chapter 20. overview oxidation-reduction reactions »balancing redox reactions half-reaction method...
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![Page 1: Chapter 20. Overview Oxidation-Reduction reactions »Balancing Redox Reactions Half-Reaction method Acidic Solution Basic Solution Voltaic Cells Cell EMF--standard](https://reader036.vdocuments.us/reader036/viewer/2022062320/56649d265503460f949fcaef/html5/thumbnails/1.jpg)
Chapter 20
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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
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• Effect of Concentration» Nernst Equation» Equilibrium Constants
• Commercial Voltaic Cells• Electrolysis
» Quantitative Aspects» Electrical Work
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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
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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
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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+
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• 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
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» 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
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• 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
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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
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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
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• 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
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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
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• 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
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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
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• 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
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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
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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
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• 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
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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
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• 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
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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
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• 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
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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
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• Ni-Cd
» NiO2 + 2H2O + 2e- Ni(OH)2 + 2OH-
Cd + 2OH- Cd(OH)2 + 2e-
• Fuel cells
» 4e- + O2 + 2H2O 4OH-
2H2 + 4OH- 4H2O
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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
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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)
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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
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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