Oxidation - Reduction Reactions

Redox Terminology—REVIEW CHAPTER 4!!!redox reaction -- electron transfer process

e.g., 2 Na + Cl2 --> 2 NaCl

Overall process involves two Half Reactions:oxidation -- loss of electron(s)reduction -- gain of electron(s)

e.g. Na --> Na+ + e– (oxidation)Cl2 + 2 e– --> 2 Cl– (reduction)

Related terms:

oxidizing agent = the substance that gets reduced (Cl2)

reducing agent = the substance that gets oxidized (Na)

Oxidation and reduction always occur together so that there is no net loss or gain of electrons overall.

OILRIG

Redox Reactions• The transfer of electrons between species, meaning that

one species is oxidized and one is reduced. The two processes will always occur together.

• When has a redox reaction occurred?– If there is a change in the oxidation state of any element in

the reaction, a redox reaction has happened.– Remember that if something is oxidized, something else

must be reduced! (and vice-versa)

More examples:

2 Ca(s) + O2(g) --> 2 CaO(s)

2 Al(s) + 3 Cl2(g) --> 2 AlCl3(s)

HALF-REACTION METHOD for balancing Redox Equations

1. Write unbalanced ionic equations for the two half-reactions. (Look at oxidation numbers in the “skeleton equation.”)

2. Balance atoms other than H and O.3. Add appropriate number of electrons.4. Balance O with H2O.

5. Balance H with H+.6. If in acidic solution, then skip to step 7. If in basic solution, then

add equal number of OH– to both sides to cancel all of the H+.7. Multiply balanced half-reactions by appropriate coefficients so

that the number of electrons are equal. 8. Rewrite the balanced half reactions.9. Add the half-reactions together.10. Cancel species that appear on both sides to get the balanced

Net Ionic Equation.11. If necessary, add spectator ions to get the balanced molecular

equation.

Check the Final Balance (atoms and charges)!

Example ProblemBalance the following redox equation.

Ce4+(aq) + Sn2+

(aq) --> Ce3+(aq) + Sn4+

(aq)

The reaction can be separated into a reaction involving the substance being reduced;

Ce4+(aq) + e– --> Ce3+

(aq)

And the substance being oxidized;

Sn2+(aq) --> Sn4+

(aq) + 2e–

We can see that the equations don’t balance; you must multiply the top equation by two, then add them to get

2 Ce4+(aq) + Sn2+

(aq) --> 2 Ce3+(aq) + Sn4+

(aq)

Sample ProblemThe following redox reaction occurs in basic solution. Write

complete, balanced equations for the individual half-reactions and the overall net ionic equation.

MnO4–(aq) + N2H4(aq) --> MnO2(s) + NO(g)

Oxidation Half Reaction:

Reduction Half Reaction:

Net Ionic Equation:

Sample ProblemThe following redox reaction occurs in basic solution. Write

complete, balanced equations for the individual half-reactions and the overall net ionic equation.

MnO4–(aq) + N2H4(aq) --> MnO2(s) + NO(g)

Oxidation Half Reaction:

8 OH–(aq) + N2H4(aq) --> 2 NO(g) + 6 H2O(l) + 8 e–

Reduction Half Reaction:

3 e– + 2 H2O(l) + MnO4–(aq) --> MnO2(s) + 4 OH–

(aq)

Net Ionic Equation:

3 N2H4(aq) + 8 MnO4–(aq) --> 6 NO(g) + 8 MnO2(s) + 2 H2O(l)

+

8 OH–(aq)

Sample Problems• Balance the following redox equations in acidic solution.

• Al(s) + I2(s) --> AlI3(s)

• KClO3(aq) + HNO2(aq) --> KCl(aq) + HNO3(aq)

• Cl–(aq) + MnO2(s) --> Mn2+(aq) + Cl2(g)

• Balance the above redox equations in basic solution.

Sample Problems• Balance the following redox equations in acidic solution.

