1. Fundamentals of Electrochemistry
Chapter 14 Notes
Fundamentals of Electrochemistry

2. Redox Reactions
Reactions involving the transfer of electrons from one species to another.
LEO says GER
Oxidation – Loss of electrons
Reduction – Gain of electrons

3. Definitions
A given species is said to be “reduced” when it gains electrons and “oxidized” when it loss electrons.
Example
2Fe Cu(s) ↔ 2Fe Cu2+ oxidizing reducing reduced oxidized agent agent species species

4. Electrochemistry - The study of redox chemistry
Electrochemical cells - The reactants are separated from one another, and the reaction is forced to occur via the flow of electrons through an electrical circuit.

5. But for what purpose?
produce electricity (batteries; chemical energy from spontaneous redox reactions are converted to electrical work).
To force non-spontaneous reactions to occur by supplying an external energy source.
Quantitative analysis of redox active analytes
To study the energetics and kinetics of redox processes

6. Types of electrochemical experiments
Construction of a battery (ch 14)
Potentiometry (measurement of cell voltages to extract chemical information, ex. pH meter)/ch 15
Redox titrations/ch 16
Electrogravimetric analysis (depositing analyte on an electrode/ ch 17
Coulometry (measuring the number of electrons being transferred at constant cell voltage)/ch 17
Voltammety (measuring current as a function of cell voltage, quantitative and qualitative info)/ch 17

7. Electrochemical Cells
Remind the students that they need to review the material presented in section 14.1
Electrochemical Cells
Cell voltage (E or EMF) – A measure of the spontaneity of the redox reaction.
.. DG = -nFE
E is the cell voltage (E = equilibrium)
n is the number of electrons transferred
F is Faradays constant – 9.649*104 C/mol.
Note: when the cell voltage is positive the reaction is spontaneous

8. Galvantic cells : A cell that uses a spontaneous chemical reaction to generate electricity
Ex. Fig 14.3
Cd(s)  Cd e- Oxidation
2AgCl(s) + 2e- ↔ 2Ag(s) + 2Cl- Reduction
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Cd(s) + 2AgCl(s) ↔ Cd Ag(s) + 2Cl-
E is positive, spontaneous reaction

9. Cd(s) + 2AgCl(s) ↔ Cd2+ + 2Ag(s) + 2Cl-e- e-
EMF
Cd(s)
Ag(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
AgCl +e- →Ag(s) + Cl-
Cl-
Cd2+
AgCl(s)
CdCl2(aq)
Cd(s) + 2AgCl(s) ↔ Cd Ag(s) + 2Cl-
Note: The two solids are separated and that the transfer of electrons
must flow through the external circuit.

10. Example 2 Cd(s)  Cd2+ + 2e- Oxidation 2Ag+ + 2e- ↔ 2Ag(s) Reduction
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Cd(s) + 2Ag+ ↔ Cd Ag(s)
If we set up this redox reaction in the same manner as the other one, no current flows through the circuit even though the energetics of the reaction is the same.
Why?

11. without electron having to flow through circuit.
Ag+ - + e- e-
EMF
Cd(s)
Ag(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+
Cd(NO3)2(aq)
AgNO3(aq)
Cd(s) + 2Ag+ ↔ Cd Ag(s)
The redox reaction can take place directly on the surface of the electrodes,
without electron having to flow through circuit.

12. Need Salt Bridge to avoid energy barrier to build up of excess charge- + e- e-
EMF
KNO3
Ag(s)
Cd(s)
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cd2+
Ag+ K+
NO3-
Cd(NO3)2(aq)
AgNO3(aq)
Cd(s) + 2Ag+ → Cd Ag(s)
Need Salt Bridge to avoid energy barrier to build up of excess charge

13. Cd(s)│CdCl2(aq) ║ AgNO3(aq)│Ag(s)
Line Notation:
anode║cathode
Cd(s)│CdCl2(aq) ║ AgNO3(aq)│Ag(s)

14. Nersnst Equation For the balance redox reaction: aA + bB ↔ cC + dD
The cell voltage is a function of the activities of the reactants and products (in an analogous manner in which DG is related to Q.
E = E - RT/nF ln {ACcADd /AAaABb}
at T = K (25 C)
E = E /n log {ACcADd /AAaABb},
where n = the number of moles of electrons transferred in the balanced redox reaction.

