Oxidation Reduction and Electrochemistry - chymist.com and Electrochemistry.pdf · Oxidation‐Reduction and Electrochemistry David A. Katz ... + Cu(NO 3) 2 (aq) ... Oxidation ‐reduction

Oxidation‐Reductionand

Electrochemistry

David A. KatzDepartment of ChemistryPima Community College

Oxidation‐Reduction Reactions

In an oxidation‐reduction (Redox) reaction, electrons are transferred from one species to another.For example, in a single replacement reaction

Cu (s) + 2 AgNO3 (aq) 2 Ag (s) + Cu(NO3)2 (aq)

The Cu atoms lose electrons to form Cu2+ in theCu(NO3)2 (aq) and the Ag+ gains electrons to form metallic Ag

Oxidation‐Reduction Reactions

• This can be more easily observed by writing the net ionic equation for the reaction:

Cu (s) + 2 Ag+ (aq) 2 Ag (s) + Cu2+ (aq)

• The metallic Cu atoms are uncombined, so they are considered to have an oxidation number of zero.

• The initial combined Ag+ ions are in a +1 oxidation state.• Each Cu atom will lose 2 electrons to 2 Ag+ ions• The resulting Ag atoms are considered to have an

oxidation number of zero

Oxidation‐Reduction ReactionsCu (s) + 2 Ag+ (aq) 2 Ag (s) + Cu2+ (aq)

Oxidation Numbers

In order to keep track of what loses electrons and what gains them, we list the oxidation numbers of each element.

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.


• A species is reduced when it gains electrons.– Here, each of the H+ gains an electron and they combine to form H2.


• The species that contains the element that is reduced is the oxidizing agent.– H+ oxidizes Zn by taking electrons from it.

• The species that contains the element that is oxidized is the reducing agent.– Zn reduces H+ by giving it electrons.

Assigning Oxidation Numbers

1. Elements in their elemental form have an oxidation number of 0.

2. The oxidation number of a monatomic ion is the same as its charge.


3. The oxidation number of metals depends on their position in the periodic table

• Group IA elements are +1• Group IIA elements are +2• Group IIIA elements are +3• Group IVA metals are usually +2 or +4• Group VA metals are usually +3 or +5


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

– Oxygen always has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1.

– Hydrogen is always −1 when bonded to a metal

– Hydrogen is +1 when bonded to a nonmetal.


4. Nonmetals (continued).– Fluorine always has an oxidation number of

−1.– The halogens (Cl, Br, and I)have an oxidation

number of −1 when they are negative– The halogens (Cl, Br, and I) will have

positive oxidation numbers in oxyanions(ClO‐, ClO2

‐, ClO3‐, etc.)


5. The sum of the oxidation numbers in a neutral compound is 0.

6. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

Balancing Oxidation‐Reduction Equations

Oxidation‐reduction equations may be difficult to balance.Generally, the easiest way to balance the equation of an oxidation‐reduction reaction is via the half‐reaction method.This involves treating 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.

Balancing Redox Equations by the Half‐Reaction Method

1. Assign oxidation numbers to the elements in the equation

2. Determine what is oxidized and what is reduced.

3. Write the oxidation and reduction half‐reactions.


4. Balance each half‐reaction.a. Balance elements other than H and O.b. Balance O by adding H2O.c. Balance H by adding H+.d. Balance charge by adding electrons.

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


6. Add the half‐reactions, subtracting things that appear on both sides.

7. Make sure the equation is balanced according to mass.

8. Make sure the equation is balanced according to charge.


Consider the reaction between MnO4− and C2O4

2− :MnO4

−(aq) + C2O4

2−(aq) Mn2+(aq) + CO2(aq) (acidic solution)


First, assign oxidation numbers (remember oxygen is ‐2)

MnO4− + C2O4

2‐ Mn2+ + CO2

+7 +3 +4+2

The oxidation number of manganese goes from +7 to +2, it is reduced.The oxidation number of carbon goes from +3 to +4, it is oxidized.


