Oxidation–Reduction Reactions
Oxidation Loss of electrons
Reduction Gain of electrons
Zn( s ) + 2 H + ( aq ) → Zn +2 ( aq ) + H 2 ( g )
Zn is oxidized
H+ is reduced
Zn is the reducing agent
H+ is the oxidizing agent
reductant
oxidant
Oxidation Numbers
1. For an atom in its elemental form the
oxidation number is zero.
2. For any monoatomic ion the oxidation
number equals the charge on the ion.
K+, O-2, Al+3, etc
3. Nonmetals usually have negative
oxidation numbers.
a. the oxidation number of O is usually -2
b. the oxidation number of H is +1 when it is
bonded to a non metal (H2O, NH3, H3PO4,
etc. ) and -1 when it is bonded to a metal
(CaH2, NaH, etc.)
c. the oxidation number of F is always -1.
The other halogens are usually -1 but when
combined with oxygen in oxycations it is +1
( OCl-1, O2Cl-2, etc. )
“If this were an ionic compound, what
would the charges be?”
Identify the oxidation numbers
H 2O
NiO2
−
4
MnO
Ni( OH )2
Al( OH )4−
−2
4
C2 O
NH 3
−2
7
Cr2O
ClO3−
Balancing Oxidation Reduction Reactions
Half-Reaction Method
Sn +2 ( aq ) + Fe +3 ( aq ) → Sn +4 ( aq ) + Fe +2 ( aq )
Sn +2 ( aq ) → Sn +4 ( aq ) + 2e −
Fe +3 ( aq ) + e − → Fe +2 ( aq )
2 Fe +3 ( aq ) + 2e − → 2 Fe+2 ( aq )
Sn +2 ( aq ) → Sn +4 ( aq ) + 2e −
Sn +2 ( aq ) + 2 Fe+3 ( aq ) → Sn +4 ( aq ) + 2 Fe+2 ( aq )
MnO4− ( aq ) + C2O4−2 ( aq ) → Mn +2 ( aq ) + CO2 ( g ) Acidic solution
2 MnO4− ( aq ) + 5C2O4−2 ( aq ) + 16 H + ( aq ) → 2 Mn +2 ( aq ) + 10CO2 ( g ) + 8 H 2O( l )
2 MnO4− ( aq ) + 5C2O4−2 ( aq ) + 16 H + ( aq )
→
2 Mn +2 ( aq ) + 10CO2 ( g ) + 8H 2O( l )
MnO4− ( aq ) → Mn +2 ( aq ) + 4 H 2O( l )
8 H + ( aq ) + MnO4− ( aq ) → Mn +2 ( aq ) + 4 H 2O( l )
5e − + 8 H + ( aq ) + MnO4− ( aq ) → Mn +2 ( aq ) + 4 H 2O( l )
C2O4−2 ( aq ) → 2CO2 ( g )
C2O4−2 ( aq ) → 2CO2 ( g ) + 2e −
10e − + 16 H + ( aq ) + 2 MnO4− ( aq ) → 2 Mn +2 ( aq ) + 8 H 2O( l )
5C2O4−2 ( aq ) → 10CO2 ( g ) +10e −
2MnO4− ( aq ) + 5C2O4−2 ( aq ) + 16 H + ( aq )
→
2Mn +2 ( aq ) + 10CO2 ( g ) + 8H 2O( l )
Rules (Acidic Solution)
1. Divide the reaction into half-reactions
2. Balance each half-reaction
a. First balance elements other than O & H
b. Balance O atoms using H2O
c. Balance H atoms with H+
d. Balance charge with e3. Multiply each half reaction by an integer
so each has the same number of electrons
4. Add half reactions and simplify
5. Check!
Balancing in a basic solution
Use OH- & H2O
Given
CN − ( aq ) + MnO4− ( aq ) → CNO − ( aq ) + MnO2 ( s )
Balanced Equation
3CN − ( aq ) + 2MnO4− ( aq ) + H 2O( l )
→
3CNO − ( aq ) + 2MnO2 ( s ) + 2OH − ( aq )
A Fundamental Observation
When a strip of Zn is placed in a solution
containing Cu+2(aq) ions the Cu+2(aq) ions
react with the Zn and form Cu metal while
the Zn dissolves, forming Zn+2(aq) ions.
Cu +2 ( aq ) + Zn( s ) → Cu( s ) + Zn +2 ( aq )
Voltaic Cell
A device in which a spontaneous oxidationreduction reaction occurs with the passage
of electrons through an external circuit.
Electrodes
Two solid metals connected
by the external circuit
Anode
Electrode at which oxidation
occurs
Cathode
Electrode at which reduction
occurs
Salt Bridge
Electrical contact between
the two solutions that
prevents them from mixing
but permits ion flow
Electrodes need not enter into reaction
Beaker A
H2O, K2Cr2O7, & H2SO4
Beaker B
HI & H2O
Put Pt electrode in both beakers
Connect electrodes with wire
Connect beakers with salt bridge
Spontaneous reaction
Cr2O7−2 ( aq ) + 14 H + ( aq ) + 6 I − ( aq )
→
2Cr +3 ( aq ) + 3I 2 ( s ) + 7 H 2O( l )
Cell EMF
The potential energy of the electrons is
higher at the anode
so they flow spontaneously through the
external circuit to the cathode.
