Redox Reactions, Applications Of Chemistry

Oxidatio n

✓an increase in oxidation state ✓loss of electron ✓in organic chemistry, it is the loss of H atom/s or the gain of O atom/s.

Oxidizing agent ✓an electron acceptor ✓a reactant that accepts electron from another reactant. ✓the reactant that is reduced in a redox

Reduction ✓a decrease in oxidation state ✓gain of electron ✓in organic chemistry, it is the loss of O atom/s or the gain of H atom/s.

Reducing agent ✓an electron donor ✓a reactant that gives up electron/s to another reactant thus decreasing the oxidation state of one of its atom. ✓the reactant that is

Oxidation Number For 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 Number ✓Example: NaCl the oxidation number of Na and Cl are +1 and –1 +

–

NaCl → Na + Cl and it can be said that sodium and chlorine are in a +1 and –1 oxidation

Oxidation State

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

Fundamental ➊ The sum ofRules the oxidation state

for all atoms in the formula for an electrically neutral compound is zero. ➋ The oxidation state for any element in the free or uncombined state is zero. ➌ The oxidation state for an ion is

Special ➊ In allConventions hydrogen compounds, the oxidation state for H is +1. Exception: in hydrides where H is – 1

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

Special ➌ In allConventions halides, the oxidation state for the halogens is –1. ➍ In all sulfides, the oxidation state for sulfur is –2. ➎ In binary compounds, the element with the greatest attraction for electrons is assigned a negative oxidation state equal to its charge

Some Important Concepts ➊ Corrosion: When a metal undergoes corrosion it is oxidized forming cations: Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g) ❋ Metals are also oxidized by acids to

form salts:

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) ❋ Metals can also be oxidized by other

salts:

Some Important Concepts ➋ Activity series: A list of

metals arranged in decreasing ease of oxidation. Some metals are easily oxidized whereas others are not. ❋ The higher the metal on the activity series, the more active that metal. ❋ Any metal can be oxidized by the ❋

Some Important Concepts ➌ Solution Composition: ❋ Solution - solute dissolved in a solvent ❋ Aqueous solution - solution in which water is the dissolving medium or solvent (ex. HCl(aq), H+(aq), Ni2+(aq)) Solute - a substance dissolved in a solvent to form a solution

❋

Some Important Concepts ➌ Solution Composition: ❋ Concentration - amount or measure of solute in a solution (generally measured as molar concentration) ❋ Molarity - moles of solute per volume of solution in liters (or simply moles/liter)

Balancing Redox Reaction

Oxidation States Method ➊ Assign the oxidation states of all the atoms. ➋ Determine which element is oxidized and which is reduced. Take note also of the change (increase/decrease) in

Oxidation States Method ➌ Choose the coefficients for the compound containing the element oxidized and the element reduced such that the total increase in oxidation state equals the total decrease in oxidation state. ➍ Balance the remainder of

Practice Problem Balance the following redox reaction using the oxidation states method: CH3OH (l) + O2

(g)

➔ CO2 (g) + H2O (g)

MnO2 (s) + Al (s) ➔ Mn

(s)

+ Al2O3 (s)

CH3OH (l) + O2 (g) ➔ CO2 (g) + H2O C: -2 O: 0 C: +4 H: +1 H: +1 O: -2 O: -2 O: -2 C: -2 ➔ +4 & O: 0 ➔ -2 C: loss 6 e- & O: gain 2 e(g)

2CH3OH + 6O2 ➔ CO2 + H2O 2CH3OH + 6O2 ➔ 2CO2 + 4H2O

CH3OH (l) + O2 (g) ➔ CO2 (g) + H2O (g)

2CH3OH + 2CH3OH +

6O2 6O2

➔ ➔

2CO2 + 4H2O 2CO2 + 4H2O

Recall: O: 0 to -2 ➔ 2e-/O not 2e-/O2

2CH3OH + 3O2 ➔ 2CO2 + 4H2O CH3OH: loss 12 e- from O: 2e/O O2: gain 12 e- from O: 6e-/C 2