• Al(s) + I2(s) --> AlI3(s)

2 Al(s) + 3 I2(s) --> 2 AlI3(s)

• KClO3(aq) + HNO2(aq) --> KCl(aq) + HNO3(aq)

KClO3(aq) + 3 HNO2(aq) --> KCl(aq) + 3 HNO3(aq)

• Cl–(aq) + MnO2(s) --> Mn2+(aq) + Cl2(g)

4H+(aq) + 2 Cl–(aq) + MnO2(s) --> Cl2(g) + Mn2+

(aq) + 2 H2O(l)

• Balance the above redox equations in basic solution.– First doesn’t change.

– Second: use NO2– and NO3

– rather than the acids.

– Third: Add 4 OH- to each side,2 H2O(l) + 2 Cl–(aq) + MnO2(s) --> Cl2(g) + Mn2+

(aq) + 4 OH–(aq)

Electrochemistry TerminologyElectrochemical Cell -- a device that converts electrical energy into chemical energy or vice versa

• Two Types:Electrolytic cell

Converts electrical energy into chemical energy Electricity is used to drive a non-spontaneous reaction

Galvanic (or voltaic) cell Converts chemical energy into electricity (a

battery!) A spontaneous reaction produces electricity

• Conduction:Metals: metallic (electronic) conduction

-- free movement of electronsSolutions: electrolytic (ionic) conduction (or molten salts) -- free movement of ions

Galvanic CellsGalvanic Cells (batteries) -- produce electrical energy

e.g. a spontaneous reaction:Cu(s) + Ag+

(aq) --> Cu2+(aq) + Ag(s)

(Ag metal will be deposited on a Cu wire dipped into aqueous AgNO3 solution)

In a galvanic cell, the half reactions are occurring in separate compartments (half-cells)

Electrochemical ConventionsCharges on the electrodes

Galvanic cell: (+) cathode ~ reduction

(–) anode ~ oxidationElectrolytic cell: (+) anode ~ oxidation

(–) cathode ~ reduction

Cell Notation -- summary of cell descriptione.g.

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

(aq) | Ag(s)

anode cathode

Cell PotentialCell potential ~ Eºcell (an electromotive force, emf)

Units of Eºcell are volts: 1 volt = 1 joule/coulomb

Eºcell is a measure of the relative spontaneity of a cell reaction

Positive (+) Eºcell --> spontaneous reaction

Eºcell depends on:– Nature of reactants– Temperature -- superscript º means 25 ºC– Concentrations -- superscript º means all conc are at

1.00 M and gases are at 1.00 atm

but, Eºcell is independent of amounts of reactants

Standard PotentialsStandard Reduction Potential:

the potential of a half-cell relative to a standard reference

(Pt)

Sample Problem; Write the cell notation for this cell.

Standard Cell PotentialStandard Reduction Potentials Eº (volts)F2(g) + 2 e– --> 2 F–

(aq) + 2.87

Ag+(aq) + e– --> Ag(s) + 0.80

Cu2+(aq) + 2 e– --> Cu(s) + 0.34

2 H+(aq) + 2 e– --> H2(g) 0.00

Zn2+(aq) + 2 e– --> Zn(s) – 0.76

Li+(aq) + e– --> Li(s) – 3.05

• Ease of Reduction ~ increases with Eºe.g. F2 is easiest to reduce, Li+ is the hardest

• Standard Cell Potential ~ Eºcell can be determined from standard reduction potentials:

Eºcell = Eºred – Eºoxid

= [reduction potential of substance reduced] – [reduction potential of substance oxidized]

Standard Potential Example ProblemIs the following a galvanic or an electrolytic cell? Write the

balanced cell reaction and calculate Eºcell.

Zn(s) | Zn2+(aq) || Cu2+

(aq) | Cu(s)

Cathode: Cu2+ + 2 e– --> Cu Eº = 0.34 VAnode: Zn --> Zn2+ + 2 e– Eº = – 0.76 VCell rxn: Zn(s) + Cu2+

(aq) --> Zn2+(aq) + Cu(s)

Eºcell = Eºred – Eºox = 0.34 V – (– 0.76 V)

= + 1.10 V ∴ galvanic.

Another Example ProblemGiven the standard reduction potentials:

O2(g) + 4 H+(aq) + 4 e– --> 2 H2O(l) Eº = 1.23 V

Cl2(g) + 2 e– --> 2 Cl–(aq) Eº = 1.36 V

Write a balanced equation and calculate Eºcell for a galvanic cell based on these half reactions. Sketch the galvanic cell.