The difference in potential energy is
measured in volts
A volt is one joule per coulomb of charge
1V = 1
J
C
Charge on one electron is 1.602 x 10-19 C
1Coulomb corresponds to the charge
on 6.24 x 1018 electrons.
Potential difference is called the
electromotive force (EMF) or cell voltage
0
Zn( s ) + Cu +2 ( aq,1M ) → Zn +2 ( aq,1M ) + Cu( s ) Ecell
= +1.10V
Standard Reduction (Half-Cell) Potentials
A→ B
AB
Ecell
= E( B ) − E( A )
X→A
XA
= E( A ) − E( X )
Ecell
X →B
XB
= E( B ) − E( X )
Ecell
AB
XB
XA
Ecell
= Ecell
( B ) − Ecell
( A)
X is the reference electrode
XB
XA
Ecell
( B )& Ecell
( A)
are called the standard reduction potentials
XB
0
Ecell
( B ) = Ered
(B)
0
XA
Ecell
( A ) = Ered
( A)
Reference Electrode is the Standard
Hydrogen Electrode
0
2 H + ( aq,1M ) + 2e − → H 2 ( g ,1atm ) Ered
= 0V
EMF Example
Determine the half-reactions that occur at
the anode and cathode and the cell EMF for
the half cells
+3
Al( s ) / Al ( aq )
I 2 ( s ) / I −1( aq )
Cr2O7−2 ( aq ) / Cr +3 ( aq )
−1
I 2 ( s ) / I ( aq )
+2
Co ( aq ) / Co( s )
AgCl( s ) / Ag( s )
EMF, Free Energy & the Nernst Equation
∆G = −nFEcell
F = the Faraday
the charge on one mole of electrons
(6.02214 x 10 23 e/mol)(1.60218 x 10 -19 C/e)=96500C/mol
n = the number of electrons transferred in the reaction
∆G = ∆G + RT ln Q
0
−nFEcell = −nFE
0
cell
Ecell = E
0
cell
Ecell = E
0
cell
+ RT ln Q
RT
−
ln Q
nF
0.0592
−
log Q
n
Example of Nernst Equation
Zn( s ) + Cu +2 ( aq ) → Zn +2 ( aq ) + Cu( s )
Ecell
0.0592V
[ Zn +2 ( aq )]
log
= 1.10V −
2
[ Cu +2 ( aq )]
[ Cu +2 ] = 5.0 M , [ Zn +2 ] = 0.05M
Ecell
0.0592V
0.05
= 1.10V −
log
2
5 .0
Ecell
0.0592V
= 1.10V −
log 10−2
2
Ecell = 1.10V + 0.0592V = 1.16V
∆G Examples
Given the reaction
3Ni +2 ( aq ) + 2Cr( OH )3 ( s ) + 10OH − ( aq )
→
3Ni( s ) + 2CrO4−2 ( aq ) + 8H 2O( l )
What is the value of n?
What is
∆G 0 ?
Same question for
1
2 Ag( s ) + O2 ( g ) + 2 H + ( aq ) → 2 Ag + ( aq ) + H 2O( l )
2
Nernst Equation Example
Zn( s ) + 2 H + ( aq ) → Zn +2 ( aq ) + H 2 ( g )
If the voltage in this cell is 0.45V at 250C
when [Zn+2] =1.0M and PH2=1.0 atm, what
is the concentration of H+?
Concentration Cell Examples
Two Hydrogen electrodes
Electrode 1
PH 2 = 1.00atm, [H + (aq)]=?
Electrode 2
PH 2 = 1.00atm, [H + (aq)]=1.00M
T=298K Ecell = 0.211 V
Current flow: electrode 1 to electrode 2
pH = ?
Consider two Zn(s)-Zn+2(aq) half cells
[ Zn +2 ( aq )]( cell 1 ) = 1.35M
[ Zn +2 ( aq )]( cell 2 ) = 3.75 x10−4 M
Which is anode?
What is emf of cell?
Strengths of Oxidizing/Reducing Agents
Standard reduction potentials can be used to
understand aqueous reaction chemistry
0
The more positive Ered
the greater
the tendency for the reactant of the
half reactant to be reduced and,
therefore, to oxidize another species.
