Chapter 20 Electrochemistry

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Chapter 20 Electrochemistry
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Oxidation and Reduction

• Oxidation (loss of e-) Na Na e-
• Reduction (gain of e-) Cl e- Cl-
• Oxidation-reduction (redox) reactions occur when
  electrons are transferred from an atom that is
  oxidized to an atom that is reduced.
• Electron transfer can produce electrical energy
  spontaneously, but sometimes electrical energy is
  needed to make them occur (nonspontaneous).

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Oxidation-Reduction (Redox) Reactions

• BOTH reduction and oxidation must occur.
• A substance that gives up electrons is oxidized
  and is called a reducing agent or reductant
  (causes another substance to be reduced).
• A substance that accepts electrons is reduced
  and is therefore called an oxidizing agent or
  oxidant (causes another substance to be
  oxidized).

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Oxidation-reduction (Redox) Reactions

• Is it a redox reaction??
• Zn(s) 2H(aq) Zn2(aq)
  H2(g)
• Quick hint

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Oxidation Number Guidelines

• Atoms in elemental form, oxidation number is
  zero.
• (Cl2, H2, P4, Ne are all zero)
• Monoatomic ion, the oxidation number is the
  charge on the ion. (Na 1 Al3 3 Cl- -1)
• O is usually -2. But in peroxides (like H2O2 and
  Na2O2) it has an oxidation number of -1.
• H is 1 when bonded to nonmetals and -1 when
  bonded to metals.
• (1 in H2O, NH3 and CH4 -1 in NaH, CaH2 and
  AlH3)
• The oxidation number of F is -1.
• The sum of the oxidation numbers for the molecule
  is the charge on the molecule (zero for a neutral
  molecule).

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Example
Determine the oxidation state of all elements in
ammonium thiosulfate (NH4)2(S2O3).
(NH4)2(S2O3)
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Determining Oxidation States

• What is the oxidation state of Mn in MnO4-?
• Answer 7

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Balancing oxidation-reduction equations

• We know Balancing chemical equations follows
  law of conservation of mass.
• AND, for redox reactions, gains and losses of
  electrons must also be balanced.

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Half-Reactions

• Separate oxidation and reduction processes in
  equation,
• Sn2(aq) 2Fe3(aq) Sn4(aq)
  2Fe2(aq)
• Oxidation
• Reduction
• Overall, the number of electrons lost in the
  oxidation half reaction must equal the number
  gained in the reduction half reaction.

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Balancing Equations by the Method of
Half-Reactions Acidic

• Consider
• MnO4-(aq) C2O42-(aq) Mn2(aq)
  CO2(g)
• Unbalanced half-reactions MnO4-(aq)
  Mn2(aq)
• 
  C2O42-(aq) CO2(g)
• First, balance everything EXCEPT hydrogen and
  oxygen.
• Deal with half-reactions SEPARATELY.

(acidic)
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Balancing Equations by the Method of
Half-Reactions Acidic

• MnO4-(aq) C2O42-(aq) Mn2(aq)
  CO2(g)
• To balance O
• Add 4H2O to products to balance oxygen in
  reactants.
• To balance H
• Add 8H to reactant side to balance the 8H in
  water.

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Balancing Equations by the Method of
Half-Reactions Acidic

• Balance charge Add up charges on both sides.
• Add 5 electrons to reactant side.
• 5e- 8H(aq) MnO4-(aq) Mn2(aq)
  4H2O(l)
• Mass balance of C in oxalate half-reaction.
• C2O42-(aq) 2CO2(g)
• Balance charge by adding two electrons to the
  products.
• Last step Cancel electrons and add reactions
  together.

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Balancing Equations by the Method of
Half-Reactions Acidic

• 5e- 8H(aq) MnO4-(aq) Mn2(aq)
  4H2O(l)
• C2O42-(aq)
  2CO2(g) 2e-
• Top reaction times 2. Bottom reaction times 5.
  ALL PARTS!!

