1
Lecture 9 Electrochemistry
Text: Chapter 12
Oxidation-Reduction Reactions
Loss of an electron - oxidation (LEO)
charge increases
Gain of an electron - reduction (GER)
charge decreases
Zn(s) + HCl(aq)  ZnCl(aq) + H2(g)

Zn s  2H aq   Zn
2
aq   H2 g
net ionic equation
Oxidation - the value of the positive charge increases
Zn s  Zn
2
aq
Reduction - the positive charge decreases (it is reduced in size)

2 H aq  H2 g
Must review Section K in Fundamentals section of text and know how to assign
oxidation numbers to atoms!!!
Half Reactions:
Zn(s)  Zn2+(aq) + 2e2e- + 2H+  H2 (g)
__________________________
oxidation half cell
reduction half cell
Zn(s) + 2H+  Zn2+(aq) + H2 (g)
overall reaction
Electrochemistry 1
2
Balancing Equations
Must balance the atoms and the total charge on both sides of the Chem Eqn.
Use the method of half reactions.
Treat oxidation as a separate reaction.
Balance its mass and charge then treat reduction as a separate reaction, balancing
its mass and charge, then add the reactions together.
Problem solving Tips
Two methods for balancing Redox equations. (Choose the one you prefer.)
Method One (See textbook)
Example:
Step 1. Verify that the reaction is an oxidation-reduction process by inspecting
oxidation numbers
Step 2. Separate the overall reaction into two half-reactions
Step 3. Balance each half-reaction for mass.
a. First balance the elements other than H and O
b. Next balance the O atoms by adding H2O.
c. Then balance the H atoms by adding H+ when in acidic solution (or by
adding OH- when in basic solution)
Step 4. Balance the half reactions for charge.
Electrochemistry 2
3
Step 5. Multiply the balanced half-reactions so the number of electrons lost in one
half-reaction equals the number of electrons gained in the other.
Step 6. Add the balanced half-reactions and simplifying by canceling species
appearing on both sides of the equation.
Step 7. Check the final result for mass and charge balance.
See Example 12.1, page 485.
Method Two (The “ECHO” method)
Set up half reactions, then balance the: Electrons, Charge, H2O
MnO4(aq) + H2C2O4(aq)  Mn2(aq) + CO2(g) in Acid solution
1(x) + 4(-2)=-1
2(1)+2(y)+4(-2)=0
1(+2)
1(z)+2(-2)=0
MnO4(aq) + H2C2O4(aq)  Mn2(aq) + CO2(g)
x= +7
y = +3
+2
Step 1) E: balance ox numbers and atoms with electrons
5e
+ 1(+7)
= 1(+2)
MnO4(aq)  Mn2(aq)
2(+3)
= 1(+4)
H2C2O4(aq)  CO2(g)
2(+3)
= 2(+4)

+2e
H2C2O4(aq)  2CO2(g)
Electrochemistry 3
z = +4
4
Result of E: find common denominator of electrons
2[ 5e + MnO4(aq)  Mn2(aq)
]
5[
H2C2O4(aq)  2CO2(g) + 2 e ]
___________________________________
10e + 2MnO4(aq) + 5H2C2O4(aq)  2Mn2(aq)+10CO2(g) + 10e
Step 2) C: balance charge with H+ ions (because in acid conditions)
2MnO4(aq) + 5H2C2O4(aq)  2Mn2(aq)+10CO2(g)
2 (-)  2(+2)
Balance charge: 6H+ + 2 (-)  2(+2)
Result:
6H+ + 2MnO4(aq) + 5H2C2O4(aq)  2Mn2(aq)+10CO2(g)
Step 3) Balance (either H or O) with H2O
6H
0H
+ 10H
 0H
16 H  0 H
+0H
add 8H2O
Resultant balanced equation:
6H+ + 2MnO4(aq) + 5H2C2O4(aq)  2Mn2(aq)+10CO2(g) + 8H2O
Verify !!!
Electrochemistry 4
5
Example 2 Balance in basic solution with ECHO method
1(x) + 4(-2)=-1
1(-1)
1(y)+2(-2)=0
1(z)+3(-2)=-1
MnO4(aq) + Br(aq)  MnO2 (s) + BrO3(aq) in Basic soln
x= +7
-1
+4
z = +5
Step 1) E: balance ox numbers and atoms with electrons
3e
+ 1(+7)
=
1(+4)
MnO4(aq)  MnO2 (s)
+6e
1(-1) = 1(+5)