F2 ( g ) + 2e − → 2 F − ( aq )
0
Ered
= +2.87 V
The half-reaction with the smallest
0
(algebraically) Ered
is most easily reversed as an oxidation
Li + ( aq ) + e − → Li( s )
0
Ered
= −3.05 V
Rank the following ions in order of
increasing strength as oxidizing agents
NO3− ( aq )
Ag + ( aq )
Cr2O7-2 ( aq )
Rank the following ions in order of
increasing strength as reducing agents
I − ( aq )
Fe(s)
Al(s)
Batteries
Portable, self-contained electrochemical
power source that consists of one or more
voltaic cells
Primary Cell
Cannot be recharged
Secondary Cell
Can be recharged
Lead-Acid Battery
Cathode:
PbO2 (s)+HSO4− ( aq ) + 3H + ( aq ) + 2e−
→
0
PbSO4 ( s ) + 2 H 2O( l ) Ered
= +1.685 V
Anode:
Pb( s ) + HSO4- ( aq )
→
PbSO4 ( s ) + H + ( aq ) + 2e −
0
= −0.356 V
Ered
Cell Reaction:
PbO2 (s)+Pb(s)+2HSO4- ( aq ) + 2 H + ( aq )
→
2 PbSO4 ( s ) + 2 H 2O( l )
0
Ecell
= E 0 ( cathode ) − E 0 ( anode )
0
Ecell
= ( +1.685 V ) - (-0.356 V ) = + 2.041 V
12 V lead-acid battery
consists of 6 voltaic cells and is
rechargeable
Alkaline Battery
Cathode:
2MnO2 (s)+2H 2O(l)+2e→
2 MnO( OH )( s ) + 2OH − ( aq )
Anode:
Zn(s)+2OH - (aq) → Zn(OH)2 (s)+2e0
Ecell
= 1.55V
Fuel Cells
Cathode :
0
O2 ( g ) + 2 H 2O( l ) + 4e − → 4OH − ( aq ) Ered
= +0.40V
Anode :
2 H 2 ( g ) + 4OH − ( aq ) → 4 H 2O( l ) + 4e−
0
Ered
= −0.83V
Re action :
0
2 H 2 ( g ) + 2O2 ( g ) → 2 H 2O( aq ) Ecell
= +1.23V
Batteries in series
Corrosion
Corrosion occurs when a metal undergoes a
spontaneous redox reaction and is converted
to an unwanted substance.
Oxidation is usually the thermodynamically
favored process for a metal
Corrosion of Iron
Preventing Iron from Corroding
Cathodic Protection
Electroplating
Electrolysis with an active electrode
Ni( s ) → Ni +2 ( aq ) + 2e −
0
Ered
= −0.28V
2 H 2O( l ) → O2 ( g ) + 4 H + ( aq ) + 4e −
Ni +2 ( aq ) + 2e − → Ni( s )
0
Ered
= +1.23V
0
Ered
= −0.28V
0
2 H 2O( l ) + 2e − → H 2 ( g ) + 2OH − ( aq ) Ered
= −0.83V
Quantitative Aspects of Electrolysis
Stoichiometry of a half-reaction shows how
many electrons are needed for the
electrolysis
+
−
Na + e → Na
Cu +2 + 2e − → Cu
+3
−
Al + 3e → Al
Current measured in amperes (A)
1A = 1
Coulomb
C
=1
sec ond
s
Coulombs of charge=magnitude of current x duration of current
Calculate the number of grams of Al
produced in 1.00 hour by the electrolysis of
molten AlCl3 if the current is 10.0A.
The half-reaction for the formation of
magnesium metal upon electrolysis of
molten MgCl2 is
Mg +2 + 2e − → Mg
Calculate the mass of magnesium formed
upon the passage of a current of 60.0A for
4.00x103 s
Electrical Work
Free energy change in a process is equal to
the maximum useful work that can be
extracted from the process.
∆G = wmax = − nFE
wmax = nFEexternal > 0
The unit of electrical power is the Watt(W)
Joule
J
1Watt = 1
=1
sec ond
s
Unit of energy employed by utility
companies is kilo-Watt-hour (kWh)
1kWh = 3.6 x10 J
6
Calculate the number of kilowatt-hours required
to produce 1.0 x 103 kg of Al by electrolysis of
Al+3 if the applied voltage is 5.50V
w = nFE
Cell EMF and Chemical Equilibrium
∆G = ∆G 0 + RT ln Q
At equilibrium
∆G = 0 & Q = K eq
0
∆G 0 = − RT ln K eq = −nFEcell
0
nFEcell
ln K eq =
RT
0
nEcell
; T=298K
log K eq =
0.0592
Calculate the equilibrium constant for the
oxidation of Fe+2 by O2 in an acidic solution
Fe +2 ( aq ) + O2 ( g ) → Fe +3 ( aq ) + H 2O( l )
Fe +2 → Fe +3 + e −
0
Ered
= +0.77V
O2 ( g ) → 2 H 2O( l )
4 H + ( aq ) + O2 ( g ) → 2 H 2O( l )
0
4 H + ( aq ) + O2 ( g ) + 4e − → 2 H 2O( l ) Ered
=+1.23V
0
n = 4 ; Ecell
= +1.23V − 0.77V = 0.46V
log K eq =
4 x 0.46V
= 31
0.0592V
K eq = 1.0 x1031