MnO4− + C2O4

2‐ Mn2+ + CO2

+7 +3 +4+2

The MnO4‐ is the oxidizing agent

The C2O42‐ is the reducing agent

Determine the oxidizing and reducing agents


The oxidation half‐reaction is

C2O42− CO2

Balance the half reaction

C2O42− 2 CO2

This balances both the carbon atoms and the oxygen atoms


Balance the charge by adding 2 electrons to the right side.

C2O42− 2 CO2 + 2 e−


The reduction half‐reaction is

MnO4− Mn2+

The manganese is balancedIn order to balance the 4 oxygens, we add 4 water molecules to the right side.

MnO4−Mn2+ + 4 H2O


MnO4− Mn2+ + 4 H2O

The addition of water on the right side of the equation included hydrogen atoms.To balance the 8 hydrogens, add 8 H+ to the left side.

8 H+ + MnO4−Mn2+ + 4 H2O


8 H+ + MnO4−Mn2+ + 4 H2O

Balance the charges, add 5 e− to the left side.

5 e− + 8 H+ + MnO4−Mn2+ + 4 H2O


Before the two half‐reactions can be added together, the number of electrons lost must be equal to the electrons gained:

C2O42− 2 CO2 + 2 e− (2 e‐ lost)

5 e− + 8 H+ + MnO4−Mn2+ + 4 H2O (5 e‐ gained)

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

5 C2O42− 10 CO2 + 10 e−

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


5 C2O42− 10 CO2 + 10 e−

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

Add these half‐reactions

10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2−

2 Mn2+ + 8 H2O + 10 CO2 +10 e−

Cancel out the electrons from both sides to get

16 H+ + 2 MnO4− + 5 C2O4

2− 2 Mn2+ + 8 H2O + 10 CO2

Balancing Redox Equationsin Basic Solution

• If a reaction occurs in basic solution, use OH‐

and H2O instead of the H+ and H2O used in acid solution

• The easiest method is to balance the equation 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 results in water on both sides of the equation, subtract water from each side as a final step.


Complete and balance the following redox equation that takes place in basic solutionCN‐

(aq) + MnO4‐(aq) CNO‐

(aq) + MnO2 (s) (basic solution)

First, assign oxidation numbers (remember oxygen is ‐2)

+2 ‐3 +7 ‐2 +4 ‐3 ‐2 +4 ‐2

CN‐(aq) + MnO4

‐(aq) CNO‐

(aq) + MnO2 (s)

Note: In an anion such as CN‐, C, which comes first, would be positive and N would be negative. If negative, N would be ‐3.


+2 ‐3 +7 ‐2 +4 ‐3 ‐2 +4 ‐2

CN‐(aq) + MnO4

‐(aq) CNO‐

(aq) + MnO2 (s)

The oxidation number of carbon goes from +2 to +4, it is oxidized. (C lost 2 e‐)The oxidation number of manganese goes from +7 to +4, it is reduced. (Mn gained 3 e‐)

MnO4‐ is the oxidizing agent

CN‐ is the reducing agent


The reduction half‐reaction is

MnO4− MnO2

Balance the half reaction

MnO4− MnO2

Balance the oxygen atoms, add H+ and H2O

4 H+ + MnO4− MnO2 + 2 H2O


Balance the charges, add 3 e‐ to the left side

3 e‐ + 4 H+ + MnO4− MnO2 + 2 H2O


The oxidation half‐reaction is

CN− CNO‐

Since the C atoms are balanced, balance the oxygen.

