Electrochemistry

Electrochemistry Terminology #1

Oxidation – A process in which an element attains a more positive oxidation state

Na(s) Na+ + e-

Reduction – A process in which an element attains a more negative oxidation state

Cl2 + 2e- 2Cl-

Electrochemistry Terminology #2

Gain Electrons = Reduction

An old memory device for oxidation and reduction goes like this… LEO says GER

Lose Electrons = Oxidation

Electrochemistry Terminology #3

Oxidizing agent**The substance that is reduced is

the oxidizing agent Reducing agent**

The substance that is oxidized is the reducing agent

**Terminology of reducing agent and oxidizing agent have been excluded from the course.

Electrochemistry Terminology #4

Anode The electrode

where oxidation occurs

CathodeThe electrode

where reduction occurs

Memory device:

Reductionat the

Cathode

Table of Reduction Potentials

Measured against

the StandardHydrogenElectrode

Measuring Standard Electrode Potential

Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.

Galvanic (Electrochemical) Cells

Spontaneous redox processes

have:A positive cell potential, E0

A negative free energy change, (-G)

Zn - Cu Galvanic

Cell

Zn2+ + 2e- Zn E = -0.76V

Cu2+ + 2e- Cu E = +0.34V

From a table of reduction potentials:

Zn - Cu Galvanic

Cell

Cu2+ + 2e- Cu E = +0.34V

The less positive, or more negative reduction potential becomes the oxidation… Zn Zn2+ + 2e- E =

+0.76VZn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Line Notation

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

An abbreviated representation of an electrochemical cell

Anodesolution

Anodematerial

Cathodesolution

Cathodematerial| |||

Calculating G0 for a Cell

0 (2 )(96485 )(1.10 )coulombs JoulesG mol emol e Coulomb

G0 = -nFE0

n = moles of electrons in balanced redox equationF = Faraday constant = 96,485 coulombs/mol e-

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

0 212267 212G Joules kJ

The Nernst Equation

0 ln( )RTE E QnF

Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M.

R = 8.31 J/(molK) T = Temperature in K

n = moles of electrons in balanced redox equationF = Faraday constant = 96,485 coulombs/mol e-

Excluded from the AP Chemistry Course!

Nernst Equation Simplified

0 0.0591log( )E E Qn

At 25 C (298 K) the Nernst Equation is simplified this way:

Equilibrium Constants and Cell Potential

At equilibrium, forward and reverse reactions occur at equal rates, therefore:1. The battery is “dead”

2. The cell potential, E, is zero volts

0 0.05910 log( )volts E Kn

Modifying the Nernst Equation (at 25 C):

Zn + Cu2+ Zn2+ + Cu E0 = + 1.10 V

Calculating an Equilibrium Constant from a Cell Potential

0.05910 1.10 log( )2

volts K

(1.10)(2) log( )0.0591

K

37.2 log( )K

37.2 3710 1.58 10K x

Concentration Cell

Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.

Both sides have the same

components but at different

concentrations.

???

Concentration Cell

Both sides have the same

components but at different

concentrations.

The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentrationZn2+ (1.0M) + 2e- Zn (reduction)

Zn Zn2+ (0.10M) + 2e- (oxidation)

???

CathodeAnode

Zn2+ (1.0M) Zn2+ (0.10M)

Concentration Cell

0 0.0591log( )E E Qn

Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C).

Both sides have the same

components but at different

concentrations.

???

CathodeAnode

Zn2+ (1.0M) Zn2+ (0.10M)

Concentration Cell

0 0.0E Volts (0.10)(1.0)

Q2n

0.0591 0.100.0 log( ) 0.0302 1.0

E Volts

Nernst CalculationsZn2+ (1.0M) Zn2+ (0.10M)

0 0.0591log( )E E Qn

Electrolytic

Processes

A negative cell potential, (-E0)

A positive free energy change, (+G)

Electrolytic processes are NOT spontaneous. They have:

Electrolysis of Water

2 22 4 4H O O H e

2 24 4 2 4H O e H OH

In acidic solution

Anode rxn:Cathode rxn:

-1.23 V-0.83 V

-2.06 V2 2 22 2H O H O

Electroplating of Silver

Anode reaction:Ag Ag+ + e-

Electroplating requirements:1. Solution of the plating metal

3. Cathode with the object to be plated2. Anode made of the plating metal

4. Source of current

Cathode reaction:Ag+ + e- Ag

Solving an Electroplating Problem

Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO3 using a 20.0 Ampere current?

5.0 g

Ag+ + e- Ag

1 mol Ag107.87 g

1 mol e-

1 mol Ag96 485 C1 mol e-

1 s20.0 C

= 2.2 x 102 s