + H2O  SO42– + 2 e–
MnO4– + 3 e–  MnO2 + 2 H2O
Balance H atoms by adding H+ ions:
SO32– + H2O  SO42– + 2 e– + 2 H+

Balance H atoms by adding H+ ions:
MnO4– + 3 e– + 4 H+  MnO2 + 2 H2O
Check to see that atoms and charges are balanced for each half-reaction,
8
Chemistry 102
Chapter 18
Step 4: Equalize loss and gain of electrons.
Oxidation half-reaction
Reduction half-reaction
[SO32– + H2O  SO42– + 2 e– + 2 H+] x
3
[MnO4– + 3 e– + 4 H+  MnO2 + 2 H2O] x
2
3 SO32– + 3 H2O  3 SO42– + 6 e– + 6
H+
2 MnO4– + 6 e– + 8 H+  2 MnO2 + 4 H2O
Step 5: Add up half-reactions and cancel electrons.
3 SO32 + 3 H2O + 2 MnO4 + 8 H+

3 SO42 + 6 H+ + 2 MnO2 + 4 H2O
Note that the H+ ions and H2O molecules may be canceled and the equation becomes:
3 SO32 + 2 MnO4 + 2 H+
3 SO42 + 2 MnO2 + H2O
Step 6: Add OH– to cancel H+ ion by forming H2O molecules
3 SO32 + 2 MnO4 + 2 H+ + 2 OH
3 SO42 + 2 MnO2 + H2O + 2 OH
2 H2 O
3 SO32 + 2 MnO4 +

Cancel one more time the H2O molecules:
3 SO32 + 2 MnO4 +

3 SO42 + 2 MnO2 + H2O + 2 OH
2 H2 O
H2 O
3 SO42 + 2 MnO2 + 2 OH
Check to see that atoms and charges are balanced for each half-reaction,
NOTE:
For an alternate method of balancing basic solutions see “Redox Review” on my website
9
Chemistry 102
Chapter 18
Examples:
Balance the following redox reactions in basic solution:
1.
Cr(OH)4– (aq) + H2O2 (aq)  CrO42– + H2O (l)
2.
MnO4– (aq) + Br– (aq)  MnO2 (s) + BrO3– (aq)
10
Chemistry 102
Chapter 18
ELECTROCHEMICAL CELLS


Electrochemical cells are systems consisting of electrodes that dip into an electrolyte and in which a
chemical reaction either uses or generates an electric current
There are of two types of electrochemical cells:
1. Voltaic Cells (galvanic cells, or battery cells)
 These are electrochemical cells in which a spontaneous reaction generates an electric current
 Example: An ordinary battery
2. Electrolytic Cells
 These are electrochemical cells in which an electric current drives a non-spontaneous reaction.
 Example: Electroplating of Metals
Voltaic Cells
 Consist of two half-cells that are electrically connected
 Each half-cell is made from a metal strip that dips into a solution of its metal ion
Zinc – Zinc Ion Half-Cell
(zinc electrode)
Copper – Copper Ion Half-Cell
(copper electrode)
Zinc strip
Copper strip
Zn2+
Cu2
11
Chemistry 102
Chapter 18
VOLTAIC CELLS
12
Chemistry 102
Chapter 18
VOLTAIC CELLS


How does the voltaic cell function?
Basic Idea: Zinc tends to lose electrons more readily than Copper (Zn is more active than Cu)


Oxidation half-reaction
Reduction half-reaction
Zn (s)  Zn2+ (aq) + 2 e–
Cu2+ (aq) + 2 e–  Cu (s)
The electrons flow away from the
zinc strip into the external circuit.
The Zinc strip is used up.


The electrons come from the external
circuit and deposit onto the Cu strip.
The Cu2+ ions plate out as solid copper.
 Net Result: An electric current is generated through the external circuit.

When sketching and labeling a voltaic cell, follow the following guidelines:
Anode:
 is the negative electrode
 is the more active metal
Cathode:
 is the positive electrode
 is the less active metal
Electron Flow:

Away from the anode

Towards the cathode
Cell Reaction:

Oxidation half-reaction

Reduction half-reaction
The Electrodes:
13
Chemistry 102
Chapter 18
SHORTHAND NOTATION FOR VOLTAIC CELLS