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Balancing Equations by the Method of
Half-Reactions Acidic Summary

• 1. Divide equation into two incomplete
  half-reactions.
• 2. Balance each half-reaction
• (a) balance elements other than H and O.
• (b) balance O atoms by adding H2O.
• (c) balance H atoms by adding H (basic
  conditions will require further work at this
  step).
• (d) balance charge by adding e- to the side with
  greater overall positive charge.

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Balancing Equations by the Method of
Half-Reactions Summary

• Multiply each half-reaction by an integer so that
  the number of electrons lost in one half-reaction
  equals the number gained in the other
  half-reaction.
• 4. Add the two half-reactions and cancel out
  all species appearing on both sides of the
  equation.
• 5. Check equation to make sure there are same
  number of atoms of each kind and the same total
  charge on both sides. Errors can be caught!!

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Balancing Equations by the Method of
Half-Reactions Basic

• Balancing process is started using H and H2O,
  then adjusting with OH- to uphold reaction
  conditions.
• (H does not exist in basic solutions.)
• Balance the following reaction
• H2O2(aq) ClO2(aq) ClO2-(aq) O2(g)

(basic)
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Basic Redox Reactions
Balancing Equations by the Method of
Half-Reactions Basic

• Split into two half-reactions.
• H2O2(aq) O2(g)
• ClO2(aq) ClO2-(aq)
• Balance elements, then oxygen by adding H2O.
  Then, add H to balance H, just like an acidic
  redox reaction.
• H2O2(aq) O2(g) 2H
• ClO2(aq) ClO2-(aq)

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Basic Redox Reactions
Balancing Equations by the Method of
Half-Reactions Basic

• Add OH- to both sides, enough to neutralize all
  H (basic reactions cannot support H).
• Combine H and OH- to form H2O.

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Basic Redox Reactions
Balancing Equations by the Method of
Half-Reactions Basic

• Balance charge by adding e-.
• 2OH- H2O2(aq) O2(g) 2H2O
• ClO2(aq) ClO2-(aq)
• Multiply each reaction so both have same e-. Then
  add them together, cancelling where possible.
• 2OH- H2O2(aq) 2ClO2(aq)
• O2(g) 2H2O 2ClO2-(aq)
• Double-check your answer!

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Balance this redox equation

• Cu(s) NO3-(aq) Cu2(aq) NO2(g)
  (acidic)
• Ans Cu(s) 2NO3-(aq) 4H(aq) Cu2(aq)
  2NO2(aq) 2H2O(l)

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Balance this redox equation

• NO2-(aq) Al(s) NH3(aq)
  Al(OH)4-(aq) (basic)
• 2Al(s) NO2-(aq) OH-(aq) 5H2O(aq)
  2Al(OH)4-(aq) NH3(aq)

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Voltaic Cells
Voltaic (aka galvanic) cells electrochemical
reactions in which electron transfer occurs via
an external circuit. The energy released in a
voltaic cell reaction can be used to perform
electrical work. Reactions are spontaneous.
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Voltaic Cells Components
For example Anode Zn(s) ? Zn2(aq) 2e-
( half-reaction) Cathode Cu2(aq)
2e- ? Cu(s) ( half reaction)
Salt bridge cations move from anode to
cathode, anions move from cathode to
anode. External circuit (wire) electrons move
from anode to cathode (between two solid metal
electrodes). Electron transfer can naturally
occur in OTHER forms besides a wire (direct
electron transport between solution and metal).
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Voltaic Cells Ion Flow
Anions and cations move through a porous barrier
or salt bridge. Cations move into the cathodic
compartment to neutralize the excess negatively
charged ions. Anions move into the anodic
compartment to neutralize the excess Zn2 ions
formed by oxidation.
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Simplified Voltaic Cell
In any voltaic cell, the electrons flow from the
anode through the external circuit to the
cathode. Anode labeled with negative sign (-)
and cathode with positive sign (). Anions
migrate toward anode. Cations toward the cathode.
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Voltaic Cells Electron Flow
As oxidation occurs, Zn(s) is converted to Zn2
and 2e-. And Cu2 is converted to Cu(s).