Add H2O to the left side and H+ to the right side.H2O + CN− CNO‐+ 2 H+


H2O + CN− CNO‐ + 2 H+

Balance the charge, add 2 e‐ to the right side

H2O + CN− CNO ̶ + 2 H+ + 2 e‐


Before the two half‐reactions can be added together, the number of electrons lost must be equal to the electrons gained:H2O + CN− CNO‐ + 2 H+ + 2 e‐ (2 e‐ lost)

3 e‐ + 4 H+ + MnO4− MnO2 + 2 H2O (3 e‐ gained)

To attain the same number of electrons on each side, multiply the first equation by 3 and the second by 2.3 H2O + 3 CN− 3 CNO‐ + 6 H+ + 6 e‐

6 e‐ + 8 H+ + 2 MnO4− 2 MnO2 + 4 H2O


Add the half‐reactions3 H2O + 3 CN− + 8 H+ + 2 MnO4

−

3 CNO‐ + 6 H+ + 2 MnO2 + 4 H2O

Cancel out the H+ and H2O to get2 H+ + 3 CN− + 2 MnO4

− 3 CNO‐ + 2 MnO2 + H2O


This reaction, however, takes place in basic solution2 H+ + 3 CN− + 2 MnO4

− 3 CNO‐ + 2 MnO2 + H2O

Add 2 OH‐ to both sides to cancel out the H+

2 OH‐ +2 H+ + 3 CN− + 2 MnO4− 3 CNO‐ + 2 MnO2 + H2O + 2 OH‐

(Remember that H+ + OH‐ H2O)

2 H2O + 3 CN− + 2 MnO4− 3 CNO‐ + 2 MnO2 + H2O + 2 OH‐

The balanced equation is

3 CN− + 2 MnO4− + H2O 3 CNO‐ + 2 MnO2 + 2 OH‐

A History of Electricity/Electrochemistry• Thales of Miletus (640‐546 B.C.) is

credited with the discovery thatamber when rubbed with cloth orfur acquired the property ofattracting light objects.

• The word electricity comes from"elektron" the Greek word foramber.

• Otto von Guericke (1602‐1686)invented the first electrostaticgenerator in 1675. It was made ofa sulphur ball which rotated in awooden cradle. The ball itself wasrubbed by hand and the chargedsulphur ball had to betransported to the place wherethe electric experiment wascarried out.

Thales of Miletus Otto von Guericke

• Eventually, a glass globe replaced the sulfur sphere used by Guericke

• Later, large disks were used

• Ewald Jürgen von Kleist (1700‐1748), invented the Leyden Jar in 1745 to store electric energy. The Leyden Jar contained water or mercury and was placed onto a metal surface with ground connection.

• In 1746, the Leyden jar was independently invented by physicist Pieter van Musschenbroek (1692‐1761) and/or his lawyer friend Andreas Cunnaeus in Leyden/the Netherlands

• Leyden jars could be joined together to store large electrical charges

• In 1752, Benjamin Franklin (1706‐1790) demonstrated that lightning was electricity in his famous kite experiment

• In 1780, Italian physician and physicist Luigi Aloisio Galvani (1737‐1798) discovered that muscle and nerve cells produce electricity. Whilst dissecting a frog on a table where he had been conducting experiments with static electricity, Galvani touched the exposed sciatic nerve with his scalpel, which had picked up an electric charge. He noticed that the frog’s leg jumped.

Count Alessandro Giuseppe Antonio Anastasio Volta (1745 – 1827) developed the first electric cell, called a Voltaic Pile, in 1800.A voltaic pile consist of alternating layers of two dissimilar metals, separated by pieces of cardboard soaked in a sodium chloride solution or sulfuric acid.

Volta determined that the best combination of metals was zinc and silver

Volta’s electric pile (right)

A Voltaic pile at the Smithsonian Institution, (far right)

• In 1800, English chemist William Nicholson (1753–1815) and surgeon Anthony Carlisle (1768‐1840) separated water into hydrogen and oxygen by electrolysis.

