Electrochemistry

The first of the

BIG FOUR

Introduction of Terms

Electrochemistry- using chemical changes to produce an electric current or using electric current to produce a chemical change

Current- movement of electrons or a transfer of electrons

Conductor- contains charge carriers1. Metallic conductor- wire where the metal ions

float around the electrons2. Electro - ions present from ionization or

dissociation of a dissolved electrolyte.

Voltaic CellsA. Definition: An electrochemical cell in

which a spontaneous reaction generates an electric current.

B. This type of cell is spontaneous.

C. Consists of two half cells; a portion of the cell contains one half of the reaction

D. Oxidation - ANODE “an ox”

Reduction - CATHODE “red cat”

Illustration

Galvanic Cell Animation

Notation for Voltaic CellsA. Anode || Cathode (the double line represents the

salt bridgeB. Reactants, products || reactants, productsC. | - represents a phase boundaryD. Examples:

1. Zn(s) | Zn2+(aq) || Cu2+ (aq)| Cu(s)

2. Zn(s) | Zn2+(aq)|| H+ (aq) | H2 (g) | Pt

3. Tl + Sn2+ → Tl 3++ Sn4. Zn + Fe3+ → Zn2+ + Fe2+

E. A gas is involved, Pt is usually the electrode

1. || Cl2 (g) | Cl- | Pt

Notation for Voltaic Cells

2 H+(aq) + 2e- H2(g)

Zn(s) Zn2+ + 2e-

Tl(s) │Tl+(aq) ║Sn2+ │Sn(s)

Zn(s) │Zn2+(aq) ║Fe3+

(aq), Fe2+(aq) │Pt

Electromotive Force (EMF)A. Definition- The maximum potential difference

between the electrodes of a voltaic cell.

B. Wmax = ∆G = -nFEcell

where F is a faraday = 9.6485 x 104 C

∆G is Gibb’s Free Energy (thermodynamics)

C. The maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell.

D. Example 1: Cu2+ (aq) + Fe (s) → Cu (s) + Fe2+

Answer: -1.5 x 105 J

Example 2

The emf of a voltaic cell with the reaction: Hg2

2+ (aq) + H2 ↔ 2 Hg(l) + 2 H+ (aq)

is 0.650 V. Calculate the maximum electrical work of this cell when 0.500 g of H2 is consumed.

Answer: -2 x 96, 480 x 0.650 = -1.25 x 105 J

.500g (1/2.02)(-1.25 x 105J/mol)= -3.09 x 104J

Standard Cell EMF’s

A. A cell emf is a measure of the driving force of the cell reaction.

B. Anode + Cathode1. Ox potential- a measure in V of the tendency

for a species to lose electrons in the oxidation half reaction

2. Red. Potential- a measure in V of the tendency for a species to gain electrons in the reduction half reaction.

Calculating Ecell

1. Equation: Ecell = Eox + Ered

2. Turning Ered into a Eox

a) Reverse the half cell reaction (to a ox. Rxn)

i. Sn2+ + 2e- → Sn

Sn → Sn2+ + 2e-

b) Reverse the sign on the potential valuei. +0.14 V

Strengths of Oxidizing & Reducing Agents

1. Oxidizing Agents – these are the substances reduced, so the strongest oxidizing agents have the highest positive Ered. (F2 is the strongest oxidizing agent while Li+ is the weakest oxidizing agent).

2. Reducing Agents- these are the substances being oxidized so they appear on the right side of the yield sign. The strongest reducing agents correspond with the highest negative Ered. (Li is the strongest reducing agent and F- is the weakest.

Examples

3. Fe3+, Cl2, H2O2

4. Cu, H2, Al

Nernst Equation1. Calculate cell potentials when

conditions are not standard

2. Calculate cell potentials of half reactions

3. Predict whether the cell voltage will decrease or increase depending on Q.

The Equation

cellEE Qn

log0592.0

Controls the voltage of the cell

Q and E are inversely relatedQ>1 : then the log value is positive and E decreases

Q<1 : then the log value is negative and E increases

Electrolytic Cells

A. Definition: An electrochemical cell in which an electric current drives an otherwise nonspontaneous reaction.

B. Electrolysis- the process by which a direct electrical current causes a chemical reaction to occur.

1. Electrolysis of MOLTEN Salts (Down’s Cell)

Anode Cathode

Withdrawing electrons so it’s positive

Withdrawing electrons so it’s positive

ReductionOxidation

The battery acts as an electron pump, pulling electrons from the anode and pushing them to the cathode.

Adding electrons so it’s negative

Adding electrons so it’s negative

2 Cl- → Cl2 + 2 e- 2 Na+ + 2 e- → 2 Na

Overall cell reaction:

2Na+ + 2Cl- → 2Na + Cl2

Overall cell reaction:

2Na+ + 2Cl- → 2Na + Cl2

Examples:

a) KCl

Anode: 2 Cl- → Cl2 + 2 e-

Cathode: 2 e- + 2 K+ → 2 K

KOHAnode: 4 OH- → O2 + 2H2O + 4 e-

Cathode: 4 K+ + 4 e- → 4 K

2. Electrolysis of Aqueous Solutions

a. There are two reactions possible at each electrode for an aqueous solution.

