1 Chapter Eighteen Electrochemistry. 2 Electrochemical reactions are oxidation-reduction reactions. The two parts of the reaction are physically separated

1

Chapter Eighteen

Electrochemistry

2

Electrochemistry• Electrochemical reactions are oxidation-reduction

reactions.

• The two parts of the reaction are physically separated.– oxidation occurs at one cell– reduction occurs in the other cell

• There are two kinds electrochemical cells.

• Electrolytic cells - nonspontaneous chemical reactions

• Voltaic or galvanic cells - spontaneous chemical reactions

3

Electrical Conduction• Ionic or electrolytic conduction• Ionic motion transports the electrons• Positively charged ions, cations, move toward the negative electrode.• Negatively charged ions, anions, move toward the positive electrode.

4

Electrodes

• Conventions for electrodes:

• Cathode - electrode at which reduction occurs

• Anode - electrode at which oxidation occurs

• Inert electrodes do not react with the liquids or products of the electrochemical reaction.

• Graphite and Platinum are common inert electrodes.

5

Voltaic or Galvanic Cells• Electrochemical cells in which a spontaneous chemical reaction produces

electrical energy.• Cell halves are physically separated so that electrons (from redox reaction)

are forced to travel through wires and creating a potential difference.• Examples include:

Auto batteries

Flashlight batteries

Other batteries

6

The Construction of Simple Voltaic Cells

• Half-cell contains the oxidized and reduced forms of an element (or other chemical species) in contact with each other.

• Simple cells consist of:– two pieces of metal immersed in solutions of their ions– wire to connect the two half-cells– salt bridge to

• complete circuit• maintain neutrality• prevent solution mixing

7

The Zinc-Copper Cell

• Cell components:Cu strip immersed in 1.0 M copper (II) sulfate

Zn strip immersed in 1.0 M zinc (II) sulfate

wire and a salt bridge to complete circuit

• Initial voltage is 1.10 volts

8


9


Anode Zn Zn e

Cathode Cu e Cu

Cell rxn. Zn Cu Zn Cu

Spontaneous rxn. E V

0 2+

2+ 0

0 2+ 2+ 0

cell0

2

2

110.

• In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode).

10


• Short hand notation for voltaic cells– Zn-Cu cell example

Zn/Zn2+(1.0 M) || Cu2+(1.0 M)/Cu

species (and concentrations)in contact with electrode surfaces

electrode surfaces

salt bridge

11

The Copper - Silver Cell


Ag strip immersed in 1.0 M silver (I) nitrate



12


13


Anode Cu Cu 2e

Cathode 2 Ag + e Ag

Cell rxn. Cu + 2 Ag Cu + 2 Ag

spontaneous rxn. E V

0 2

+ - 0

0 + 2+ 0

cell0

0 46.• Compare the Zn-Cu cell to the Cu-Ag cell

Cu electrode is cathode in Zn-Cu cell

Cu electrode is anode in Cu-Ag cell

• Whether a particular electrode behaves as an anode or as a cathode depends on what the other electrode of the cell is.

14

The Copper - Silver Cell• Demonstrates that Cu2+ is a stronger oxidizing agent than Zn2+

Cu2+ oxidizes metallic Zn to Zn2+

• Ag+ is is a stronger oxidizing agent than Cu2+

Ag+ oxidizes metallic Cu to Cu2+

• Arrange these species in order of increasing strengths

Zn < Cu < Ag Zn > Cu > Ag

strength as oxidizing agent strength as reducing agent

2+ 2+ + 0 0 0

15

Standard Electrode Potential

• Establish an arbitrary standard to measure potentials of a variety of electrodes

• Standard Hydrogen Electrode (SHE)– assigned an arbitrary voltage of 0.000000… V

16

Standard Electrode Potential

17

The Zinc-SHE Cell• Cell components:

Zn strip immersed in 1.0 M zinc (II) sulfate

other electrode is a SHE



18

The Zinc-SHE Cell

19

The Zinc-SHE Cell

E

Anode Zn Zn + 2 e 0.763 V

Cathode 2 H + 2 e H 0.000 V

Cell rxn. Zn + 2H Zn + H E V

0

0 2+ -

+ -2 g

0 + 2+2 g cell

0

0 763.

