Transcript Chapter 14 Notes - University of Massachusetts Boston

Chapter 14 Notes
Fundamentals of Electrochemistry
Redox Reactions
• Reactions involving the transfer of
electrons from one species to another.
• LEO says GER
• Oxidation – Loss of electrons
• Reduction – Gain of electrons
Definitions
• A given species is said to be “reduced”
when it gains electrons and “oxidized”
when it loss electrons.
Example
• 2Fe3+ + Cu(s) ↔ 2Fe2+ +
Cu2+
oxidizing reducing
agent
agent
reduced
species
oxidized
species
• 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.
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
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
• 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 = 0 @ 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
Galvantic cells : A cell that uses a spontaneous
chemical reaction to generate electricity
Ex. Fig 14.3
• Cd(s)  Cd2+ + 2e-
Oxidation
2AgCl(s) + 2e- ↔ 2Ag(s) + 2ClReduction
___________________________________
Cd(s) + 2AgCl(s) ↔ Cd 2+ + 2Ag(s) + 2Cl• E is positive, spontaneous reaction
e-
+
e-
EMF
Cd(s)
Ag(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cd2+
Cl-
Cathode –where
reduction occurs
AgCl +e- →Ag(s) + Cl-
CdCl2(aq)
AgCl(s)
Cd(s) + 2AgCl(s) ↔ Cd2+ + 2Ag(s) + 2ClNote:
The two solids are separated and that the transfer of electrons
must flow through the external circuit.
Example 2
Cd(s)  Cd2+ + 2eOxidation
2Ag+ + 2e- ↔ 2Ag(s)
Reduction
____________________________________
Cd(s) + 2Ag+ ↔ Cd2+ + 2Ag(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?
e-
+
EMF
e-
Cd(s)
Ag(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+
Ag+
Cd(NO3)2(aq)
AgNO3(aq)
Cd(s) + 2Ag+ ↔ Cd2+ + 2Ag(s)
The redox reaction can take place directly on the surface of the electrodes,
without electron having to flow through circuit.
e-
+
EMF
e-
KNO3
Ag(s)
Cd(s)
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cd2+
Ag+
NO3-
Cd(NO3)2(aq)
K+
AgNO3(aq)
Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)
Need Salt Bridge to avoid energy barrier to build up of excess charge
Line Notation:
anode║cathode
Cd(s)│CdCl2(aq) ║ AgNO3(aq)│Ag(s)
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 = 298.15 K (25 C)
• E = E - 0.05916 /n log {ACcADd /AAaABb},
where n = the number of moles of electrons
transferred in the balanced redox reaction.
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 = + 1.507 V
Positive voltage means the rxn with H2 is spontaneous at
unit activities, implies Ag+ is a strong oxidizing agent.
• Cd2+ + 2e- ↔ Cd(s)
E0 = -0.402 V
Negative voltage means rxn between H+ and Cd(s) is
spontaneous at unit activities, implies Cd2+ is a very
weak oxidizing agent.
Using the Nernst Equation for example 2
Cd(s)  Cd2+ + 2eOxidation
2Ag+ + 2e- ↔ 2Ag(s)
Reduction
__________________________________
Cd(s) + 2Ag+ ↔ Cd2+ + 2Ag(s)
Ecell = Ecathode – Eanode
The Nernst equation for a half-cell reaction is
always written as a reduction!!!!!
Cd(s) + 2Ag+ ↔ Cd2+ + 2Ag(s)
EMF = (E+ - E-) - (0.05916/2)log([Cd2+]/[Ag+]2)
= Ecell - (0.05916/2)log ([Cd2+] / [Ag+]2)
[Cd2+] / [Ag+]2 = Q (the reaction quotient)
Nernst Eq as half reactions
EMF = E+ - EE+ = E+ – (0.05916)log(1/[Ag+]}
E- = E- – (0.05916/2)log(1/[Cd2+])}
The nernst Eq for both ½ rxn written as reductions
Cd2+ + 2e-  Cd(s)
2Ag+ + 2e- ↔ 2Ag(s)
anode ½ rxn
cathode ½ rxn
Gives same answer. The math is equivalent.
