session 5

G. Bandarage , CMU 1220/CME 3220 — Basic principles of chemistry, Volume V – Electrochemistry © 2008 Open University of Sri Lanka

SESSION 5 Electrode Potentials Contents Introduction 5.1

Significance and recording of standard emf

5.2

Electrode potential and standard electrode potential

5.2.1

What is the reaction corresponding to a standard electrode potential?

5.3

Working with electrode potentials

5.3.1 5.3.2 5.3.3

An enlightening exercise Electrode potentials and Nernst equation How do you deal with oxidation potentials?

Summary Objectives Introduction

In the previous session, you learned what the standard emf of a cell is. It is an important constant of a cell reaction which could be tabulated (as melting points of pure compounds). In this session you discuss an efficient way of constructing such a table. You introduce the concept of electrode potential and develop rules (shortcuts) that could be used in problem solving. (However, these rules should not be used in rigorous derivations.)

5.1 Significance and recording of standard emf In the previous session, you learned that Nernst equation can be used to calculate the emf of a cell at any concentration of ionic species if you know the standard emf of the cell at the same temperature. Moreover, the standard emf of a cell (or cell reaction) is a constant at constant temperature. So, keeping a record (or table) of E 0 values for all possible cell reactions (for a given set of electrodes) appears to be a good idea. However you do not do this in practice due to practical difficulties. Q46

How many different cells can you construct if you are given four different electrodes.

A46

Denote the electrodes by 1, 2, 3 and 4. You can construct a cell by making electrical contact between 1 and 2 [which may be represented by (1,2)] using a salt bridge. Similarly you can have (1,3), (1,4), (2,3), (2,4) and (3,4) cells. So, altogether you have six different cells.

CMU 1221/CME 3221 — Basic principles of chemistry — Unit V

In general n different electrodes will give us

n

C2 = 12 n (n − 1) different cells.

So a complete table of E 0 values of cells involving n electrodes will have 1 2 n ( n − 1) entries. This number is much larger than n itself for large n: e.g. there are 435 E 0 values when n = 30. Q47

Is there a way to reduce the number of entries, in an E 0 table, for n electrodes?

A47

How about recording the potential difference at the interface of each electrode, under standard conditions? Then, for n electrodes you have n entries. You could determine the standard emf of a cell by using this table and the IUPAC convention, E = Φ R − Φ L .

Thus you see that it is more efficient to record interfacial potential difference of electrodes, under standard conditions, than recording the standard emfs of all cells that can be constructed out of them.

5.2 Electrode potential and standard electrode potential From the above discussion it is clear that you must construct a table of interfacial potential difference of electrodes under standard conditions. However, you immediately see a problem; you cannot (experimentally) measure this potential difference (of an electrode, which you call the test electrode) without using another electrode to make electrical contact with the electrolytic phase. This additional electrode is called a reference electrode. Thus you end up measuring the emf of a cell. However, you can have a table with n entries for n electrodes (as mentioned in Q47) by keeping the reference electrode constant. Conventionally, standard hydrogen electrode (i.e. hydrogen gas electrode under standard conditions) is used as the reference. Each potential value in this table may be interpreted as the interfacial potential difference of the test electrode relative to the standard hydrogen electrode. This relative potential difference is called the electrode potential of the test electrode. To be specific, it is defined as follows.

•

The electrode potential of an electrode is defined as the emf assigned to the cell diagram where the standard hydrogen electrode appears on the left hand side and the electrode of interest appears on the right hand side.

An electrode potential is represented by letter E. You may use a subscript to indicate the electrode; e.g. E 2+ . Cu

(aq) Cu (s)

Here is an example. Assume that you have found (say by experiment) that the emf assigned to the cell diagram Pt (s) H 2 (g ) (1 bar ) H (aq) ( a = 1) Mg +

60

2+

(

(aq) c = 0.001 moldm

−3

) Mg(s)

Electrochemistry

to be − 2.445 V at 298 K and 1 bar. Then you say that the electrode potential of Mg(s) Mg 2 + (aq) electrode is − 2.445 V at 298 K and 1 bar , when the

magnesium ion concentration is 0.001 mol dm −3 . Electrode potential of an electrode under standard conditions is called the standard electrode potential. Standard electrode potential of an electrode is represented by E 0 . You may indicate the electrode as a subscript e.g. E 0 2+ . Cu

(aq) Cu (s)

Q48

The emf assigned to the cell diagram Mg(s) Mg 2+ (aq) ( a = 1) H + (aq) ( a = 1) H 2 (g) ( P = 1bar ) Pt(s) is 2.356 V at 298 K. What is E 0

Mg 2 + (aq) Mg(s)

E0

A48

Mg 2 + (aq) Mg(s)

at 298 K and 1 bar?

= − 2.356 V at 298 K. Note that you get the negative

sign since the standard hydrogen electrode appears on the left hand side of the cell diagram corresponding to an electrode potential.

Thus the table you discussed in the beginning of this session is constructed with standard electrode potentials. A good book of constants has such a table. Activity 30 1. State the significance of the standard emf of a reaction. 2. Describe an advantage of using a common reference electrode in measuring the interfacial potential difference of an electrode under standard conditions. 3. Define the electrode potential and standard electrode potential of an electrode. 4. At 298 K, the emf assigned to the cell diagrams

(

Pt(s) H 2 (g) (1 bar) H + (aq) a

(

Ag(s) AgCl(s) Cl − (aq) a

Cl−

H+

)

(

= 1 Al3+ (aq) a

)

(

= 1 H + (aq) a

H+

Al 3+

)

= 1 H 2 (g) (1bar) Pt(s)

are −1.67 V and −0.2221 V respectively. Deduce E

E0

Ag(s) AgCl(s) Cl− (aq)

)

= 1 Al(s) and

0 Al 3 + (aq) Al(s)

and

at 298 K .

