Chapter 14

Electrode Potentials

14-1 Redox Chemistry & Electricity

Oxidation: a loss of electrons to an oxidizing agentReduction: a gain of electrons from a reducing agent

Reduction-Oxidation reaction (redox reaction)

ex:

Half-reactions:re: Fe3+ + e- Fe2+

ox: V2+ V3+ + e-

1. Chemistry & Electricity

• Electrochemistry: the study of the interchange of chemical & electrical energy.

• Electric charge (q) is measured in coulombs(C). • The magnitude of the charge of a single electron (or proto

n) is 1.602×10 － 19 C. A mole of electrons therefore has a charge of (1.602×10 － 19 C)(6.022×1023 /mol)= 9.649×104 C/mol, which is called the Faraday constant, F.

Example at p.310

2. Electric current is proportional to the rate of a redox reaction

I (ampere; A) = electric current

= a flow of 1 coulomb per second = 1C/s

Example at p. 310:

Sn4+ + 2e- Sn2+

at a constant rate of 4.24 mmole/h.

How much current flows into the solution?

3. Voltage & Electrical Work

wire q I E

hose H2O VH2O PH2O

The difference in electric potential between two points measures the work that is needed (or can be done) when electrons move from one point to another.

Ask yourself at p.312

• Consider the redox reaction

14.2 Galvanic Cells

Chemical reaction spontaneously occurs to produce electrical energy.

Ex: lead storage battery

When the oxidizing agent & reducing agent are physically separated, e transfer through an external wire.

generates electricity.

A cell in action

Electrodes: the redox rxn occur

anode: oxidation occur

cathode: reduction occur

Salt bridge: connect two solns.

External wire

cathodeanode e

Cell representation: Line Notation

Example ：： Interpreting Line Diagrams of Cells

Figure 14-4 Another galvanic cell.

14-3 Standard PotentialsCell potential ( Ecell)

a)The voltage difference between the electrodes.

electromotive force (emf)

b)can be measured by voltmeter.

c)emf of a cell depends on

The nature of the electrodes & [ions]

Temp.

14-3 Standard Potentials

S.H.E. (standard hydrogen electrode )

It is impossible to measure Ecell of a half-rxn directly, need a reference rxn.

standard hydrogen electrode:

1atmP 1M,H

0E )( H2e2H

2H

0red cell2

gaq)(

The standard reduction potential (E0) for each half-cell is measured by an experiment shown in idealized form in Fig.14-6.

Table 14-1 & Appendix C

( 於 1953, the 17th IUPAC meeting 決定半反應以「還原反應」來表示 )

Standard Reduction Potentials for reaction

0.34V

0CuCu

0.76V

0ZnZn

1.10V

0cell

(s)(aq2

(aq)2

(s)

0.76V

0ZnZn

0

0HH

0.76V

0cell

2(g)(aq)2

(s)(aq)

0red

0ox

0cell

22

22

EEE

Cu)ZnCuZn :rxn

EEE

HZnZn 2H:rxn

EEE

0.34V0.77VEEE

0.77VE 2Fe2e2Fe

0.34VEE 2eCuCu

0.77VE FeeFe:red

FeCuCue F

0ox

0re

0cell

0re

23

0re

0ox

2

0re

23

(aq)2

(aq)2

(s)(aq)3

2

rxn:

Standard Reduction Potentials for reaction

• AgCl (s) + e- Ag (s) + Cl-

0.222 V

0.197 V in saturated KCl (formal potentional)

E0 = 0.222V

S.H.E.║ Cl- (aq, 1M) | AgCl (s) | Ag(s)

E0’ (formal potential) = 0.197 V (in saturated KCl)

S.H.E.║ KCl (aq, saturated) | AgCl (s) | Ag(s)

Formal potential

Formal Potential

Ex: Ce4+ + e- Ce3+ E°=1.6V

with H+A- E°≠1.61V

Formal potential: (E°’)

The potential for a cell containing a [reagent] ≠1M.

Ex: Ce4+/Ce3+ in 1M HCl E°’=1.28V

E is the reduction potential at the specified concentrations n: the number of electrons involved in the half-reactionR: gas constant (8.3143 V coul deg-1mol-1)T: absolute temperatureF: Faraday constant (96,487 coul eq-1) at 25°C 2.3026RT/

F=0.05916

14-4 The Nernst EquationThe net driving force for a reaction is expressed by the Nernst eqn.

Nernst Eqn for a Half-Reaction

where

a

b

[A]

[B]log

n

0.05916EE

bBneaA

E = E0 when [A] = [B] = 1M

Q (Reaction quotient ) =1 E = E0

Where, Q = [B]b / [A]a

Nernst equation for a half-reaction at 25ºC

[C] & Ecell

standard conditions: [C]=1M

what if [C]≠1M?