• SHE is the cathodeZn reduces H+ to H2

• Zn is the anode

20

The Copper-SHE Cell


other electrode is a SHE



21

The Copper-SHE Cell

22

The Copper-SHE Cell

• SHE is the anodeCu2+ ions oxidize hydrogen to H+

• Cu is the cathode

E

Anode H 2 H + 2 e 0.000 V

Cathode Cu + 2e Cu 0.337 V

Cell rxn. H + Cu 2H + Cu E V

0

2 g+ -

2+ - 0

2 g2+ + 0

cell0

0 337.

23

The Electromotive (Activity) Series of the Elements

• Electrodes that force the SHE to act as an anode are assigned positive standard reduction potentials.

• Electrodes that force the SHE to act as the cathode are assigned negative standard reduction potentials.

• Table 18.1 in Text lists the Activity of some of the elements

24

Uses of the Electromotive Series

• Standard electrode (reduction) potentials tell us the tendencies of half-reactions to occur as written.

• For example, the half-reaction for the standard potassium electrode is:

• The large negative value tells us that this reaction will occur only under extreme conditions.

K e K E = -2.925 V0 0

25

Uses of the Electromotive Series• Compare the potassium half-reaction to fluorine’s half-reaction:

• The large positive value denotes that this reaction occurs readily as written.• Positive E0 values tell us that the reaction tends to occur to the right

– larger the value, greater tendency to occur to the right

• Opposite for negative values

F + 2 e 2 F E = +2.87 V20 - - 0

26


• Use standard electrode potentials to predict whether an electrochemical reaction at standard state conditions will occur spontaneously.

• Steps for obtaining the equation for the spontaneous reaction.

27


1 Choose the appropriate half-reactions from a table of standard reduction potentials.

2Write the equation for the half-reaction with the more positive E0 value first, along with its E0 value.

3Write the equation for the other half-reaction as an oxidation with its oxidation potential, i.e. reverse the tabulated reduction half-reaction and change the sign of the tabulated E0.

28


4 Balance the electron transfer.

5Add the reduction and oxidation half-reactions and their potentials. This produces the equation for the reaction for which E0

cell is positive, which indicates that the forward reaction is spontaneous.

29

Electrode Potentials for Other Half-Reactions

• Example: Will permanganate ions, MnO4-, oxidize iron (II) ions to iron

(III) ions, or will iron (III) ions oxidize manganese ions to permanganate ions in acidic solution?

• Follow the steps outlined in the previous slides.E0 values are not multiplied by any stoichiometric relationships in this procedure.

30


• Example: Will permanganate ions, MnO4-, oxidize iron (II)

ions to iron (III) ions, or will iron (III) ions oxidize manganese ions to permanganate ions in acidic solution?

V 1.51 + O)4HMn5e+8H1(MnOn Red'

E

22--

4

0

31




V) 0.771(- )e + Fe5(Fen Oxd'


E

-+32

22--

4

0

32




V +0.74=E 5Fe+O4HMn8HFe5MnO rxn. Cell

V) 0.771(- )e + Fe5(Fen Oxd'


E

0cell

+32

22-4

-+32

22--

4

0

33


• Example: Will permanganate ions, MnO4-, oxidize iron (II) ions to iron (III) ions, or will iron (III) ions oxidize manganese ions to permanganate ions in acidic

solution?

• Thus permanganate ions will oxidize iron (II) ions to iron (III) and are reduced to manganese (II) ions in acidic solution.

V +0.74=E 5Fe+O4HMn8HFe5MnO rxn. Cell

V) 0.771(- )e + Fe5(Fen Oxd'


E

0cell

+32

22-4

-+32

22--

4

0

34

Effect of Concentrations (or Partial Pressures) on Electrode Potentials -

Nernst Equation

• Standard electrode potentials are determined at thermodynamic standard conditions.1 M solutions

1 atm of pressure for gases

liquids and solids in their standard states

temperature of 250 C

35


Nernst Equation

• Potentials change if conditions are nonstandard.• Nernst equation describes the electrode potentials

at nonstandard conditions.