15. Standard Reduction Potentials (E) analogous to DG
The cell voltage when the activities of all reactants and products are unity.
Standard half cell reduction potentials (Table in back of book) –
E for 2H+ + 2e- ↔ H2 is arbitrarily set to 0 V.
MnO4- + 8H+ + 5e- ↔ Mn2++ 4H2O
E0 = V
Positive voltage means the rxn with H2 is spontaneous at unit activities, implies Ag+ is a strong oxidizing agent.
Cd e- ↔ Cd(s)
E0 = V
Negative voltage means rxn between H+ and Cd(s) is spontaneous at unit activities, implies Cd2+ is a very weak oxidizing agent.

16. Using the Nernst Equation for example 2
Cd(s)  Cd e- Oxidation
2Ag e- ↔ 2Ag(s) Reduction
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Cd(s) + 2Ag+ ↔ Cd Ag(s)
Ecell = Ecathode – Eanode

17. [Cd2+] / [Ag+]2 = Q (the reaction quotient)
The Nernst equation for a half-cell reaction is always written as a reduction!!!!!
Cd(s) + 2Ag+ ↔ Cd Ag(s)
EMF = (E+ - E-) - ( /2)log([Cd2+]/[Ag+]2)
= Ecell - ( /2)log ([Cd2+] / [Ag+]2)
[Cd2+] / [Ag+]2 = Q (the reaction quotient)

18. Nernst Eq as half reactions
EMF = E+ - E-
E+ = E+ – ( )log(1/[Ag+]}
E- = E- – ( /2)log(1/[Cd2+])}
The nernst Eq for both ½ rxn written as reductions
Cd e-  Cd(s) anode ½ rxn
2Ag e- ↔ 2Ag(s) cathode ½ rxn
Gives same answer. The math is equivalent.

19. - + e- e-
EMF
KNO3
Ag(s)
Cd(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+ K+
NO3-
0.100 M Cd(NO3)2(aq)
0.100 M
AgNO3(aq)
Cd(s) + 2Ag+ → Cd Ag(s)

20. Calculating Ecell using the Nerst Eq
EMF =
(Ec - Ea) - ( /2)log([Cd2+]/[Ag+]2)
= (0.7993–(-0.402)) – (( /2)log(0.100/(0.100)2))
= V

21. - + e- e-
EMF = V
KNO3
Ag(s)
Cd(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+ K+
NO3-
0.100 M Cd(NO3)2(aq)
0.100 M
AgNO3(aq)
Cd(s) + 2Ag+ → Cd Ag(s)

22. Determination of the anode and cathode
On paper
When you write an overall redox reaction in a given direction and solve the Nernst Eq accordingly, if the EMF is positive, then the reaction is spontaneous as written.
If it is negative it is spontaneous in the opposite direction. Reduction always occurs at the cathode and oxidation always occurs at the anode.

23. Determination of the anode and cathode
In the Lab
The voltmeter has a positive and a negative lead.
If you connect the negative lead to the anode and the positive lead to the cathode, the EMF will be positive.
If the EMF reads negative, it means that you have connected the negative lead to the cathode and the positive lead to the anode.
The sign of the EMF on the voltmeter display indicates the directions in which the electrons flow.

