1 Electrochemistry Electrochemistry Generating Voltage (Potential)
Electrochemistry
description
Transcript of Electrochemistry
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