MnO2 (s) + Al (s) ➔ Mn (s) + Al2O3 (s) Mn: +4 +3 O: -2

Al: 0

Mn: 0

Al: O: -2

Mn: +4 ➔ 0 & Al: 0 ➔ +3 Mn: gain 4 e- & Al: loss 3 e3MnO2 + 4Al ➔ Mn + Al2O3 3MnO2 + 4Al ➔ 3Mn + 2Al2O3

MnO2 (s) + Al (s) ➔ Mn (s) + Al2O3 (s)

3MnO2 + 4Al ➔ 3Mn + 2Al2O3 MnO2: gain 12 e- from Mn: 4e/Mn Al: loss 12 e- from Al: 3e-/Al

Half-Reaction Method

➊ Identify the elements that is oxidized and reduced. ➋ Write the equations for the oxidation and reduction half reactions

Half-Reaction ✓parts of a redox reaction ✓2 types (oxidation half reaction and reduction half

Example ✓Ce4+(aq) + Sn2+(aq)➔ Ce3+(aq) + Sn4+(aq) ✓Reduction half-reaction: Ce4+(aq) ➔ Ce3+(aq) ✓Oxidation half-reaction:

Half-Reaction Method

➌ For each half-reaction: ✓Balance all the elements except H and O. ✓Balance O using H2O ✓Balance H using H+ ✓Balance the charge using

Example ✓Ce4+(aq) + Sn2+(aq) ➔ Ce3+(aq) + Sn4+(aq) ✓Reduction half-reaction: Ce4+(aq) + e- ➔ Ce3+(aq) ✓Oxidation half-reaction: Sn2+(aq) ➔ Sn4+(aq) + 2e-

Half-Reaction Method

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

Example ✓Reduction half-reaction: Ce4+(aq) + e- ➔ Ce3+(aq) ✓Oxidation half-reaction: Sn2+(aq) ➔ Sn4+(aq) + 2eReduction half-rxn lacks 1 e2 [ Ce4+(aq) + e- ➔ Ce3+(aq) ]

Half-Reaction Method

➎ Add the two halfreactions and cancel identical species. ➏ Check to ensure that the elements and charges are balance.

Example 2Ce4+(aq) + 2e- ➔ 2Ce3+(aq) Sn2+(aq) ➔ Sn4+(aq) + 2e_____________________________________________

2Ce4+(aq) + Sn2+ ➔ 2Ce3+(aq) Sn4+(aq)

Practice Problem Balance the following redox reaction using the halfreaction method: MnO4

–

(aq) + Fe

2+

acid

(aq)

→ Fe3+(aq) + Mn2+(aq) base

Ag(s) + CN–(aq) + O2 → Ag(CN)2– (aq)

Half-Reaction Method (Basic)

➊ Follow the same step as in the acidic solution. ➋ In the final equation, + identify the H and – add OH equal to the +

Half Reaction Method (Basic) + –

➌ Combine the H and OH ions by forming H2O.

➍ Eliminate the number of H2O that appears on both sides of the equation. ➎ Check to ensure that the equation is balance.

Electrochemistry  the study of the interchange of chemical and electrical energy  primarily concerned with current generation by chemical reaction and the use of current to produce a

Electrochemical Cell ➊ galvanic cell or voltaic cell ➋ a device that converts chemical energy into electrical energy ➌ utilizes a spontaneous redox reaction to produce current that can be used to

Electrochemical Cell ➊ can be divided into to half cell, one cell containing Zn metal and ZnSO4 while the other one Cu and CuSO4 ➋ Zn metal is oxidized (to 2+ Zn ) into solution while Cu2+ are reduced forming

Electrochemical Cell ➌ a point of contact between the metal and the solution is called an electrode ➍ the reaction in the cell occurs at the interface between the electrode and the solution (its also the place where electron