• Johann Wilhelm Ritter (1776‐1810) repeated Nicholson’s separation of water into hydrogen and oxygen by electrolysis. Soon thereafter, Ritter discovered the process of electroplating. He also observed that the amount of metal deposited and the amount of oxygen produced during an electrolytic process depended on the distance between the electrodes

• Humphrey Davy (1778‐1829) utilized the voltaic pile, in 1807, to isolate elemental potassium by electrolysis which was soon followed by sodium, barium, calcium, strontium, magnesium.

William Nicholson

Johann Wilhelm Ritter

Humphrey Davy

• Michael Faraday (1791‐1867) began his career in 1813 as Davy's Laboratory Assistant.

• In 1834, Faraday developed the two laws of electrochemistry: • The First Law of Electrochemistry

The amount of a substance deposited on each electrode of an electrolytic cell is directly proportional to the amount of electricity passing through the cell.

• The Second Law of ElectrochemistryThe quantities of different elements deposited by a given amount of electricity are in the ratio of their chemical equivalent weights.

• Faraday also defined a number of terms:The anode is therefore that surface at which the electric current, according to our present expression, enters: it is the negative extremity of the decomposing body; is where oxygen, chlorine, acids, etc., are evolved; and is against or opposite the positive electrode.

The cathode is that surface at which the current leaves the decomposing body, and is its positive extremity; the combustible bodies, metals, alkalies, and bases are evolved there, and it is in contact with the negative electrode.

Many bodies are decomposed directly by the electric current, their elements being set free; these I propose to call electrolytes....

Finally, I require a term to express those bodies which can pass to theelectrodes, or, as they are usually called, the poles. Substances are frequently spoken of as being electro‐negative or electro‐positive, according as they go under the supposed influence of a direct attraction to the positive or negative pole...I propose to distinguish such bodies by calling those anions which go to the anode of the decomposing body; and those passing to the cathode, cations; and when I have occasion to speak of these together, I shall call them ions.

…the chloride of lead is an electrolyte, and when electrolyzed evolves the two ions, chlorine and lead, the former being an anion, and the latter a cation.

• John Frederic Daniell (1790‐1845), professor of chemistry at King's College, London.

• Daniell's research into development of constant current cells took place at the same time (late 1830s) that commercial telegraph systems began to appear. Daniell's copper battery (1836) became the standard for British and American telegraph systems.

• In 1839, Daniell experimented on the fusion of metals with a 70‐cell battery. He produced an electric arc so rich in ultraviolet rays that it resulted in an instant, artificial sunburn. These experiments caused serious injury to Daniell's eyes as well as the eyes of spectators.

• Ultimately, Daniell showed that the ion of the metal, rather than its oxide, carries an electric charge when a metal‐salt solution is electrolyzed.

Left: An early Daniell Cell

Right:Daniell cells used by Sir William Robert Grove, 1839.

Voltaic Cells

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

Voltaic Cells

• If the reaction is separated into two parts, we can use that energy to do work by making the electrons flow through an external device.

• This type of setup is called a voltaic cell.

Voltaic Cells• This is a typical voltaic

cell• A strip of zinc metal is

immersed in a solution of Zn(NO3)2

• A strip of copper metal is immersed in a solution of Cu(NO3)2

• The two solutions are connected by a salt bridge containing NaNO3

• The oxidation occurs at the anode (Zn)

• The reduction occurs at the cathode (Cu)

Voltaic Cells• The salt bridge is used to

prevent electrons flowing directly from the zinc to the copper

• The salt bridge consists of a U‐shaped tube that contains a salt solution, sealed with porous plugs, or an agar gel solution of the salt

• The salt bridge keeps the charges balanced and forces the electron to move through the wire – Cations move toward the

cathode.– Anions move toward the

anode.

Voltaic Cells

• In the cell, 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.

• Eventually, if the cell is used for a long time, the anode (zinc) will dissolve

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.

• Eventually, if the cell is used for a long time, all the copper ions will plate onto the copper cathode

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.

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 Ecell.

Cell Potential

Cell potential is measured in volts (V).

