ELECTROCHEMISTRY

By Ethan Foreman & Janani Raman

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)

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.

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

SAMPLE PROBLEM TIME!

What are the oxidation numbers of each element in the following molecular compound and polyatomic ion?

Pb(NO3)2

Cr2O72-

SAMPLE PROBLEM ANSWER TIME!

Pb(NO3)2

Pb: +2N: +5O: -2

Cr2O72-

Cr: +6O: -2

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.

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

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-

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

Oxidation/Reduction Reaction(in basic solution)

Fe(OH)2+ CrO42- Fe2O3 + Cr(OH)4

-

Balance This!

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-

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

-

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

Look at the Picture!!!

From: www.sparknotes.com

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

Is the reaction spontaneous?

EO ΔG K

SPONTANEOUS + - >>1

NON-SPONTANEOUS - + <<1

EQUILIBRIUM 0 0 1

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!

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

WHIN? (What Help is Needed)

© Juan Mazzini & Liza Cohen

© Mr. John Charles Bruss

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