“galvanic” implies a positive value for Eºcell

so, Cl2/Cl– should be the reduction half reaction,

since 1.36 – 1.23 = + 0.13 V∴ reverse the O2 half reaction and make it the oxidation

Multiply the Cl2 reaction by 2, to make e–’s cancel, hence:

2 Cl2(g) + 2 H2O(l) --> 4 Cl–(aq) + O2(g) + 4 H+(aq)

Note: When a half-reaction is multiplied by a coefficient, the Eº IS NOT MULTIPLIED by the coefficient.

Diagram of cell

voltmeter

salt bridgecathode

anode

(reduction)

(oxidation)

Pt electrode

Pt electrode

supporting electrolyte

Cl–

Cl–

Cl–

NO3–

NO3–

NO3–

K+

K+

K+

K+

K+

K+

e–e–

O2 gas

Cl2 gas

Sample ProblemUnder standard conditions, is the following a galvanic or an

electrolytic cell? Support your conclusion with appropriate calculations and balanced chemical reactions.

Ag(s), AgBr(s) | Br–(aq) || Au3+

(aq) | Au(s)

--From table, Au3+

(aq) + 3e – Au(s) Eº = 1.50 V

AgBr(s) + e – Ag(s) + Br –(aq) Eº = 0.071 V

Sample ProblemUnder standard conditions, is the following a galvanic or an

electrolytic cell? Support your conclusion with appropriate calculations and balanced chemical reactions.

Ag(s), AgBr(s) | Br–(aq) || Au3+

(aq) | Au(s)

--From table, Au3+

(aq) + 3e – Au(s) Eº = 1.50 V

AgBr(s) + e – Ag(s) + Br –(aq) Eº = 0.071 V

3 Ag(s) + 3 Br–(aq) + Au3+

(aq) --> 3 AgBr(s) + Au(s), Eºcell = + 1.43 V, so galvanic.

Reactions of Metals with Acids

Generally: metal + acid --> salt + H2(g)

e.g. Zn(s) + H2SO4(aq) --> ZnSO4(aq) + H2(g)

Metal is oxidized to yield the metal cationH+ from the acid is reduced to yield H2

Some acids (especially HNO3) are “oxidizing acids” -- the NO3

– ion is reduced to form NO2 or NO depending on concentration of the acid.

e.g. reactions of Cu metal with HNO3

Cu(s) + concentrated HNO3 --> NO2(g)

Cu(s) + dilute HNO3 --> NO(g)

21

Cell Potential and ThermodynamicsFree Energy Change (G)G = – Wmax (maximum work)

• For an electrochemical cell:W = nFEcell where n = # moles e–

F = 96,500 coulombs/mole e–

Units: volt =joule

coulomb

joules = (moles e–)coulombsmole e–

joulecoulomb

∴ free energy is related to Ecell as follows:

G = – nFEcell

under standard conditions: use Gº and Eºcell

Equilibrium Constantsince Gº = – RTlnKc = – nFEºcell can rearrange to:

Eºcell = ln Kc

at 25 ºC, use values for R, T, and F, then convert to base-10 log:

Eºcell = log Kc

RTnF

0.0592n

(Use 8.314 J mol-1 K-1 for R!)

Overall,

Sig Fig Review: Log, LnRules are the same for log and ln!

When you report the log or ln of a number, the significant digits are the ones after the decimal point, e.g.

ln (24.3) = 3.190 [3 sig fig!]log (0.068) = –1.17 [2 sig fig!]

Antilog is the opposite; when you convert back to natural numbers, you will “appear” to lose significant digits, e.g. Antilog (23.32) = 1023.32 = 2.1 x 1023 [2 sig fig!] Anti(natural log) of 23.32 = e23.32 = 1.3 x 1010 [2 sig fig!]

Sample ProblemGiven the following standard reduction potentials, calculate

the solubility product constant (Ksp) for lead sulfate, PbSO4.

PbSO4(s) + 2 e– --> Pb(s) + SO42–

(aq) Eº = – 0.360 V

Pb2+(aq) + 2 e– --> Pb(s) Eº = – 0.130 V

Sample ProblemGiven the following standard reduction potentials, calculate

the solubility product constant (Ksp) for lead sulfate, PbSO4.

