Chemistry 30
Unit 7:
REDOX Reactions
Teacher
A Simple Redox Experiment
When a copper wire is placed in an aqueous solution of silver nitrate, a spectacular reaction results.
Within minutes a large number of shiny metallic crystals form on the wire, and the solution turns pale
blue. When we shake off the crystals, we observe that much of the copper wire has disappeared. The
crystals are silver crystals and the blue colour of the solution indicates the presence of hydrated
copper ions. Copper metal has displaced silver from its salt.
 Cu(NO3)2(aq) + 2 Ag(s)
Cu(s) + 2 AgNO3(aq) 
 Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq) 
Copper atoms in the wire each lose two electrons and go into solution as hydrated copper ions. The
electrons are transferred to hydrated silver ions, which crystallize as metallic silver.
Any reaction in which electrons are lost is called an oxidation reaction. A reaction that involves the
gain of electrons is called a reduction reaction. In the copper-silver nitrate reaction, the copper metal
is oxidized to copper ions and the silver ions are reduced to silver metal. Because oxidation and
reduction reactions always take place concurrently, we refer to these reactions as redox reactions.
Electronegativity and Oxidation Numbers





The oxidation number tells us how many electrons an atom in a compound would gain or lose.
In assigning oxidation numbers we assume that the electrons of each bond are transferred to
the more electronegative atom in the bond.
A positive oxidation number indicates a loss of electrons and a negative number indicates a
gain of electrons.
The oxidation number of an atom in its pure form, as an element, is zero.
The oxidation number of a monatomic ion is equal to the charge on the ion, since the charge
indicates the number of electrons lost or gained.
Example
What are the oxidation numbers of each element in calcium chloride?
CaCl2
Ca2+ + 2 Cl−
Ca = +2 and Cl = −1
(+2) + 2(−1) = 0
Example
What are the oxidation numbers of iron in the two iron oxides, FeO and Fe2O3? Assume that oxygen
has an oxidation number of −2 in each case.
The sum of the oxidation numbers in each compound will be zero.
FeO = Fe2+ + O2−
Fe2O3 = 2 Fe3+ + 3 O2−
Fe = +2 and O = −2
Fe = +3 and O = −2
(+2) + (−2) = 0
2(+3) + 3(−2) = 0
-2-
Oxidation Number Rules
1. The oxidation number of an atom in its pure form, as an element, is zero.
2. The oxidation number of a monatomic ion is equal to the charge on the ion.
3. The algebraic sum of the oxidation numbers in a neutral polyatomic compound is zero.
4. The algebraic sum of the oxidation numbers in a polyatomic ion is equal to the charge on the
ion.
5. The most common oxidation number for hydrogen is +1; in metal hydrides it is −1.

In most of its compound, hydrogen has the lowest electronegativity value and an oxidation
number of +1. But in metal hydrides, such as sodium hydride, NaH, the hydrogen is bonded
to a less electronegative metal and has an oxidation number of −1.
6. The most common oxidation number for oxygen is −2; in peroxides it is −1, and in combination
with fluorine it is +2.
7. In combinations of nonmetals, the oxidation number of the more electronegative atom is
negative. The oxidation number of the less electronegative atom is positive.
Example
H
He
2.01
-
What are the oxidation numbers of oxygen and
fluorine in oxygen difluoride, OF2?
Li
0.98 1.57
2.04 2.55 3.04 3.44 3.98 -
The electronegativity values of oxygen and
fluorine are 3.44 and 3.98 respectively. The
fluorine atoms attract electrons more strongly
than oxygen; therefore it is assigned the
negative oxidation number, and the oxygen,
the positive oxidation number.
Na
Al
OF2 = O2+ + 2 F−
O = +2 and F = −1
(+2) + 2(−1) = 0
Be
Mg
B
C
Si
Ar
Ga
Ca
Ge
As
Se
Br
Kr
0.82 1.00
1.81 2.01 2.18 2.55 2.96 3.0
Rb
In
Sr
Sn
Sb
Te
I
Xe
0.82 0.95
1.78 1.96 2.05 2.1
2.66 2.6
Cs
Tl
Po
At
Rn
2.04 2.33 2.02 2.0
2.2
-
Ba
b) ethyne, C2H2
b)
Cl
Ne
K
Pb
What is the oxidation number of carbon in each of the following compounds:
C2H6 = 2 C3− + 6 H+
S
F
1.61 1.90 2.19 2.58 3.16 -
Example
a)
P
O
0.93 1.31
0.79 0.89
a) ethane, C2H6
N
C2H2 = 2 C− + 2 H+
C = −3 and H = +1
C = −1 and H = +1
2(−3) + 6(+1) = 0
2(−1) + 2(+1) = 0
-3-
Bi
Oxidation Numbers of Atoms in Polyatomic Ions or Molecules
Example
What are the oxidation numbers of the elements in the fluorosulfate ion, FSO3−?
FSO3− = F− + S6+ + 3 O2−
F = −1, S = +6, and O = −2
(−1) + (+6) + 3(−2) = −1
Example
What are the oxidation numbers of each of the elements in sodium perxenate, Na4XeO6?
Na
Xe
O
+1
Xe
−2
4(+1) + (Xe) + 6(−2) = 0
4 + (Xe) + (−12) = 0
Xe = +8
Na = +1, Xe = +8, and O = −2
Example
What are the oxidation numbers of each of the elements in ammonium carbonate, (NH4)2CO3?
NH4+ =
N
H
C
O
N
+1
C
−2
(N) + 4(+1) = +1
CO32− = (C) + 3(−2) = −2
(N) + 4 = +1
(C) + −6 = −2
N = −3
C = +4
N = −3, H = +1, C = +4, and O = −2
Determine Oxidation Numbers Assign
Redox Reactions





