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Redox reaction( 氧化還原反應 ), Oxidation number and Electrolysis( 電解 )

International Junior Science Olympiad (IJSO)

Dr. Yu-San Cheungyscheung@cuhk.edu.hk

Department of ChemistryThe Chinese University of Hong Kong

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Oxidation: Reaction of an element or a compound with O2 to give an oxide

e.g., 4 Na(s) + O2(g) 2 Na2O(s) (sodium oxide)

2 H2(g) + O2(g) 2 H2O(l) (“hydrogen oxide”?)

Reduction: Reverse of oxidatione.g., 2 CuO(s) 2 Cu(s) + O2 (g)

Oxidation( 氧化 ) and Reduction( 還原 )

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Reduction: Reverse of oxidatione.g., 2 CuO(s) 2 Cu(s) + O2 (g)

CuO(s) can also be converted to Cu(s) with hydrogen:e.g., CuO(s) + H2(g) Cu(s) + H2O(l)

Therefore, we may also say that:Reduction is a reaction of an element or a compound with H2.

e.g., H2(g) + Cl2(g) 2 HCl(g) [Cl2(g) is reduced]

Here, we consider that H2 and O2 have opposite properties.

Oxidation( 氧化 ) and Reduction( 還原 )

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In the following reaction, copper is oxidized and loses electrons to have a positive charge:e.g., 2 Cu(s) + O2(g) 2 CuO(s)

Therefore, we may also say that:In an oxidation, an element or a compound loses electron(s) to have a positive charge.

Similarly: in an reduction, an element or a compound gains electron(s) to have a positive charge.e.g., Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s)

Which ion is oxidized? __________Which ion is reduced? __________

Oxidation ( 氧化 ) and Electron Transfer( 電子轉移 )

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Consider: H+(aq) + OH–(aq) H2O(l)

Is OH–(aq) oxidized or reduced? (Gaining “hydrogen” but with charge increased)

Oxidation Number/Oxidation State: Numbers assigned to elements, ions, and compounds to help us to tell whether they are oxidized or reduced in a reaction.

It is also assigned to individual atoms in ions and molecules.

Oxidation Number( 氧化數 )

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From high to low priority:

• O.N. for atoms in elements: 0

Overall O.N. for neutral molecules and compounds: 0

e.g., H2(g), Na(s), NaCl(s), CO2(g)

• Overall O.N. for ions: equal to the charges

e.g., Na+ & NH4+(O.N.= +1), SO4

2–(O.N.= –2)

Rules of Assigning Oxidation Number (O.N.)

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For individual atoms in neutral molecules and compounds:

• F: –1

• Group 1A metals (Li, Na, K, …): +1

• Group 2A metals (Be, Mg, Ca, …): +2

• H: +1 (except in metal hydride, see below)

• O: –2

• Cl: –1

• Br, I: –1; N, P: –3; S: –2

Rules of Assigning Oxidation Number (con’t)

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SO4 2–: overall: –2 O.N. for O: –2 O.N. for S = +6

SO2: overall: 0 O.N. for O: –2 O.N. for S = +4

SF2: overall: 0 O.N. for F: –1 O.N. for S = +2

H2S: overall: 0 O.N. for H: +1 O.N. for S = –2

PCl5: overall: 0 O.N. for Cl: –1 O.N. for P = +5

PO4 3–: overall: –3 O.N. for O: –2 O.N. for P = +5

PO3 3–: overall: –3 O.N. for O: –2 O.N. for P = +3

PH3: overall: 0 O.N. for H: +1 O.N. for S = –3

H2O: overall: 0 O.N. for O: –2 O.N. for H = +1

CaH2: containing Ca2+ and H– O.N. for H = –1

Examples:

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ClO4–: overall: –1 O.N. for O: –2 O.N. for Cl =

+7 ClO3

–: overall: –1 O.N. for O: –2 O.N. for Cl = +5ClO2

–: overall: –1 O.N. for O: –2 O.N. for Cl = +3ClO–: overall: –1 O.N. for O: –2 O.N. for Cl = +1

MnO4–: overall: –1 O.N. for O: –2 O.N. for Mn =

+7 Note: Mn has bonding with oxygen atoms. It does not exist as Mn7+ ion.

