redox reactions, applications of chemistry

OxidatioOxidationnan increase in oxidation an increase in oxidation

statestateloss of electronloss of electronin organic chemistry, it is in organic chemistry, it is the loss of H atom/s or the the loss of H atom/s or the gain of O atom/s.gain of O atom/s.

Oxidizing agentOxidizing agentan electron acceptoran electron acceptora reactant that accepts a reactant that accepts electron from another electron from another reactant.reactant.

the reactant that is the reactant that is reduced in a redox reduced in a redox reactionreaction

ReductionReduction

a decrease in oxidation a decrease in oxidation statestate

gain of electrongain of electronin organic chemistry, it is in organic chemistry, it is the loss of O atom/s or the the loss of O atom/s or the gain of H atom/s.gain of H atom/s.

Reducing agentReducing agentan electron donoran electron donora reactant that gives up a reactant that gives up electron/s to another electron/s to another reactant thus decreasing the reactant thus decreasing the oxidation state of one of its oxidation state of one of its atom.atom.

the reactant that is oxidized the reactant that is oxidized in a redox reactionin a redox reaction

Oxidation Oxidation NumberNumberFor an element or

radical, it is the same as the charge of the ion formed from an atom of the element or the ion of the same composition as the radical

Oxidation Oxidation NumberNumberExample: NaClExample: NaCl

the oxidation number of the oxidation number of Na and Cl are +1 and –1Na and Cl are +1 and –1

NaCl NaCl Na Na++ + Cl + Cl––

and it can be said that and it can be said that sodium and chlorine are in a sodium and chlorine are in a +1 and –1 +1 and –1 oxidation statesoxidation states

Oxidation Oxidation StateState

A concept that provides a way to keep track of electrons in redox reaction according to certain rules.

Fundamental Fundamental RulesRules The sum of the oxidation state The sum of the oxidation state

for all atoms in the formula for an for all atoms in the formula for an electrically neutral compound is electrically neutral compound is zerozero..

The oxidation state for any The oxidation state for any element in the free or element in the free or uncombined state is uncombined state is zerozero..

The oxidation state for an ion is The oxidation state for an ion is the same as its charge.the same as its charge.

Special Special ConventionsConventions In all hydrogen compounds, the In all hydrogen compounds, the

oxidation state for H is +1.oxidation state for H is +1.Exception: in hydrides where H is –1Exception: in hydrides where H is –1

In all oxygen compounds, the In all oxygen compounds, the oxidation state for O is –2.oxidation state for O is –2.

Exception: in peroxides where O is –1Exception: in peroxides where O is –1

Special Special ConventionsConventions In all halides, the oxidation In all halides, the oxidation

state for the halogens is state for the halogens is –1.–1. In all sulfides, the oxidation In all sulfides, the oxidation

state for sulfur is state for sulfur is –2.–2. In binary compounds, the element In binary compounds, the element

with the greatest attraction for with the greatest attraction for electrons is assigned a negative electrons is assigned a negative oxidation state equal to its charge in oxidation state equal to its charge in its ionic compound.its ionic compound.

Some Important ConceptsSome Important Concepts Corrosion: When a metal Corrosion: When a metal

undergoes corrosion it is oxidized undergoes corrosion it is oxidized forming cations:forming cations:

MgMg((ss)) + 2 H + 2 H++((aqaq)) Mg Mg2+2+

((aqaq)) + H + H2(2(gg))

Metals are also oxidized by acids to form Metals are also oxidized by acids to form salts:salts:

MgMg((ss)) + 2 HCl + 2 HCl((aqaq)) MgCl MgCl2(2(aqaq)) + H + H2(2(gg))

Metals can also be oxidized by other Metals can also be oxidized by other salts:salts:

FeFe((ss)) + Ni + Ni2+2+((aqaq)) Fe Fe2+2+

((aqaq)) + Ni + Ni((ss))

Some Important ConceptsSome Important Concepts Activity series: A list of Activity series: A list of

metals arranged in decreasing metals arranged in decreasing ease of oxidation.ease of oxidation. Some metals are easily oxidized Some metals are easily oxidized

whereas others are not.whereas others are not. The higher the metal on the activity The higher the metal on the activity

series, the more active that metal.series, the more active that metal. Any metal can be oxidized by the Any metal can be oxidized by the

ions of elements below it.ions of elements below it.