Electrochemical Cell ➎ oxidation occurs at the anode while reduction at the cathode ➏ oxidation of Zn causes a transfer of 2 electrons to occur ➐ electron flow is measured

Voltmeter

An instrument that measures cell potential by drawing electric current through a known resistance

Electrochemical Cell ➑ a U-tube filled with electrolyte is responsible for the maintaining a zero net charge for the cells. ➒ the electrolyte-filled U-tube is called a salt bridge and allows the ion flow without

Electrolyte ➊ substances whose solution conducts electric current ➋ a material that dissolves in water to give a solution that conducts electric current ➌ a solution that conducts

Electrolytic Cell ➊ uses electrical energy to produce a chemical change that would otherwise not occur spontaneously ➋ a device that converts electrical energy into

Electrolytic Cell ➊ utilizes a power source to force a chemical reaction ➋ electrons from the battery are forced towards the Zn anode causing electrolysis ➌ the salt bridge maintains

Electrolytic Cell ➍ Zn metal is formed from the solution of ZnSO4 while the Cu metal is oxidized ➎oxidation occurs in the anode and reduction at the cathode ➏ migration of cations is

Cell Notation ✓ a shortcut to indicate the parts of an electrochemical cell ✓also called “line cell notation” ✓the salt bridge is symbolized by a double

Cell Notation ✓the 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 Notation

Representation of electrochemical cells Zn(s) / Zn (1M) // Cu (1M) / Cu(s) 2+

2+

Cell Notation The half reactions indicate the oxidation of magnesium: Mg(s) → Mg2+ + 2eand the reduction of H+: + 2H + 2e → H2(g) Since there is no conductive material in the cathode, we

Cell Potential, Ecell ➊ electromotive force (emf) ➋ the “pull” or driving force on the electrons from the reducing agent towards the oxidizing agent through a wire ➌ measured in volts (V) and

Cell 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 Potentials ➊ emf or the “pull” or driving force on the electrons ➋ measured in volts ➌ intensive property ➍ based on the assignment of 0 volts to “2H+ + 2e- ➙ H2” process

Standard Hydrogen Electrode

➊ consists of a platinum electrode for its cathode ➋ Pt electrode is in contact with 1M H+ ions ➌ H2 gas at 1 atm passes through the Pt electrode

2H+ + 2e0.00

➙

H2

E° =

Standard Hydrogen Potential

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

0.00 V

- 0.76 V

0.34 V

Table of Standard Reduction Potentials

➊ half-reactions are given as reduction processes ➋ if the half-reaction is reversed, the sign of E° is also reversed ➌ if the half-reaction is multiplied by an integer, E°

Practice Problem Using the standard reduction potential table, calculate the cell potential based on the following galvanic cell reactions: ➊ Zn + Cu2+ ➜ Zn2+ + Cu ➋ Cu2+ + Fe ➜ Cu + Fe2+ ➌ 2Al + 3Mn2+ ➜ 2Al3+ + 3Mn

Galvanic Cell

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

Galvanic Cell

➊ Dry

batteriesanare selfCell electrochemic contained and need al cell that no salt bridge

requires a ➋ Mercury Battery

➌ ➍

continuous supply of Lead Storage Battery reactants to keep functioning. Hydrogen-Oxygen Fuel

Anode: Zn(s)  Zn2+(aq) + 2eCathode: 2NH4+ (aq) + 2MnO2(s) + 2e-  Mn2O3(s) + 2NH3(aq) + H2O(l) +

erall: Zn(s) + 2NH4 (aq) + 2MnO2(s)  Zn2+(aq) + Mn2O3(s) + 2NH3(aq) + H2

Anode: Zn(s) + 2OH-(aq)  ZnO(s) + H2O(l) + 2eCathode: HgO + H O + 2e  Hg (s) 2(l)ZnO (l) + rall: Zn(s) + HgO + Hg (s) (s) (l) 2OH-(aq) Because the overall reaction doesn’t have any aqueous solutions, conc. doesn’t change and