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One-Pot Voltaic Cells
If a strip of Zn is placed in a solution of
CuSO4, Cu is deposited on the Zn and the Zn
dissolves, forming Zn2. Zn is spontaneously
oxidized to Zn2 by Cu2 (Cu is the oxidizing
agent). The Cu2 is spontaneously reduced to Cu0
by Zn (Zn is the reducing agent). The entire
process is spontaneous.
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The Atomic Level

A Cu2(aq) ion comes into contact with a Zn(s)
atom on the surface of the electrode. Two
electrons are directly transferred from the
Zn(s), forming Zn2(aq), to the Cu2(aq) forming
Cu(s).
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Cell Electromotive Force (EMF)
Electromotive force (emf) the force required to
push electrons through the external
circuit. Potential difference difference per
electrical charge between two electrodes. Units
Volts (V). One volt is the potential required to
impart one joule of energy to a charge of one
coulomb
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Cell EMF
Cell potential (Ecell) the emf of a cell. Aka
cell voltage. Positive for spontaneous cell
reactions. Ecell strictly depends on reactions
that occur at the cathode and anode, the
concentration of reactants and products, and the
temperature and is described by the
equation Ecell Ered(cathode) -
Ered(anode)
For 1M solutions at 25C and 1 atm (standard
conditions), the standard emf (standard cell
potential), Ecell, is written as Ecell.
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Standard Reduction Potentials

• Potential associated with each half-reaction is
  chosen to be the potential for reduction to occur
  at that electrode. Hence, Standard Reduction
  Potentials.
• Standard reduction potentials, Ered, are
  measured relative to the standard hydrogen
  electrode (SHE).

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Standard Hydrogen Electrode (SHE)
SHE is assumed as the cathode (for consistency).
Pt electrode in a tube containing 1M H
solution. H2(g) is bubbled through the tube and
equilibrium is established.
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Determining Standard Reduction Potentials
For the SHE 2H(aq, 1M) 2e- ? H2(g, 1 atm)
Ered . Standard reduction
potentials can then be calculated using the SHE
(E 0V) as Ered(cathode). Each calculated
E is rewritten as a reduction and tabulated.
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Finding the Standard Reduction Potential

• Ecell is measured, 0V (SHE) is used for
  E(cathode), and the reduction potential for Zn
  is found.

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Zn Standard Reduction Potential
We measure Ecell relative to the SHE Ecell
Ered(cathode) - Ered(anode) 0.76V 0V -
Ered(anode). Therefore, Ered(anode)
-0.76V And we find that -0.76V can be assigned
to reduction of zinc.
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Zn Standard Reduction Potential
Since Ered -0.76 V (negative!) we conclude
that the reduction of Zn2 in the presence of the
SHE is not spontaneous. However, the oxidation
of Zn with the SHE is spontaneous. Reactions
with Ered gt 0 are spontaneous reductions
relative to the SHE. Ered lt 0 are spontaneous
oxidations. Changing the stoichiometric
coefficient does not affect Ered. 2Zn2(aq)
4e- ? 2Zn(s) Ered -0.76 V
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Full list in Appendix E
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EMF Trends
The larger the difference between Ered values,
the larger Ecell. Spontaneous voltaic cell
Ered(cathode) is more positive than
Ered(anode), resulting in a positive Ecell.
Recall
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Calculating Ered from Ecell

• Zn(s) Cu2(aq, 1M) Zn2(aq, 1M)
  Cu(s) Ecell 1.10V
• Given Zn2 2e- Zn(s)
  Ered -0.76V
• Calculate the Ered for the reduction of Cu2 to
  Cu.
• What about Ered for the oxidation of Cu(s)
  reverse reaction?

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Oxidizing and Reducing Agents

• We can use Ered values to understand aqueous
  reaction chemistry.
• e.g. Zn(s) Cu2(aq) Zn2(aq)
  Cu(s)
• Since Cu2 is responsible for the oxidation of
  Zn(s), Cu2 is called the
  .
• Since Zn(s) is responsible for the reduction of
  Cu2, Zn(s) is called the
  .
• Ered for Cu2 (0.34V) indicates that, compared
  to Ered for Zn (-0.76V), it will be reduced.