E = E -2.303RT

nF log Q0

36


Nernst Equation

E = E -2.303RT

nF log Q

E = potential under condition of interest

E potential under standard conditions

R = universal gas constant = 8.314 Jmol K

T = temperature in K

n = number of electrons transferred

F = the faraday = (96,487 C / mol e 1 JC V

J / V mol e

Q = reaction quotient

0

0

-.

-

)

,96 487

37


Nernst Equation

2.303 RTF

so

E = E -0.0592

n log Q0

0 0592.

• Substitution of the values of the constants into the Nernst equation at 250 C gives:

38


Nernst Equation

Cu e Cu E = +0.153 V2+ - +

• For a typical half-reaction:

• The corresponding Nernst equation is

E = E -

0.05921

log Cu

Cu0

+

2

39


Nernst Equation• Substituting E0 into the above expression gives

E = 0.153 V -

0.05921

log Cu

Cu

+

2

40


Nernst Equation• If [Cu2+] and [Cu+] are both 1.0 M, standard conditions, then E = E0 because concentration term is zero.

• Since log 1= 0, we have

E = 0.153 V -0.0592

1 log

11

E 0.153 V

41

Effect of Concentrations (or Partial Pressures) on Electrode Potentials - Nernst Equation

• Example: Calculate the potential for the Cu2+/ Cu+ electrode at 250C when the concentration of Cu+ ions is three times that of Cu2+ ions.

Cu + e Cu Q =

Cu

Cu

Cu

Cu2+ - +

+

2+

2+

2+

33

42


• Example: Calculate the potential for the Cu2+/ Cu+ electrode at 250C when the concentration of Cu+ ions is three times that of Cu2+ ions.

0.125 E

3 log 1

0.0592-V 0.153=Q log

1

0.0592-E=E 0

43


• The Nernst equation can also be used to calculate the potential for a cell that consists of two nonstandard electrodes.

• Example: Calculate the initial potential of a cell that consists of an Fe3+/Fe2+ electrode in which [Fe3+]=1.0 x 10-2 M and [Fe2+]=0.1 M connected to a Sn4+/Sn2+ electrode in which [Sn4+]=1.0 M and [Sn2+]=0.10 M . A wire and salt bridge complete the circuit.

44


• Calculate the E0 cell by the usual procedure.

E

Red'n 2 Fe + e Fe 0.771 V

Oxd'n 1 Sn Sn + 2 e - 0.15 V

Cell rxn. 2 Fe Sn 2 Fe Sn E V

0

3+ 2+

2+ 4+ -

3+ 2+ 2+ 4+cell

0 0 62.

45


• Substitute the ion concentrations into Q to calculate Ecell.

E = E -0.0592

2Q

= 0.62 V -0.0592

2

Fe Sn

Fe Sn

cell cell0

2 4

3 2

log

log

2

2

46


• Substitute the ion concentrations into Q to calculate Ecell.

E = 0.62 V -0.0592

2

0.62 V -0.0592

2 V

= 0.53 V

cell log. .

. .

. . .

010 10

10 10 010

3 00 0 62 0 09

2

2 2

47

Relationship of E0cell to G0 and K

• From previous chapters we know the relationship of G0 and K for a reaction.

K log RT 303.2Gor lnK -RTG 00

48


• The relationship between G0 and E0cell is also a

simple one.

e ofnumber n

e mol J/V 96,487 F where

E F-n G

-

-

0cell

0

49


• Combine these two relationships into a single relationship to relate E0

cell to K.

RT

E Fn K ln

or

lnK RT -E Fn -

0cell

0cell

50


• Example: Calculate the standard Gibbs free energy change, G0 , at 250C for the following reaction.

Cu PbCuPb 22

51


• Calculate E0cell using the appropriate half-reactions.

V 463.0E CuPbPbCu

V 0.126-- e 2PbPb

V 0.337 Cue 2Cu

E

0cell

022

2

02

0

52


• Now that we know E0cell , we can calculate G0 .

• The negative value tells us that the reaction is spontaneous as written.

kJ -84.9

)rxn" of mol"(per J 1049.8

V 463.0e mol V

J 500,96e mol 2

E Fn G

4

-

0cell

0

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1 Chapter Eighteen Electrochemistry. 2 Electrochemical reactions are oxidation-reduction reactions. The two parts of the reaction are physically separated