24. The EMF reads positive,
Because the negative
Terminal is connected to the
Anode and the positive
cathode. - + e- e-
EMF = V
KNO3
Cd(s)
Ag(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+ K+
NO3-
0.100 M Cd(NO3)2(aq)
0.100 M
AgNO3(aq)
Cd(s) + 2Ag+ → Cd Ag(s)

25. Need Salt Bridge to avoid energy barrier to build up of excess charge
If the leads are switched,
the voltage reads negative. + - e- e-
EMF = V
KNO3
Ag(s)
Cd(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+ K+
NO3-
0.100 M Cd(NO3)2(aq)
0.100 M
AgNO3(aq)
Cd(s) + 2Ag+ → Cd Ag(s)
Need Salt Bridge to avoid energy barrier to build up of excess charge

26. Prob 14-19 Calculate EMF for the following cell
Pb (s)│PbF2(s)│NaF(aq)(0.10 M)║ NaF(aq)(0.10 M)│AgCl(s)│Ag(s)
2 ways to approach it (the first one is less work)
PbF2(s) + 2e- → Pb (s) + 2F- (aq) and AgCl(s) + e- → Ag(s) + Cl- (aq) Or
Pb (s) → Pb2+ + 2e- and Ag+ + e- → Ag (s)

27. Anode –where
oxidation occurs
Pb(s) + 2I- → PbI2(s) + 2e- or
Pb(s) → Pb2+ + 2e- - + e- e-
EMF
Pb(s)
Ag(s)
Cathode –where
reduction occurs
AgCl +e- →Ag(s) + Cl- or
Ag+ + e- → Ag(s)
Cl-
Cd2+
AgCl(s)
CdCl2(aq)
PbI2(s)
2I- + Pb(s) + 2AgCl(s) ↔ PbI2(s) + + 2Ag(s) + 2Cl-
Pb(s) + 2Ag+ → 2Ag(s) + Pb2+

28. PbF2(s) + 2e- → Pb (s) + 2F- (aq) E0 = -0.350 V
AgCl(s) + e- → Ag(s) + Cl- (aq) E0 = V
E+ = log([Cl-]) = V
E- = /2 log([F-]2) = V
E = – (-0.291) = V

29. Second way E+ = 0.7993 -0.05916 log(1/[Ag+])
Pb2+ + 2e- → Pb(s) E0 = V
Ag+ + e- → Ag(s) E0 = V
E+ = log(1/[Ag+])
E- = /2 log(1/[Pb2+])
[Ag+] = Ksp(AgCl)/[Cl-] = 1.8E-10/0.10 = 1.8E-9
[Pb2+] = Ksp(PbF2)/[F-]2 =3.6E-8/0.01 = 3.6E-6
E = – (-0.287) = V

30. Adding half reactions
Adding together two half reactions to obtain a new half reaction. To do this it is best to convert to Gs
Problem 14.22
You must determine which reactions you must to add together to obtain the reaction in question.
HOBr  Br 2(aq)
Br2(aq)  2Br-(aq)
HOBr  2Br -(aq)

31. Balance Redox Rxns
Balance reaction between HOBr and Br- in an acidic solution
balancing the half-cell reaction:
step 1: add H+ to the reactant side and water to the product side
HOBr + H+  Br- + H2O
Step 2: Stoichiometrically balance the reaction.
It already is in this case!
Step 3: balance the charge by adding electrons
HOBr + 2e- + H+  Br- + H2O
Repeat for the reactions that you must add to obtain the above rxn.

32. When you add to half-rxns to obtain a third half-rxn, the safe thing to do is to add the DGs.
HOBr + H+ + e- 1/2Br2 + H2O DG1 = -F(1.584)
1/2Br2 + e-  Br DG2 = -F(1.098)
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HOBr + 2e- + H+  Br- + H2O DG3 = -2FE30
DG3 = DG1 + DG2 = -2F(E30)
= -F(1.584) + -F(1.098) = -2FE30
E30 = ( )/2 = V

33. Problem 14-29
You are forming a half-reaction. It is safest to add together multiply Ks (or add DGs).
Pd(OH)2(s)  Pd+2 + 2OH- Ksp = 3·10 -28
Pd+2 + 2e-  Pd(s) K1 = 10(nE/.05916) = 8.9·1030
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Pd(OH)2(s) +2e-  Pd (s) + 2OH- K = KspK1
K = 3·103
Log K = -nE/
E = -( /2)log(3·103) = V