1 V = 1 JC

Where J = JoulesC = Coulombs

Recall that 1 electron has a charge of 1.6 x 10‐19 C

Standard Reduction Potentials

The cell potential is the difference between two electrode potentials.By convention, electrode potentials are written as reductionsReduction potentials for most common electrodes are tabulated as standard reduction potentials.

Standard Hydrogen Electrode

• Electrode potentials 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)

Standard Cell Potentials

The cell potential at standard conditions is calculated

Ecell = Ered (cathode) − Ered (anode)

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

Substance reduced Substance oxidized

Cell Potentials

Oxidation: E°red = -0.76 V Reduction: E°red = +0.34 V

Cell Potentials

Ecell = Ered (cathode) − Ered (anode)

= +0.34 V − (−0.76 V)= +1.10 V

Generally, most of the common cells used, on the average, generate approximately 1.5 V

Oxidizing and Reducing Agents

• The strongest oxidizers have the most positive reduction potentials.

• The strongest reducers have the most negative reduction potentials.

Oxidizing and Reducing Agents

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

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 transferredF is a constant, the Faraday.

1 F = 96,485 C/mol = 96,485 J/V‐molE° = The standard cell potential in V

Under standard conditions,G = −nFE

Nernst Equation• Remember that

G = G + RT ln Q• This means

−nFE = −nFE + RT ln Q

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

or, using base‐10 logarithms,

E = E −RTnF ln Q

E = E −2.303 RT

nF log Q

Nernst EquationAt room temperature (298 K), and R = 8.314 J/mol KF = 96,485 J/V‐mol

The final form of the Nernst Equation becomes

E = E − 0.0592n

log Q

2.303 RTF = 0.0592 V

Walther Hermann Nernst (1864 ‐1941) • Nernst's early studies in electrochemistry were inspired

by Arrhenius' dissociation theory of ions in solution. • In 1889 he elucidated the theory of galvanic cells by

assuming an "electrolytic pressure of dissolution" which forces ions from electrodes into solution and which was opposed to the osmotic pressure of the dissolved ions.

• Also, in 1889, he showed how the characteristics of the current produced could be used to calculate the free energy change in the chemical reaction producing the current. This equation, known as the Nernst Equation, relates the voltage of a cell to its properties.

• Independently of Thomson, he explained why compounds ionize easily in water. The explanation, called the Nernst‐Thomson rule, holds that it is difficult for charged ions to attract each other through insulating water molecules, so they dissociate.

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, E°cell would be 0, but Q would not.

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

Applications of Oxidation‐Reduction Reactions

Why Study Electrochemistry?Why Study Electrochemistry?• Batteries• Corrosion• Industrial production

of chemicals such as Cl2, NaOH, F2and Al

• Biological redoxreactions

The heme group

BATTERIESPrimary, Secondary, and Fuel Cells

BatteriesMost commercial batteries produce 1.5 V. To get a higher voltage, batteries are joined together.

Dry Cell Battery

Anode (-)

Zn Zn2+ + 2e-

Cathode (+)

2 NH4+ + 2e-2 NH3 + H2

Primary battery — uses redox reactions that cannot be restored by recharge.

Nearly same reactions as in common dry cell, but under basic conditions.

Alkaline Battery

Anode (‐): Zn + 2 OH‐ ZnO + H2O + 2e‐Cathode (+): 2 MnO2 + H2O + 2e‐ Mn2O3 + 2 OH‐

Alkaline Batteries

Lead Storage Battery

• Secondary battery

• Uses redoxreactions that can be reversed.

• Can be restored by recharging

Lead Storage BatteryAnode (-) Eo = +0.36 VPb + HSO4

- PbSO4 + H+ + 2e-Cathode (+) Eo = +1.68 VPbO2 + HSO4

- + 3 H+ + 2e- PbSO4 + 2 H2O

Ni-Cad BatteryAnode (-)Cd + 2 OH- Cd(OH)2 + 2e-Cathode (+) NiO(OH) + H2O + e- Ni(OH)2 + OH-

Fuel Cells: H2 as a Fuel•Fuel cell ‐ reactants are supplied continuously from an external source.•Cars can use electricity generated by H2/O2 fuel cells.•H2 carried in tanks or generated from hydrocarbons.

