chapter 20. overview oxidation-reduction reactions »balancing redox reactions half-reaction method...

Chapter 20

Overview• Oxidation-Reduction reactions

» Balancing Redox Reactions• Half-Reaction method• Acidic Solution • Basic Solution

• Voltaic Cells• Cell EMF--standard reduction potentials• Oxidizing & Reducing reagents• Spontaneity of Redox reactions

• Effect of Concentration» Nernst Equation» Equilibrium Constants

• Commercial Voltaic Cells• Electrolysis

» Quantitative Aspects» Electrical Work

Redox Reactions• Involve a transfer of electrons

» Oxidation loss of one or more electron(s)• oxidation state will increase

» Reduction gain of one or more electron(s)• oxidation state will decrease

» Must occur simultaneously

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

(aq) + Cu(s)

Zn Zn2+(aq) + 2e- oxidation ½ rxn

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

oxidation

reduction ½ rxn

You must know oxidation states:(Review: Section 8.10)• What are the oxidation states of each

atom in the following:

» H2

» CO

» ClO2-

» HC2H3O2

H 0C +2, O -2Cl +3, O -2

H +1, C1 +3, C2 -3, O -2

Balancing Redox Reactions• Mass balance must be observed• e--transfer must be balanced• Simple reactions:

» Sn2+ + Fe3+ Sn4+ + Fe2+ • Sn2+ Sn4+ + 2e-

• Fe3+ + e- Fe2+

oxidation ½ rxn

reduction ½ rxnx 2

2Fe3+ + 2e- 2Fe2+

Sn2+ + 2Fe3+ Sn4+ + 2Fe2+

• Reactions involving H & O in acid:» MnO4

- + C2O42- Mn2+ + CO2

» write both ½ reactions• MnO4

- Mn2+

• C2O42- CO2

» mass balance (all except H & O)• MnO4

- Mn2+

• C2O42- 2CO2

» add H2O & H+ to balance O & H

• 8H+ + MnO4- Mn2+ + 4H2O

• C2O42- 2CO2

» balance charge by adding electrons• 5e- + 8H+ + MnO4

- Mn2+ + 4H2O

• C2O42- 2CO2 + 2e-

» balance electrons transferred• 10e- + 16H+ + 2MnO4

- 2Mn2+ + 8H2O

• 5C2O42- 10CO2 + 10e-

» add half reactions• 16H+ + 2MnO4

-+ 5C2O42- 10CO2 + 2Mn2+ + 8H2O

»check the balance

• Reactions in base: MnO4- + CN- CNO- + MnO2

» use exactly the same process• CN- CNO-

• MnO4- MnO2

3e- +

H2O +

+ 2H2O4H+ +

+ 2H+ + 2e-

» since H+ cannot exist in basic solution, add OH-

• 2OH- + CN- CNO- + H2O + 2e-

• 3e- + 2H2O + MnO4- MnO2 + 4OH-

» balance electrons transferred & sum• 6OH- + 3CN- 3CNO- + 3H2O + 6e-

• 6e- + 4H2O + 2MnO4- 2MnO2 + 8OH-

» 3CN- + H2O + 2MnO4- 2MnO2

+ 3CNO- +2OH-

»check balance

Voltaic Cells• A spontaneous redox reaction that

does work• Anode

» electrode at which oxidation occurs» loses mass» electrons released, sign is negative

• Cathode» electrode at which reduction occurs» gains mass» electrons consumed, sign is positive

Cell EMF• Difference in potential energy of

electrons at the anode and cathode» Diff. in potential energy per electrical charge

measured in volts » 1 V = 1 J

C

• Potential difference = EMF, electromotive force

• Ecell = cell potential = cell voltage

» Eºcell = cell potential under std. conditions• 1 M, 1 atm, 25 ºC

• Standard reduction potentials

» E ºred in tables

» E ºcell = E ºred (cathode) - E ºred (anode)

• Based on “standard hydrogen electrode”

» 2H+(aq, 1M) + 2e- H2(g, 1atm) E ºred = 0 V

» Zn(s) + 2H+(aq) Zn2+

(aq) + H2(g) E ºcell = 0.76 V

» 0.76 V = 0 V - E ºred (anode)

• Zn2+(aq, 1M) + 2e- Zn(s) E ºred (anode) = -0.76 V

Problem:

• Calculate Eºcell for

• 2Al(s) + 3I2(s) 2Al3+(aq) + 6I-(aq)

» Anode: 2Al 2Al3+ + 6e-

» Cathode: 3I2 + 6e- 6I-

• Eºcell = E ºred (cathode) - E ºred (anode)

• E ºcell = 0.54 V - (-1.66 V)

• E ºcell = 2.20 V

• Note: stoichiometric coefficient does not affect

the value of the E ºred (it is an intensive property)

• E ºox = - E ºred

• 2Al(s) + 3I2(s) 2Al3+(aq) + 6I-(aq)

• 2Al 2Al3+ + 6e- E ºox = +1.66 V

• 3I2 + 6e- 6I- E ºred = +0.54 V

• E ºcell = E ºox + E ºred = 2.20V

• The more positive the E ºcell the more driving force

for the reaction

Oxidizing/Reducing Agents

• Oxidizing agents cause oxidation» oxidizing agents are reduced

» the more (+) the E ºred the better the ox.

agent

• Reducing agents cause reduction» reducing agents are oxidized

» the more (-) the E ºred the better the red.

agent

• Which is the better oxidizing agent?

