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Chapter 20�Electrochemistry

John D. Bookstaver

St. Charles Community College

Cottleville, MO

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Electrochemical Reactions

In electrochemical reactions, electrons are transferred from one species to another.

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Oxidation Numbers

In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

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Oxidation and Reduction

A species is oxidized when it loses electrons.

Here, zinc loses two electrons to go from neutral zinc metal to the Zn²⁺ ion.

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Oxidation and Reduction

A species is reduced when it gains electrons.

Here, each of the H⁺ gains an electron, and they combine to form H₂.

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Oxidation and Reduction

What is reduced is the oxidizing agent.

H⁺ oxidizes Zn by taking electrons from it.

What is oxidized is the reducing agent.

Zn reduces H⁺ by giving it electrons.

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Assigning Oxidation Numbers

Elements in their elemental form have an oxidation number of 0.
The oxidation number of a monatomic ion is the same as its charge.

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Assigning Oxidation Numbers

Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.

Oxygen has an oxidation number of −2, except in the peroxide ion, which has an oxidation number of −1.
Hydrogen is −1 when bonded to a metal and +1 when bonded to a nonmetal.

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Assigning Oxidation Numbers

Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.

Fluorine always has an oxidation number of −1.
The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.

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Assigning Oxidation Numbers

The sum of the oxidation numbers in a neutral compound is 0.
The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

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Balancing Oxidation-Reduction Equations

Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.

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Balancing Oxidation-Reduction Equations

This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.

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The Half-Reaction Method

Assign oxidation numbers to determine what is oxidized and what is reduced.
Write the oxidation and reduction half-reactions.

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The Half-Reaction Method

Balance each half-reaction.

Balance elements other than H and O.
Balance O by adding H₂O.
Balance H by adding H⁺.
Balance charge by adding electrons.

Multiply the half-reactions by integers so that the electrons gained and lost are the same.

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The Half-Reaction Method

Add the half-reactions, subtracting things that appear on both sides.
Make sure the equation is balanced according to mass.
Make sure the equation is balanced according to charge.

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The Half-Reaction Method

Consider the reaction between MnO₄⁻ and C₂O₄²⁻:

MnO₄⁻₍_aq₎ + C₂O₄²⁻₍_aq₎  Mn²⁺₍_aq₎ + CO_{2 (}_aq₎

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The Half-Reaction Method

First, we assign oxidation numbers.

MnO₄⁻ + C₂O₄^2-  Mn²⁺ + CO₂

+7

+3

+4

+2

Since the manganese goes from +7 to +2, it is reduced.

Since the carbon goes from +3 to +4, it is oxidized.

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Oxidation Half-Reaction

C₂O₄²⁻  CO₂

To balance the carbon, we add a coefficient of 2:

C₂O₄²⁻  2 CO₂

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Oxidation Half-Reaction

C₂O₄²⁻  2 CO₂

The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.

C₂O₄²⁻  2 CO₂ + 2 e⁻

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Reduction Half-Reaction

MnO₄⁻  Mn²⁺

The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.

MnO₄⁻  Mn²⁺ + 4 H₂O

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Reduction Half-Reaction

MnO₄⁻  Mn²⁺ + 4 H₂O

To balance the hydrogen, we add 8 H⁺ to the left side.

8 H⁺ + MnO₄⁻  Mn²⁺ + 4 H₂O

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Reduction Half-Reaction

8 H⁺ + MnO₄⁻  Mn²⁺ + 4 H₂O

To balance the charge, we add 5 e⁻ to the left side.

5 e⁻ + 8 H⁺ + MnO₄⁻  Mn²⁺ + 4 H₂O

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Combining the Half-Reactions

Now we evaluate the two half-reactions together:

C₂O₄²⁻  2 CO₂ + 2 e⁻

5 e⁻ + 8 H⁺ + MnO₄⁻  Mn²⁺ + 4 H₂O

To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.

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Combining the Half-Reactions

5 C₂O₄²⁻  10 CO₂ + 10 e⁻

10 e⁻ + 16 H⁺ + 2 MnO₄⁻  2 Mn²⁺ + 8 H₂O

When we add these together, we get:

10 e⁻ + 16 H⁺ + 2 MnO₄⁻ + 5 C₂O₄²⁻ 

2 Mn²⁺ + 8 H₂O + 10 CO₂ +10 e⁻

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Combining the Half-Reactions

10 e⁻ + 16 H⁺ + 2 MnO₄⁻ + 5 C₂O₄²⁻ 

2 Mn²⁺ + 8 H₂O + 10 CO₂ +10 e⁻

The only thing that appears on both sides are the electrons. Subtracting them, we are left with:

16 H⁺ + 2 MnO₄⁻ + 5 C₂O₄²⁻ 

2 Mn²⁺ + 8 H₂O + 10 CO₂

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Balancing in Basic Solution

If a reaction occurs in basic solution, one can balance it as if it occurred in acid.
Once the equation is balanced, add OH⁻ to each side to “neutralize” the H⁺ in the equation and create water in its place.
If this produces water on both sides, you might have to subtract water from each side.

