3. 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
4. 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
5. Applications
Moving electrons is electric current.
8H++MnO4
-+ 5Fe+2 +5e-
® 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
6. Connected this way the reaction starts
Stops immediately because charge builds
up.
H+
MnO4
e-e-
e-e-e-
- Fe+2
11. 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
12. Zn+2 SO4
-
2
1 M HCl
Anode
0.76
1 M ZnSO4
H+
Cl-
H2 in
Cathode
13. Standard Hydrogen Electrode
This is the reference
all other oxidations
are compared to
Eº = 0
º indicates standard
states of 25ºC,
H+
1 atm, 1 M
solutions.
Cl-
1 M HCl
H2 in
14. Cell Potential
Zn(s) + Cu+2 (aq) ® Zn+2(aq) + Cu(s)
The total cell potential is the sum of the
potential at each electrode.
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.
15. 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)
Fe+3(aq) + e-® Fe+2(aq) Eº = 0.77 V
Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V
Cu(s) ® Cu+2(aq)+2e- Eº = -0.34 V
2Fe+3(aq) + 2e-® 2Fe+2(aq) Eº = 0.77 V
16. 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
17. Line Notation
solid½Aqueous½½Aqueous½solid
Anode on the left½½Cathode 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.
18. For the last reaction
Cu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s)
Cu2+ Fe+2
19. 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
20. 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+
21. Galvanic Cell
The reaction always runs
spontaneously in the direction that
produced a positive cell potential.
Four things for a complete description.
1) Cell Potential
2) Direction of flow
3) Designation of anode and cathode
4) Nature of all the components-electrodes
and ions
22. 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
23. 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.
-w = q E
Faraday = 96,485 C/mol e-
q = nF = moles of e- x charge/mole e-
w = -qE = -nFE = DG
24. Potential, Work and DG
DGº = -nFEº
if Eº 0, then DGº 0 spontaneous
if Eº 0, then DGº 0 nonspontaneous
In fact, reverse is spontaneous.
Calculate DGº for the following reaction:
Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq)
Fe+2(aq) + e-® Fe(s) Eº = 0.44 V
Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V
25. Cell Potential and
Concentration
Qualitatively - Can predict direction of
change in E from LeChâtelier.
2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)
Predict 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
if [Al+3] = 1.5 M and [Mn+2] = 1.5 M
26. The Nernst Equation
DG = DGº +RTln(Q)
-nFE = -nFEº + RTln(Q)
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.
27. 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
28. Batteries are Galvanic Cells
Car batteries are lead storage batteries.
Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O
32. Corrosion
Rusting - spontaneous oxidation.
Most structural metals have reduction
potentials that are less positive than O2 .
Fe ® Fe+2 +2e- Eº= 0.44 V
O2 + 2H2O + 4e- ® 4OH-Eº= 0.40 V
Fe+2 + O2 + H2O ® Fe2O3 + H+
Reactions happens in two places.
33. Salt speeds up process by increasing
conductivity
Water
Rust
Iron Dissolves-
Fe ® Fe+2
e-
Fe2+
O2 + 2H2O +4e- ® 4OH-Fe2+
+ O2 + 2H2O ® Fe2O3 + 8 H+
34. 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.
35. Electrolysis
Running a galvanic cell backwards.
Put a voltage bigger than the potential
and reverse the direction of the redox
reaction.
Used for electroplating.
36. 1.10
e- e-
Zn Cu
1.0 M
Zn+2
1.0 M
Cu+2
Anode Cathode
37. A battery
1.10V
e- e-
Zn Cu
1.0 M
Zn+2
1.0 M
Cu+2
Cathode Anode
38. 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+
39. Calculating plating
1. Current x time = charge
2. Charge ∕Faraday = mole of e-
3. Mol of e- to mole of element or
compound
4. Mole to grams of compound
Or the reverse if you want time to plate
40. 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.
41. 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?
42. 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.
43. 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)
44. 6. Combination Oxidizing agent of one
element will react with the reducing agent
of the same element to produce the free
element.
I- + IO- + H+ ® I+ HO
3
2 27. Decomposition.
a) peroxides to oxides
b) Chlorates to chlorides
c) Electrolysis into elements.
d) carbonates to oxides
45. 45
Examples
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.
46. 46
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.
47. A stream of chlorine gas is passed through
47
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.
48. 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.
48
49. A way to remember
An 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
50. 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.)
51. 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.)
52. 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.
53. 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?