Some Important ConceptsSome Important Concepts Solution Composition:Solution Composition:

Solution - solute dissolved in a Solution - solute dissolved in a solventsolvent

Aqueous solution - solution in which Aqueous solution - solution in which water is the dissolving medium or water is the dissolving medium or solvent (ex. HClsolvent (ex. HCl(aq)(aq), H, H++

(aq)(aq), Ni, Ni2+2+(aq)(aq)))

Solute - a substance dissolved in a Solute - a substance dissolved in a solvent to form a solutionsolvent to form a solution

Solvent - the dissolving medium in a Solvent - the dissolving medium in a solutionsolution

Some Important ConceptsSome Important Concepts Solution Composition:Solution Composition:

Concentration - amount or Concentration - amount or measure of solute in a solution measure of solute in a solution (generally measured as molar (generally measured as molar concentration)concentration)

Molarity - moles of solute per Molarity - moles of solute per volume of solution in liters (or volume of solution in liters (or simply moles/liter)simply moles/liter)

Balancing Balancing Redox Redox

ReactionReaction

Oxidation States Oxidation States MethodMethod Assign the oxidation states Assign the oxidation states

of all the atoms.of all the atoms. Determine which element is Determine which element is oxidized and which is oxidized and which is reduced. Take note also of reduced. Take note also of the change the change (increase/decrease) in (increase/decrease) in oxidation state.oxidation state.

Oxidation States Oxidation States MethodMethod Choose the coefficients for the Choose the coefficients for the

compound containing the compound containing the element oxidized and the element oxidized and the element reduced such that the element reduced such that the total increase in oxidation state total increase in oxidation state equals the total decrease in equals the total decrease in oxidation state.oxidation state.

Balance the remainder of the Balance the remainder of the equation by inspectionequation by inspection

Practice ProblemPractice Problem

Balance the following redox Balance the following redox reaction using the oxidation reaction using the oxidation states method:states method:

CHCH33OHOH (l) (l) + O + O22 (g)(g) CO CO2 (g)2 (g) + H + H22OO (g) (g)

MnOMnO2 (s)2 (s) + Al + Al (s) (s) Mn Mn (s)(s) + Al + Al22OO3 (s)3 (s)

C: -2C: -2 O: 0 O: 0 C: +4 C: +4 H: H: +1+1H: +1H: +1 O: -2 O: -2 O: -2O: -2O: -2O: -2

C: -2 C: -2 +4 & O: 0 +4 & O: 0 -2 -2

C: loss 6 eC: loss 6 e-- & O: gain 2 e- & O: gain 2 e-

2CH2CH33OH + 6OOH + 6O22 CO CO22 + H + H22OO

2CH2CH33OH + 6OOH + 6O22 2CO 2CO22 + 4H + 4H22OO

CHCH33OH OH (l)(l) + O + O22 (g)(g) CO CO22 (g)(g) + H + H22O O

(g)(g)

CHCH33OH OH (l)(l) + O + O22 (g)(g) CO CO22 (g)(g) + H + H22O O

(g)(g)

2CH2CH33OH + 3OOH + 3O22 2CO 2CO22 + 4H + 4H22OO

2CH2CH33OH + 6OOH + 6O22 2CO 2CO22 + 4H + 4H22OO

2CH2CH33OH + 6OOH + 6O22 2CO 2CO22 + 4H + 4H22OO

Recall: O: 0 to -2 Recall: O: 0 to -2 2e 2e--/O not 2e/O not 2e--/O/O22

CHCH33OH: loss 12 eOH: loss 12 e-- from O: from O: 2e2e--/O/OOO22: gain 12 e: gain 12 e-- from O: 6e from O: 6e--/C/C

MnOMnO22 (s)(s) + Al + Al (s)(s) Mn Mn (s)(s) + Al + Al22OO33

(s)(s)Mn: +4Mn: +4 Al: 0 Al: 0 Mn: 0 Mn: 0 Al: Al: +3+3 O: -2O: -2 O: -2O: -2

Mn: +4 Mn: +4 0 & Al: 0 0 & Al: 0 +3 +3

Mn: gain 4 eMn: gain 4 e-- & Al: loss 3 e- & Al: loss 3 e-

3MnO3MnO22 + 4Al + 4Al Mn + Al Mn + Al22OO33

3MnO3MnO22 + 4Al + 4Al 3Mn + 2Al 3Mn + 2Al22OO33

MnOMnO22 (s)(s) + Al + Al (s)(s) Mn Mn (s)(s) + Al + Al22OO33

(s)(s)3MnO3MnO22 + 4Al + 4Al 3Mn + 2Al 3Mn + 2Al22OO33

MnOMnO22: gain 12 e: gain 12 e-- from Mn: from Mn: 4e4e--/Mn/MnAl: loss 12 eAl: loss 12 e-- from Al: 3e from Al: 3e--/Al/Al

Half-Reaction MethodHalf-Reaction Method

Identify the elements Identify the elements that is oxidized and that is oxidized and reduced.reduced.

