Oxidation-Reduction (Redox)
Electrochemistry involves the study of the conversions between chemical and electrical energy.
Voltaic (galvanic) cells use chemical reactions to produce an electric current.
Batteries (electrochemical cells) utilize oxidation-reduction reactions to provide an electric current at a
constant voltage.
Electrolysis involves the use of electrical energy to produce chemical changes.
OXIDATION-REDUCTION reactions involve the transfer of electrons between atoms.
Oxidation involves the loss of electrons which makes the oxidized substance more positive, thus
increasing its ability to combine with oxygen.
Reduction involves the gain of electrons which makes the reduced substance more negative thus
reducing its ability to combine with oxygen.
The substance that is oxidized gives up electrons – allowing another substance to gain electrons
(reducing it). Thus the oxidized substance is classified as the reducing agent.
The substance that is reduced gains electrons – allowing another substance to lose electrons
(oxidizing it). Thus the reduced substance is classified as the oxidizing agent.
In a chemical reaction, oxidation cannot occur without an accompanying reduction.
Oxidation numbers (a.k.a. oxidation states) are a “bookkeeping” tool to help determine whether or
not electrons have been transferred during a chemical reaction.
Oxidation numbers are assigned on a per atom basis. The following general “rules” are adapted
mainly from pp. 138-139 of General Chemistry (6th Ed.), Whitten, Davis, and Peck, Saunders
College Publishing will help with the assignment of oxidation numbers. In order of precedence the
rules are:
1.
2.
3.
4.
5.
6.
7.
The atoms of free, uncombined elements have oxidation numbers of zero. This includes
the diatomic elements: H2, N2, O2, F2, Cl2, Br2, and I2 as well as O3, P4, S8, and C60.
The oxidation number of an element in a monatomic ion is equal to the charge on the ion.
The sum of the oxidation numbers of all the atoms in a compound is zero.
The sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on
the ion.
Fluorine’s oxidation number is 1- in all of its compounds.
Hydrogen’s oxidation number is 1+ in most of its compounds, except its oxidation number is
1- in metallic hydrides.
Oxygen’s oxidation number is 2- in its compounds, except for:
(a)
peroxides, O22–, where each O has a 1-. Examples: H2O2, Na2O2, and CaO2.
(b)
superoxides, O2–, where each O has a 1/2–. Examples: KO2 and RbO2.
(c)
In OF2, O has a 2+.
Page 1 of 10
8.
9.
The element’s position on the periodic table often helps with the assignment of oxidation
numbers:
(a)
Group 1 metals have an oxidation number of 1+ in all of their compounds.
(b)
Group 2 metals have an oxidation number of 2+ in all of their compounds.
(c)
Group 3 metals have an oxidation number of 3+ in all of their compounds with a few
rare exceptions.
(d)
Group 14 can an oxidation number of 2+ or 4+, except that C and Si often have
other charges.
(e)
Group 15 elements have an oxidation number of 3– in binary compounds with
metals, hydrogen, or in ammonium. They can also have an oxidation number of 3+
or 5+ in compounds with more electronegative elements.
(f)
Group 16 (except O) elements have 2– in binary compounds with metals or
hydrogen, or with ammonium. They can also have an oxidation number of 4+ or 6+
with O and the lighter halogens.
(g)
Group 17 elements have 1– in binary compounds with metals or hydrogen, or with
ammonium. Cl, Br, and I can have an oxidation number of 1+, 3+, 5+, or 7+ with O
or a lighter halogen.
The oxidation number for the transition (Groups 3-12) and the inner-transition metals must
be determined based on what they are combined with – or from the Roman numeral in their
compound name. Exceptions: Ag forms 1+, Cd and Zn form 2+,
Example:
Zn(s) + 2 HBr(aq)  ZnBr2(aq) + H2(g)
Zn(s) and H2(g) have oxidation numbers of zero
H+(aq) has a 1+ oxidation number
Br–(aq) is a spectator ion and has an oxidation number of 1–.
Zn2+(aq) has an oxidation number of 2+.
