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 reduced oxidized agent agent species 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.• .. G = -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) + 2Cl- Reduction

___________________________________

Cd(s) + 2AgCl(s) ↔ Cd 2+ + 2Ag(s) + 2Cl-

• E is positive, spontaneous reaction

CdCl2(aq)

- +

EMF

AgCl(s)

Ag(s)Cd(s)

Cd2+Cl-

e- e-

Cd(s) + 2AgCl(s) ↔ Cd2+ + 2Ag(s) + 2Cl-

Note: The two solids are separated and that the transfer of electrons must flow through the external circuit.

Anode –where oxidation occursCd(s) → Cd2+ + 2e- Cathode –where

reduction occursAgCl +e- →Ag(s) + Cl-

Example 2

Cd(s) Cd2+ + 2e- Oxidation

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?

Cd(NO3)2(aq)

- +

EMF

Ag(s)Cd(s)

Cd2+Ag+

e- e-

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.

AgNO3(aq)

Ag+

Anode –where oxidation occursCd(s) → Cd2+ + 2e-

Cathode –where reduction occursAg+ +e- →Ag(s)

Cd(NO3)2(aq)

- +

EMF

Ag(s)Cd(s)

Cd2+Ag+

e- e-

Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)

AgNO3(aq)

Need Salt Bridge to avoid energy barrier to build up of excess charge

KNO3

K+NO3-

Anode –where oxidation occursCd(s) → Cd2+ + 2e-

Cathode –where reduction occursAg+ +e- →Ag(s)

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 G is related to Q.

• E = E - RT/nF ln {ACcAD

d /AAaAB

b}at T = 298.15 K (25 C)• E = E - 0.05916 /n log {AC

cADd /AA

aABb},

where n = the number of moles of electrons transferred in the balanced redox reaction.

Standard Reduction Potentials (E) analogous to G

• 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++ 4H2OE0 = + 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 VNegative 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+ + 2e- Oxidation

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)= Ecell - (0.05916/2)log ([Cd2+] / [Ag+]2)

[Cd2+] / [Ag+]2 = Q (the reaction quotient)

Nernst Eq as half reactions

EMF = E+ - E-

E+ = 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) anode ½ rxn2Ag+ + 2e- ↔ 2Ag(s) cathode ½ rxn

Gives same answer. The math is equivalent.

0.100 M Cd(NO3)2(aq)

- +

EMF

Ag(s)Cd(s)

Cd2+Ag+

e- e-

Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)

0.100 MAgNO3(aq)

KNO3

K+NO3-

Anode –where oxidation occursCd(s) → Cd2+ + 2e-

Cathode –where reduction occursAg+ +e- →Ag(s)

Calculating Ecell using the Nerst Eq

EMF =

(Ec - Ea) - (0.05916/2)log([Cd2+]/[Ag+]2)

= (0.7993–(-0.402)) – ((0.05916/2)log(0.100/(0.100)2))

= 1.172 V

0.100 M Cd(NO3)2(aq)

- +

EMF = 1.172 V

Ag(s)Cd(s)

Cd2+Ag+

e- e-

Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)

0.100 MAgNO3(aq)

KNO3

K+NO3-

Anode –where oxidation occursCd(s) → Cd2+ + 2e-

Cathode –where reduction occursAg+ +e- →Ag(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.

0.100 M Cd(NO3)2(aq)

- +

EMF = 1.172 V

Ag(s)Cd(s)

Cd2+Ag+

e- e-

Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)

Anode –where oxidation occursCd(s) → Cd2+ + 2e- Cathode –where

reduction occursAg+ +e- →Ag(s)

0.100 MAgNO3(aq)

KNO3

K+NO3-

The EMF reads positive, Because the negative Terminal is connected to theAnode and the positive Terminal is connected to the cathode.

0.100 M Cd(NO3)2(aq)

-+

EMF = -1.172 V

Ag(s)Cd(s)

Cd2+Ag+

e- e-

Cd(s) + 2Ag+ → Cd2+ + 2Ag(s)

0.100 MAgNO3(aq)

Need Salt Bridge to avoid energy barrier to build up of excess charge

KNO3

K+NO3-

If the leads are switched, the voltage reads negative.

Anode –where oxidation occursCd(s) → Cd2+ + 2e-

Cathode –where reduction occursAg+ +e- →Ag(s)

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)

CdCl2(aq)

- +

EMF

AgCl(s)

Ag(s)Pb(s)

Cd2+Cl-

e- e-

2I- + Pb(s) + 2AgCl(s) ↔ PbI2(s) + + 2Ag(s) + 2Cl-

Anode –where oxidation occursPb(s) + 2I- → PbI2(s) + 2e-orPb(s) → Pb2+ + 2e-

Cathode –where reduction occursAgCl +e- →Ag(s) + Cl-

orAg+ + e- → Ag(s)

PbI2(s)

Pb(s) + 2Ag+ → 2Ag(s) + Pb2+

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 wayPb2+ + 2e- → Pb(s) E0 = -0.126 V

Ag+ + e- → Ag(s) 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 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)

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 sideHOBr + 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 half-rxn, the safe thing to do is to add the Gs.

HOBr + H+ + e- 1/2Br2 + H2O G1 = -F(1.584)

1/2Br2 + e- Br- G2 = -F(1.098)

____________________________________

HOBr + 2e- + H+ Br- + H2O G3 = FE3

G3 = G1 + G2 = -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 Gs).

Pd(OH)2(s) Pd+2 + 2OH-Ksp = 3·10 -28

Pd+2 + 2e- Pd(s) K1 = 10(nE/.05916) = 8.9·1030

________________________________________________________

Pd(OH)2(s) +2e- Pd (s) + 2OH- K = KspK1

K = 3·103

Log K = -nE/0.05916E = -(0.05916/2)log(3·103) = 0.103 V