Write the equations for Write the equations for the oxidation and the oxidation and reduction half reactionsreduction half reactions

Half-ReactionHalf-Reaction

parts of a redox parts of a redox

reactionreaction

2 types (oxidation half 2 types (oxidation half

reaction and reduction reaction and reduction

half reaction) half reaction)

ExampleExample

CeCe4+4+(aq)(aq) + Sn + Sn2+2+

(aq)(aq) Ce Ce3+3+(aq) (aq) + Sn+ Sn4+4+

(aq)(aq)

Reduction half-reaction:Reduction half-reaction:

CeCe4+4+(aq)(aq) Ce Ce3+3+

(aq)(aq)

Oxidation half-reaction:Oxidation half-reaction:

SnSn2+2+(aq)(aq) Sn Sn4+4+

(aq)(aq)

Half-Reaction MethodHalf-Reaction Method

For each half-reaction:For each half-reaction:Balance all the elements Balance all the elements except H and O.except H and O.

Balance O using HBalance O using H22O O Balance H using HBalance H using H++ Balance the charge using Balance the charge using ee––

ExampleExampleCeCe4+4+

(aq)(aq) + Sn + Sn2+2+(aq)(aq) Ce Ce3+3+

(aq)(aq) + Sn + Sn4+4+(aq)(aq)

Reduction half-reaction:Reduction half-reaction:

CeCe4+4+(aq)(aq) + e + e-- Ce Ce3+3+

(aq)(aq)

Oxidation half-reaction:Oxidation half-reaction:

SnSn2+2+(aq)(aq) Sn Sn4+4+

(aq)(aq) + 2e + 2e--

Half-Reaction MethodHalf-Reaction Method

Check if the electrons Check if the electrons transferred in the two transferred in the two half- reactions are equal. half- reactions are equal. If not, multiply one or If not, multiply one or both equation by an both equation by an integer to balance the einteger to balance the e–– transfer.transfer.

ExampleExampleReduction half-reaction:Reduction half-reaction:

CeCe4+4+(aq)(aq) + e + e-- Ce Ce3+3+

(aq)(aq)

Oxidation half-reaction:Oxidation half-reaction:

SnSn2+2+(aq)(aq) Sn Sn4+4+

(aq)(aq) + 2e + 2e--

Reduction half-rxn lacks 1 eReduction half-rxn lacks 1 e--

2 [ Ce2 [ Ce4+4+(aq)(aq) + e + e-- Ce Ce3+3+

(aq)(aq) ] ]

2Ce2Ce4+4+(aq)(aq) + 2e + 2e-- 2Ce 2Ce3+3+

(aq)(aq)

Half-Reaction MethodHalf-Reaction Method

Add the two half-Add the two half-reactions and cancel reactions and cancel identical species.identical species.

Check to ensure that Check to ensure that the elements and the elements and charges are balance.charges are balance.

ExampleExample

2Ce2Ce4+4+(aq)(aq) + 2e + 2e-- 2Ce 2Ce3+3+

(aq)(aq)

SnSn2+2+(aq)(aq) Sn Sn4+4+

(aq)(aq) + 2e + 2e--

__________________________________________________________________________________________

2Ce2Ce4+4+(aq)(aq) + Sn + Sn2+2+ 2Ce 2Ce3+3+

(aq)(aq) Sn Sn4+4+(aq)(aq)

Balance the following redox Balance the following redox reaction using the half-reaction using the half-reaction method:reaction method:

MnOMnO44––(aq) (aq) + Fe+ Fe2+2+

(aq)(aq) Fe Fe3+3+(aq)(aq) + Mn + Mn2+2+

(aq)(aq)

AgAg(s)(s) + CN + CN––(aq)(aq) + O + O22 Ag(CN) Ag(CN)22

–– (aq) (aq)

Practice ProblemPractice Problem

base

acid

Half-Reaction Method Half-Reaction Method (Basic)(Basic) Follow the same step Follow the same step

as in the acidic as in the acidic solution.solution.

In the final equation, In the final equation, identify the Hidentify the H++ and and addadd OHOH–– equal to the equal to the number of Hnumber of H++ ions. ions.

