Electrochemistry
Applications of Redox
Review
 Oxidation
reduction reactions involve a
transfer of electrons.
 OIL- RIG
 Oxidation Involves Loss
 Reduction Involves Gain
 LEO-GER
 Lose Electrons Oxidation
 Gain Electrons Reduction
Solid lead(II) sulfide reacts with oxygen in
the air at high temperatures to form
lead(II) oxide and sulfur dioxide. Which
substance is a reductant (reducing
agent) and which is an oxidant
(oxidizing agent)?
A. PbS, reductant; O2, oxidant
B. PbS, reductant; SO2, oxidant
C. Pb2+, reductant; S2- oxidant
D. PbS, reductant; no oxidant
E. PbS, oxidant; SO2, reductant
Applications
 Moving
electrons is electric current.
+
+2 +5e 8H +MnO4 + 5Fe
 Mn+2 + 5Fe+3 +4H2O
 Helps to break the reactions into half
reactions.
 8H++MnO4-+5e-  Mn+2 +4H2O
 5(Fe+2  Fe+3 + e- )
 In the same mixture it happens without
doing useful work, but if separate
 Connected
this way the reaction starts
 Stops immediately because charge builds
up.
H+
MnO4-
Fe+2
Galvanic Cell
Salt
Bridge
allows
current
to flow
H+
MnO4-
Fe+2
 Electricity
travels in a complete circuit
H+
MnO4-
eFe+2
Instead
of a salt bridge
H+
MnO4-
Porous
Disk
Fe+2
e-
e-
e-
e-
Anode
e-
Reducing
Agent
Cathode
e-
Oxidizing
Agent
Cell Potential
 Oxidizing
agent pulls the electron.
 Reducing agent pushes the electron.
 The push or pull (“driving force”) is called
the cell potential Ecell
 Also
called the electromotive force (emf)
 Unit is the volt(V)
 = 1 joule of work/coulomb of charge
 Measured with a voltmeter
0.76
H2 in
Anode
Zn+2
SO4-2
1 M ZnSO4
Cathode
H+
Cl1 M HCl
Standard Hydrogen Electrode
 This
is the reference
all other oxidations
are compared to
 Eº = 0
º
indicates standard
+
H
states of 25ºC,
1 atm, 1 M
Clsolutions.
1 M HCl
H2 in
Cell Potential
+ Cu+2 (aq)  Zn+2(aq) + Cu(s)
 The total cell potential is the sum of the
potential at each electrode.
 Zn(s)
 Eºcell = EºZn Zn+2 + EºCu+2 Cu
 We
can look up reduction potentials in a
table.
 One of the reactions must be reversed,
so change it sign.
Cell Potential
 Determine
the cell potential for a galvanic
cell based on the redox reaction.
 Cu(s) + Fe+3(aq)  Cu+2(aq) + Fe+2(aq)
Eº = 0.77 V
 Cu+2(aq)+2e- Cu(s)
Eº = 0.34 V
 Cu(s) Cu+2(aq)+2eEº = -0.34 V
 2Fe+3(aq) + 2e- 2Fe+2(aq) Eº = 0.77 V
 Fe+3(aq) +
e- Fe+2(aq)
Reduction potential
 More
negative Eº
– more
easily electron is added
– More easily reduced
– Better oxidizing agent
 More
positive Eº
– more
easily electron is lost
– More easily oxidized
– Better reducing agent
Line Notation
solidAqueousAqueoussolid
 Anode on the leftCathode on the right
 Single line different phases.
 Double line porous disk or salt bridge.
 If all the substances on one side are
aqueous, a platinum electrode is
indicated.

For
the last reaction
Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)
Cu2+
Fe+2
In a galvanic cell, the electrode that
acts as a source of electrons to the
solution is called the __________;
the chemical change that occurs at
this electrode is called________.
a. cathode, oxidation
b. anode, reduction
c. anode, oxidation
d. cathode, reduction
Under standard conditions, which of
the following is the net reaction that
occurs in the cell?
Cd|Cd2+ || Cu2+|Cu
a. Cu2+ + Cd → Cu + Cd2+
b. Cu + Cd → Cu2+ + Cd2+
c. Cu2+ + Cd2+ → Cu + Cd
d. Cu + Cd 2+ → Cd + Cu2+
Galvanic Cell