Water can be oxidized to O2 at the anode:

Anode: 2 H2O → O2 + 4 H+ + 4 e-

♦ Water can be reduced to H2 at the cathode:

Cathode: 2 H2O + 2 e- → H2 + 2 OH-

b. Chloride SolutionsAt the anode there are 2 possible half reactions:

2 Cl- → Cl2 + 2 e- -1.36 V 2 H2O → O2 + 4 H+ + 4 e- -1.23 V

At the cathode there are 2 possible half reactions:

2 Na+ + 2 e- → 2 Na -2.71 V

2 H2O + 2 e- → H2 + 2 OH- -0.83 V

Overall Balanced Reaction:

2 H2O + 2 Cl- → Cl2 + H2 + 2 OH-

Since O2 has the more positive potential, we would expect to see O2 produced at the anode. This does not happen. There is a much higher voltage needed to produce the O2, in excess of the expected value (overvoltage). The overvoltage is greater than the production of Cl2 so the chlorine is produced at the anode.

Since O2 has the more positive potential, we would expect to see O2 produced at the anode. This does not happen. There is a much higher voltage needed to produce the O2, in excess of the expected value (overvoltage). The overvoltage is greater than the production of Cl2 so the chlorine is produced at the anode.

c. Sulfuric acid solutions/ SO42- ions

Possible Anode Reactions:

2 SO42- → S2O8

2- + 2 e- -2.71 V

2 H2O → O2 + 4 H+ + 4 e- -1.23 V

Possible Cathode Reactions:

2 H+ + 2 e- → H2

2 H2O + 2 e- → H2 + 2 OH-

Overall Balanced Reaction: 2 H2O → O2 + 2 H2

In this case, the sulfate is not as easily oxidized as chloride in the previous example, so this time the water is oxidized as expected.

In this case, the sulfate is not as easily oxidized as chloride in the previous example, so this time the water is oxidized as expected.

Examples1. Half reactions of CuSO4 (aq)

Anode:

2 SO42- → S2O8

2- + 2 e- -2.71 V

2 H2O → O2 + 4 H+ + 4 e- -1.23 V

Cathode:

Cu2+ + 2e- → Cu +0.34 V 2 H2O + 2 e- → H2 + 2 OH- -0.83V

Overall Balanced Reaction:

2 Cu2+ + 2 H2O → 2 Cu + O2 + 4H+

(acidic)

Examples2. Aqueous AgNO3 (NO3

- is not oxidized)

Anode:

2 H2O → O2 + 4 H+ + 4 e- -1.23 V

Cathode:

Ag+ + e- → Ag +0.80 V

2 H2O + 2 e- → H2 + 2 OH- -0.83V

Overall Balanced Reaction:

4 Ag+ + 2 H2O → Ag + O2 + 4 H+ (acidic)

*NOTE:It is difficult to reduce elements

in group IA (alkali metals) in the presence of water (II A too!). You have the formation of H2 instead.

Answers to Homework Anode: 2 Cl- Cl2 + 2e-

Cathode: 2e- + Ca2+ Ca

Anode: 4 OH- → O2 + 2H2O + 4 e-

Cathode: 4e- + 4 Cs+ 4 Cs

Anode: 2 H2O → O2 + 4 H+ + 4 e-

Cathode: 2 H2O + 2 e- → H2 + 2 OH-

Overall Balanced Reaction:

2 H2O → O2 + 2 H2

Answers to Homework

Anode: 2 Br- Br2 + 2e-

Cathode: 2 H2O + 2 e- → H2 + 2 OH-

Overall Reaction:2 Br- + 2 H2O Br2 + H2 + 2 OH-

Stoichiometry of Electrolysis

A. Electrical Units1. ampere- base unit of current (rate of flow)

2. Coulomb – SI unit of electrical charge (amp•sec): a current of 1 amp flowing for 1 second.

3. Faraday- 1 mole of electrons = 6.02x1023 e-

1F = 96,500 C (96,480 book #) = 96,500 amp•sec

Examples

1.

2.

3.

Example problems:

1. How many grams of copper is plated out when a current of 10.0 amps is passed for 30.0 minutes?

mol

gX

molE

molCuX

C

molEXXX

amps

1

546.63

0.2

1

500,96

1

min1

sec0.60

1

min0.30

1

0.10

2. How long must a current of 5.00 A be applied to a solution of Ag+ to produce 10.5 g silver metal?

AX

C

ampX

molE

CX

molAg

molEX

g

molAgX

g

00.5

1

1

sec1

1

500,96

1

1

868.107

1

1

5.10