PbSO4(s) + 2 e– --> Pb(s) + SO42–

(aq) Eº = – 0.360 V

Pb2+(aq) + 2 e– --> Pb(s) Eº = – 0.130 V

Kc = 2 x 10–8

Effect of Concentration on Cell Potential

For a general reaction:aA + bB cC + dD

G = Gº + RTln Qwhere Q = “concentration quotient” or “reaction

quotient”= [C]c[D]d / [A]a[B]b (use given concentrations)

Since G = – nFE :– nFE = – nFEº + RTln Q

Rearrange to get the “Nernst Equation”

E = Eº – [RT/nF]ln Q

(review fromCh. 14-16)

The Nernst Equation

E = Eº – [RT/nF]ln Q

The Nernst Equation shows the relationship between the standard cell potential (Eº) and the cell potential (E) under actual, non-standard conditions.

this can be simplified at 25 ºC to:

E = Eº – (0.0592/n)log Q

Major use of the Nernst Equation:• Determine concentrations from standard reduction

potentials

• Use actual concentrations (i.e. Q) to calculate Ecell

Sample ProblemA silver wire coated with AgCl is sensitive to the chloride ion

concentration because of the following half-cell reaction. AgCl(s) + e– --> Ag(s) + Cl–(aq) Eº = 0.2223 V

Calculate the molar concentration of Cl– when the potential of this half-cell is measured to be 0.4900 volts relative to a standard hydrogen electrode.

Sample ProblemA silver wire coated with AgCl is sensitive to the chloride ion

concentration because of the following half-cell reaction. AgCl(s) + e– --> Ag(s) + Cl–(aq) Eº = 0.2223 V

Calculate the molar concentration of Cl– when the potential of this half-cell is measured to be 0.4900 volts relative to a standard hydrogen electrode.

3 x 10–5 M

Quantitative Aspects of Electrochem ~ the Faraday

e.g. consider a cell with the cathode reaction:Cr3+

(aq) + 3 e– --> Cr(s)

Stoichiometry: 1 mole Cr3+ ~ 3 moles e– ~ 1 mole Cr– If a current equivalent to 3 moles of electrons is passed

through the solution of Cr3+, then 1 mole of Cr metal will be produced

By definition: 1 Faraday (F) = 1 mole electronsbut, how much electrical current is this?

1 F = 9.65 x 104 coulombs and

1 coulomb = 1 amp sec

or, 1 amp = 1 coulomb

1 sec

Batteries• Dry cells (paste)

– Metal anode is oxidized, solid MnO2 in paste is reduced

– Supporting electrolyte is either• acid (can corrode more easily, cannot

recharge), or• alkaline (base, lasts longer)

• Lead storage (car batteries!)– Lead is oxidized, PbO2 is reduced

– Acid electrolyte (H2SO4)

– Rechargeable, heavy

Light-Weight Rechargeable Batteries• NiCAD

– Cadmium is oxidized, NiO2 is reduced

– Supporting electrolyte is KOH

• NiMH– Hydrogen is oxidized (stored as metal

hydride), NiO2 is reduced

– Supporting electrolyte is KOH

• Lithium ion – “Concentration” cell– Migration of lithium ions from graphite

anode to metal oxide cathode– Supporting electrolyte is KOH

Fuel Cells• like batteries in which reactants are constantly

being added– so it never runs down!

• Pt electrodes, KOH electrolyte

Simpler diagram

Electrolysis

-- chemical reactions that occur during electrolytic conduction

e.g. molten NaCl

reduction:Na+

(l) + e– --> Na(s)

Na+ ions migrate toward the negative electrode and are reduced

oxidation:2 Cl–(l) --> Cl2(g) + 2e–

Cl– ions migrate toward the positive electrode and are oxidized

Electrolysis Cell• Electrodes:

cathode -- where reduction occursanode -- where oxidation occurs

• Net Cell Reaction: add anode & cathode half-reactions so that # of

electrons cancel

2 Cl– --> Cl2 + 2 e– (anode ~ oxidation)

2 [ Na+ + e– --> Na ] (cathode ~ reduction)2 Cl–(l) + 2 Na+

(l) --> Cl2(g) + 2 Na(s) (Cell Reaction)