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Voltaic Cells�(Galvanic Cells)

In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

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Voltaic Cells

We can use that energy to do work if we make the electrons flow through an external device.
We call such a setup a voltaic cell.

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Voltaic Cells

A typical cell looks like this.
The oxidation occurs at the anode.
The reduction occurs at the cathode.
Red Cat, An Ox

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Voltaic Cells

Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

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Voltaic Cells

Therefore, we use a salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced.

Cations move toward the cathode.
Anions move toward the anode.

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Voltaic Cells

In the cell, then, electrons leave the anode and flow through the wire to the cathode.
As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.

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Voltaic Cells

As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode.
The electrons are taken by the cation, and the neutral metal is deposited on the cathode.

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Example

A nickel-cadmium (nicad) battery (a rechargeable dry-cell) uses the following reaction:
Cd_(s) + NiO_2(s) + 2H₂O_(l) → Cd(OH)_2(s) + Ni(OH)_2(s)
ID the substances that are oxidized & reduced & ID the oxidizing agent & reducing agent.

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Complete & balance via half-reactions

Cr₂O₇^2-_(aq) + Cl^-_(aq) → Cr³⁺_(aq) + Cl_2(g)
(acidic solution)

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Redox Cell

Cr₂O₇^2- + 14H⁺ + 6I^- → 2Cr^3- + 3I_2(s) + 3H₂O_(l)
A sln of potassium dichromate & sulfuric acid is poured into a container
A sln of potassium iodide is poured into another.
Inert electrodes (such as Pt) are placed in each sln & connected to a volt-meter, along with a salt-bridge. A current is detected.
Indicate:

Rxn at anode
Rxn at cathode
Direction of electron migration
Direction of ion migration
Signs of the electrodes

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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.

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Cu²⁺

Fe²⁺

Fe³⁺

For the last reaction
Cu(s)½Cu⁺²(aq)½½Fe⁺²(aq),Fe⁺³(aq)½Pt(s)

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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

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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⁺

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Electromotive Force (emf)

Water only spontaneously flows one way in a waterfall.
Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

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Electromotive Force (emf)

The potential difference between the anode and cathode in a cell is called the electromotive force (emf).
It is also called the cell potential and is designated E_cell.

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Cell Potential

Emf is work /charge.

Cell potential is measured in volts (V).

1 V = 1

J

C

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Standard Reduction Potentials

Reduction potentials for many electrodes have been measured and tabulated.

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Standard Hydrogen Electrode

Their values are referenced to a standard hydrogen electrode (SHE).
By definition, the reduction potential for hydrogen is 0 V:

2 H⁺ (aq, 1M) + 2 e⁻  H₂ (g, 1 atm)

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Standard Cell Potentials

The cell potential at standard conditions can be found through this equation:

E_cell



= E_red (cathode) − E_red (anode)



Because cell potential is based on the potential energy per unit of charge, it is an intensive property.

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Cell Potentials

For the oxidation in this cell,

For the reduction,

E_red = −0.76 V



E_red = +0.34 V



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Cell Potentials

E_cell



=

E_red



(cathode) −

E_red



(anode)

= +0.34 V − (−0.76 V)

= +1.10 V

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Oxidizing and Reducing Agents

The strongest oxidizers have the most positive reduction potentials.
The strongest reducers have the most negative reduction potentials.

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Oxidizing and Reducing Agents

The greater the difference between the two, the greater the voltage of the cell.

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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

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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

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Rank the following

From strongest oxidizing agent to weakest
NO₃^-_(aq)
Ag⁺_(aq)
Cr₂O₇^2-_(aq)

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Free Energy

G for a redox reaction can be found by using the equation

G = −nFE

where n is the number of moles of electrons transferred, and F is a constant, the Faraday.

1 F = 96,485 C/mol = 96,485 J/V-mol

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Free Energy

Under standard conditions,

G = −nFE

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Spontaneous or Not?

Cu²⁺ + H₂ → Cu + 2H⁺

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Nernst Equation

Remember that

G = G + RT ln Q

This means

−nFE = −nFE + RT ln Q

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Nernst Equation

Dividing both sides by −nF, we get the Nernst equation:

E = E −

RT

nF

ln Q

or, using base-10 logarithms,

E = E −

2.303 RT

nF

log Q

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Nernst Equation

At room temperature (298 K),

Thus the equation becomes

E = E −

0.0592

n

log Q

2.303 RT

F

= 0.0592 V

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Concentration Cells

Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.