Zn(s)  Zn2+(aq) + 2 e–
2 H+(aq) + 2 e–  H2(g)
is the oxidation, Zn becomes more +
is the reduction, H+ becomes less +
Zn(s) is the reducing agent and H+(aq) is the oxidizing agent
Common Oxidizing and Reducing Agents, Kotz & Treichel, Chemistry & Chemical Reactivity,
3rd Ed., p. 193
Oxidizing Agent
Reaction Product
Reducing Agent
Reaction Product
H+, hydrogen ion or H
2–
O2, oxygen
O , oxide
H2, hydrogen
combined in H2O
Mn+, metal ions such
–
–
Halogens: F2, Cl2, Br2,
Halide ion: F , Cl ,
M, metals such as Na,
as Na+, K+, Fe2+ or
or I2
Br–, or I–
K, Fe, and Al
Fe3+, or Al3+
Nitrogen oxides, such
C, carbon (used to
HNO3, nitric acid
CO and CO2
as NO and NO2
reduce metal oxides)
Cr2O72–, dichromate
Cr3+, chromium (III) ion
ion
(in acid solution)
2+
Mn
, manganese (II)
MnO4–
ion (in acid solution)
Page 2 of 10
Balancing redox equations
Step 1:
Assign oxidation numbers to the various atoms to determine whether or not the reaction
involves an electron transfer.
Step 2:
Separate the reaction into reduction and oxidation half-reactions.
Step 3:
Balance each half-reaction by mass (number of atoms of each element). Note: The
procedure differs depending on whether the solution is acidic or basic.
Step 4:
Balance each half-reaction by charge. Add electron(s) to whichever side has the greater
positive net charge.
Step 5:
Multiple each half-reaction by appropriate factors so that each contains the same number of
electrons.
Step 6:
Add the half-reactions to obtain a balanced overall equation.
Step 7:
Cancel common reactants and products.
Step 8:
Verify that the resulting net equation is balanced both by mass and charge.
EXAMPLE 1:
Balance the following acidic redox reaction:
C2H5OH(aq) + Cr2O72–(aq)  CH3CO2H(aq) + Cr3+(aq)
2–
6+
1+ 2 – 1+
2– 2–
0
1+
0 2–
1+
Step 1:
3+
C2 H 5 O H (aq) !!!+!Cr2 O7 (aq) !!!!C H 3 C O 2 H (aq) !!+!Cr(aq)
Step 2:
Oxidation:
C2 H 5 O H (aq) !!!!!!!C H 3 C O 2 H (aq) !
Reduction:
3+
Cr2 O7 (aq) !!!!Cr(aq)
Step 3:
2–
6+
1+ 2 – 1+
0
1+
0 2–
1+
2– 2–
First we balance the various elements (except H and O). Then, since this is acidic, we add
H2O to the oxygen deficient side and H+ to balance the hydrogens.
+
C2 H 5 O H (aq) !+!H 2O(l ) !!!!!!C H 3 C O 2 H (aq) !!+!!4!H (aq)
!
+
3+
Cr2O72 – (aq) !+!14!H (aq)
!!!2!Cr(aq)
!!+!!7!H 2O(l )
Step 4:
+
C2 H 5 O H (aq) !+!H 2O(l ) !!!!!!C H 3 C O 2 H (aq) !!+!!4!H (aq)
!+!4!e –
+
3+
6!e – !+!Cr2O72 – (aq) !+!14!H (aq)
!!!2!Cr(aq)
!!+!!7!H 2O(l )
Page 3 of 10
Step 5:
Since the two half-reactions contain a different number of electrons, we must find the least
common multiple – in this case, it is 12, so we triple the first equation and double the second
equation:
+
3![C2 H 5 O H (aq) !+!H 2O(l ) !!!!!!C H 3 C O 2 H (aq) !!+!!4!H (aq)
!+!4!e – ]
+
3+
2![6!e – !+!Cr2O72 – (aq) !+!14!H (aq)
!!!2!Cr(aq)
!!+!!7!H 2O(l ) ]
Steps 6 & 7:
+
–
3+
3!C2 H 5OH (aq) !+!16!H (aq)
!!+!2!Cr2O72!(aq)
!!!3!CH 3CO2 H (aq) !+!4!Cr(aq)
!+!11!H 2O(l )
Step 8:
The charges and number of atoms are equal on both sides.
EXAMPLE 2:
Balance the following basic redox reaction:
–
2"
SnO22!(aq)
!+!Bi(OH )3!(s ) !!!Bi(s ) !+!SnO3!(aq)
2+ 2–2–
3+ 2 – 1–
4+ 2–2"
0
Step 1:
Sn O 2 !(aq) !+!Bi(O H )3!(s ) !!!Bi (s ) !+!Sn O 3!(aq)
Step 2:
Oxidation:
Reduction:
Step 3:
–
2"
SnO22!(aq)
!!!!!SnO3!(aq)
Bi(OH )3!(s ) !!!Bi(s )
Balance each element (except for H and O). Since this is a basic solution, we balance the
oxygens using OH– and H2O. Add OH–- to the oxygen deficient side and, if necessary, add
H2O to the other side.