Br(aq)  BrO3(aq)
Result of E: find common denominator of electrons
2[ 3e + MnO4(aq)  MnO2 (s) ]
Br(aq)  BrO3(aq) + 6 e
__________________________________
6 e + 2MnO4(aq) + Br(aq)  2MnO2(s) + BrO3(aq) + 6 e
Step 2) C: balance charge with OH ions (because in basic soln.)
2MnO4(aq) + Br(aq)  2MnO2(s) + BrO3(aq)
2(-)
+ 1(-)
=
0
1(-)
+ 2 OH
Result:
2MnO4(aq) + Br(aq)  2MnO2(s) + BrO3(aq) + 2 OH
Electrochemistry 5
6
Step 3) Balance (either H or O) with H2O
2MnO4(aq) + Br(aq)  2MnO2(s) + BrO3(aq) + 2 OH
0H  2H
2H2O + 2MnO4(aq) + Br(aq)2MnO2(s) + BrO3(aq) + 2 OH
Verify !!!
Chemical Change Leading to an Electrical Current
Galvanic (or Voltaic) cell: a spontaneous redox reaction to do electrical work.
Direction of electron flow
electrode (-)
electrode (+)
salt bridge
reduced species
-
e
oxidized
species
oxidized
species
e-
cations
ANODE compartment
OXIDATION occurs
reduced
anions species
CATHODE comparment
REDUCTION occurs
Electrochemistry 6
7
electrode - a device for conducting electrons into and out of solutions in
electrochemical cells.
Electric neutrality is maintained by migration of ions through a porous glass disc or
through a salt bridge.
salt bridge allows flow of non-reacting charged species between compartments.
Usually, saturated KCl
Electrochemistry 7
8
Supposing that
Zn(s)  Zn2+(aq) + 2e2e- + 2H+(aq)  H2 (g)
__________________________
oxidation half cell
reduction half cell
Zn(s) + 2H+ (aq)  Zn2+(aq) + H2 (g)
overall reaction
Cl-(aq) migrate out of the salt bridge to balance the new Zn2+(aq) in the oxidation
cell, and K+ migrate out to replace the H+(aq) used up in the reduction cell.
Electrochemical Cells and Potentials
electromotive force (emf) : electrons move from an electrode of higher potential to
one of lower potential (potential difference, in Volts).
Electrical work = charge x potential difference
charge in coulombs, C
charge of one electron : 1.6022 x 10-19 C
Volt = Ошибка!
or
Work (Joule) = 1 volt x 1 coulomb
standard conditions,
•
pure solid, or
• 1.0 M solution, or
• gas at 1 bar., give standard electrode potential, E°
Electrochemistry 8
9
Calculating the Potential E° of an Electrochemical Cell
Cell potentials for product-favored electrochemical reactions have positive values.
Ecell  Ered  Eox
standard half-cell, standard hydrogen electrode (SHE)


2 H3 O aq, 1 M  2e  H2 g, 1 bar   H2 Ol E  0.00 V
Tables list standard reduction potentials
Look at Zn, Cu cell
Zn s  Cu
2
aq, 1M  Zn 2 aq  Cus Ecell  1.10V
Zn is oxidized, use opposite of reduction potential in table
Zn
2
aq   2e   Zn s Ecell  0.76 V
Zn s  Zn
2
aq  2e  Ecell  0.76 V
Cu is reduced, use reduction potential from table
Cu
2
aq  2e   Cus Ecell  0.34 V
E cell  0.76 V  0.34 V  1.10 V
Electrochemistry 9
10
Oxidizing and Reducing Agents
The more positive the Ered value for a half-reaction, the greater the tendency for
the reactant of the half-reaction to be reduced, and therefore, to oxidize another
species.
Look at the half-reactions in the table of reduction potentials.
Most positive values of Ered
strongest
oxidizing
agent
2F–(aq)
•
•
•
•
•
•
2H+ (aq)
H 2(g)
•
•
•
•
•
•
Li+ (aq) +e –
Li(s)
Most negative values of Ered
Increaseing strength of reducing agent
Increasing strength of oxidizing agent
F2(g) + 2e –
strongest
reducing
agent
F2 is most easily reduced and consequently the strongest oxidizing reagent.


F2 g  2e  2F aq Ered  2.87 V
Electrochemistry 10
11
+
Lithium ion, Li , is the most difficult to reduce and is therefore the poorest
oxidizing reagent.