Half Reaction Method Half Reaction Method (Basic)(Basic) Combine the HCombine the H++ and OH and OH––

ions by forming Hions by forming H22O. O. Eliminate the number of Eliminate the number of HH22O that appears on both O that appears on both sides of the equation.sides of the equation.

Check to ensure that the Check to ensure that the equation is balance.equation is balance.

ElectrochemistryElectrochemistry

the study of the the study of the interchange of chemical and interchange of chemical and electrical energyelectrical energy

primarily concerned with primarily concerned with current generation by current generation by chemical reaction and the chemical reaction and the use of current to produce a use of current to produce a chemical changechemical change

Electrochemical CellElectrochemical Cell

galvanic cell or voltaic cellgalvanic cell or voltaic cell a device that converts a device that converts chemical energy into chemical energy into electrical energyelectrical energy

utilizes a spontaneous redox utilizes a spontaneous redox reaction to produce current reaction to produce current that can be used to do workthat can be used to do work

Electrochemical CellElectrochemical Cell

can be divided into to half can be divided into to half cell, one cell containing Zn cell, one cell containing Zn metal and ZnSOmetal and ZnSO44 while the while the other one Cu and CuSOother one Cu and CuSO44

Zn metal is oxidized (to Zn metal is oxidized (to ZnZn2+2+) into solution while Cu) into solution while Cu2+2+ are reduced forming Cu metalare reduced forming Cu metal

Electrochemical CellElectrochemical Cell

a point of contact between a point of contact between the metal and the solution is the metal and the solution is called an electrodecalled an electrode

the reaction in the cell occurs the reaction in the cell occurs at the interface between the at the interface between the electrode and the solution (its electrode and the solution (its also the place where electron also the place where electron transfer occurs)transfer occurs)

Electrochemical CellElectrochemical Cell

oxidation occurs at the anode oxidation occurs at the anode while reduction at the cathodewhile reduction at the cathode

oxidation of Zn causes a oxidation of Zn causes a transfer of 2 electrons to occur transfer of 2 electrons to occur

electron flow is measured and electron flow is measured and monitored by a voltmetermonitored by a voltmeter

VoltmeterVoltmeter

An instrument that An instrument that measures cell potential measures cell potential

by drawing electric by drawing electric current through a known current through a known

resistanceresistance

Electrochemical CellElectrochemical Cell a U-tube filled with electrolyte a U-tube filled with electrolyte

is responsible for the is responsible for the maintaining a zero net charge maintaining a zero net charge for the cells. for the cells.

the electrolyte-filled U-tube is the electrolyte-filled U-tube is called a salt bridge and allows called a salt bridge and allows the ion flow without extensive the ion flow without extensive mixing of the solution in the cellmixing of the solution in the cell

ElectrolyteElectrolyte

substances whose solution substances whose solution conducts electric currentconducts electric current

a material that dissolves in a material that dissolves in water to give a solution that water to give a solution that conducts electric currentconducts electric current

a solution that conducts a solution that conducts electricityelectricity

Electrolytic CellElectrolytic Cell

uses electrical energy to uses electrical energy to produce a chemical change produce a chemical change that would otherwise not that would otherwise not occur spontaneouslyoccur spontaneously

a device that converts a device that converts electrical energy into electrical energy into chemical energychemical energy

Electrolytic CellElectrolytic Cell

utilizes a power source to utilizes a power source to force a chemical reactionforce a chemical reaction

electrons from the battery electrons from the battery are forced towards the Zn are forced towards the Zn anode causing electrolysisanode causing electrolysis

the salt bridge maintains a the salt bridge maintains a zero net charge in each cellzero net charge in each cell

Electrolytic CellElectrolytic Cell

Zn metal is formed from the Zn metal is formed from the solution of ZnSOsolution of ZnSO44 while the Cu while the Cu metal is oxidizedmetal is oxidized

oxidation occurs in the anode oxidation occurs in the anode and reduction at the cathodeand reduction at the cathode

migration of cations is migration of cations is towards the Zn cathodetowards the Zn cathode

Cell NotationCell Notation a shortcut to indicate the parts of an electrochemical cell

also called “line cell notation”the salt bridge is symbolized by a double slash (//) that separates the anode and the cathode reaction

Cell NotationCell Notationthe anode and the cathode reaction are separated from the electrode by a slash (/)

the line format goes:Anode electrode/Anode rxn//Cathode rxn/Cathode

electrode

Cell NotationCell Notation

Representation of Representation of electrochemical cellselectrochemical cells

ZnZn(s)(s) / Zn / Zn2+2+(1M) // Cu(1M) // Cu2+2+(1M) / (1M) / CuCu(s)(s)

