Download - Electrochemistry
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Electrochemistry
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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-
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Electrochemistry Terminology #2
Gain Electrons = Reduction
An old memory device for oxidation and reduction goes like this… LEO says GER
Lose Electrons = Oxidation
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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.
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Electrochemistry Terminology #4
Anode The electrode
where oxidation occurs
CathodeThe electrode
where reduction occurs
Memory device:
Reductionat the
Cathode
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Table of Reduction Potentials
Measured against
the StandardHydrogenElectrode
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Measuring Standard Electrode Potential
Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.
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Galvanic (Electrochemical) Cells
Spontaneous redox processes
have:A positive cell potential, E0
A negative free energy change, (-G)
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Zn - Cu Galvanic
Cell
Zn2+ + 2e- Zn E = -0.76V
Cu2+ + 2e- Cu E = +0.34V
From a table of reduction potentials:
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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
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Line Notation
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
An abbreviated representation of an electrochemical cell
Anodesolution
Anodematerial
Cathodesolution
Cathodematerial| |||
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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
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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!
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Nernst Equation Simplified
0 0.0591log( )E E Qn
At 25 C (298 K) the Nernst Equation is simplified this way:
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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):
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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
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Concentration Cell
Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.
Both sides have the same
components but at different
concentrations.
???
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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)
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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
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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
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Electrolytic
Processes
A negative cell potential, (-E0)
A positive free energy change, (+G)
Electrolytic processes are NOT spontaneous. They have:
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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
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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
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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