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Example Problem Cell emf

• Calculate the standard emf for the following
• Ni(s) 2Ce4(aq) Ni2(aq) 2Ce3(aq)
• Ask yourselfWhat are the half-reactions?
• Which one is oxidized? Reduced?
• Ni2(aq) 2e- Ni(s) Ered
  -0.28V
• Ce4(aq) 1e- Ce3(aq) Ered
  1.61V
• is the oxidizing agent.

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Example Problem Cell emf

• Which reaction will undergo reduction? What
  reaction occurs at the anode?
• Sn4(aq) 2e- Sn2(aq)
  Ered 0.154V
• MnO4-(aq) 8H(aq) 5e- Mn2(aq)
  4H2O(l) Ered 1.51V

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Oxidizing and Reducing Agents EMF
The more positive Ered, the greater the tendency
for the reactant in the half-reaction to be
reduced. Therefore, the stronger the oxidizing
agent. The more negative Ered, the greater the
tendency for the product in the half-reaction to
be oxidized. This means the product is tends to
be a reducing agent.
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• A species at higher left of the table of
  standard reduction potentials will spontaneously
  oxidize a species that is lower right in the
  table.
• F2 will oxidize H2 or Li.
• Ni2 will oxidize Al(s).

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Example Redox Agents

• Which is the strongest reducing agent? Oxidizing
  agent?
• Ce4, Br2, H2O2, or Zn?
• Ered 1.61V for Ce4 (aq) 1.065V for
  Br2 (l)
• 1.776 for H2O2 (aq) -0.763V for Zn(s)

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Spontaneity of Redox Reactions
In a spontaneous voltaic cell, Ered(cathode)
must be more positive than Ered(anode). A
positive Ecell indicates a spontaneous voltaic
cell process. A negative Ecell indicates a
non-spontaneous process. It all relates back to
Gibbs Free Energy.
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Spontaneity of Redox Reactions
Cell emf and free-energy change are related by
?G is the change in free-energy, n is the
number of moles of electrons transferred, F is
Faradays constant, and E is the emf of the cell.
?G units are J (assumed per mole). 1F 96,500
C/mol 96,500 J/Vmol If standard conditions
?G -nFEcell If Ecell gt 0 then
, both of which indicate spontaneous processes.
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Example Cell spontaneity

• Calculate ?G for the following reaction. Is it
  spontaneous?
• Cl2(g) 2I-(aq) 2Cl-(aq) I2(s)
  Ecell 0.823V

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Batteries
Defn Self-contained electrochemical power source
with one (or more) complete voltaic cell. When
multiple cells or multiple batteries are
connected in series, greater emfs can be
achieved. Cathode labeled with a plus sign the
anode with a minus sign.
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How Batteries Work

• Spontaneous redox reactions, as in voltaic cells,
  serve as the basis for battery operation.
• Specific reactions at anode and cathode determine
  the voltage of the battery, and the usable life
  of the battery depends on the quantity of these
  substances.
• cannot be
  recharged.
• are rechargeable
  using an external power source after its emf has
  dropped below a usable level.