Examples:

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O.N. for O is usually –2, except (including but not limited to):

Oxidation Number( 氧化數 )for Oxygen

• In fluorine-oxygen compoundse.g.,: OF2

Neutral molecule: overall O.N. = 0Then assign O.N. = –1 for FSo, O.N. = +2 for O

How about FIO3?

O.N. = –1 for F, O.N. = –2 for O, finally we have +7 for I

• In some ions:Peroxide: O2

2– O.N. for O atom: –1

Superoxide: O2– O.N. for O atom: –1/2

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Redox reaction( 氧化還原反應 )(Oxidation-Reduction)

Oxidation of an atom: O.N. increasesReduction of an atom: O.N. decreases

In a reaction, O.N. increase of an atom must be accompanied with O.N. decrease of another atom

e.g., Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag (s)

O.N. change: Fe: +2 +3 Ag: +1 0

We say that: Fe2+(aq) is oxidized to Fe3+(aq) Ag+(aq) is reduced to Ag(s)

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O.N. and Electron Transfer( 電子轉移 )

e.g., Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag (s)

Ag+: gaining electron, being reduced Fe3+: losing electron, being oxidized

In general:- gaining electron, being reduced - losing electron, being oxidized

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Oxidizing Agent( 氧化劑 ) and Reducing Agent( 還原劑 )

Oxidizing agent (oxidant): - oxidizing another species- being reduced (O.N. decreased)

Reducing agent (reductant) : - reducing another species- being oxidized (O.N. increased)

e.g., Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag (s)Oxidizing Agent: Ag+(aq) Reducing Agent: Fe2+(aq)

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Common Oxidizing Agents ( 氧化劑 ) Ions of metal at low position in the reactivity series (e.g., Ag+ Ag)

MnO4– & Cr2O7

– in acidic medium:

MnO4– Mn2+; Cr2O7

– Cr3+

Conc. HNO3(aq): NO3– NO2(g)

Conc. H2SO4(aq): SO42– SO2(g)

Cl2(g) 2 Cl–; Br2(l) 2 Br–

O2(g) O2–; H2O2(aq) H2O(l)

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Common Reducing Agents( 還原劑 )Metal at high position in the reactivity series (e.g., Na Na+)

C(s) CO(g) or CO2(g)

CO(g) CO2(g)

SO32– SO4

2–

Fe2+ Fe3+

2I– I2

H2(g) 2 H+

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Some chemical can act as both oxidizing agent and reducing agent.

e.g., SO2(g) S(s) oxidation / reduction SO2(g) SO4

2–(aq) oxidation / reduction

Species with high O.N. atom has higher chance to be an oxidizing agent.

Similarly, species with low O.N. atom has higher chance to be a reducing agent.

Remarks:

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Acid-Base Reaction & Redox Reaction

H+(aq) + OH–(aq) H2O(l)

O.N.: H +1 +1 +1O –2 –2

No atom has its O.N. change.

In general, acid-base reaction is NOT a redox reaction.

ExerciseVerify the conclusion for the following:HCl(aq) + NaHCO3(aq) NaCl(aq) + CO2(g) + H2O(l)

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Concentration( 濃度 )Effect on Oxidizing/Reducing Power

Example

Dilute nitric acid (HNO3) reacts with magnesium but not copper. Concentrated nitric acid reacts with copper.

• Dil. HNO3(aq) with Mg: 2 H+(aq) + Mg(s) H2(g) + Mg2+(aq)

• Dil. HNO3(aq) with Cu: no reaction

• Conc. HNO3(aq): (unbalanced equation) NO3

–(aq) + Cu(s) NO2(g) + Cu2+(aq)

• Dil. HNO3(aq) acts as an acid, conc. HNO3(aq) can act as an oxidizing agent. Similarly for H2SO4(aq).

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Oxidation Number ( 氧化數 ) in Chemical Naming

Roman number is sometimes used for O.N./O.S.

e.g., “The Oxidation State of Mn in MnO4– is +VII.”

