electrochemistry by ethan foreman & janani raman
TRANSCRIPT
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ELECTROCHEMISTRY
By Ethan Foreman & Janani Raman
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Basic Constants and Equations to Remember!
• F=Faraday Constant=96,485 Coulombs/mol• E=Joules/Coulomb• ΔGo=-nFEo
• R=8.314 J/(mol*K)• ΔG=ΔGo+ RT ln(Q)• E=Eo – (RT)/(nF)*lnQ• Current=I=Charge/Time=q/t• Measured in Amps (1A=1Coulomb/Sec)
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What are the oxidation numbers (ox#) for the following categories?
• Atoms in elemental form: 0– Example: H atom in H2 molecule has ox# 0
• Monatomic Ions: oxidation number is the charge– Ex. Na+ (ox#=+1), Cl- (ox=-1)
• Nonmetals: generally negative ox#– Oxygen: ox#=-2 except with peroxides, which have O2
2-, in which case each oxygen has an ox# of -1
– Hydrogen: +1 when bonded to nonmetals, but -1 when bonded to metals
– Halogens: Fluorine is always -1, and the other halogens are -1 except when combined with oxygen, in which case they have positive oxidation states.
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The Super Duper Oxidation Rule!
• A neutral compound’s oxidation number must be 0, and this must be the sum of all the oxidation numbers of the atoms in the compound
• A polyatomic ion’s oxidation number must equal the charge of the ion, and this must be the sum of the oxidation numbers of the atoms in the compound
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SAMPLE PROBLEM TIME!
What are the oxidation numbers of each element in the following molecular compound and polyatomic ion?
Pb(NO3)2
Cr2O72-
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SAMPLE PROBLEM ANSWER TIME!
Pb(NO3)2
Pb: +2N: +5O: -2
Cr2O72-
Cr: +6O: -2
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Changes in Ox# (during RXNs)
• An ox# increase indicates a loss of electrons, meaning the substance is being oxidized; this substance is the reducing agent.
• An ox# decrease indicates a gain of electrons, meaning the substance is being reduced; this substance is the oxidizing agent.
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Oxidation/Reduction Half Reactions (in acidic solution)
MnO4- + _______ Mn2+ + _______
MnO4- + _______ Mn2+ + 4H2O
MnO4- + 8H+ +___ Mn2+ + 4H2O
MnO4- + 8H+ + 5e- Mn2+ + 4H2O
Ox# of Mn Before Reaction: +7Ox # of Mn After Reaction: +2
Ox# decrease indicates a GAIN of electrons; therefore, Mn undergoes REDUCTION
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Oxidation/Reduction Half Reactions(in acidic solution)
C2O42- 2CO2
+ _______
Ox# of C Before Reaction: +3Ox # of C After Reaction: +4
Ox# increase indicates a LOSS of electrons; therefore, C undergoes OXIDATION
C2O42- 2CO2 + 2e-
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Oxidation/Reduction Net Reaction(in acidic solution)
+[C2O42- 2CO2 + 2e-] * 5
[MnO4- + 8H+ + 5e- Mn2+ + 4H2O] * 2
2MnO4- + 16H+ + 10e- +5C2O4
2- 2Mn2+ + 8H2O + 10CO2 + 10 e-
Net Reaction:2MnO4
- + 16H+ 5C2O42- 2Mn2+ + 8H2O + 10CO2
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Oxidation/Reduction Reaction(in basic solution)
Fe(OH)2+ CrO42- Fe2O3 + Cr(OH)4
-
Balance This!
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Oxidation/Reduction Reaction(in basic solution)
Fe(OH)2 Fe2O3
2Fe(OH)2 Fe2O3 + H2O + 2H+ + 2e-
CrO42- Cr(OH)4
-
CrO42-
+ 4H+ + 3e- Cr(OH)4-
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Oxidation/Reduction Reaction(in basic solution)
[2Fe(OH)2 Fe2O3 + H2O + 2H+ + 2e-] *3 +[CrO4
2- + 4H+ + 3e- Cr(OH)4-] *2
2OH- + 6Fe(OH)2 + 2CrO42- + 8H+ + 6e-
3Fe2O3 + 3H2O + 6H+ + 6e- + 2Cr(OH)4- + 2OH-
Net Reaction:6Fe(OH)2 + 2CrO4
2- 3Fe2O3 + 1H2O + 2OH- + 2Cr(OH)4
-
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Electricity• Anode: substance being oxidized• Cathode: substance being reduced
– (A hint we picked up from last year’s group: remember VOWELS WITH VOWELS)
• Electrons flow from anode to cathode; ions flow from cathode to anode
• A reaction with greater potential is reduced• A reaction with smaller potential has the potential sign changed
(don’t multiply by moles because voltage is independent of moles)• E◦cell= E◦ox + E◦red• Oxidized means mass decreases• Reduced means mass increases
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Look at the Picture!!!
From: www.sparknotes.com
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Which reaction is oxidized and which is reduced?
Cu2+ +2e- Cu E=+.34 VAl3+ + 3e- Al E=-1.66 V
Cu2+ +2e- Cu E=+.34 V REDUCED Al Al3+ + 3e- E=+1.66 VOXIDIZED
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Is the reaction spontaneous?
EO ΔG K
SPONTANEOUS + - >>1
NON-SPONTANEOUS - + <<1
EQUILIBRIUM 0 0 1
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One Last Problem!
• A battery pushes 20.0 mA through a Ni-Cd battery for 4.00 hours. How much Cd metal is deposited on the electrode?
• Cd2+ +2e- Cd
GO!
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Solution to Last Problem!
• 4.00 hr * (3600 sec / 1 hr)= 14400 sec• 14400 sec * (.020 C / 1 sec)= 288 C• 288 C * (1 mol e- / 96485 C)= 0.0029849… e-
• .00298… e- (1 mol Cd / 2 mol e-)= .00149… mol Cd• .00149…mol Cd (112.41 g Cd / 1 mol Cd)= .1677…g Cd• ≈ .168 g Cd
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WHIN? (What Help is Needed)
© Juan Mazzini & Liza Cohen
© Mr. John Charles Bruss
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Picture Citations
• Voltorb, thanks to www.geocities.com• Zapdos, thanks to www.vaporeoncave.fanspace.com• Harry Potter, thanks to www.weeklyreader.com• Pikachu, thanks to www.gifninja.com• Superman, thanks to www.contrib.andrew.cmu.edu• John Bruss, thanks to www.dist113.org