Hydrogen—Air Fuel Cell

Hydrogen Fuel Cells

H2 as a Fuel

Comparison of the volumes of substances required to store 4 kg of hydrogen relative to car size.

Storing H2 as a Fuel

One way to store H2 is to adsorb the gas onto a metal or metal alloy.

ElectrolysisUsing electrical energy to produce chemical change.

Sn2+(aq) + 2 Cl‐(aq) Sn(s) + Cl2(g)

Electrolysis of Aqueous NaOH

Anode (+)4 OH- O2(g) + 2 H2O + 4e-

Cathode (-) 4 H2O + 4e- 2 H2 + 4 OH-

Eo for cell = -1.23 V

Electric Energy fChemical Change

Anode Cathode

ElectrolysisElectric Energy f Chemical Change

BATTERY

+

Na+Cl-

Anode Cathode

electrons

BATTERY

+

Na+Cl-

Anode Cathode

electrons• Electrolysis of molten NaCl.

• Here a battery “pumps” electrons from Cl‐ to Na+.

•NOTE: Polarity of electrodes is reversed from batteries.

Electrolysis of Molten NaCl

See Figure 20.18

Electrolysis of Molten NaClAnode (+)

2 Cl- Cl2(g) + 2e-Cathode (-)

Na+ + e- Na

BATTERY

+

Na+Cl-

Anode Cathode

electrons

BATTERY

+

Na+Cl-

Anode Cathode

electrons

Eo for cell (in water) = E˚c ‐ E˚a= ‐ 2.71 V – (+1.36 V)= ‐ 4.07 V (in water)

External energy needed because Eo is (‐).

Electrolysis of Aqueous NaClAnode (+) 2 Cl- Cl2(g) + 2e-

Cathode (-) 2 H2O + 2e- H2 + 2 OH-


Note that H2O is more easily reduced than Na+.

BATTERY

+

Na+Cl-

Anode Cathode

H2O

electrons

BATTERY

+

Na+Cl-

Anode Cathode

H2O

electrons

Also, Cl‐ is oxidized in preference to H2O because of kinetics.Also, Cl‐ is oxidized in preference to H2O because of kinetics.

Electrolysis of Aqueous NaClCells like these are the source of NaOH and Cl2.In 1995: 25.1 x 109 lb Cl2 and 26.1 x 109 lb NaOH

Also the source of NaOCl for use in bleach.

Electrolysis of Aqueous NaI

Anode (+): 2 I- I2(g) + 2e-Cathode (-): 2 H2O + 2e- H2 + 2 OH-


Electrolysis of Aqueous CuCl2Anode (+)

2 Cl- Cl2(g) + 2e-

Cathode (-)

Cu2+ + 2e- Cu


Note that Cu is more easily reduced than either H2O or Na+.

BATTERY

+

Cu2+Cl-

Anode Cathode

electrons

H2O

BATTERY

+

Cu2+Cl-

Anode Cathode

electrons

H2O

Electrolytic Refining of Copper

Impure copper is oxidized to Cu2+ at the anode. The aqueous Cu2+ ions are reduced to Cu metal at the cathode.The copper at the cathode is over 99% pure

Producing Aluminum2 Al2O3 + 3 C f 4 Al + 3 CO2

Charles Hall (1863‐1914) developed electrolysis process. Founded Alcoa.

Corrosion and…

…Corrosion Prevention

Oxidation Reduction and Electrochemistry - chymist.com and Electrochemistry.pdf · Oxidation‐Reduction and Electrochemistry David A. Katz ... + Cu(NO 3) 2 (aq) ... Oxidation ‐reduction

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