An increase in the oxidation number of an atom indicates that it has lost electrons and has
undergone oxidation.
A decrease in the oxidation number of an atom means an increase in the number of
electrons and indicates reduction.
Since electrons are neither created nor destroyed in chemical reactions, the total number of
electrons will always remain unchanged. This means that the loss of electrons must always
be accompanied by the opposite process – the gain of electrons. In other words, oxidation
and reduction are processes that always take place together.
The element that undergoes oxidation is the reducing agent.
The element that undergoes reduction is the oxidizing agent.
-4-
Example
Which of the following reactions is a redox reaction?
 2 NaCl(aq) + Br2(aq)
a) 2 NaBr(aq) + Cl2(aq) 
 2 NaCl(aq) + H2S(g)
b) 2 HCl(aq) + Na2S(aq) 
 2 NaCl(aq) + Br2(aq)
a) 2 NaBr(aq) + Cl2(aq) 
+1 −1
0
+1 −1
 2 NaCl(aq) + H2S(g)
b) 2 HCl(aq) + Na2S(aq) 
0
+1 −1
Bromide to bromine increases from −1 to 0
(oxidation)
+1 −2
+1 −1
+1 −2
No change
Chlorine to chloride decreases from 0 to −1
(reduction)
Activity Series of Metals
Example
Use the activity series to predict whether a
reaction takes place when the following
substances are mixed. If there is a reaction,
complete and balance the equation, and
then write the net ionic equation.

a) Fe(s) + ZnSO4(aq) 

b) Al(s) + H2SO4(aq) 

c) Sn(s) + CuSO4(aq) 
 no reaction
a)Fe(s) + ZnSO4(aq) 
 Al2(SO4)3(aq) + 3
b)2 Al(s) + 3 H2SO4(aq) 
H2(g)
 2 Al3+(aq) + 3 H2(g)
2 Al(s)+ 6 H+(aq) 
The Activity Series
Lithium
Potassium
These metals displace hydrogen from water
Barium
Calcium
Sodium
Magnesium
Aluminium
Zinc
Chromium
These metals displace hydrogen from acids.
Iron
Cadmium
Nickel
Tin
Lead
Hydrogen
Copper
These metals do not displace hydrogen from
Mercury
Acids or water.
Silver
Gold
 SnSO4(aq) + Cu(s)
c) Sn(s) + CuSO4(aq) 
 Sn2+(aq) + Cu(s)
Sn(s) + Cu2+(aq) 
-5-
Oxidizing and Reducing Agents





A good oxidizing agent must be a good electron acceptor and a good reducing agent must be a
good electron donor.
The cheapest and most widely-used oxidizing agent is atmospheric oxygen.
Elemental chlorine is also a good oxidizing agent.
The most commonly-used reducing agent is hydrogen gas.
Reactive metals such as magnesium and aluminium are also good reducing agents.
Example
Consider the addition or removal of oxygen atoms, as well as the change in oxidation numbers, to
determine which substance is oxidized and which is reduced in the following reaction:

CH3OH(aq) + 2 HClO(aq) 
HCO2H(aq) + 2 HCl(aq) + H2O(l)
methanol
formic acid
hypochlorous acid
Carbon increases from −2 to +2 (oxidized)
Chlorine decreases from +1 to −1 (reduced)
Balancing Redox Equations Using Oxidation Numbers
In a balanced redox equation, the total increase in oxidation number of the element that is oxidized
must equal the total decrease in oxidation number of the element that is reduced.
Procedure:
1. Assign oxidation numbers to all the atoms in the equation.
2. Identify which atoms undergo a change in oxidation number.
3. Determine the ratio in which these atoms must react so that the total increase in oxidation
numbers equals the decrease.
4. Balance the redox participants in the equation.
5. Balance the other atoms by the inspection method.
Example: Balance the following redox reaction
HNO3 + H2S
Step 1:
+1+5−2
+1−2

 NO + S + H2O
+2−2
0
+1−2
Step 2:
Only nitrogen and sulfur undergo a change
Step 3:
Nitrogen: +5  +2, a decrease of 3
Sulfur: −2  0, an increase of 2
Step 4:
 2 NO + 3 S + H2O
2 HNO3 + 3 H2S 
Step 5:
 2 NO + 3 S + 4 H2O
2 HNO3 + 3 H2S 
-2-
Example: Balance the following redox reaction
 Mn2+ + HSO4− + H2O
MnO4− + H2SO3 + H+ 
Step 1:
+7 −2
+1+4−2
+1
+2
+1+6−2
Step 2:
Only manganese and sulfur undergo a change
Step 3:
Manganese: +7  +2, a decrease of 5
+1−2
Sulfur: +4  +6, an increase of 2
Step 4:
 2 Mn2+ + 5 HSO4− + H2O
2 MnO4− + 5 H2SO3 + H+ 
Step 5:
 2 Mn2+ + 5 HSO4− + 3 H2O
2 MnO4− + 5 H2SO3 + H+ 
Examples
a)
 HIO3 + NO2 + H2O
I2 + HNO3 
0
+1+5−2
+1+5−2
+4−2
+1−2
 2 HIO3 + NO2 + H2O
I2 + HNO3 
Only iodine and nitrogen undergo a change
Iodine: 0 → +10, an increase of 10
Nitrogen: +5 → +4, a decrease of 1
 2 HIO3 + 10 NO2 + H2O
I2 + 10 HNO3 
 2 HIO3 + 10 NO2 + 4 H2O
I2 + 10 HNO3 
b)
 Br2 + H2O
HBr + HBrO3 
+1−1
+1+5 −2
0
+1 −2
Each bromide undergoes a change to bromine
Bromine: −1 → 0, an increase of 1
Bromine: +5 → 0, a decrease of 5
 3 Br2 + H2O
5 HBr + HBrO3 
 3 Br2 + 3 H2O
5 HBr + HBrO3 
Balancing Using Oxidation #s Assign
-3-
Half-Reactions
By separating a redox reaction into two half-reactions, we can get a better understanding of how
redox reactions take place.
 MgCl2(s)
Mg(s) + Cl2(g) 
 Mg2+(s) + 2 e−
Mg(s) 
 2 Cl−(s)
Cl2(g) + 2 e− 
 Cu(NO3)2(aq) + 2 Ag(s)
Cu(s) + 2 AgNO3(aq) 
 Cu2+(aq) + 2 NO3−(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq) + 2 NO3−(aq) 
 Cu2+(aq) + 2 Ag(s)
Cu(s) + 2 Ag+(aq) 
 Cu2+(aq) + 2 e−
Cu(s) 
 2 Ag(s)
2 Ag+(aq) + 2 e− 
Example
Write the half-reactions for the reaction:
 2 AlCl3(aq) + 3 H2(g)
2 Al(s) + 6 HCl(aq) 
 2 Al3+(aq) + 3 H2(g)
2 Al(s) + 6 H+(aq) 
Oxidation:
 2 Al3+(aq) + 6 e−
2 Al(s) 
Reduction:
 3 H2(g)
6 H+(aq) + 6 e− 
Ion-Electron Method
We can use half-reactions to balance redox equations. When we do so, we are using the ion-electron
method.
1. Write balanced oxidation and reduction half-reactions.
2. Change the coefficients in the balanced half-reactions so that the number of electrons
produced in the oxidation half-reaction equals the number of electrons consumed in the
reduction half-reaction.
3. Add the two half-reactions to obtain a balanced net ionic equation for the redox
reaction.
-4-
In acidic solution: To write a balanced half-reaction for the oxidation of ethanol, C2H5OH to acetic
acid, CH3CO2H, in acidic solution, we would follow the steps outlined below.
1. Write the skeletal equation:
 CH3CO2H
C2H5OH 
2. Balance for species other than oxygen and hydrogen:
 CH3CO2H (no change)
C2H5OH 
3. Balance for oxygen using one water molecule for each oxygen you require:
 CH3CO2H
C2H5OH + H2O 
4. Balance for hydrogen using a hydrogen ion for each hydrogen you require:
 CH3CO2H + 4 H+
C2H5OH + H2O 
5. Balance for charge by adding electrons to either the product or reactant side:
 CH3CO2H + 4 H+ + 4 e−
C2H5OH + H2O 
In basic solution: We will use the reduction of the permanganate ion to manganese(IV) oxide in basic
solution as an example.
1. Write the skeletal equation:
 MnO2
MnO4− 
2. Balance for species other than oxygen and hydrogen:
 MnO2 (no change)
MnO4− 
3. Balance for oxygen atoms by adding one water molecule for each oxygen
that you require:
 MnO2 + 2 H2O
MnO4− 
4. Balance for hydrogen atoms by adding one hydrogen ion for every
hydrogen atom:
 MnO2 + 2 H2O
MnO4− + 4 H+ 
5. For every hydrogen ion add one hydroxide ion to both sides of the equation:
 MnO2 + 2 H2O + 4 OH−
MnO4− + 4 H+ + 4 OH− 
6. Combine hydrogen and hydroxide ions to form water:
 MnO2 + 2 H2O + 4 OH−
MnO4− + 4 H2O 
7. Balance for charge by adding electrons:
 MnO2 + 2 H2O + 4 OH−
MnO4− + 4 H2O + 3 e− 
8. Finally, cancel water molecules:
 MnO2 + 4 OH−
MnO4− + 2 H2O + 3 e− 
-5-
Example
Write a balanced half-reaction for the oxidation of aluminum metal to the aluminate ion (AlO2−) in
basic solution.
 AlO2−(aq)
Al(s) 
 AlO2−(aq)
Al(s) + 2 H2O(l) 
 AlO2−(aq) + 4 H+(aq)
Al(s) + 2 H2O(l) 
 AlO2−(aq) + 4 H+(aq) + 3 e−
Al(s) + 2 H2O(l) 
 AlO2−(aq) + 4 H+(aq) + 3 e− + 4 OH−(aq)
Al(s) + 2 H2O(l) + 4 OH−(aq) 
 AlO2−(aq) + 4 H2O(l) + 3 e−
Al(s) + 2 H2O(l) + 4 OH−(aq) 
 AlO2−(aq) + 2 H2O(l) + 3 e−
Al(s) + 4 OH−(aq) 
Balancing Half Reactions Assign
Combining Half-Reactions
We can combine the half-reactions for the reduction of iron(III) to iron(II) with the oxidation of
hydrogen gas to hydrogen ion:
 Fe2+(aq)
Fe3+(aq) + e− 
 2 H+(aq) + 2 e−
H2(g) 
 2 Fe2+(aq)
2 Fe3+(aq) + 2 e− 
(multiply by 2)
 2 H+(aq) + 2 e−
H2(g) 
___________________________________________________________________________________________________________________________
 2 Fe2+(aq) + 2 H+(aq)
2 Fe3+(aq) + H2(g) 
The Full Ion-Electron Method
A reaction between the purple permanganate ion and iron(II) ion in acid solution to produce the
colourless manganese(II) ion and the iron (III) ion.
 Mn2+ + Fe3+
MnO4− + Fe2+ 
 Mn2+ + 4 H2O
MnO4− 
 Mn2+ + 4 H2O
MnO4− + 8 H+ + 5 e− 
 Fe3+ + e−
Fe2+ 
 5 Fe3+ + 5 e−
5 Fe2+ 
(multiply by 5)
 Mn2+ + 4 H2O
MnO4− + 8 H+ + 5 e− 
 5 Fe3+ + 5 e−
5 Fe2+ 
___________________________________________________________________________________________________________________________________________
 Mn2+ + 4 H2O + 5 Fe3+
MnO4− + 8 H+ + 5 Fe2+ 
-6-
Example
Use the ion-electron method to write a balanced equation for the reaction between copper metal and
concentrated nitric acid to produce copper(II) ions and nitrogen dioxide gas.
 Cu2+ + NO2
Cu + HNO3 
 NO2 + H2O
HNO3 
 NO2 + H2O
HNO3 + H+ 
 NO2 + H2O
HNO3 + H+ + e− 
 Cu2+
Cu 
 Cu2+ + 2 e−
Cu 
 2 NO2 + 2 H2O
2 HNO3 + 2 H+ + 2 e− 
(multiply by 2)
 Cu2+ + 2 e−
Cu 
__________________________________________________________________________________________________________________________________
 2 NO2 + 2 H2O + Cu2+
2 HNO3 + 2 H+ + Cu 
See Ion Electron Assign
Electrochemical Cells