(ex)

a)[Al3+]=2.0M, [Mn2+]=1.0M Ecell<0.48V

b)[Al3+]=1.0M, [Mn2+]=3.0M Ecell>0.48V

0.48VE

3Mn2Al3MnAl 2

:ε

0cell

(s)(aq)3

(aq)2

(s)

0cell

Dependence of potential on pH

0.059pHE'E

]AsO[H

]AsO[Hlog

2

0.0592-0.0592pHE

]AsO[H

]AsO[Hlog

2

0.0592-]H0.0592log[E

][H ]AsO[H

]AsO[Hlog

2

0.0592-EE

OHAsOH2e2HAsOH :Ex

43

33

43

33

243

33

233-

43

Many redox reactions involved protons, and their potentials are influenced greatly by pH.

Nernst Equation for a Complete Reaction Equation for a Complete Reaction

• 1. Write reduction half-reactions for both half-cells and find E0 for each in Appendix C.

• 2. Write Nernst equation for the half-reaction in the right half-cell.

• 3. Write Nernst equation for the half-reaction in the left half-cell.

• 4. Fine the net cell voltage by subtraction: E=E ＋－ E － .

• 5. To write a balanced net cell reaction.

P.321

Nernst Equation for a complete reaction

]3.6

1log 18 [log

2

0.0592-201.1

]10[3.6

[0.01]log

2

0.0592--0.402)](-799.0[

][Ag

][Cdlog

2

0.0592-EE

2

210

2

2

Example at p. 321 Rxn: 2Ag+ (aq) + Cd (s) Ag (s) + Cd 2+ (aq)

2Ag+ + 2e- Ag (s) E0+ = 0.799

Cd 2+ + 2e- Cd (s) E0- = -0.402

Electrons Flow Toward More Positive Potential• Electrons always flow from left to right in a diagra

m like Figure 14-7.

14-5 E0 and the Equilibrium Constant

Qlog n

0.059 E

]][Cl[Fe

][Felog

n

0.059EE

]log[Cln

0.059E

][Fe

][Felog

n

0.059E

(EEEEE

VE AgClFeAgFe

VE ClAgeAgCl

VE FeeFe

3

2

3

2

AgAgCl,Fe,Fecell

23

--

2-3

23

)(

))(

.)s()s(

.)s()s(

.

0

5490

2220

7710

14-5 E0 and the Equilibrium Constant

Klog n

0.05916E

Klog n

0.05916E0

Qlog n

0.05916EE

• E = 0 and Q = K

• E0 > 0 K > 1, • E0 < 0, K < 1

At equilibrium

Ex:• One beaker contains a solution of 0.020 M KMnO4, 0.005

M MnSO4, and 0.500 M H2SO4; and a second beaker

contains 0.150 M FeSO4 and 0.0015 M Fe2 (SO4)3. The 2

beakers are connected by a salt bridge and Pt electrodes are placed one in each. The electrodes are connected via a wire with a voltmeter in between.

• What would be the potential of each half-cell (a) before reaction and (b) after reaction?

• What would be the measured cell voltage (c) at the start of the reaction and (d) after the reaction reaches eq.?

• Assume H2SO4 to be completely ionized and equal

volumes in each beaker.

Ans:

5Fe+2 + MnO4- + 8H+ 5Fe+3 + Mn+2 + 4H2O

Pt | Fe+2(0.15 M), Fe+3(0.003 M)║MnO4-(0.02 M), Mn+2(0.005 M), H+(1.00 M) | Pt

(a) EFe = EoFe(III)/Fe(II) – (0.059/1) log [Fe+2]/[Fe+3]

= 0.771 – 0.059 log (0.150)/(0.0015 × 2) = 0.671 V

EMn = EoMnO4-/Mn+2 – (0.059/5)log [Mn+2]/[MnO4

-][H+]8

  = 1.51 – 0.059/5 log (0.005)/(0.02)(1.00) 8 = 1.52 V

(b) At eq., EFe = EMn, 可以含鐵之半反應來看，先找出平衡時兩個鐵離子的濃度，得EFe = 0.771 – 0.059 log (0.05)/(0.103) = 0.790 V

(c) Ecell = EMn - EFe = 1.52 – 0.671 = 0.849 V

(d) At eq., EFe = EMn, 所以 Ecell = 0 V

Determinea) e- flow direction?

b) anode? cathode?

c) E =? at 25℃

Concentration Cells

Ag Ag

1MAgNO 3(aq)

1M NaCl(s)

& AgCl(s)

1.0Msp

0

ClAgK

1.0

Aglog

1

0.05920

0.58Vε

0?ε

(ex) Calculate Ksp for AgCl at 25℃ ε=0.58V

soln:

Ex: Systems involving ppt

14-6 Reference Electrodes

Indicator electrode: responds to analyte concentration

Reference electrode: maintains a fixed potential

]log[Cl0592.0E][Fe

][Felog0592.0EEEE

3

2

cell

• Silver-Sliver Chloride• AgCl + e- Ag(s) +Cl-

E0 = 0.222 V

E (saturated KCl) = 0.197 V

• Calomel • Hg2Cl2 + 2e- 2Hg(l) +2Cl-

E0 = 0.268 V

E (saturated KCl) = 0.241 V saturated calomel electrode (S.C.E.)

Reference Electrodes

Voltage conversion between different reference scales

• The potential of A ?

?