For such a cell,

would be 0, but Q would not.

E_cell

^

Therefore, as long as the concentrations are different, E will not be 0.

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Practice

Using standard reduction potentials at 298K
4Ag + O₂ + 4H⁺ → 4Ag⁺ + 2H₂O_(l)
Determine the free energy change & K

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Voltaic Cell under Non-Standard Conditions

Cr₂O₇^2- + 14H⁺ + 6I^- → 2Cr^3- + 3I_2(s) + 3H₂O_(l)
[Cr₂O₇^2-] = 2.0M
[H⁺] = 1.0M
[I^-] = 1.0M
[Cr³⁺] = 1.0x10^-5M

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Finding Concentrations

Zn + 2H⁺ → Zn²⁺ + H₂
The voltage of this cell is .45V at 298K when [Zn²⁺] = 1.0M and P_H2 = 1.0atm.
What is [H⁺]
Solution:

Write the std cell potentials for red & ox
Determine #e- transferred
Use Nernst equation to solve for Q
Use Q to determine [H⁺]

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Applications of Oxidation-Reduction Reactions

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Batteries

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Alkaline Batteries

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Hydrogen Fuel Cells�http://www.fueleconomy.gov/feg/fuelcell8.swf

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Corrosion

Undesirable spontaneous redox reactions
Thin coating can protect some metals (like aluminum) – forms a hydrated oxide)
Iron - $$$$$

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Protection

Higher pH
Paint surface
Galvanize (zinc coating) – why?
Zinc is a better anode
Called cathodic protection – sacrificial metal

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Electrolysis

Cells that use a battery or outside power source to drive an electrochemical reaction in reverse
Example NaCl → Na⁺ + Cl^-
Reduction at the cathode, oxidation at the anode
Voltage source pumps electrons to cathode.

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Diagram

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Solutions

High temperatures necessary for previous electrolysis (ionic solids have high MP)
Easier for solutions, but water must be considered
Example: NaF
Possible reductions are:

Na⁺ + e^- → Na(s) (E_red = -2.71 V)
2H₂O + 2 e^- → H₂(g) + 2 OH^- (E_red = -.83 V)

Far easier to reduce water!

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Continued

Look at possible oxidations:

2F- → F₂(g) (E_red = 2.87 volts)
2H₂O → O₂(g) + 4H⁺ + 4e^-(E_red = 1.23 volts)
Far easier to oxidize water, or even OH-!

So for NaF, neither electrode would produce anything useful, and doesn’t by experiment
With NaCL, neither electrode is favored over water. However, the oxidation of Cl- is kinetically favored, and thus occurs upon experimentation!
Use E_red values of two products to find Ecell (minimum amount of energy that must be provided to force cell to work)

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Active electrodes

If electrode is not inert, it can be coated with a thin layer of the metal being reduced, if its reduction potential is greater than that of water.
This is called electroplating
Ecell = 0, so a small voltage is needed to push the reaction.

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Applications of Electrolytic Cells

Aluminium refining: The major ore of aluminium is bauxite, Al₂O₃.

Anhydrous Al₂O₃ melts at over 2000°C. This is too high to permit its use as a molten medium for electrolytic formation of free aluminium. The electrolytic process commercially used to produce aluminium is known as the Hall process, named after its inventor, Charles M. Hall. Al₂O₃ is dissolved in molten cryolite, Na₃AlF₆, which has a melting point of 1012^oC and is an effective conductor of electric current. Graphite rods are employed as anodes and are consumed in the electrolysis process. The cell electrolytic reaction is:

2Al₂O₃ + 3C → 4Al(l) + 3CO₂(g)

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Electrolysis of brine

Chlorine and sodium hydroxide are both manufactured by electrolysis of brine (aqueous sodium chloride) using inert electrodes. Chlorine is evolved at the anode, Cl^- →1/2Cl₂ + e

Hydrogen is evolved at the cathode: H⁺ + e →1/2H₂

The removal of chloride ions and hydrogen ions leaves sodium ions and hydroxide ions in solution.

Chlorine is used to disinfect municipal water supplies and water in swimming pools. It is used to manufacture household bleaches and disinfectants. It is used to manufacture plastics (e.g. PVC), pesticides, anaesthetics, CFCs etc.

Sodium hydroxide is used in the manufacture of synthetic fibres, soaps and detergents.

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Quantitative relationship

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