MgMg(s)(s) | Mg | Mg2+ 2+ || H|| H++ | H | H2(g) 2(g) | Pt| Pt

Cell NotationCell Notation The half reactions indicate the The half reactions indicate the oxidation of magnesium:oxidation of magnesium:

MgMg(s)(s) Mg Mg2+2+ + 2e + 2e--

and the reduction of Hand the reduction of H++::

2H2H++ + 2e + 2e-- H H2(g)2(g)

Since there is no conductive Since there is no conductive material in the cathode, we must material in the cathode, we must use an inert electrode such as Ptuse an inert electrode such as Pt

Cell Potential, Cell Potential, EEcellcell

electromotive force (emf)electromotive force (emf) the “pull” or driving force on the “pull” or driving force on the electrons from the the electrons from the reducing agent towards the reducing agent towards the oxidizing agent through a oxidizing agent through a wirewire

measured in volts (V) and is measured in volts (V) and is equal to 1 joule per coulombequal to 1 joule per coulomb

Cell PotentialCell Potential

Just as a redox reaction can be divided up into two half reactions, the cell potential can be divided up into two half cell potentials.

Standard Cell PotentialsStandard Cell Potentials

emf or the “pull” or driving emf or the “pull” or driving force on the electronsforce on the electrons

measured in voltsmeasured in volts intensive propertyintensive property based on the assignment of 0 based on the assignment of 0

volts to “2Hvolts to “2H++ + 2e + 2e- - HH22” process” process

symbol is symbol is E°E°cellcell

Standard Hydrogen Standard Hydrogen ElectrodeElectrode consists of a platinum consists of a platinum

electrode for its cathodeelectrode for its cathode Pt electrode is in Pt electrode is in contact with 1M Hcontact with 1M H++ ions ions

HH22 gas at 1 atm passes gas at 1 atm passes through the Pt electrodethrough the Pt electrode

2H+ + 2e- H2 E° = 0.00

Standard Hydrogen PotentialStandard Hydrogen Potential

It is the reference potential against which all half-reaction potentials are assigned

0.00 V0.00 V

- 0.76 V- 0.76 V 0.34 V0.34 V

Table of Standard Reduction PotentialsTable of Standard Reduction Potentials

half-reactions are given as half-reactions are given as reduction processesreduction processes

if the half-reaction is if the half-reaction is reversed, the sign of reversed, the sign of E°E° is is also reversedalso reversed

if the half-reaction is if the half-reaction is multiplied by an integer, multiplied by an integer, E°E° remains the sameremains the same

Practice ProblemPractice Problem

Using the standard Using the standard reduction potential table, reduction potential table, calculate the cell potential calculate the cell potential based on the following based on the following galvanic cell reactions:galvanic cell reactions: Zn + CuZn + Cu2+2+ Zn Zn2+2+ + Cu + Cu CuCu2+2+ + Fe + Fe Cu + Fe Cu + Fe2+2+

2Al + 3Mn2Al + 3Mn2+2+ 2Al 2Al3+3+ + 3Mn + 3Mn

Galvanic CellGalvanic Cell

A galvanic cell runs spontaneously in the direction that gives a positive cell potential.

Galvanic CellGalvanic Cell

Dry CellDry Cell

Mercury BatteryMercury Battery

Lead Storage BatteryLead Storage Battery

Hydrogen-Oxygen Fuel CellHydrogen-Oxygen Fuel Cell

batteries are self-contained and need no salt bridge

an electrochemical cell that requires a continuous supply of reactants to keep functioning.

Anode: Zn(s) Zn2+(aq) + 2e-

Cathode: 2NH4+

(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + H2O(l)

Overall: Zn(s) + 2NH4+

(aq) + 2MnO2(s) Zn2+

(aq) + Mn2O3(s) + 2NH3(aq) + H2O(l)

Overall: Zn(s) + HgO(s) ZnO(s) + Hg(l)

Because the overall

reaction doesn’t have any aqueous

solutions, conc. doesn’t change and

the voltage is constant

Anode: Zn(s) + 2OH-(aq) ZnO(s) + H2O(l) + 2e-

Cathode: HgO(s) + H2O(l) + 2e- Hg(l) + 2OH-(aq)

Anode: Pb(s) + SO42-

(aq) PbSO4(s) + 2e-

Cathode: PbO2(s) + 4H+(aq) + SO4

2-(aq) + 2e- PbSO4(s) + 2H2O(l)

Overall: Pb(s) + PbO2(s) + 2SO42-

(aq) + 4H+(aq) 2PbSO4(s) + 2H2O(l)