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Lead-Acid Batteries
A 12V car battery consists of 6 cathode/anode
pairs, connected in series, each producing
2V. Cathode PbO2(s) on a metal grid in
sulfuric acid PbO2(s) SO42-(aq) 4H(aq)
2e- ? PbSO4(s) 2H2O(l) Anode Pb(s) in
sulfuric acid Pb(s) SO42-(aq) ? PbSO4(s)
2e- The overall electrochemical reaction is
PbO2(s) Pb(s) 2SO42-(aq) 4H(aq) ?
2PbSO4(s) 2H2O(l)
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Lead-Acid Batteries
E?cell E?red(cathode) - E?red(anode) (1.685
V) - (-0.356 V) 2.041 V. Wood or fiberglass
spacers prevent electrode contact. H2SO4
consumed during discharge, (voltage may vary with
use).
Recharging reverses forward reaction PbO2(s)
Pb(s) 2SO42-(aq) 4H(aq) ? 2PbSO4(s) 2H2O(l)
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Alkaline Batteries
Most common nonrechargeable (primary) battery
(100 billion produced annually). Anode Powdered
Zn in gel in contact with cKOH solution Zn(s)
2OH-(aq) ? Zn(OH)2(s) 2e- Cathode MnO2 and
C paste 2MnO2(s) 2H2O(l) 2e- ?
2MnO(OH)(s) 2OH-(aq)
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Rechargeable Batteries
Lightweight batteries for use in cell phones,
notebook computers, etc. Nickel-cadmium (NiCad)
battery still common. Cadmium oxidized at anode,
nickel oxyhydroxide NiO(OH)(s) reduced at
cathode. Cathode 2NiO(OH)(s) 2H2O(l)
2e- ? 2Ni(OH)2(s) 2OH-(aq) Anode Cd(s)
2OH-(aq) ? Cd(OH)2(s) 2e-
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Rechargeable Batteries
Solid reaction products adhere to the
electrodes, permitting electrode reactions to be
reversed during recharging. Drawbacks of NiCad
battery Toxicity of Cd (disposal problems),
relatively heavy.
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NiMH and Li-ion Batteries
Cathode reaction in nickel metal hydride (NiMH)
battery is same as for Ni-Cd battery.
2NiO(OH)(s) 2H2O(l) 2e- ? 2Ni(OH)2(s)
2OH-(aq) Anode consists of a metal alloy, e.g.
ZrNi2, capable of absorbing hydrogen atoms.
During oxidation, the hydrogen atoms lose
electrons (used by cathode), and the resulting H
ions react with OH- to form water (drives cathode
reaction to the right). Another type Li-ion
battery - lightweight with very high energy
density.
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Fuel Cells
Direct, efficient production of electricity. NOT
batteries as they are not self-contained (some
products must be released). On Apollo moon
flights, the H2-O2 fuel cell was the primary
source of electricity. Cathode 2H2O(l) O2(g)
4e- ? 4OH-(aq) Anode 2H2(g) 4OH-(aq) ?
4H2O(l) 4e- Whats the overall reaction?
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Corrosion
Generally the result of an undesirable redox
reaction. Spontaneous! A metal is attacked by
some substance in its environment to form an
unwanted compound. For many metals oxidation is
thermodynamically favored at RT. Corrosion can
form an insulating protective coating, preventing
further oxidation. (e.g., Al forms a hydrated
form of Al2O3) 20 of iron produced annually in
USA is used to replace rusted items!!
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Corrosion of Iron
Rusting of iron requires both oxygen and
water. Other factors that can accelerate rusting
of iron - - - - Rusting of iron
is electrochemical and iron itself conducts
electricity.
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Corrosion of Iron
Recall Ered Fe2 (-0.44V) lt Ered O2 (1.23V),
so iron can be oxidized by oxygen. Cathode
O2(g) 4H(aq) 4e- ? 2H2O(l) Ered
1.23V Anode Fe(s) ? Fe2(aq) 2e- Ered
-0.44V Dissolved oxygen in water usually causes
Fe oxidation. Fe2 initially formed can be
further oxidized to Fe3 which forms rust, Fe2O3
. xH2O(s).
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Corrosion of Iron
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Corrosion of Iron
Oxidation occurs more readily at the site with
the greatest concentration of O2 (water-air
junction). Salts produce the necessary
electrolyte required to complete the electrical
circuit. Preventing Corrosion of Iron Corrosion
can be prevented by coating the iron with paint
or another metal. Galvanized iron (commonly used
for nails) is coated with a thin layer of zinc.
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Corrosion Prevention Galvanization
Zinc protects iron since Zn is more easily
oxidized (EredZn is more negative than
EredFe). Anode Zn2(aq) 2e- ? Zn(s)
Ered -0.76 V Cathode Fe2(aq) 2e- ? Fe(s)
Ered -0.44 V Zn is oxidized INSTEAD OF
IRON, preventing corrosion. Protects even if the
surface coat is broken.
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Preventing Corrosion of Iron
Corrosion Prevention Galvanization
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Corrosion Prevention Sacrificial Anodes
To protect underground pipelines. The water pipe
is the cathode and a more active metal is used as
the anode (cathodic protection). Often, Mg is
used as the sacrificial anode Mg2(aq) 2e- ?
Mg(s) Ered -2.37 V Fe2(aq) 2e- ? Fe(s)
Ered -0.44 V Since Ered of Mg is more
negative, it will oxidize first.
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Corrosion Prevention Sacrificial Anodes
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Electrolysis

• Defn Use of electrical energy to drive a
  non-spontaneous redox reaction.
• In voltaic and electrolytic cells
• reduction occurs at the cathode, and
• oxidation occurs at the anode.
• However, in electrolytic cells, electrons are
  forced to flow from the anode to cathode.
• Anode is now designated as POSITIVE().

