Determine the oxidation number for each atom in the following molecules

1.H2S

2.P2O5

3.S8

4.SCl2

5.Na2SO3

6. SO4-2

7. NaH 8. Cr2O7

-2

9. SnBr4

10. Ba(OH)2

A

Intersection 14

12/05/06Electrochemistry

19.9-19.13 p 941-955

A

December in Studio

S M Tu W Th F S12/5Exam 3

12/6Studio

12/8 Polymers; check out

12/11 Poster session, paper due

12/12 final IS

12/13 In-class assignment

12/17 Review session 7-9 pm

12/19 Final exam 8-10 am

A

Watershed Poster Session

• Monday, December 11 in USB 2165• Board (4 ft x 4ft), easel, pins• Set up by 1:10 and 3:10• One person stationed at poster; others

evaluate• Rubric available• Paper due same time

A

Last In-Class Assignment

• Wednesday, December 13th in studio• Available on-line• Read papers before coming to class; bring

them with you. • May make any notes you like on the papers• Goal: to evaluate scientific method and

data

A

Outline• Ed’s demos• Balancing Redox Reactions• Electrochemistry

– Electrochemical cells and Standard Hydrogen Electrodes

– Nernst– Quantifying current

A

Ed’s Demos

A

Oxidation States of Vanadium: Reduction of V5+ to V+2

• Reaction 1– Zn (s) + 2 VO3

- (aq) + 8 H3O+ (aq) ↔ 2 VO2+ (aq) + Zn+2 (aq) + 12 H2O (l)

• Reaction 2– Zn (s) + 2 VO2

+ (aq) + 8 H3O+ (aq) ↔ 2 V3+ (aq) + Zn+2 (aq) + 6 H2O (l)

• Reaction 3– Zn (s) + 2 V3+ (aq) ↔ 2 V2+ + Zn+2 (aq)

V+5 (aq) → V+4 (aq) yellow to greenV+4 (aq) → V+3 (aq) green to blueV+3 (aq) → V+2 (aq) blue to violet

A

Oxidation States of Manganese: Mn+7, Mn+6, Mn+4, and Mn+2

• +7 (purple) to +2 (colorless)– 2 MnO4

- (aq) + H+ (aq) + 5 HSO3- (aq) ↔ 2 Mn+2 (aq) + 5 SO4

-2 (aq) + 3 H2O(l)

• + 7 (purple) to +4 (brown)– OH- + 2 MnO4

- (aq) + 3 HSO3- (aq) ↔ 2 MnO2 (s) + 3 SO4

-2 (aq) + 2 H2O(l)

• + 7 (purple) to + 6 (green)– 2 MnO4

- (aq) + 3 OH- + HSO3- (aq) ↔ 2 MnO4

-2(aq) + SO4-2 (aq) + 2 H2O(l)

A

Thinking back….

• What happened when Na(s) was added to water?

Na(s) + H2O(l) Na+ (aq) + H2(g) + OH-

(aq)

• Determine the oxidation state of each reactant and product

• What was oxidized?• What was reduced?

A

Balancing Redox Reactions

When KMnO4 (potassium permanganate) is mixed with Na2C2O4 (sodium oxalate) under acidic conditions, Mn+2(aq) ions and CO2(g) form.

The unbalanced chemical equation is:KMnO4(aq) + Na2C2O4(aq) Mn+2

(aq) + CO2(g) + K+(aq) + Na+

(aq)

K+ and Na+ are spectator ions, so we can ignore them at this point.MnO4

- (aq) + C2O4

-2(aq) Mn+2

(aq) + CO2(g)

M

Half-Reactions

• Reduction reaction

• Oxidation reaction

MnO4- (aq) + C2O4

-2(aq) Mn+2

(aq) + CO2(g)

M

Reduction reaction

Step 1: Balance all elements other than oxygen and hydrogen. Step 2: Balance the oxygens by adding water. Step 3: Balance the hydrogens using H+

Step 4: Balance the electrons Mn+7 on reactants sideMn+2 on products side

Step 5: Check charge balance and elemental balance

MnO4- Mn+2

M

Oxidation reaction

C2O4-2 CO2

M

Combine Half Reactions

5 e- + 8H+ + MnO4- Mn+2 + 4 H2O

C2O4-2 2 CO2 + 2e-

We are assuming the reaction takes place under acidic conditions!