–
Bi(OH )3!(s ) !!!Bi(s ) !+!3!OH (aq)
!
–
–
2"
2!OH (aq)
!+!SnO22!(aq)
!!!!!SnO3!(aq)
!+!H 2O(l )
Step 4:
–
3!e – !+!Bi(OH )3!(s ) !!!Bi(s ) !+!3!OH (aq)
!
–
–
2"
2!OH (aq)
!+!SnO22!(aq)
!!!!!SnO3!(aq)
!+!H 2O(l ) !+!2!e –
Step 5:
Since the half-reactions have a different number of electrons, we find the least common
multiple – 6. So we double reduction half-reaction and triple the oxidation half-reaction.
–
2![3!e – !+!Bi(OH )3!(s ) !!!Bi(s ) !+!3!OH (aq)
!]
–
–
2"
3![2!OH (aq)
!+!SnO22!(aq)
!!!!!SnO3!(aq)
!+!H 2O(l ) !+!2!e – ]
Steps 6 & 7:
–
2–
3!SnO22!(aq)
!+!2!Bi(OH )3!(aq) !!!2!Bi(s ) !+!3!SnO3!(aq)
!+!3!H 2O(l )
Step 8:
The charges and the number of atoms are the same on both sides.
NOTE:
Basic redox reactions could be balanced using the same process as for acidic redox reactions
then the H+ ions could be “cancelled out” by adding the appropriate number of OH– ions to both sides.
Page 4 of 10
Electrochemical cells (a.k.a as galvanic or voltaic cells – named after Luigi Galvani and Allessandro Volta
who constructed early versions) use spontaneous redox reactions to produce electricity.
In an electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode. A salt bridge
containing inert electrolytes is used to connect the two solutions so that anions and cations can flow between
the electrodes to complete the circuit – if the two solutions are mixed in the same container, the redox
reaction would occur without producing useful electrical work. Electrons flow from the anode to the
cathode due to the difference in electrical potential energy between the two electrodes. A voltmeter is used
to measure the cell potential, E, (a.k.a. cell voltage or electromotive force – emf). The cell voltage
depends on the nature of the electrodes, the type and concentration of the ions, and the temperature.
A cell diagram can be used to represent an electrochemical cell. If the electrochemical cell consists of the
following half-cell reactions:
Zinc electrode (anode)
Copper electrode (cathode)
Zn(s)  Zn2+(aq) + 2 e–
Cu2+(aq) + 2 e–  Cu(s)
Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu(s)
This notation indicates that oxidation half-reaction consists of a zinc metal anode is immersed in a 1 M
solution of Zn2+ (ZnSO4 for example) which is connected with a salt bridge (or other suitable conductor) to
the reduction half-reaction consisting of a copper metal cathode immersed in a 1 M solution of Cu2+ (CuSO4
for example).
Standard reduction potentials (voltages) are used to indicate cell potentials. Standard reduction potentials
are experimentally determined relative to the hydrogen “standard electrode”:
2 H+ (1 M) + 2 e–  H2 (g) (1 atm, 25 oC)
By definition, this “standard electrode” is assigned an Eo value of 0.00 V
For an electrochemical cell, the standard emf, Eocell, is determined (using the standard reduction potentials
for both electrodes) by the formula:
Eocell = Eocathode – Eoanode
Using the Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu(s) cell as an example, using Eo values from a Standard
Reduction Potentials table, we would have:
Zn2+(aq) + 2 e–  Zn(s)
Eo = – 0.76 V
Cu2+(aq) + 2 e–  Cu(s)
Eo = + 0.34 V
NOTE:
Eocell = Eocathode – Eoanode
Eocell = + 0.34 V – (– 0.76 V) = + 1.10 V
Page 5 of 10
This is written as the reverse of the
actual oxidation.
When deciding trying to determine which half-reaction will be oxidation and which will be reduction, refer
to the standard potentials. The half-reaction with the more positive standard reduction potential will occur
at the cathode (reduction).
Example 3:
If an electrochemical cell consists of an aluminum electrode in 1.0 M Al(NO3)3 and a barium
electrode in 1.0 M Ba(NO3)2: (a) identify the anode and cathode, (b) draw a cell diagram,
and (c) determine the cell’s standard emf.
Referring to a standard reduction potentials table:
Al3+(aq) + 3 e-  Al(s)
Ba2+(aq) + 2 e–  Ba(s)
Eo = – 1.66 V
Eo = – 2.90 V
(a)
Since Al3+ has the more positive (less negative) standard reduction potential, it will be reduced at the
cathode. Ba(s) will be oxidized at the anode.