Electrochemical cells are often represented by a compact notation as shown below:
Zn (s)
Zn2+ (aq)
Cu2+ (aq)
Cu (s)
Anode terminal
Cathode terminal
Phase boundary
Phase boundary
Anode electrolyte
Cathode electrolyte
Salt bridge
The Anode (Negative electrode)
(the more active metal)
The Oxidation half-cell
The Cathode (Positive electrode)
(the less active metal)
The Reduction half-cell
Examples:
1. Write the cell notation for the reaction shown below:
2 Al (s) + 3 Zn2+ (a)  2 Al3+ (aq) + 3 Zn (s)
2. Write the half-reactions and the overall reaction for the redox reaction that occurs in the
voltaic cell shown below:
Sc (s)  Sc3+ (aq)  Ag+ (aq)  Ag (s)
14
Chemistry 102
Chapter 18
HALF-REACTIONS INVOLVING GASES
electrical terminal
and
electrode surface
An inert material (ex: Platinum) serves as:
A Hydrogen electrode
(catalyzes the half-reaction, but otherwise is not involved in it)
Acidic Solution
2 H+ (aq)
Half Reaction:
+
2 e–
Cathode Reaction
H2 (g)
Anode Reaction

Notation for the hydrogen electrode:
As an Anode
Pt
H2 (g)
As a Cathode
H+ (aq)
H+(aq)
Anode Reaction:
H2(g)
H2 (g)
Pt
Cathode Reaction:
2 H+(aq) + 2 e–
2 H+ (aq) + 2e–
15
H2 (g)
Chemistry 102
Chapter 18
Example:
1. Consider the voltaic cell shown below. (a) Identify and label the anode and the cathode,
(b) determine the direction of the electron flow and (c) write a balanced equation for the
overall reaction.
2. Write the notation for a cell in which the electrode reactions are:
2 H+(aq) +
Zn (s)
2 e–
H2(g)
2+
Zn (aq) + 2 e–
3. Give the overall cell reaction for the voltaic cell with the notation shown below:
Cd (s)  Cd2+ (aq) 
16
H+(aq)  H2(g)  Pt
Chemistry 102
Chapter 18
ELECTROMOTIVE FORCE

Through the external wire, the electrons are forced to flow:
From the negative
(Anode)
(high electric potential)

electrons
To the positive electrode
(Cathode)
(low electric potential)
The Electrical Work needed to move an electrical charge (the electron) through a conductor
depends on:
 the total electrical charge moved, and
 the potential difference

This difference in electrical potential (voltage) is measured in volts, V (SI unit of potential difference)

It follows:
Electrical Work = Electrical Charge x Potential Difference
Joules

=
coulombs
x
volts
The magnitude of the electrical charge is conveniently measured as the electrical charge carried by
1 mole of electrons:
Electrical charge carried by one electron
= 1.602 x 1019 coulombs/e
Number of electrons in 1 mole of electrons = 6.022 x 1023 electrons
Electrical Charge
carried by
=
1 mole of electrons
Electrical Charge
carried by
1 mole of electrons

19
1.602 x 10
coulombs

electron
9.65 x 104 coulombs
17
x

6.022 x 1023 electrons
96,500 coulmbs
Chemistry 102

Chapter 18
The Electrical Charge carried by 1 mole of electrons:
 is equal to 9.65 x 104 coulombs (96,500 coulmbs)
 is referred to as the Faraday constant, F
(in honor of Michael Faraday, who laid the foundations of Electrochemistry)
Let:
w = the electrical work done by the voltaic cell in moving 1 mole of electrons from one
electrode to another
w = Electrical Work = (Electrical Charge) x (Potential Difference)
w =  ( F ) (Potential Difference)
negative sign indicates that the
voltaic cell loses energy, as it does
work on the surroundings
Maximum
Electrical Work
obtainable
=
Electromotice Force
emf
=
Ecell
Maximum Voltage
=
Maximum Electrical
Work
obtainable
=
Maximum Potential Difference
Maximum Voltage
=
Maximum
Potential Difference
between the electrodes
measured with
a voltmeter

In the normal operation of a voltaic cell:
The Actual Voltage  The Electromotive Force
Reason: it takes energy to drive a current through the cell itself
Conclusion:
Maximum Work Obtainable
=
wmax = – n F Ecell
Cell emf
Faraday constant = 9.65 x 105 C
Number of electrons transferred
in the overall cell equation
18
Chemistry 102
Chapter 18
Examples:
1. What is the maximum electrical work that can be obtained from 6.54 g of zinc metal that
reacts in a Daniell cell (Zn as an anode and Copper as a cathode), whose emf is 1.10 V.
The cell reaction is :
Zn (s) + Cu2+aq)
Zn2+(aq) + Cu (s)
Wmax = – n F Ecell
To determine “n”, write out the half-reactions:
Anode (–) half-reaction
Cathode (+) half-reaction
Zn (s)  Zn2+ (aq)
+ 2 e–
Note that:
n=2
Cu2+ (aq) + 2 e– 
Cu (s)
wmax = – n F Ecell = – (2) (9.65 x 104 C) (1.10 V) = – 2.12 x 105 J/mol
The maximum work for 6.54 g of Zn(s) is:
1 mol Zn – 2.12 x 105 J
6.54 g Zn x  x  = – 2.12 x 104 J
65.39 g Zn
1 mol Zn
2. Calculate the free energy change (G0) for the reaction shown below with standard
potential of 0.92 V at 25 C:
Al (s) + Cr3+ (aq)  Al3+ (aq) + Cr (s)
(Hint: recall that G = maximum work)
19
Chemistry 102
Chapter 18
STANDARD CELL Emf’s AND STANDARD ELECTRODE POTENTIALS