M

Balancing in Base

5 e- + 8H+ + MnO4- Mn+2 + 4 H2O

C2O4-2 2 CO2 + 2e-

Change H+ to water by adding OH- to each side

M

M

Electrochemical Cells

A

Definitions• Electrochemical cell: A combination of anode,

cathode, and other materials arranged so that a product-favored redox reaction can cause a current to flow or an electric current can cause a reactant-favored redox reaction to occur

• Voltaic cell (battery): An electrochemical cell or group of cells in which a product-favored redox reaction is used to produce an electric current.

• Galvanic cell: A cell in which an irreversible chemical reaction produces electrical current

• Electrolytic cell: electrochemical reactions are produced by applying electrical energy

A

A Copper-Zinc battery – What Matters?Consider reduction potentials:

Cu+2 + 2e- → Cu(s) 0.3419 VZn+2 + 2e- → Zn(s) -0.7618 V

Place Zn electrode in copper sulfate solution – What happens?

Cu+2 + 2e- → Cu(s) 0.3419 VZn(s) → Zn+2 + 2e- 0.7618 V

Cu+2 + Zn(s) → Zn+2 + Cu(s) 1.1 V E > 0, spontaneous

Note, no need for electron to flow external to cell for reaction to occur!!

Copper is plated on Zn electrode

A

A Copper-Zinc battery – What Matters?Consider reduction potentials:

Cu+2 + 2e- → Cu(s) 0.3419 VZn+2 + 2e- → Zn(s) -0.7618 V

Place Cu electrode in zinc sulfate solution – What happens?

Cu(s) → Cu+2 + 2e- -0.3419 VZn+2 + 2e- → Zn(s) -0.7618 V

Zn+2 + Cu(s) → Cu+2 + Zn(s) -1.1 V E < 0, not spontaneous

No reaction occurs !!

Zn doesn’t plate on copper electrode?!

A

Fig. 19-3, p.918

A

                                                                             

              

What are the ½ reactions?What is the overall reaction?

Identify the oxidation, reduction, anode, and cathode

A

Fig. 19-7, p.922

SHE: Standard Hydrogen Electrode

2 H3O+(aq, 1.00 M) + 2e- <-> H2(g, 1 atm) + 2H2O(l)

Eo = 0V

Standard conditions:1M, 1atm, 25oC

A

Measuring Relative Potentials

                                                            

               

Table of Standard Reduction Potentials

A

Standard Reduction Potentials

What is the standard potential of a Au+3/Au/Mg+2/Mg cell?

A

The half-reaction with the more positive standard reduction potential occurs at the cathode as reduction.

The half-reaction with the more negative standard reduction potential occurs at the anode as oxidation.  

A

Picture from: www.corrosion-doctors.org/ Biographies/images/

Nernst

M

Is potential always the same?

Standard conditions: 1 atm, 25oC, 1 M

What will influence the potential of a cell?

M

Mathematical Relationships: Nernst

The Nernst Equation:  Eo = standard potential of the cellR = Universal gas constant = 8.3145 J/mol*KT = temperature in Kelvinn = number of electrons transferredF = Faraday’s constant = 96,483.4 C/mol Q = reaction quotient (concentration of anode divided by the concentration of the cathode)

E = Eo - RT ln Q nF

Cu+2 + Zn(s) → Zn+2 + Cu(s) Q =

M

Applying the Nernst Equation

This cell is operating at 25oC with 1.00x10-5M Zn2+ and 0.100M Cu2+?