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Electrolysis of Molten Salts
Used industrially to produce metals such as Na
and Al. Requires high temperatures. Example
Electrolysis of molten NaCl NaCl(s) ? Na(l)
Cl-(l) Two reactants Na and Cl- Cathode
2Na(l) 2e- ? 2Na(l) Ered -2.71V Anode
2Cl-(l) ? Cl2(g) 2e- Ered -1.359V This
is clearly a non-spontaneous reaction. Ecell
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Electrolysis of Molten Solutions
Electrons still flow from anode to cathode, but
are forced.
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Electrolysis of Aqueous Solutions
Aqueous solution electrolysis is complicated by
water. Ask Is water oxidized (to form O2) or
reduced (to form H2) in reference to the other
components? e.g. For an aqueous solution of
NaF in an electrolytic cell, possible reactants
are Na, F- and H2O. Both Na and H2O can be
reduced but not F-. Thus possible reactions at
cathode are Na(aq) e- Na(s)
Ered -2.71 V 2H2O(l)
2e- H2(g) 2OH-(aq) Ered
-0.83 V
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Electrolysis of Aqueous Solutions
More positive the Ered favors reduction
reactions. Reduction of H2O (Ered -0.83V)
occurs at cathode with H2 gas produced. At
anode, either F- or H2O must be oxidized since
Na cannot lose additional electrons. 2F-(aq)
F2(g) 2e-
Ered 2.87 V 4OH-(aq) O2(g)
2H2O(l) 4e- Ered 0.40 V Thus
oxidation of H2O occurs at anode (more negative
Ered favors oxidation reactions).
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Electrolysis of Aqueous Solutions
Cathode 4H2O(l) 4e- 2H2(g)
4OH-(aq) Ered -0.83 V Anode
4OH-(aq) O2(g) 2H2O(l) 4e-
Ered -0.40 V Process is
non-spontaneous (as expected).
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Electroplating
Active electrodes - take part in
electrolysis. Example electrolytic
plating. Defn Electrolysis used to deposit a
thin layer of one metal on another in order to
improve beauty or resistance to corrosion. e.g.
electroplating nickel on a piece of steel.
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Electroplating
Consider an active Ni electrode and another
metallic electrode placed in an aqueous solution
of NiSO4. Anode Ni(s) ? Ni2(aq) 2e- Cathode
Ni2(aq) 2e- ? Ni(s) With voltage supplied,
Ni plates on the steel electrode. Electroplating
is important in protecting objects from corrosion
(chrome).
Ecell 0V, so outside source of voltage needed!
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Quantitative Aspects of Electrolysis

• How much material can we obtain with
  electrolysis?
• Consider the reduction of Cu2 to Cu(s).
• Cu2(aq) 2e- ? Cu(s).
• 2 mol of electrons will plate 1 mol of Cu
• The charge of 1 mol of electrons is 96,500 C
  (1F).
• Using Q It
• The amount of Cu can be calculated from the
  current used (I) and time (t) allowed to plate.

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Calculate the number of grams of aluminum
produced in 1.00h by the electrolysis of molten
AlCl3 if the electrical current is 10.0 A.
Example
Ans. 3.36g
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Electrical Work
Free-energy is a measure of the maximum amount of
useful work that can be obtained from a
system. We know If work is negative, then
work is performed by the system and E is positive
(which means its spontaneous)
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Electrical Work
The emf can be thought of as a measure of the
driving force for a redox reaction to
proceed. In order to drive non-spontaneous,
electrolytic reactions, the external, applied emf
must be greater than Ecell. If Ecell -1.35 V,
then external emf must be at least 1.36
V. Note work can be expressed in Watts (W) 1 W
1 J/s.