Anode: Pb(s) + SO42-(aq)  PbSO4(s) + 2eCathode: PbO2(s) + 4H+(aq) + SO42-(aq) + 2e-  2+ + 2H O l:PbSO Pb(s) + PbO + 2SO + 4H 4(s) 2(s)2 (l) 4 (aq) (aq)  2PbSO4(s) +

Overall: 2H2(g) + O2(g)  2H2O(l)

Spontaneity of Reactions electrical energy = volts x F = Faraday Constant coulombs = 96,485 C/mol = the charge on a mole of electrons with 1 mole A = = synonymous Ampere of e= 1 coulomb of charge per

For Chem 16, we will only be considering this:

ΔG = -nFEcell or ΔG° = -nFE°

Spontaneity of Reactions CONDITIONS

Reaction under standard state

ΔG°

K

E°cell

<0

>1

>0

Spontaneous

0

=1

0

At equilibrium

<0

Reverse reaction is spontaneous

>0

<1

Practice Problem Calculate the ∆G° for the following reaction and predict if it is spontaneous: Cu2+ + Fe ➜ Cu + Fe2+ -o1o.5o1 x 1o5 J

Au(s) + NO3–(aq) + 4H+(aq) ➜ Au3+(aq) + NO(g) + 2H2O(l)

Solution Cu2+ + 2e- → Cu Fe2+ + 2e- → Fe

0.34V -0.44V

Cu2+ + Fe → Cu + Fe2+ 0.78V Au3+ + 3e- → Au 1.50V NO3- + 4H+ + 3e- → NO + 2H2O 0.96V Au + NO3- + 4H+ → Au3+ NO + 2H2O -0.54V

Solution 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 J 5

Solution 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 10 J 5

Electrolysis a method of separating elements and compounds by passing electric current through them a process of forcing electrons through a chemical cell thus causing a

QUANTITATIVE ASPECTS OF ELECTROLYSI

•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

QUANTITATIVE ASPECTS OF ELECTROLYSI

•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.

Practice Problem

Thirty minutes of electrolysis of a solution of CuSO4 produced 3.175g Cu at the cathode. How many Faradays and how many Coulombs passed through the cell? What is the

Practice Problem 3.175g Cu

1 mol Cu 63.5g Cu

=

0.05 mol Cu

0.05 mol Cu

2 mol e1 mol Cu

=

0.10 mol e-

0.10 mol e-

96,485 C 1 mol e-

=

9,648.5 C

9,648.5 C 30 min 60 sec 1 min

=

5.36 A

Take Home Problem ➊ Calculate the amounts of Cu produced in 1.0 h at inert electrodes in a solution of CuBr2 by a current of 2.50 A. ➋ If a steady current was passed through molten Al2O3 and 15.0 g of Al was collected. How many coulombs of

Metallurgy the science and technology of extracting metals ➊ minerals. Pyrometallurgy from ➋ Hydrometallurgy

Principal Mineral Sources of Some Common Metals METAL

MINERAL

Composition

Aluminum Chromium Copper

Bauxite Chromite Chalcocite Chalcopyrite Malachite Hematite Magnetite Galena Pyrolusite Cinnabar Molybdenite Cassiterite Rutile Ilmenite Sphalerite

Al2O3 FeCr2O4 Cu2S CuFeS2 Cu2CO3(OH)2 Fe2O3 Fe3O4 PbS MnO2 HgS MoS2 SnO2 TiO2 FeTiO3 ZnS

Iron Lead Manganese Mercury Molybdenum Tin Titanium Zinc

Five Important ➊Mining (getting the ore out of the Steps Refining is the process

ground) during which a crude, ➋Concentrating (preparing it for impure metal is converted further treatment) into a(topure ➌Reduction obtainmetal. the free metal in the zero oxidation state) ➍Refining (to obtain the pure metal) ➎Mixing with other metals (to form