Overvoltage, or overpotential — in some cases additional voltage must be applied in order to get an electrolytic reaction to occur (kinetic reasons)

Electrolysis of WaterIn aqueous solution electrolysis of water may occur:

• oxidation (H2O --> O2) -- at anode

2 H2O --> O2 + 4 H+ + 4e– Eº = 1.23 V

• reduction (H2O --> H2) -- at cathode

in neutral or basic solution2 H2O + 2 e– --> H2 + 2 OH– Eº = -0.83 V

• reduction (H+ --> H2) -- at cathode

in acidic solution2 H+ + 2 e– --> H2 Eº = 0.00 V

Higher Eº easier to reduce!Lower Eº easier to oxidize!

Some Industrial Applications of Electrolysis

Preparation of Aluminum from molten Al2O3

(can’t use Al3+ in solution since H2O is easier to reduce)

anode: 3 [O2– --> 1/2 O2 + 2 e– ]

cathode: 2 [ Al3+ + 3 e– --> Al ] (Eº = –1.66 V)

Cell: 2 Al3+(l) + 3 O2–

(l) --> 3/2 O2(g) + 2 Al(s)

Electroplating reduction of metal ions (from solution) to pure metals (Ag, Cr, Cu, etc.)e.g.

Cr3+(aq) + 3 e– --> Cr(s) (chrome plating)

Example ProblemHow long (in hours) would it take to produce 25 g of Cr from a solution of

Cr3+ by a current of 2.75 amp? Cr3+

(aq) + 3 e– --> Cr(s)

1 mole Cr ~ 3 moles e– ~ 3 F ~ 3 x 9.65 x 104 coulombs

25 g Cr x1 mole Cr

52 g Crx

3 F

1 mole Cr= 1.44 F

1.44 F x96,500 coulombs

F= 139,000 coulombs

139,000 amp sec

2.75 ampx

1 hr

3600 sec = 14 hr

= 139,000 amp sec

Review Problems1. A large electrolytic cell that produces metallic aluminum

from Al2O3 ore is capable of making 250 kg of aluminum in 24 hours. Determine the current (in amps) that is required for this process. Include appropriate chemical reactions.

2. An aqueous solution of NaBr was electrolyzed with a current of 2.50 amps for 15.0 minutes. What volume (in mL) of 0.500 M HCl would be required to neutralize the resulting solution? (Hint: H2 is produced at the cathode and Br2 at the anode.)

Review Problems1. A large electrolytic cell that produces metallic aluminum

from Al2O3 ore is capable of making 250 kg of aluminum in 24 hours. Determine the current (in amps) that is required for this process. Include appropriate chemical reactions.

3.1 x 104 amps (Al3+(aq) + 3 e– --> Al(s) is ½-reaction,

overall reaction is 2 Al2O3(l) 3 O2(g) + 4 Al(s))

2. An aqueous solution of NaBr was electrolyzed with a current of 2.50 amps for 15.0 minutes. What volume (in mL) of 0.500 M HCl would be required to neutralize the resulting solution? (Hint: H2 is produced at the cathode and Br2 at the anode.)

46.6 mL HCl soln

Review Problems, cont.3. Consider the following electrochemical cell in which the

volume of solution in each half-cell is 100 mL.Zn(s) | Zn2+ (1.00 M) || Ag+ (1.00 M) | Ag(s)

(a) Write balanced chemical equations for the anode, cathode, and overall cell reactions. (b) Determine Eºcell, Gº, and the equilibrium constant (Kc) for the cell reaction.

(c) If current is drawn from this cell at a constant rate of 0.10 amp, what will the cell potential be after 15.0 hours?

Answers3. (a) Anode: Zn(s) --> Zn2+

(aq) + 2 e–

Cathode: Ag+(aq) + e– --> Ag(s)

Overall: Zn(s) + 2 Ag+(aq) --> Zn2+

(aq) + 2 Ag(s)

(b) Eºcell = + 1.56 V

G = - 301 kJ/mol

Kc = 6 x 1052

(c) 1.54 V