Predict if the voltage will be higher or lower than the standard potential

E = Eo - RT ln Q nF

Cu+2 + Zn(s) → Zn+2 + Cu(s)

M

Eo = standard potential of the cell

R = Universal gas constant = 8.3145 J/mol*K

T = temperature in Kelvin

n = number of electrons transferred

F = Faraday’s constant = 96,483.4 C/mol

Q = reaction quotient (concentration of anode divided by the concentration of the cathode)

E = Eo - RT ln Q nF Zn+2 + 2e- -> Zn -0.76 V

Cu+2 + 2e- -> Cu 0.34 V

25oC + 273 = Kn =

1.00x10-5M Zn2+ and 0.100M Cu2

Cu+2 + Zn(s) → Zn+2 + Cu(s)

Q = [Zn+2]/[Cu+2]

M

Were your predictions correct?

                                                          

   

M

Picture from: musee-ampere.univ-lyon1.fr/

Ampere

A

The Units of Electrochemistry

• Coulomb– 1 coulomb equals 2.998 x 109 electrostatic units (eu)– eu is amount of charge needed to repel an identical charge 1

cm away with unit force– Charge on one electron is -1.602 x 10-19 coulomb

Problem: An aluminum ion has a +3 charge. What is this value in coulombs?

magnitude of charge is same at that of e-, opposite sign3 x 1.602 x10-19 = 4.806 x 10-19 coulomb

Key Point: electrons or ions charges can be measured in coloumbs

A

The Units of Electrochemistry• Ampere

– Amount of current flowing when 1 coulomb passes a given point in 1 second

– Units of Amperes are Coulombs per second– Current (I) x time (C/s x s) gives an amount of charge.

Problem: How much current is flowing in a wire in which 5.0 x 1016 electrons are flowing per second?

The charge transferred each second = (5.0 x 1016 electrons/sec) x (1.602 x 10-19 coulomb/electron)= 8.0 x 10-3 coulombs/sec = amps

A

The Units of Electrochemistry• Ampere

– Amount of current flowing when 1 coulomb passes a given point in 1 second

– Units of Amperes are Coulombs per second– Current (I) x time (C/s x s) gives an amount of charge.– We can express electron or ION flow in amps!

Problem: If 1 mol Al+3 ions passes a given point in one hour, what is the current flow?

1 hour = 3600 seconds; Al+3 charge is 4.806 x 10-19 coulomb

1 mol Al +3 ions 6.022 x 1023 Al +3 ions 4.806 x 10-19 coulomb 1 hour

Hour 1 mol Al +3 ions 1 Al+3 ion 3600 sec= 80 C/s = 80 A

A

ElectrolysisThe process of using electrical current to drive a redox reaction that is not spontaneous. Balance the reaction in acid.

At the anode (where oxidation occurs):

H2O (l) → O2 (g)

At the cathode (where reduction occurs):

H3O+ (aq) → 2 H2 (g)

Overall reaction:

M

Overall reaction: 2 H2O (l) → 2 H2(g) + O2 (g)

p.951Overall reaction: 2 H2O (l) → 2 H2(g) + O2 (g)

M

Standard Reduction Potentials

What is the standard potential for the electrolysis of water?

M

Electrolysis ReactionsHow many coulombs of charge are required to electrolyze 1 mol of water?

Note that 6 water are consumed at the anode, but 4 are produced at the cathode.At the anode (where oxidation occurs): 6 H2O (l) → O2 (g) + 4 H3O+ (aq) + 4e-

At the cathode (where reduction occurs): 4 H3O+ (aq) + 4 e- → 2 H2 (g) + 4 H2O (l)

The net amount of water consumed per every 4 electrons is 2 molecules.

1.0 mol H2O 4 mol e- 96,487 coulombs

2 mol H2O 1 mol e-

= 190000 C

M

Electrolysis Reactions

A power supply puts out a maximum current of 10 amps. How long will it take to electrolyze 18 g (1 mole) water?

190,000 coulombs

sec hour

10 coulombs 3600 secs= 5.4 hours

M

ElectrolysisWhat similarities do you find between voltaic

cells and electrolysis?

What differences?

Overvoltage: However, the potential that must be applied to electrolysis cell is always greater than that calculated from standard reduction potentials.  This excess potential is called an overvoltage. This additional voltage is needed to overcome limitations in electron transfer rate at the interface between electrode and solution.

M