Overall: 2H2(g) + O2(g) 2H2O(l)

Spontaneity of ReactionsSpontaneity of Reactions

F = Faraday Constant

= 96,485 C/mol

= the charge on a mole of electrons

= synonymous with 1 mole of e-

electrical energy = volts x coulombs

A = Ampere

= 1 coulomb of charge per second

For Chem 16, we will only be considering this:For Chem 16, we will only be considering this:

ΔG = -nFEcell

or

ΔG° = -nFE°cell

Spontaneity of ReactionsSpontaneity of Reactions

CONDITIONS Reaction under standard state

ΔG° K E°cell

<0 >1 >0 Spontaneous

0 =1 0 At equilibrium

>0 <1 <0 Reverse reaction is spontaneous

Practice ProblemPractice Problem

Calculate the Calculate the G° for the G° for the following reaction and predict following reaction and predict if it is spontaneous: if it is spontaneous:

CuCu2+2+ + Fe + Fe Cu + Fe Cu + Fe2+2+-o1o.5o1 x 1o-o1o.5o1 x 1o55 J J

AuAu(s)(s) + NO + NO33––(aq)(aq) + 4H + 4H++

(aq)(aq)

AuAu3+3+(aq)(aq) + NO + NO(g)(g) + 2H + 2H22OO(l)(l)

1o.5o6 x 1o1o.5o6 x 1o55 J J

SolutionSolution

Cu2+ + 2e- Cu 0.34VFe2+ + 2e- Fe -0.44V

CuCu2+2+ + Fe + Fe Cu + Fe Cu + Fe2+2+ 0.78V0.78V

Au3+ + 3e- Au 1.50VNO3

- + 4H+ + 3e- NO + 2H2O 0.96V

Au + NO3- + 4H+ Au3+ NO + 2H2O -

0.54V

SolutionSolution

ΔG° = -nFE°cell

ΔG° = -(2 mol e-)(96,485 C/mol e-)(0.78V)

ΔG° = -(2 mol e-)(96,485 C/mol e-)(0.78J/C)

ΔG° = -1.51 x 10-1.51 x 1055 J

SolutionSolution

ΔG° = -nFE°cell

ΔG° = -(3 mol e-)(96,485 C/mol e-)(-0.54V)

ΔG° = -(3 mol e-)(96,485 C/mol e-)(-0.54J/C)

ΔG° = 1.56 x 101.56 x 1055 J

ElectrolysisElectrolysisa method of separating a method of separating elements and compounds elements and compounds by passing electric current by passing electric current through themthrough them

a process of forcing a process of forcing electrons through a electrons through a chemical cell thus causing chemical cell thus causing a chemical reactiona chemical reaction

QUANTITATIVE ASPECTS OF ELECTROLYSIS

• The number of electrons transferred in the half reactions can be used as in stoichiometry.

• A common problem would involve calculating the amount (in grams) of a substance that would be formed during electrolysis.

• Ex. In electrolysis, one can calculate how many grams of Na would be formed if given the information to obtain the number of electrons transferred.

Na+ + e- Na• 1 mole of e- makes one mole of Na

QUANTITATIVE ASPECTS OF ELECTROLYSIS

• Current is measured in Amperes (A)• 1C = 1 A x 1 s or 1 A = 1 C/1 s• Therefore if given the current and time, one can

calculate the charge of electrons that passed through the electrolytic cell.

• Once you know the charge (in C), you can use the Faraday constant (96,500 C/mol e-) to calculate the number of electrons that are passed through the electrolytic cell.

• Now you can calculate the moles of substance formed by using the balanced half reaction.

Practice ProblemPractice Problem

Thirty minutes of Thirty minutes of electrolysis of a solution of electrolysis of a solution of CuSOCuSO44 produced 3.175g Cu produced 3.175g Cu at the cathode. How many at the cathode. How many Faradays and how many Faradays and how many Coulombs passed through Coulombs passed through the cell? What is the current? the cell? What is the current?

Practice ProblemPractice Problem

3.175g Cu 63.5g Cu

1 mol Cu= 0.05 mol Cu

0.05 mol Cu

= 0.10 mol e-

0.10 mol e-

= 9,648.5 C

9,648.5 C

1 mol Cu

2 mol e-

1 mol e-

96,485 C

1 min

60 sec = 5.36 A30 min

Take Home ProblemTake Home Problem

Calculate the amounts of Cu Calculate the amounts of Cu produced in 1.0 h at inert produced in 1.0 h at inert electrodes in a solution of CuBrelectrodes in a solution of CuBr22 by a current of 2.50 A. by a current of 2.50 A.