An electrochemical cell uses energy released from a spontaneous redox reaction to generate an
electric current.
The current is derived from the flow of electrons through metal, called metallic conduction, and
the movements of ions in solution, called electrolytic conduction.
A battery consists of a single electrochemical cell or a number of cells connected in series.
How Electrochemical Cells Work
When a piece of zinc metal is placed in an aqueous solution of copper(II) sulfate, a rapid
redox reaction takes place. Electrons are transferred from the zinc metal to the copper
ions. The reaction is spontaneous since zinc oxidizes more readily than copper. It appears
above copper in the activity series:
 Zn2+(aq) + Cu(s)
Zn(s) + Cu2+(aq) 

Oxidation:
 Zn2+(aq) + 2 e−
Zn(s) 
Reduction:
 Cu(s)
Cu2+(aq) + 2 e− 
The electrical circuit of an electrochemical cell consists of two parts, an external circuit and an
internal circuit.
-7-
Electrode: an electrically-conducting medium, usually a solid, that is used to make electrical contact
Anode: the site of oxidation and electron release
Cathode: the site of reduction and electron consumption
Potential difference: the force that pushes the electrons from the anode to the cathode (a.k.a. emf)
Volts: a unit used to measure potential difference
Salt bridge: a device filled with a salt solution that connects the half-cells to allow the passage of ions
and to maintain the electrical neutrality of the two solutions
As the reaction continues, the voltage decreases and finally becomes zero when the reaction reaches
equilibrium. The cell will then be “dead.”
Notation for an Electrochemical Cell
Zn( s ) Zn 2 ( aq ) (1 M ) Cu 2 ( aq ) (1 M ) Cu( s )
By convention the anode appears on the left, the cathode on the right. The single vertical
lines indicate the contact boundaries between phases in each half-cell. The double
vertical lines represent the salt bridge or separator. An electrochemical cell is referred
to as a standard cell when the reactant and product ions or molecules are present in
concentrations of 1.0 M.
-8-
Example
When the metal electrodes of an electrochemical cell are connected by a voltmeter, the meter indicates
a potential difference of 0.46 V.
a)
using the activity series (page 8), identify the half-reaction taking place at each
electrode.
From the position of the metals in the activity series we conclude that copper oxidizes more
readily than silver and that silver ions reduce more readily than copper ions. The half-reactions
are therefore:
b)
Anode (oxidation):
 Cu2+(aq) + 2 e−
Cu(s) 
Cathode (reduction):
 Ag(s)
Ag+(aq) + e− 
Write the shorthand notation for the cell, identifying the anode and cathode, and
indicate the direction of movement of electrons and ions.
The copper electrode is the anode; the silver electrode is the cathode. The correct
notation for this standard cell is
Cu( s ) Cu 2  ( aq ) Ag  ( aq ) Ag ( s )
 (Cu(s) to Ag(s))
e− 
 cathode
cations (+) 
 anions (−)
anode 
See Cell Assign
-9-
Standard Electrode Potentials