The Romanic number system is also used in the “Stock system” to distinguish different compounds.

e.g.,Cu2O, containing Cu+ ion: copper(I) oxide

CuO, containing Cu2+ ion: copper(II) oxide

SO42–, sulphate(VI) ion (more common: sulphate)

SO32–, sulphate(IV) ion (more common: sulphite)

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O.N. of Atoms( 原子 ) in Various Species

Exercise: verify the O.N.

Oxidation number

Sulphur Nitrogen

Carbon Iron Copper

Manganese

Chromium

+7 KMnO4

+6 H2SO4 K2MnO4 K2Cr2O7

+5 HNO3

+4 SO2 NO2 CaCO3 MnO2

+3 HNO2 FeCl3 Mn2O3 CrCl3

+2 SCl2 NO CO FeSO4 CuSO4 MnSO4 CrCl2

+1 N2O CuCl

0 S N2 C Fe Cu Mn Cr

-1 NH2OH C2H2

-2 H2S N2H4 C2H4

-3 NH3 C2H6

-4 CH4

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Electrolysis( 電解 )

Charging and discharge of rechargeable battery

• Charging: applying voltage new substances (electrical energy chemical energy)

• Discharge: giving out electrical energy (chemical energy electrical energy)

Electrolysis: chemical reaction by applying voltage

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Electrolysis( 電解 ) of molten PbBr2

Anode (“+”, attracting anions): Br - Br + e-

Cathode (“-”, attracting cations): Pb2+ + 2e- Pb

“An Ox”: anode - oxidation

“Red Cat”: reduction - cathode

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Electrolysis( 電解 ) of CuSO4 solution

Ions attracted to anode: OH-(aq) & SO42-(aq)

Oxidation: 4OH-(aq) O2(g) + 2H2O(l) + 4e-

SO42-(aq): no reaction

Ions attracted to cathode: Cu2+(aq) & H+(aq)

Reduction: Cu2+(aq) + 2e- Cu(s)H+(aq): no reaction

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Preference of ion discharge

(1) Electrochemical series

Na Na+ + e- (More easily) Ag Ag+ + e- (More difficultly)

Therefore, we can infer that sodium ions gain electrons to form

atoms more difficultly than silver ions.

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Different species have different abilities to gain electrons to be reduced.  These abilities are summarized in the electrochemical series.

Example: http://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29

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Characteristics • The half-reactions are reductions (i.e., gaining electrons on the left- hand-side).

• The half-reactions are equilibriums.

• The abilities of gaining electrons are quantified by “standard electrode potentials”. The smaller the potentials (near the top) are, the more difficultly the reductions occur.

Compare Na(s) Na+(aq) + e- Eo = +2.71 V Ag(s) Ag+(aq) + e- Eo = -0.80 V

(Note that when the reactions are reversed, the signs of Eo are changed.)

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• Higher concentration or pressure: more favorable to go to the opposite side. “Standard” values: 1 M for concentration and 1 atm pressure for pressure.

e.g., Ca2+(aq) + 2e- Ca(s) Eo = -2.87 V

If increasing Ca2+ concentration, more favorable to the right-hand- side, larger E (less negative or even positive).

e.g., NO3-(aq) + 2H+(aq) + e- NO2(g) + H2O(l) Eo = +0.78

V

If increasing NO2(g) pressure, more favorable to the left-hand-side, smaller E.

Characteristics

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Characteristics

•A complete reaction consists of a reduction and an oxidation.

e.g., anode: 4OH-(aq) O2(g) + 2H2O(l) + 4e- (1) cathode: Cu2+(aq) + 2e- Cu(s) (2)

They are combined to form a complete reaction, in which no electron shows up.

(1):(2)x2:

4OH-(aq) O2(g) + 2H2O(l) + 4e-

2Cu2+(aq) + 4e- 2Cu(s)

Adding:

4OH-(aq) + 2Cu2+(aq) O2(g) + 2H2O(l) + 2Cu(s) (3)

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• Voltage of an electrochemical cell:

(1): 4OH-(aq) O2(g) + 2H2O(l) + 4e- -0.40 V (2)x2: 2Cu2+(aq) + 4e- 4Cu(s) +0.34 V

IMPORTANT: Eo does not change when the equation is “doubled”.