1)
2)
3)
4)
The reaction always runs
spontaneously in the direction that
produced a positive cell potential.
Four things for a complete description.
Cell Potential
Direction of flow
Designation of anode and cathode
Nature of all the componentselectrodes and ions
Practice
 Completely
describe the galvanic cell
based on the following half-reactions
under standard conditions.
 MnO4- + 8 H+ +5e-  Mn+2 + 4H2O
Eº=1.51 V
 Fe+3
+3e-  Fe(s)
Eº=0.036V
Potential, Work and DG
 emf
= potential (V) = work (J) / Charge(C)
E = work done by system / charge
E = -w/q
 Charge
is measured in coulombs.
=qE
 Faraday = 96,485 C/mol e q = nF = moles of e- x charge/mole e -w
w
= -qE = -nFE = DG
Potential, Work and DG

DGº = -nFEº
Eº > 0, then DGº < 0 spontaneous
 if Eº< 0, then DGº > 0 nonspontaneous

if
 In
fact, reverse is spontaneous.
 Calculate DGº for the following reaction:
 Cu+2(aq)+ Fe(s) Cu(s)+ Fe+2(aq)
+2(aq) +
 Fe
e-Fe(s)
 Cu+2(aq)+2e- Cu(s)
Eº = 0.44 V
Eº = 0.34 V
Cell Potential and
Concentration
 Qualitatively
- Can predict direction of
change in E from LeChâtelier.
 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
if Ecell will be greater or less than
Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M
 if [Al+3] = 1.0 M and [Mn+2] = 1.5M
+3
+2
 if [Al ] = 1.5 M and [Mn ] = 1.5 M
 Predict
 DG
The Nernst Equation
= DGº +RTln(Q)
E = -nFEº + RTln(Q)
 -nF
E = Eº - RTln(Q)
nF
 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
Eº = 0.48 V
 Always have to figure out n by balancing.
 If concentration can gives voltage, then
from voltage we can tell concentration.
The Nernst Equation
 As
reactions proceed concentrations of
products increase and reactants
decrease.
 Reach equilibrium where Q = K and
Ecell = 0
 0 = Eº - RTln(K)
nF
Eº = RTln(K)
nF

nF Eº = ln(K)
RT
Batteries are Galvanic Cells
 Car
batteries are lead storage batteries.
 Pb +PbO2 +H2SO4 PbSO4(s) +H2O
Batteries are Galvanic Cells
 Dry
Cell
Zn + NH4+ +MnO2 
Zn+2 + NH3 + H2O + Mn2O3
Batteries are Galvanic Cells
 Alkaline
Zn +MnO2  ZnO+ Mn2O3 (in base)
Batteries are Galvanic Cells
 NiCad