(b)
Ba(s) | Ba2+ (1 M) || Al3+ (1 M) | Al(s)
(c)
Eocell = Eocathode – Eoanode
Eocell = – 1.66 V – (–2.90 V) = + 1.24 V
Since the Eocell > 0, the reaction will occur spontaneously as written. If Eocell < 0, the reverse reaction
would be spontaneous.
NOTE:
Do not adjust the half-reaction voltages to reflect the coefficients in the balanced
equation.
Spontaneity of Redox Reactions
ΔGo = – nFEocell = –RT ln K
where ΔGo represents the standard Gibbs Free Energy
n represents the number of moles of electrons that flows through the circuit
C
J ! mol
F is the Faraday constant, the charge on one mole of electrons (1 F = 96,485.3
= 96,485.3
)
mol
V
J
R = 8.314
mol ! K
K represents the equilibrium constant
Since ΔGo = – nFEocell = –RT ln K
J
(8.314!
)(298.15!K )
RT
0.0257!V
o
mol
!
K
ln!K !=!
(ln!K )!=!
!(ln!K )
E cell =
J
!
mol
nF
n
n(93500!
)
V
ΔGo
–
0
+
K
>1
1
<1
Eocell
+
0
–
Reaction under Standard-State Conditions
Spontaneous
At Equilibrium
Nonspontaneous. Reverse reaction is spontaneous.
Page 6 of 10
Reactant and product concentrations effect an electrochemical cell’s emf according to the following
relationships:
ΔG = –nFE and ΔGo = –nFEo, so ΔG = ΔGo + RT ln Q can be rearranged to obtain
RT
E!=!E o !–!
!ln!Q
nF
0.0257!V
The Nernst Equation, E = Eo –
!ln!Q , allows for the calculation of cell emf when the
n
concentrations are not 1.0 M.
For the reaction: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
E = 1.10 V –
Example 4:
0.0257!V
[Zn 2 + ]
!ln!
2
[Cu 2 + ]
What is the cell emf at 25 oC for the reaction
Fe(s) + Cd2+(aq)  Fe2+(aq) + Cd(s) when:
(a)
(b)
[Fe2+] = 0.010 M and [Cd2+] = 1.00 M
[Fe2+] = 1.00 M and [Cd2+] = 0.010 M
0.0257!V
[Fe2 + ]
0.0257!V 0.010
!ln!
= +0.04!V !–!
!ln
!=!+0.10!V
2+
2
[Cd ]
2
1.00
Since E > Eo, product formation is favored.
(a)
E = Eo –
(b)
E = 1.10 V –
0.0257!V
[Fe2 + ]
0.0257!V
1.00
!ln!
= +0.04!V !–!
!ln
!=!–0.02!V
2+
2
[Cd ]
2
0.010
Since E < Eo, reactant formation is favored. The reverse reaction occur under these
conditions.
Since the cell emf is effected by concentration, it is possible to create a concentration cell using the same
species in both anode and cathode half-cells using different concentrations.
E = Eo –
0.0257!V
[!]dilute
!ln!
n
[!]concentrated
What would be the emf for a concentration cell consisting of Ni metal electrodes in 0.001 M Ni2+ and 1.0 M
Ni2+? Since the reaction leads to an equilibrium concentration, oxidation of the Ni electrode (the anode)
will occur in the dilute solution and Ni2+ will be reduced at the Ni electrode (the cathode) in the
concentrated solution.
0.0257!V
[!]dilute
!ln!
n
[!]concentrated
0.0257!V
0.001!M
E=0–
!ln!
!=!+!0.0888!V
2
1!M
E = Eo –
Page 7 of 10
Batteries consist of one or more electrochemical cells that product direct electric current at a constant
voltage.
“Dry cell” batteries contain electrolytes in a paste, MnO2, a graphite cathode, and a zinc anode. Dry cells
batteries produce about 1.5 V.
Alkaline batteries consist of a powdered zinc metal anode immobilized in a gel surrounded by a KOH paste
sealed in a steel case. The cathode is a mixture of MnO2(s) and graphite. Alkaline batteries produce 1.55 V
at room temperature and offer better performance than to conventional dry cells.
Mercury batteries provide more consistent voltage (1.35 V) and capacity.
Rechargeable Cells
Nickel-Cadmium (Nicad) batteries are rechargeable and produce 1.30 V. Since Nicad batteries lose their
ability to be recharged, the toxic cadmium poses a major waste disposal problem.