Liaq  e  Lis  Ered  3.05 V
However, the reverse reaction is favorable, and makes Li(s) the strongest reducing
agent on the table.
–
F (aq) is the weakest reducing agent on the table.
Free Energy Change and Equilibrium Constants
Grxn  nFE
Grxn  RT ln K
ln 𝐾 =
𝑛𝐹𝐸 𝑜
𝑅𝑇
𝐾 = 𝑒 𝑛𝐹𝐸
𝑜 /𝑅𝑇
Faraday's constant, F, 96,485 C/mol (J/V mol), charge on one mole of electrons



F  6.022  1023 mol 1 1.6022  10 19 C  96485 C mol 1
n, number of electrons transferred in the reaction
Electrochemistry 11
12
Effect of Concentration on Cell EMF: The Nernst Equation
G  G  RTlnQ
G  nFE
nFE  nFE  RTlnQ
RT
EE 
lnQ
nF
G  nFE  RTlnK
RT
EE 
ln Q
nF
E
E 
RT
lnK
nF
RT  K 
ln 
nF  Q 
EE 
ln K 
0.0257V
ln Q
n
nE
0.0257V
RT
 0.0257_ V
F
K  enFE
Examples 12.9 and 10 Self-tests 12.11 and 12
Suppplementary material: Nerve cells
Electrochemistry 12
RT
 e nE
0.0257_V
13
Electrolysis - use of an electrical current to bring about chemical change.
Direction of electron flow
–
+
Battery
electrode (+)
electrode (–)
salt bridge
reduced species
oxidized
species
-
e
oxidized
species
e-
cations
reduced
anions sp ecies
ANODE compartment
OXIDATION occurs
CATHODE comparment
REDUCTION occurs
1) Counting Electrons
Current, I (amperes, A) = Ошибка!
• Current times time, in seconds, gives charge in coulombs,
Q  It
• Divide charge by Faraday's constant to get moles of e• Divide moles of electrons by moles of electrons per mole of metal reduced
• Divide moles of metal by the atomic weight of the metal to get grams of metal
1_ Faraday 1_ mol _ e  mol _ metal
g_ metal
C 
current   time(s) 



 g_ metal

s 
96485_ C
Faraday
mol _ metal
mol_ e
m
QM
nF
Electrochemistry 13
14
m = Ошибка!
Calculate the mass of aluminum produced in 1.00 hr by the electrolysis of molten
AlCl3 if the electrical current is 10.0 A.
Summary
A diagram of the galvanic cell (left) connected to an electrolytic cell (right)
containing hydrogen and iodide ions is drawn below. With arrowheads describe the
flow of ions to metal electrodes in the cells, the KCl ions in a salt bridge, and
electrons in the wires. Label the anodes and cathodes. (Will be on test)
e?
-ode
-ode
?
-ode
-ode
e-
K+Cl-
?
?
H+
?
I?
Cr3+
Cr
?
Sn2+(aq) + 2e–  Sn(s)
Cr3+(aq) + 3e–  Cr(s)
Sn2+
Sn
?
E° = –0.136 V
E° = –0.74 V
Pt
Pt
2 H+ + 2e– 
2 I-  + 2e–
Solution;
Step 1) The galvanic cell does the work. So its cell voltage must be positive.
1)To achieve this, choose the half-cell with the greatest absolute value of Eo.
Electrochemistry 14
15
Cr3+(aq) + 3e–  Cr(s)
E° = –0.74 V
2a) If this Eo > 0, switch the other equation and its Eo.
2b) If this Eo < 0, switch this equation and its Eo.
Cr3+(aq) + 3e–  Cr(s)
Cr(s)  Cr3+(aq) + 3e–
E° = –0.74 V
E° = +0.74 V
3) Add two equations (common denominator for masses, E independent of mass)
2[Cr(s)  Cr3+(aq) + 3e– ]
E° = +0.74 V
oxidation=anode
–
2+
3[Sn (aq) + 2e  Sn(s) ]
E° = –0.136 V
reduction=cathode
__________________________________________________
2Cr(s) + 3Sn2+(aq)  2Cr3+(aq) + 3Sn(s) E° = +0.60 V
total cell
Step 2) Arrow heads
1) Cr3+(aq) formed, Sn2+(aq) lost to Sn surface
2) Salt bridge: Cl- ions meet the new Cr3+(aq), K+(aq) replace lost Sn2+(aq)
3) Electron flow: from anode (-), to cathode (+)
Step 3) Electrolytic Cell
1) Electrons arrive at cathode(-), where they are used to reduce.
2 H+ + 2e– 
2) Then the electrons leave at the anode(+), where they are produced through
oxidation 2 I-  + 2e–
Electrochemistry 15
16
Batteries and Fuel Cells
Primary Batteries (not easily reversed; at equilibrium "dead")
Secondary Batteries (easily reversed or "recharged")
car battery
Fuel Cells (reactants supplied from external source)
Corrosion
Oxidation of most metals by oxygen is favorable.
Iron:
Cathode:
O2 (g) + 4 H+(aq) + 4e- 
2 H2O(l) Ered  0.14_V
Anode:
Fe 
Fe2+ (aq) + 2e-
Ered  0. 44_ V
anodic inhibition - coat with inert material, paint or layer of oxide of metal
protecting (Al2O3)
cathodic inhibition - make Fe the cathode, use Zn or Mg as the anode
(sacrificial metal)
Electrochemistry 16