• NO3- + 4H+ + 3e- NO + 2H2O E ºred 0.96 V

• Ag+ + e- Ag E ºred 0.80 V

• Cr2O72- + 14H+ + 6e- 2Cr3+ + H2O E ºred 1.33 V

• Which is the strongest reducing agent?

• I2 + 2e- 2I- Eºred +0.54 V

• Fe2+ + 2e- Fe Eºred -0.44 V

• MnO4- + 8H+ + 5e- Mn2+ + 4H2O Eºred +1.51 V

Spontaneity of Redox Reactions• Spontaneous redox rxns have positive

potentials• Non-spontaneous redox rxns have

negative potentials• Is this rxn spont. or non-spont.?

» MnO4- + 8H+ + 5Fe2+ 5Fe3+ + Mn2+ + 4H2O

• Fe2+ Fe3+ + 1e- Eºox = -0.77 v

• MnO4- + 8H+ + 5e- Mn2+ + 4H2O E ºred = +1.51

v

• E ºox + E ºred = + 0.74 vYes

EMF & Free Energy

• If both G & E are a measure of

spontaneity, they must be related

» G = - nFE

• F is Faraday’s constant 1 F = 96,500 J/v mol

e-

• remember: 1 C = 1 J/v

• n = mol e- transferred

» In the standard state Gº = - nFEº

• Calculate the standard free energy change

for

Hg + 2Fe3+ Hg2+ + 2Fe2+

» n = 2 mol electrons transferred

• Hg Hg2+ + 2e- Eox = - 0.854 v

• 2Fe3+ +2e- 2Fe2+ Ered= + 0.771 v

• Ecell = - 0.083 v

» G = - (2 mol e-)(-0.083 v)(96,500 J/v

mol e-)

» = + 16 kJ

Concentration & Cell EMF

• Nernst Equation

» relationship between G & concentrations

• G = Gº + RT ln Q Q = [prod]x/[react]y

» substitute -nFE for G

• E = Eº - (RT/nF) ln Q or

• E = Eº - (2.303 RT/nF) log Q

• 2.303 RT/F = 0.0592 v-mol e- at std. temp.

• E = Eº - (0.0592/n) log Q

• Calculate the emf that the following cell

generates when [Mn2+] = 0.10 M &

[Al3+] = 1.5 M 2Al + 3Mn2+ 2Al3+ +

3Mn • Eº = + 0.48 v

• E = (+ 0.48 v) - (0.0592 v/ 6) log [(1.5)2/(0.10)3]

• E = + 0.45 v

• when [Mn2+] = 1.5 M & [Al3+] = 0.10 M• E = (+ 0.48 v) - (0.0592 v/ 6) log [(0.10)2/(1.5)3]

• E = + 0.51 v

Equilibrium Constants

• Remember G = Gº + RT ln Q, if Q = K, then G = 0, therefore -nFE = 0 and

• 0 = Eº - (RT/nF) ln K or

• 0 = Eº - (0.0592/n) log K

• K can be calculated from cell potentials

• log K = nE º/0.0592

• Calculate the equilibrium constant, K, for

2IO3- + 5Cu + 12H+ I2 + 5Cu2+ + 6H2O

• Eº = + 0.858 v

• n = 10 mol e- transferred

• log K = nEº/0.0592

• log K = 145

• K = 1 x 10145

Voltaic Cells

• Lead storage battery» PbO2 + SO4

-2 + 4H+ + 2e- PbSO4 + H2O

Pb + SO42- PbSO4 + 2e-

• Ecell = + 2.041 v

• Dry cell» NH4

+ + 2MnO2 + 2e- Mn2O3 + 2NH3 + H2O

Zn Zn2+ + 2e- • In an alkaline cell the NH4Cl is replaced with KOH

• Ni-Cd

» NiO2 + 2H2O + 2e- Ni(OH)2 + 2OH-

Cd + 2OH- Cd(OH)2 + 2e-

• Fuel cells

» 4e- + O2 + 2H2O 4OH-

2H2 + 4OH- 4H2O

Electrolytic Cells• Redox reactions that are not spontaneous

• Must be driven by an outside source of electrical energy

• Cathode» reduction occurs

» by sign convention, is negative

• Anode» oxidation occurs

» by sign convention, is positive

Quantitative Aspects

• Redox reactions occur in stoichiometric relationship to the transfer of electrons

• Electrons put into a system through electrical energy, can be quantized» Coulomb = quantity of charge passing

through electrical circuit in 1 s at 1 ampere (A) current• Coulomb = (amp) (seconds)

Problem: Calculate the mass of Mg formed upon passage of a current of 60.0 A for a period of 4.00 x 10 3 s.

• MgCl2 Mg + Cl2

» Mg2+ + 2e- Mg 2Cl- Cl2 + 2e-

• we are concerned with the reduction

• (60.0 A)(4 x 103s)(1C/1 A-s) = 2.4 x 105 C

• (2.4 x 105 C)(1 mol e-/ 96,500 C) = 2.49 mol e-

• (2.49 mol e-)(1 mol Mg/2 mol e-) = 1.24 mol Mg

• (1.24 mol Mg)(24.3 g/mol) = 30.1 Mg

Electrical Work

• G = wmax G = - nFE wmax = - nFE

• Max work proportional to potential• wmax = - n F E

• J = (mol) (C/mol) (J/C)

• Electrical work = (watt) (time)• 1 watt (W) = 1 J/s or watt-s = J

• 1 kWh = 3.6 x 106 J