If a steady current was passed If a steady current was passed through molten Althrough molten Al22OO33 and 15.0 g and 15.0 g of Al was collected. How many of Al was collected. How many coulombs of electricity was used?coulombs of electricity was used?

MetallurgyMetallurgythe science and

technology of extracting metals from minerals. PyrometallurgyPyrometallurgy

HydrometallurgyHydrometallurgy ElectrometallurgyElectrometallurgy

METAL MINERAL Composition

AluminumChromium

Copper

Iron

LeadManganese

MercuryMolybdenum

TinTitanium

Zinc

BauxiteChromiteChalcocite

ChalcopyriteMalachiteHematiteMagnetite

GalenaPyrolusiteCinnabar

MolybdeniteCassiterite

RutileIlmenite

Sphalerite

Al2O3

FeCr2O4

Cu2SCuFeS2

Cu2CO3(OH)2

Fe2O3

Fe3O4

PbSMnO2

HgSMoS2

SnO2

TiO2

FeTiO3

ZnS

Principal Mineral Sources of Some Principal Mineral Sources of Some Common MetalsCommon Metals

Mining (getting the ore out of the Mining (getting the ore out of the ground)ground)

Concentrating (preparing it for Concentrating (preparing it for further treatment)further treatment)

Reduction (to obtain the free Reduction (to obtain the free metal in the zero oxidation state)metal in the zero oxidation state)

Refining (to obtain the pure metal)Refining (to obtain the pure metal)Mixing with other metals (to form Mixing with other metals (to form

an alloy).an alloy).

Five Important StepsRefining is the process

during which a crude, impure metal is converted

into a pure metal.

PyrometallurgyPyrometallurgy

utilizes high temperatures to obtain the free

metal

Steps involved in the processSteps involved in the processCalcination - it is heating of ore to cause Calcination - it is heating of ore to cause

decomposition and elimination of a decomposition and elimination of a volatile product:volatile product:

PbCOPbCO3(3(ss)) PbO PbO((ss)) + CO + CO2(2(gg))

Roasting - it is heating which causes Roasting - it is heating which causes chemical reactions between the ore and chemical reactions between the ore and the furnace atmosphere:the furnace atmosphere:

2ZnS2ZnS((ss)) + 3O + 3O2(2(gg)) 2ZnO 2ZnO((ss)) + 2SO + 2SO2(2(gg))

2MoS2MoS2(2(ss)) + 7O + 7O2(2(gg)) 2MoO 2MoO3(3(ss)) + 4SO + 4SO2(2(gg))

Smelting - it is a melting process that Smelting - it is a melting process that causes materials to separate into two or causes materials to separate into two or more layers.more layers.

Pyrometallurgy of ironPyrometallurgy of ironPyro

meta

llurg

y o

f Pyro

meta

llurg

y o

f ir

on

iron

The ore, limestone and coke are The ore, limestone and coke are added to the top of the blast added to the top of the blast furnace.furnace.

Coke is coal that has been Coke is coal that has been heated to drive off the volatile heated to drive off the volatile components.components.

Coke reacts with oxygen to form Coke reacts with oxygen to form CO (the reducing agent):CO (the reducing agent):2C2C((ss)) + O + O2(2(gg)) 2CO 2CO((gg)), , HH = -221 kJ = -221 kJ

Pyrometallurgy of ironPyrometallurgy of iron

Pyrometallurgy of ironPyrometallurgy of iron

CO is also produced by the reaction of CO is also produced by the reaction of water vapor in the air with C:water vapor in the air with C:

C(C(ss) + H) + H22O(O(gg) ) CO( CO(gg) + H) + H22((gg), ), HH = +131 kJ = +131 kJ

Since this reaction is endothermic, if Since this reaction is endothermic, if the blast furnace gets too hot, water the blast furnace gets too hot, water vapor is added to cool it down without vapor is added to cool it down without interrupting the chemistry.interrupting the chemistry.

At around 250At around 250C limestone is C limestone is calcinated (heated to decomposition calcinated (heated to decomposition and elimination of volatiles).and elimination of volatiles).