An almost unlimited number of half-cells can be constructed and combined into
electrochemical cells. However, the reduction potential of each half-cell, the tendency of its ions
and molecules to gain electrons, will be different. As a result, the electrodes will have different
electrical potentials.
Unfortunately we cannot measure the electrical potential of one half-cell. We can only measure
the voltage difference, or emf, between two half-cells by combining them into an
electrochemical cell. For example, a cell voltage of 1.10 V in the standard zinc-copper ion cell
tells us that the reduction potential below zinc in the activity series. It also means that the
oxidation potential (the tendency to lose electrons) is 1.10 V greater for zinc than copper.
Standard Half-cell Potentials

The standard hydrogen electrode has been assigned a standard potential (E °) of 0.00 V.
 H2(g)
2 H+(aq) + 2 e− 
1M



E ° = 0.00 V
100 kPa
T = 25°C
This electrode is then used as a reference to assign potentials to other half-cells. The measured
electromotive force of a standard electrochemical cell containing a hydrogen half-cell is the
standard potential (E °) of the second half-cell.
The standard reduction potential of a half-cell is a measure of the tendency to gain electrons
from the hydrogen half-cell.
Ions or molecules with negative reduction potentials gain electrons less easily than hydrogen
ions. Those with positive values gain electrons more easily than hydrogen ions.
o Lithium ions in aqueous solution have the weakest tendency to gain electrons (E ° =
−3.05 V); molecular fluorine has the greatest potential to gain electrons (E ° = +2.87 V).
Calculating Standard Cell Potentials

In an electrochemical cell, one half-cell undergoes reduction while the other undergoes
oxidation. Thus when we are combining half-reactions and calculating the cell potential, one of
the tabulated reduction equations will have to be reversed and the sign of the potential
changed. Let us find the potential for the following standard cell:
Fe( s ) Fe2 ( aq ) 1 M Cu 2 ( aq ) (1 M ) Cu( s )
From table:
 Fe(s)
Fe2+(aq) + 2 e− 
E ° = −0.44 V
 Cu(s)
Cu2+(aq) + 2 e− 
E ° = +0.34 V
The Fe 2 Fe half-cell is the anode. Thus we reverse the Fe 2 Fe standard half-cell reduction
reaction and the sign of E ° to give:
 Fe2+(aq) + 2 e−
Fe(s) 
E ° = +0.44 V
 Cu(s)
Cathode (reduction) Cu2+(aq) + 2 e− 
E ° = +0.34 V
Anode (oxidation)
The sum of these half-cell potentials provides the potential of the cell.
E °cell = 0.44 V + 0.34 V = 0.78 V
- 10 -
Example
Calculate the potential of the standard cell:
Zn( s ) Zn 2 ( aq ) (1 M ) Cl  ( aq ) (1 M ) Cl2( g ) (100 kPa), Pt( s )
First we determine the half-reactions from the shorthand notation of the cell. We reverse the Zn Zn2+
standard half-cell reduction and the sign of E ° given in table thus,
 Zn2+(aq) + 2 e−
Zn(s) 
E ° = +0.76 V
 2 Cl−(aq)
Cathode (reduction) Cl2(g) + 2 e− 
E ° = +1.36 V
Anode (oxidation)
E °cell = 0.76 V + 1.36 V = 2.12 V
Predicting Spontaneity of Redox Reactions
Reduction potentials can also be used to predict whether a redox reaction is spontaneous or
nonspontaneous. A redox reaction is spontaneous when the sum of the potentials of the half-reactions
is positive. The reaction is nonspontaneous when the sum is negative.
Example
Predict whether sulfur dioxide and bromine will react in aqueous solution given the equation:
 SO42−(aq) + 2 Br−(aq) + 4 H+(aq)
SO2(g) + Br2(aq) + 2 H2O(l) 
Oxidation:
 SO42−(aq) + 4 H+(aq) + 2 e−
SO2(g) + 2 H2O(l) 
E ° = −0.18 V
Reduction:
 2 Br−(aq)
Br2(aq) + 2 e− 
E ° = +1.09 V
E °cell = −0.18 V + 1.09 V = +0.91 V
Half Cell Potentials and Spontaneity Assign
- 11 -
Standard Reduction Potentials at 25oC*
 4 OH-(aq)
O2(g) + 2 H2 + 4 e- 
+0.40
Half-Reaction
E0 (V)
 2 I-(aq)
I2(s) + 2 e- 
+0.53
 Li(s)
Li+(aq) + e- 
-3.05
+0.59
 K(s)
K+(aq) + e- 
-2.93
 MnO2(s) +
MnO4-(aq) + 2 H2O + 3 e- 
4 OH (aq)
 Ba(s)
Ba2+(aq) + 2 e- 
-2.90
 H2O2(aq)
O2(g) + 2 H+(aq) + 2 e- 
+0.68
 Sr(s)
Sr2+(aq) + 2 e- 
-2.89
 Fe2+(aq)
Fe3+(aq) + e- 
+0.77
 Ca(s)
Ca2+(aq) + 2 e- 
-2.87
 Ag(s)
Ag+(aq) + e- 
+0.80
 Na(s)
Na+(aq) + e- 
-2.71
 2 Hg(l)
Hg22+(aq) + 2 e- 
+0.85
 Mg(s)
Mg2+(aq) + 2 e- 
-2.37
 Hg22+(aq)
2 Hg2+(aq) + 2 e- 
+0.92
 Be(s)
Be2+(aq) + 2 e- 
-1.85
+0.96
 Al(s)
Al3+(aq) + 3 e- 
-1.66
 NO(g) + 2
NO3-(aq) + 4 H+(aq) + 3 e- 
H2O
 2 Br-(aq)
Br2(l) + 2 e- 
+1.07
 2 H2O
O2(g) + 4 H+(aq) + 4 e- 
+1.23
 Mn2+(aq) +
MnO2(s) + 4 H+(aq) + 2 e- 
2 H2O
+1.2`3
+1.33
Mn2+(aq)
+2
e-