Nickel-Metal-Hydride batteries use an metal alloy anode that can absorb hydrogen atoms. Water is
produced at the anode. Recharging reverses the process.
Lithium batteries use a lithium anode with either TiS2 or V6O13 used as the cathode. Lithium’s large
negative reduction potential and light mass contribute to a light battery with a higher voltage (about 3 V).
Lead storage batteries consist of a series of cells with a lead anode and a lead (IV) oxide cathode in a
sulfuric acid solution.
Anode:
Cathode:
Net:
Pb(s) + SO42–(aq)  PbSO4 (s) + 2 e–PbO2 (s) + 4 H (aq) + SO42–(aq) + 2 e–-  PbSO4 (s) + 2 H2O(l)
Pb(s) + PbO2 (s) + 4 H+(aq) + SO42–(aq)  2 PbSO4 (s) + 2 H2O(l)
+
A typical “12 Volt” car battery consists of six of these cells connected in series to produce 13.6
volts.
Car batteries are “recharged” by applying external voltage to reverse the above reactions.
Fuel cells use a continuous supply of reactants to produce a constant, efficient source of electric current. A
hydrogen-oxygen fuel cell produces 1.23 V using readily available fuels (obtained from water).
Corrosion is the electrochemical deterioration (oxidation) of metals. Most metals oxidize (rust). Precious
metals such as gold, platinum, silver, and copper oxidize much more slowly and are more likely to be found
in nature in their elemental form.
Metal corrosion can be slowed (or prevented) by anodic inhibition – coating the surface of the metal with
paint to prevent oxygen and water from coming in contact with the metal. A second method, passivation,
involves the application of a thin layer of metal oxide by treating the metal with a strong oxidizing agent
such as sulfuric acid or sodium chromate. Cathodic protection involves putting the metal in contact with a
more easily oxidized metal. Iron can be protected by applying a layer of zinc that acts as a sacrificial
anode.
Page 8 of 10
Electrolysis uses electrical energy to cause a nonspontaneous chemical reaction to occur. Electrolysis is
used to split water into hydrogen and oxygen or convert salt into sodium hydroxide, hydrogen, and chlorine.
Since chloride is more easily than oxide, it is released rather than oxygen at the anode. The reaction vessel
is called an electrolytic cell.
Example 5:
What volume of gas can be collected at 30.0 oC and 99.5 kPa when 2.5 A is passed through
through a dilute sulfuric acid solution for 5.25 hours?
Anode (oxidation)
2 H2O(l)  O2 (g) + 4 H+(aq) + 4 e–
Cathode (reduction) 4 [H+(aq) + e–  1/2 H2 (g)]
Net Reaction:
2 H2O(l)  O2 (g) + 2 H2 (g)
mol!O2 !=!2.5!A!x!5.25!h!x!
3600!s
1!C
1!mol!e– 1!mol!O2
!x!
!x!
!x!
1!h
1!A ! s 96500!C 4!mol!e–
mol!O2 !=!0.12!mol
2!mol!H 2
!=!0.24!mol
1!mol!O2
total!mol!gas!=!0.12!+!0.24!=!0.36!mol
mol!H 2 !=!0.12!mol!O2 !x!
PV !=!nRT
V !=!
nRT
!=!
P
L ! atm
)(30.0!+!273.15!K )
mol ! K
1!atm
(99.5!kPa!x!
101.325!kPa)
(0.36!mol)(0.082057!
V !=!9.1!L
Example 6:
If electricity costs $0.105 per kW-h, what would it cost to produce 1,000 kg of chlorine gas
by the electrolysis of brine (salt water) using 4.6 V current?
10 3 !g
1!mol!Cl2
2!F
9.65!x!10 4 !C
1!J
1!W ! s
1000!kg!Cl2 !x!
!x!
!x!
!x!
!x!4.6!V !x!
!x!
!
1!kg
70.91!g!Cl2 1!mol!Cl2
1!F
C !V
1!J
1!kW
1!h
$0.105
x! 3 !x!!
!x!
!=!$365.15
10 !W
3600!s 1!kW ! h
NOTE:
It might be useful to remember the following conversion factor:
Page 9 of 10
1!kW ! h
3.6!x!10 6 !J
ACTIVITY SERIES OF THE ELEMENTS
METALS
lithium
potassium
calcium
sodium
magnesium
aluminum
zinc
chromium
iron
nickel
tin
lead
hydrogen
copper
mercury
silver
platinum
gold
NONMETALS
highest
activity
fluorine
chlorine
bromine
iodine
lowest
activity
Page 10 of 10