Pyrometallurgy of ironPyrometallurgy of iron

Also around 250Also around 250C iron oxides are C iron oxides are reduced by CO:reduced by CO:

FeFe33OO4(4(ss)) + 4CO + 4CO((gg)) 3Fe 3Fe((ss)) + 4CO + 4CO2(2(gg)), , HH = -15 = -15 kJkJ

FeFe33OO4(4(ss)) + 4H + 4H2(2(gg)) 3Fe 3Fe((ss)) + 4H + 4H22OO((gg)), , HH = = +150 kJ+150 kJ

Molten iron is produced lower down Molten iron is produced lower down the furnace and removed at the the furnace and removed at the bottom.bottom.

Slag (molten silicate materials) is Slag (molten silicate materials) is removed from above the molten removed from above the molten iron. iron.

Slag consists mostly of molten silicates in addition to aluminates, phosphates, fluorides, and other inorganic materials.

HydrometallurgyHydrometallurgy

the extraction of metals from ores

using water

it is the selective it is the selective dissolution of the desired dissolution of the desired mineral.mineral.

typical leaching agents typical leaching agents are dilute acids, bases, salts, are dilute acids, bases, salts, and sometimes water.and sometimes water.

LeachingLeaching

Gold can be extracted from low-Gold can be extracted from low-grade ore by cyanidation:grade ore by cyanidation:NaCN is sprayed over the crushed ore NaCN is sprayed over the crushed ore

and the gold is oxidized:and the gold is oxidized:

4Au4Au((ss)) + 8CN + 8CN--((aqaq)) + O + O2(2(gg)) + 2H + 2H22OO((ll))

4Au(CN)4Au(CN)22--((aqaq)) + 4OH + 4OH--

((aqaq))

The gold is then obtained by The gold is then obtained by reduction:reduction:2Au(CN)2Au(CN)22

--((aqaq)) + Zn + Zn((ss)) Zn(CN) Zn(CN)44

2-2-((aqaq)) + 2Au + 2Au((ss))

Hydrometallurgy of Hydrometallurgy of goldgold

Hydrometallurgy of aluminumHydrometallurgy of aluminum

Aluminum is found in bauxite as AlAluminum is found in bauxite as Al22OO33•xH•xH22O.O.Extraction involves the Bayer process:Extraction involves the Bayer process:

The crushed ore is digested in 30% The crushed ore is digested in 30% NaOH (by mass) at 150 - 230NaOH (by mass) at 150 - 230C and high C and high pressure (30 atm to prevent boiling).pressure (30 atm to prevent boiling).AlAl22OO33.H.H22OO((ss)) + 2H + 2H22OO((ll)) + 2OH + 2OH--

((aqaq)) 2Al(OH) 2Al(OH)44--((aqaq))

The aluminate solution is separated by The aluminate solution is separated by lowering the pH.lowering the pH.

The aluminate solution is calcined and The aluminate solution is calcined and reduced to produce the metal.reduced to produce the metal.

ElectrometallurgyElectrometallurgy

the process of obtaining metals

through electrolysis.

Electrometallurgy of sodiumElectrometallurgy of sodium

Two different starting materials: Two different starting materials: molten salt or aqueous solution.molten salt or aqueous solution.

Sodium is produced by electrolysis Sodium is produced by electrolysis of molten NaCl in a Downs cell.of molten NaCl in a Downs cell.

CaClCaCl22 is used to lower the melting is used to lower the melting point of NaCl from 804point of NaCl from 804C to 600C to 600C.C.

An iron screen is used to separate An iron screen is used to separate Na and Cl so that NaCl is not re-Na and Cl so that NaCl is not re-formed).formed).

At the cathode At the cathode (iron): 2Na(iron): 2Na++

((aqaq)) + + 2e2e-- 2Na 2Na((ll))

At the anode At the anode (carbon): 2Cl(carbon): 2Cl--

((aqaq)) ClCl2(2(gg)) + 2e + 2e--

Electrometallurgy of sodiumElectrometallurgy of sodium

Electrometallurgy of Electrometallurgy of aluminumaluminumHall-Heroult process Hall-Heroult process Hall electrolysis cell is used to produce Hall electrolysis cell is used to produce

AlAlAlAl22OO33 melts at 2000 melts at 2000C and it is C and it is

impractical to perform electrolysis on impractical to perform electrolysis on the molten salt.the molten salt.

Purified AlPurified Al22OO33 in molten cryolite in molten cryolite (Na(Na33AlFAlF66, m.p. 1012, m.p. 1012C) is used C) is used

Graphite rods are employed and are Graphite rods are employed and are consumed in the reaction. consumed in the reaction.

Electrometallurgy of Electrometallurgy of aluminumaluminum

Anode: C(s) + 2O2-(l) CO2

(g) + 4e-

Cathode: 3e- + Al3+(l) Al(l)