Adding: Eo = -0.40 + 0.34 = -0.06 V

Characteristics

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http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf

• An electrochemical cell consists of two

half-reactions. A single half-reaction

does not exist alone and the absolute

values of Eo for half-reactions cannot

be measured. Therefore, the Eo of

one of the half-reactions,

2H+(aq) + 2e- H2(g),

is set to zero. The electrode for this

half-reaction is shown on the right

and is called “standard hydrogen

electrode (SHE)”. The Eo of all the

others can be determined as values

relative to this “standard”.

Characteristics

Glass tubeH2(g) at 1 atm

Pt electrode

H+(aq, 1 M)

Metal wire

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e.g., 2H+(aq) + 2e- H2(g) E1o = 0

V Zn2+(aq) + 2e- Zn(s) E2

o = ?

Consider Zn(s) Zn2+(aq) + 2e- -E2o

Adding: 2H+(aq) + Zn(s) H2(g) + Zn2+(aq) Ecell

o = E1o + (-E2

o) = -E2o

Ecello of the last electrochemical cell is measured as +0.76 V.

Therefore, E2o = -0.76 V.

Characteristics

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• If the voltage is positive, the reaction occurs spontaneously. If the voltage is negative, the reaction does not occur spontaneously and an external voltage must be applied. The external voltage must be larger than the magnitude of the cell voltage.

e.g., H2(g) + Zn2+(aq) 2H+(aq) + Zn(s) Eo = –0.76 V

The external voltage applied must be at least 0.76 V.

Characteristics

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• In electrolysis, the preference of ion discharge depends on Eo of the relevant half-reaction potential. For example,

4OH-(aq) O2(g) + 2H2O(l) + 4e- -0.40 V 2SO4

2-(aq) S2O82-(aq) + 2e- -2.01 V

The first half-reaction is preferred because its Eo is larger.

Characteristics

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Increasing concentration increases the discharge tendency of an ion. For example,

4OH-(aq) O2(g) + 2H2O(l) + 4e- -0.40 V2Cl-(aq) Cl2(g) + 2e- -1.36 V

For dilute NaCl solution, OH- is discharged because the Eo value of the first half-reaction is preferred. But for concentrated NaCl solution, Cl- concentration is high enough for Cl- to be discharged.

Concentrations of species

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Commonly used graphite and platinum electrodes are inert and have no effect on the preference of ion discharge. But some may.

• Mercury electrode(汞電極 ) If graphite or platinum electrodes are used in the electrolysis of concentrated NaCl solution, only H+ is discharged at the cathode. But if mercury electrode is used for the cathode, Na+ is discharged because sodium metal forms an alloy with mercury. (This method is used in industry for the production of sodium.)

Electrodes ( 電極 )

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• Metal electrode

When an anion discharges at anode, it gives out electrons. If a metal electrode is used as the anode, the metal atoms may also give out electrons to form metal ions, i.e., the metal electrode may compete with the anion in giving out electrons. For example, if copper electrode is used as the anode in the electrolysis of copper sulfate solution, the copper electrode becomes thinner and thinner.

Electrodes ( 電極 )

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cf. Cu(s) Cu2+(aq) + 2e- -0.34 V 4OH-(aq) O2(g) + 2H2O(l) + 4e- -0.40 V

Copper metal of the anode completes with OH-. The potential of the first half-reaction is larger. Copper metal, rather than OH-, gives out electrons.

In principle, if platinum electrode is used, platinum may also give out electrons to form platinum ion. But in practice, it seldom happens due to the very negative value of Eo for this process:

Pt(s) Pt2+(aq) + 2e- -1.20 V

Electrodes( 電極 )

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Summary of common cases:

SolutionElectrodes

Main products at

Anode Cathode Anode Cathode

NaNO3 or NaSO4

Graphite

O2

H2

H2SO4

NaOH

NaCl

(Dil)

(Conc)

Cl2

(Conc)

Graphite Mercury Cl2 Na

CuSO4

GraphiteGraphite

O2 CopperCopperCopper Cu2+

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The container in which a half-reaction occurs is called a “half-cell”. In the diagram shown, the two half-cells are in the same beaker.

Salt Bridge

But in some cases they must be separated physically, because the species of the two half-cells react directly without electrons going through the external circuit.