NiO2 + Cd + 2H2O  Cd(OH)2 +Ni(OH)2
Corrosion
 Rusting
- spontaneous oxidation.
 Most structural metals have reduction
potentials that are less positive than O2 .
 Fe
 Fe+2 +2e-
 O2
+ 2H2O + 4e- 4OH-
Eº= 0.44 V
Eº= 0.40 V
+ O2 + H2O Fe2O3 + H+
 Reactions happens in two places.
 Fe+2
Salt speeds up process by increasing
conductivity
Water
Fe2+
Iron DissolvesFe  Fe+2
Rust
eO2 + 2H2O +4e-  4OH-
Fe2+ + O2 + 2H2O  Fe2O3 + 8 H+
Preventing Corrosion
 Coating
to keep out air and water.
 Galvanizing - Putting on a zinc coat
 Has a lower reduction potential, so it is
more easily oxidized.
 Alloying with metals that form oxide
coats.
 Cathodic Protection - Attaching large
pieces of an active metal like magnesium
that get oxidized instead.
Electrolysis
 Running
a galvanic cell backwards.
 Put a voltage bigger than the potential
and reverse the direction of the redox
reaction.
 Used for electroplating.
1.10
e-
e-
Zn
Cu
1.0 M
Zn+2
Anode
1.0 M
Cu+2
Cathode
e-
A battery
>1.10V
Zn
e-
Cu
1.0 M
Zn+2
Cathode
1.0 M
Cu+2
Anode
Calculating plating
 Have
to count charge.
 Measure current I (in amperes)
 1 amp = 1 coulomb of charge per second
q = I x t
 q/nF = moles of metal
 Mass of plated metal
 How long must 5.00 amp current be
applied to produce 15.5 g of Ag from Ag+
Calculating plating
1. Current x time = charge
2. Charge ∕Faraday = mole of e3. Mol of e- to mole of element or
compound
4. Mole to grams of compound
Or the reverse if you want time to plate
Calculate the mass of copper which can be
deposited by the passage of 12.0 A for
25.0 min through a solution of copper(II)
sulfate.
How long would it take to plate 5.00 g Fe
from an aqueous solution of Fe(NO3)3 at
a current of 2.00 A?
Other uses
 Electrolysis
of water.
 Separating mixtures of ions.
 More positive reduction potential means
the reaction proceeds forward.
 We want the reverse.
 Most negative reduction potential is
easiest to plate out of solution.
Redox
Know the table
2. Recognized by change in oxidation
state.
3. “Added acid”
4. Use the reduction potential table on the
front cover.
5. Redox can replace. (single replacement)
6. Combination Oxidizing agent of one
element will react with the reducing agent
of the same element to produce the free
element.
I- + IO3- + H+  I2 + H2O
7. Decomposition.
a) peroxides to oxides
b) Chlorates to chlorides
c) Electrolysis into elements.
d) carbonates to oxides
Examples
1.
2.
3.
A piece of solid bismuth is heated
strongly in oxygen.
A strip or copper metal is added to a
concentrated solution of sulfuric acid.
Dilute hydrochloric acid is added to a
solution of potassium carbonate.
44
23.
24.
25.
26.
Hydrogen peroxide solution is added
to a solution of iron (II) sulfate.
Propanol is burned completely in air.
A piece of lithium metal is dropped into
a container of nitrogen gas.
Chlorine gas is bubbled into a solution
of potassium iodide.
45
Examples
5.
6.
7.
A stream of chlorine gas is passed
through a solution of cold, dilute sodium
hydroxide.
A solution of tin ( II ) chloride is added
to an acidified solution of potassium
permanganate
A solution of potassium iodide is added
to an acidified solution of potassium
dichromate.
46
70.
71.
72.
73.
74.
Magnesium metal is burned in
nitrogen gas.
Lead foil is immersed in silver nitrate
solution.
Magnesium turnings are added to a
solution of iron (III) chloride.
Pellets of lead are dropped into hot
sulfuric acid
Powdered Iron is added to a solution of
iron(III) sulfate.
47
A way to remember
Ox – anode is where oxidation occurs
 Red Cat – Reduction occurs at cathode
 Galvanic cell- spontaneous- anode is
negative
 Electrolytic cell- voltage applied to make
anode positive
 An
A student places a copper electrode in a 1
M solution of CuSO4 and in another
beaker places a silver electrode in a 1 M
solution of AgNO3. A salt bridge
composed of Na2SO4 connects the two
beakers. The voltage measured across
the electrodes is found to be + 0.42 volt.
 (a) Draw a diagram of this cell.
 (b) Describe what is happening at the
cathode (Include any equations that may
be useful.)
A student places a copper electrode in a 1
M solution of CuSO4 and in another
beaker places a silver electrode in a 1 M
solution of AgNO3. A salt bridge
composed of Na2SO4 connects the two
beakers. The voltage measured across
the electrodes is found to be + 0.42 volt.
 (c) Describe what is happening at the
anode. (Include any equations that may
be useful.)
A student places a copper electrode in a 1
M solution of CuSO4 and in another
beaker places a silver electrode in a 1 M
solution of AgNO3. A salt bridge
composed of Na2SO4 connects the two
beakers. The voltage measured across
the electrodes is found to be + 0.42 volt.
 (d) Write the balanced overall cell
equation.
 (e) Write the standard cell notation.
A student places a copper electrode in a 1 M
solution of CuSO4 and in another beaker places
a silver electrode in a 1 M solution of AgNO3. A
salt bridge composed of Na2SO4 connects the
two beakers. The voltage measured across the
electrodes is found to be + 0.42 volt.
(f) The student adds 4 M ammonia to the
copper sulfate solution, producing the complex
ion Cu(NH3)+ (aq). The student remeasures the
cell potential and discovers the voltage to be
0.88 volt. What is the Cu2+ (aq) concentration in
the cell after the ammonia has been added?