e-
+
EMF
e-
KNO3
Ag(s)
Cd(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+
NO3-
0.100 M
Cd(NO3)2(aq)
K+
0.100 M
AgNO3(aq)
Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)
Calculating Ecell
using the Nerst Eq
EMF =
(Ec - Ea) - (0.05916/2)log([Cd2+]/[Ag+]2)
= (0.7993–(-0.402)) –
((0.05916/2)log(0.100/(0.100)2))
= 1.172 V
e-
+
EMF = 1.172 V
e-
KNO3
Ag(s)
Cd(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+
NO3-
0.100 M
Cd(NO3)2(aq)
K+
0.100 M
AgNO3(aq)
Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)
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.
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.
e-
+
EMF = 1.172 V
e-
KNO3
Cd(s)
The EMF reads positive,
Because the negative
Terminal is connected to the
Anode and the positive
Terminal is connected to the
cathode.
Ag(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+
NO3-
0.100 M
Cd(NO3)2(aq)
K+
0.100 M
AgNO3(aq)
Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)
+
e-
-
EMF = -1.172 V
e-
If the leads are switched,
the voltage reads negative.
KNO3
Ag(s)
Cd(s)
Anode –where
oxidation occurs
Cd(s) → Cd2+ + 2e-
Cathode –where
reduction occurs
Ag+ +e- →Ag(s)
Cd2+
Ag+
NO3-
0.100 M
Cd(NO3)2(aq)
K+
0.100 M
AgNO3(aq)
Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)
Need Salt Bridge to avoid energy barrier to build up of excess charge
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)
Anode –where
oxidation occurs
Pb(s) + 2I- → PbI2(s) + 2eor
Pb(s) → Pb2+ + 2e-
e-
+
e-
EMF
Pb(s)
Ag(s)
Cd2+
Cl-
Cathode –where
reduction occurs
AgCl +e- →Ag(s) + Clor
Ag+ + e- → Ag(s)
CdCl2(aq)
PbI2(s)
2I- + Pb(s) + 2AgCl(s) ↔ PbI2(s) + + 2Ag(s) + 2Cl-
Pb(s) + 2Ag+ → 2Ag(s) + Pb2+
AgCl(s)
PbF2(s) + 2e- → Pb (s) + 2F- (aq) E0 = -0.350 V
AgCl(s) + e- → Ag(s) + Cl- (aq) E0 = 0.222 V
E+ = 0.222 -0.05916 log([Cl-]) = 0.281 V
E- = -0.350 -0.05916/2 log([F-]2) = -0.291 V
E = 0.281 – (-0.291) = 0.572 V
Second way
Pb2+ + 2e- → Pb(s)
Ag+ + e- → Ag(s)
E0 = -0.126 V
E0 = 0.7993 V
E+ = 0.7993 -0.05916 log(1/[Ag+])
E- = -0.126 -0.05916/2 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 = 0.282 – (-0.287) = 0.569 V
Adding half reactions
•Adding together two half reactions to obtain a new
half reaction. To do this it is best to convert to DGs
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)
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.
• When you add to half-rxns to obtain a third halfrxn, the safe thing to do is to add the DGs.
HOBr + H+ + e- 1/2Br2 + H2O DG1 = -F(1.584)
1/2Br2 + e-  BrDG2 = -F(1.098)
____________________________________
HOBr + 2e- + H+  Br- + H2O
DG3 = -2FE30
DG3 = DG1 + DG2 = -2F(E30)
= -F(1.584) + -F(1.098) = -2FE30
E30 = (1.584 + 1.098)/2 = 1.341 V
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 + 2OHPd+2 + 2e-  Pd(s)
Ksp = 3·10 -28
K1 = 10(nE/.05916)
= 8.9·1030
________________________________________________________
Pd(OH)2(s) +2e-  Pd (s) + 2OH- K = KspK1
K = 3·103
Log K = -nE/0.05916
E = -(0.05916/2)log(3·103) = 0.103 V