Emf is a measure of the driving force of the cell half-reactions
Example:
The Daniell Cell
Anode half-reaction (Oxidation)
Cathode half-reaction (Reduction)
Zn (s)  Zn2+(aq) + 2e–

Cu2+(aq) +2e–  Cu (s)
Each of the two ions (Zn2+ and Cu2+) has a certain tendency to acquire electrons from the its
respective electrode and become reduced
Zn2+(aq) + 2e–  Zn (s)
Cu2+(aq) + 2e–  Cu (s)
 Smaller tendency to acquire electrons
 Greater tendency to acquire electrons
 Lower potential to undergo reduction
 Greater potential to undergo reduction
 Is forced to undergo Oxidation
 Is forced to undergo Reduction
Reduction Potential = the tendency of the metallic ion to become reduced
Ecell = the difference in the tendencies of the two ions to become reduced
Ecell
Reduction Potential
of the substance
that actually undergoes
reduction
Cu2+/Cu
Reduction Potential
of the substance
that is forced to undergo
oxidation
Zn2+/An

Ecell = Ered (cathode) – Ered (anode)
For the Daniell cell:
Ecell =
ECu
Reduction Potential
of substance being
reduced
20
–
EZn
Reduction Potential
of substance being
oxidized
Chemistry 102

Chapter 18
An Oxidation half-reaction is the reverse of the corresponding Reduction half-reaction:
Zn
2+
–
Reduction
(aq) + 2e
Zn(s)
Oxidation
It follows that:
Oxidation Potential for a half-reaction = – Reduction Potential for the reverse half-reaction
Ecell
=
Reduction Potential
of the substance
that actually undergoes
reduction
Cu2+/Cu
Oxidation Potential
of the substance
that is forced to undergo
oxidation
Zn2+/An
+
Examples:
1. A voltaic cell is based on the following two half-reactions, shown below. Determine the
standard potential for this cell.
Cd2+ (aq) + 2 e–  Cd (s)
E = –0.403 V
Sn2+ (aq) + 2 e–  Sn (s)
E = –0.136 V
2. Using standard cell potentials in your text, calculate the emf for a cell that operates on the
following overall reaction:
2 Al (s) + 3 I2 (s)  2 Al3+ (aq) + 6 I– (aq)
21
Chemistry 102
Chapter 18
STANDARD ELECTRODE POTENTIALS

E cell = standard emf = emf of a voltaic cell operating under standard-state conditions:
 solute concentrations are each 1M
 gas pressures are each 1 atm
 temperature is 25 C

To find E cell a table of standard electrode potentials is needed.

Experimentally it is not possible to measure the potential of a single electrode.

However, it is possible to measure emf’s of cell constructed from various electrodes connected to a
chosen “reference electrode”

The “reference electrode” chosen is arbitrarily assigned a Potential of Zero, under standard conditions

This electrode is the standard Hydrogen electrode
1 atm pressure
2 H+(aq) + 2 e–  H2(g)
Standard Potential = E = 0.00 V
25 C
1 M acidic solution

Any substance that is more easily reduced than H+ has a (+) value of E0 :
Cu2+(aq) + 2e– → Cu (s)

Any substance that is more difficult to reduce than H+ has a (-) value of E0
Zn2+(aq) +