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For example, an electrochemical cell can be constructed for the following reaction:

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

We can break it into two half-reactions:Cu(s) Cu2+(aq) + 2e– –0.34 V2Ag+(aq) + 2e– 2Ag(s) +0.80 V

If we put everything into the same beaker, it does not work as expected.

Salt Bridge

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Reason: when Ag+ is in contact with Cu, they react on the surface of the copper plate. Electrons are given out by Cu to Ag+ directly and they do not go through the external circuit.

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

Voltmeter

Salt Bridge

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Therefore, the Ag+ solution and Cu2+ solution must be separated in two beakers.

Salt-bridge:• “connecting” the two half-cells. • providing ions to keep the half-cells electrically neutral• simplest version: a strip of filter-paper soaked in KNO3 or NH4NO3

Electrical current

Modified from: http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electr

ochemistry.pdf

Salt Bridge

Voltmeter

Salt

bridge

Electron flow

Cu Ag

Cu2+ Cu2+ NO3– Ag+

NO3– K+

1 M Cu(NO3)2(aq) 1 M AgNO3(aq)

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Fuel Cell( 燃料電池 )

• Burning fuel … electrical energy: some energy is lost in heating

• Fuel cell: chemical energy electrical energy

Example: H2 fuel cell

H2 fuel cell is reversible and it can act as a rechargeable battery

Charging (storing up electrical energy):2H2O(l) 2H2(g) + O2(g)

Discharging (releasing electrical energy): 2H2(g) + O2(g) 2H2O(l)

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Half-reactions: (discharging)

H2(g) 2H+(aq) + 2e– 0.00 V

O2(g) + 4H+(aq) + 4e– 2H2O(l) +1.23 V

The cell gives an electrical potential of 1.23 V.Fuel Cell – Car & Experiment Kit Lab Manual, Thames & Kosmos (2000)

H2 fuel cell

Anode PEM Cathode PEM: proton exchange membrane

H2

O2

H2O

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Other than H2, some other compounds can also be used for fuel cell. For example: • Methanol (CH3OH)• Ethanol (CH3CH2OH)

These kinds of fuel cell may not be reversible, but easier to handle and more energy-rich.

Fuel Cell( 燃料電池 )

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http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf

Silver-Zinc( 鋅 )Battery(1.8V)

Reduction:Oxidation:

Ag2O(s) + H2O(l) + 2e- 2Ag(s) + 2OH- (aq)Zn(s) + 2OH-(aq) ZnO(s) + H2O(l) + 2e-

Overall: Zn(s) + Ag2O(s) ZnO(s) + 2Ag(s)

Half reactions of discharge:

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http://www.yorku.ca/skrylov/Teaching/Chemistry1001/electrochemistry.pdf

The Nickel-Cadmium Rechargeable Battery (1.4 V)

Reduction:Oxidation:

2NiO(OH)(s) + 2H2O(l) + 2e- 2Ni(OH)2(s) + 2OH- (aq)Cd(s) + 2OH-(aq) Cd(OH)2(s) + 2e-

Overall: Cd(s) + 2NiO(OH)(s) + 2H2O(l) 2Ni(OH)2(s) + Cd(OH)2(s)

Half reactions of discharge:

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Eo and K (Equilibrium Constant( 平衡常數 ))

nFEo = RT·ln(K)

e.g, Calculate K for the equilibrium: Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)

Solution:Cu(s) Cu2+(aq) + 2 e– –0.34 V2 Ag+(aq) + 2 e– 2 Ag(s) +0.80 V

Eo = –0.34 V + 0.80 V = +0.46 V

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n = no. of electrons in the half-reactions = 2F = 96485 C mol–1 (Faraday constant)

Put into the equation, nFEo = RT·ln(K),

(2) (96485 C mol–1) (0.46 V) = (8.314 J mol–1 K–1) (298 K) ln(K)

K = 3.6 x 1015 dm3 mol–1

Note: the unit of K is determined by the expression:

K = [Cu2+(aq)]/[Ag+(aq)]2

(mol dm–3 for concentration, atm for pressure)

Eo and K (Equilibrium Constant( 平衡常數 ))