Pyrometallurgy

utilizes high temperatures to obtain the free metal

Steps involved in the process ✓Calcination - it is heating of ore to cause decomposition and elimination of a volatile product: PbCO3(s) → PbO(s) + CO2(g) ✓Roasting - it is heating which causes chemical reactions between the ore and the furnace atmosphere: 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) 2MoS2(s) + 7O2(g) → 2MoO3(s) + 4SO2(g) ✓Smelting - it is a melting process that causes materials to separate into two or

Pyrometallurgy of iron

Pyrometallurgy of iron

Pyrometallurgy of iron ✓The ore, limestone and coke are added to the top of the blast furnace. ✓Coke is coal that has been heated to drive off the volatile components. ✓Coke reacts with oxygen to form CO (the reducing agent): 2C

+O

→ 2CO , ∆H = -221 kJ

Pyrometallurgy of iron ✓CO is also produced by the reaction of water vapor in the air with C: C(s) + H2O(g) → CO(g) + H2(g), ∆H = +131 kJ

Since this reaction is endothermic, if the blast furnace gets too hot, water vapor is added to cool it down without interrupting the chemistry. ✓At around 250°C limestone is calcinated (heated to decomposition

Pyrometallurgy of iron ✓Also around 250°C iron oxides are Slag consists mostly of reduced by CO:

molten in , ∆H = -15 Fe O + 4COsilicates → 3Fe + 4CO kJ addition to aluminates, Fe O + 4H → 3Fe + 4H O , ∆H = +150 kJ phosphates, fluorides, ✓Molten iron is produced lower down and other inorganic the furnace and removed at the materials. bottom. 3

4(s)

3

(g)

4(s)

2(g)

(s)

2(g)

(s)

2

(g)

✓Slag (molten silicate materials) is

Hydrometallurgy

the extraction of metals from ores using water

Leaching

✓ it is the selective dissolution of the desired mineral. ✓ typical leaching agents are dilute acids, bases, salts, and sometimes water.

Hydrometallurgy of gold ✓Gold can be extracted from low-grade ore by cyanidation: NaCN is sprayed over the crushed ore and the gold is oxidized: 4Au(s) + 8CN-(aq) + O2(g) + 2H2O(l) → 4Au(CN)2-(aq) + 4OH-(aq)

✓The gold is then obtained by reduction: 2Au(CN)2-(aq) + Zn(s) → Zn(CN)42-(aq) +

Hydrometallurgy of aluminum ✓Aluminum is found in bauxite as Al2O3•xH2O. ✓Extraction involves the Bayer process: The crushed ore is digested in 30% NaOH (by mass) at 150 - 230°C and high pressure (30 atm to prevent boiling). Al2O3.H2O(s) + 2H2O(l) + 2OH-(aq) → 2Al(OH)4-(aq)

The aluminate solution is separated by

Electrometallurgy

the process of obtaining metals through electrolysis.

Electrometallurgy of sodium ✓Two different starting materials: molten salt or aqueous solution. ✓Sodium is produced by electrolysis of molten NaCl in a Downs cell. ✓CaCl2 is used to lower the melting point of NaCl from 804°C to 600°C.

Electrometallurgy of sodium ✓ At the cathode (iron): 2Na+(aq) + 2e- → 2Na(l) ✓ At the anode (carbon): 2Cl-(aq) → Cl2(g) + 2e-

Electrometallurgy of aluminum ✓Hall-Heroult process ✓Hall electrolysis cell is used to produce Al ✓Al2O3 melts at 2000°C and it is impractical to perform electrolysis on the molten salt. ✓Purified Al2O3 in molten cryolite (Na3AlF6, m.p. 1012°C) is used ✓Graphite rods are employed and are

Electrometallurgy of aluminum Anode: C(s) + 2O2-(l) → CO2(g) + 4eCathode: 3e- + Al3+(l) → Al(l)