E = + 0.34 V
2e– → Zn (s)
E = – 0.76 V
To find a standard electrode potential, the electrode is connected to a standard Hydrogen electrode
and the emf is measured with a voltmeter:
22
Chemistry 102
Chapter 18
23
Chemistry 102
Chapter 18
STRENGTHS OF OXIDIZING AND REDUCING AGENTS
Oxidized Species
+
–
reduction
ne
Reduced Species
oxidation
The higher the
tendency of the
species to gain
electrons
Example:
The higher the
tendency of the
species to be
reduced
F2(g) +
2 e–
The stronger
Oxidizing Agent
the species is
2 F(aq)
The larger (more +)
the electrode reduction
potential (E) of the
species
E = + 2.87 V
Strong Oxidizing
Agent
The higher the
tendency of the
species to lose
electrons
Example:
The higher the
tendency of the
species to be
oxidized
Li+(aq) +
e–
The stronger
Reducing Agent
the species is
Li (s)
Strong Reducing
Agent
24
The smaller (more -)
the electrode reduction
potential (E) of the
species
E = – 3.04 V
Chemistry 102
Chapter 18
PREDICTING SPONTANEITY FOR OXIDATION-REDUCTION REACTIONS
Oxidized Species +
ne
–
spontaneous
Reduced Species
nonspontaneous
Stronger Oxidizing
Agent (ex: F2)
Oxidized Species +
Weaker Oxidizing
Agent (ex: 2 F)
ne
–
nonspontaneous
Reduced Species
spontaneous
Weaker Reducing
Agent (ex: Li+)
Stronger Reducing
Agent (ex: Li)
25
Chemistry 102
Chapter 18
Examples:
1. Does the following reaction occur spontaneously in the direction indicated, under standard conditions?
Cu2+(aq) + 2 I(aq)
Cu(s)
+
I2(s)
Write the half-reactions and look up the E values (Standard Reduction Potential)
I2(s)
2I (aq)
+ 2e
Cu2+ (aq) + 2 e–
E = 0.54 V
Cu (s)

E(I2)
E = 0.34 V
E(Cu2+)
 I2 has a higher tendency to be reduced than Cu2+
 I2 is a stronger oxidizing agent than Cu2+
nonspontaneous

2+
Cu (aq) + 2 I (aq)
weaker
oxidizing
agent
Cu(s) +
spontaneous
I2(s)
stronger
oxidizing
agent
2. Which is the stronger oxidizing agent, NO3(aq) in acidic solution (to NO) or Ag+(aq)?

First write the two appropriate reduction half-reactions:
NO3(aq) + 4H+(aq) + 3 e–
Ag+(aq) +

e–
NO(g) + 2 H2O(l)
E = + 0.96 V
Ag(s)
E = + 0.80 V
Compare the two reduction potentials:
E(NO3)

E(Ag+)
 NO3 has a higher tendency to be reduced than Ag+
 NO3 is a stronger oxidizing agent than Ag
26
Chemistry 102
Chapter 18
EQUILBRIUM CONSTANTS AND EMF

The free energy change, G, for a reaction equals the maximum useful work of the reaction.
G = wmax

For a voltaic cell, this work is the electrical work, –nFEcell.
G = –nFEcell

Also recall that,
G = –RTlnK

Combining the last two equations give another method for obtaining the equilibrium constant for a
reaction:
nFEcell = RTlnK
or
Ecell =
RT
2.303 RT
ln K =
log K
nF
nF
Substituting values for constants R and F at 25C gives the equation shown below:
Ecell =

0.0592
log K
n
(at 25 C)
Recall that free energy change, G, and standard free energy change, G, were related as shown
below:
G = G + RT ln Q
Therefore,
–nFEcell = –nFEcell + RT ln Q

Rearrangement leads to the Nernst equation shown below:
Ecell = Ecell -
RT
ln Q
nF
or
Ecell = Ecell -
2.303RT
log Q
nF
Substituting values for constants R and F at 25C gives the equation shown below:
Ecell = Ecell -
0.0592
log Q
n
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(at 25 C)
Chemistry 102
Chapter 18
Examples:
1. The standard emf for a Zn/Cu voltaic cell is 1.10 V. Calculate the equilibrium constant for the
reaction shown below:
Zn (s) + Cu2+ (aq)
Zn2+ (aq) + Cu (s)
0.0592
log K
n
0.0592
1.10 =
log K
2
log K = 37.2
E=
Taking antilog of both sides
K = 1037.2 = 1.6x1037
2. Calculate the equilibrium constant for the following reaction from the standard electrode potentials:
Fe (s) + Sn4+ (aq)
Fe2+ (aq) + Sn2+ (aq)
3. What is the emf for the Zn/Cu voltaic cell in example 1 when
[Zn2+] = 1.00x10–5 M
Q=
and
[Cu2+] = 0.100 M
[Zn 2+ ]
=
[Cu 2+ ]
Ecell = Ecell -
0.0592
log Q =
n
4. What is the emf for the following voltaic cell?
Zn  Zn2+ (0.200 M)  Ag+ (0.00200 M)  Ag (s)
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