 Mn(s)
 H2(g) + 2 OH-(aq)
2 H2O + 2 e- 
-1.18
-0.83

 Zn(s)
-0.76
 Cr(s)
Cr3+(aq) + 3 e- 
-0.74
 Fe(s)
Fe2+(aq) + 2 e- 
-0.44
2
Cr2O72-(aq) + 14 H+(aq) + 6 e- 
3+
Cr (aq) + 7 H2O
 Cd(s)
Cd2+(aq) + 2 e- 
-0.40
 2 Cl-(aq)
Cl2(g) + 2 e- 
+1.36
 Pb(s) + SO42-(aq)
PbSO4(s) + 2 e- 
-0.31
 Au(s)
Au3+(aq) + 3 e- 
+1.50
 Co(s)
Co2+(aq) + 2 e- 
-0.28
+1.51
 Ni(s)
Ni2+(aq) + 2 e- 
-0.25
 Mn2+(aq)
MnO4-(aq) + 8 H+(aq) + 5 e- 
+ 4 H2O
+1.61

 Sn(s)
-0.14
 Ce3+(aq)
Ce4+(aq) + e- 
+1.70

 Pb(s)
-0.13

PbO2(s) + 4 H+ + SO42-(aq) + 2 e- 
PbSO4(s) + 2 H2O
 H2(g)
2 H+(aq) + 2 e- 
0.00
 2 H2O
H2O2(aq) + 2 H+(aq) + 2 e- 
+1.77
 Sn2+(aq)
Sn4+(aq) + 2 e- 
+0.13
 Co2+(aq)
Co3+(aq) + e- 
 Cu+(aq)
Cu2+(aq) + e- 
+0.13
+0.14
 2 SO42−
S2O82− + 2 e− 
+1.82
+2.01
 O2(g) + H2O
O3(g) + 2 H+(aq) + 2 e- 
+2.07
 SO2(g) + 2
SO42-(aq) + 4 H+(aq) + 2 e- 
H 2O
+0.20
 F-(aq)
F2(g) + 2 e- 
+2.87
 Ag(s) + Cl-(aq)
AgCl(s) + e- 
+0.22
*For
 Cu(s)
Cu2+(aq) + 2 e- 
+0.34
Zn2+(aq)
Sn2+(aq)
Pb2+(aq)
+2
e-
+2
e-
+2
e-
 H2S
S + 2 H+ 2 e− 
all half-reactions the concentration is 1M
for dissolved species and the pressure is 1 atm
for gases. These are standard state values.