61


5. Deduce the relationship among E differences Φ

0 Al 3+ (aq) Al(s)

and Φ

0 Al 3 + (aq) Al(s)

and the interfacial potential

0

where Φ A is the interfacial 0

Pt (s) H 2 (g) H+ (aq)

potential difference of electrode A. 6. Deduce that the standard electrode potential of the hydrogen gas electrode is zero.

Q49

How do you calculate the standard emf of a cell using a table of standard electrode potentials?

A49

To be specific, let us consider the calculation of the standard emf, at 298 K, of the cell Pb(s) PbSO 4 (s) SO 42 − (aq) Fe3 + (aq) , Fe 2 + (aq) Pt(s) . Since lead/lead sulphate and ferrous/ferric electrodes appear on the left and right hand sides of the above cell diagram let us denote the standard electrode potentials of them by E 0L and E 0R respectively. Then by definition E 0L = Φ 0

Pb(s) PbSO4 (s) SO

2− 4 (aq)

− Φ0

Pt(s) H 2 (g) H + (aq)

This can be rearranged to give Φ0

2− Pb(s) PbSO4 (s) SO (aq) 4

= E 0L + Φ 0

(40a)

= E 0R + Φ 0

(40b)

Pt(s) H 2 (g) H+ (aq)

Similarly you have Φ0

Pt (s) Fe3 + (aq),Fe2 + (aq)

Pt(s) H 2 (g) H+ (aq)

By definition the standard emf of the above cell is

E0 = Φ0

Pt (s) Fe3 + (aq),Fe2 + (aq)

− Φ0 Pb(s) PbSO4 (s) SO

2− (aq) 4

(41)

Now substitute for interfacial potential differences by (40a) and (40b) to obtain E 0 = E R0 − E L0

(42)

You obtain the standard electrode potentials of lead/lead sulphate and ferrous/ferric electrodes from a book of constants. They are E 0R = 0.771 V and E 0L = − 0.351 V respectively. Thus, for the above

(

)

given cell E 0 = 0.771 − ( − 0.351) V = 1122 . V

Equation (42) refers to standard electrode potentials. Same equation applies to electrode potentials under non-standard conditions; viz.

62

Electrochemistry

E = ER − EL

(43)

Note that Equation (43) is similar to the equation relating interfacial potentials to the emf; you learned in session 2; viz. E = Φ R − Φ L .

Activity 31 1. Following data are given for three electrodes, at 298 K.

E0

= − 0.403 V

E0

= − 0.44 V

Cd 2 + (aq) Cd(s) Fe2 + (aq) Fe(s)

E0

Hg( l ) Hg 2Cl2 (s) Cl− (aq)

= 0.268 V .

Write down all possible non-equivalent cell diagrams that can be constructed out of these three electrodes and determine their standard emfs. 2. Electrode potentials of two electrodes A and B at 298 K were fond to be 1.00 V and 2.03 V respectively. A student prepared a cell out of A and B using a salt bridge and measured its emf using a potentiometer. What is the value he should get for the emf?

5.2.1 The reaction corresponding to a standard electrode potential. Q50

Earlier you learned that the electrode potential of Mg(s) Mg 2 + (aq) electrode, when the magnesium ion concentration is 0.001 mol dm −3 , is − 2.445 V at 298 K and 1 bar. What reaction is this emf assigned to?

A50

The cell diagram corresponding the electrode potential is Pt(s) H 2 (g)(1 bar) H (aq) ( a = 1) Mg +

2+

(

(aq) c = 0.001 moldm

−3

) Mg(s)

According to the IUPAC convention, corresponding reaction is

H 2 (g ) + Mg 2 + (aq )  → 2 H + (aq ) + Mg (s)

(44)

Note the following facts about the chemical reaction (44). 1 Hydrogen gas (a reactant) and hydrogen ions (a product) are in their standard states. This is true with the reaction associated with electrode potential of any electrode. So you might think of saving time and paper by disregarding hydrogen gas and hydrogen ions in writing down reactions corresponding to electrode potentials.

63


2 Reduction takes place at the Mg (s) Mg 2 + (aq ) electrode. In fact this is true for any electrode. i.e. in the reaction, corresponding to an electrode potential, reduction takes place at the electrode of interest. So always (unless specified otherwise) you write down the reduction reaction. (Remember, in session 3 you learned to write down the reduction reactions for electrodes.) So in a sense, electrode potential is a reduction potential. Using these two facts you develop a convention in representing the reaction corresponding to an electrode potential. RULE 1

Omit H 2 (g) and H + (aq) from the reaction corresponding to an electrode potential and write down the reduction half reaction of the electrode of interest as the reaction corresponding to the electrode potential of an electrode. Q51

Write down the reaction referred to in Q50 using the convention.

A51

The abbreviated form of reaction (44) is Mg 2 + (aq) + 2 e−

 → Mg(s)

(45)

In a sense, with rule 1 you have assigned an electrode potential to a half cell reaction. For example, you say the electrode potential of reaction (45) at 298 K and 1 bar when the magnesium ion concentration is 0.001 mol dm −3 , is − 2.445 V. So hereafter, unless specified otherwise, you write the reactions corresponding to electrode potentials according to RULE 1.

Activity 32 1. State the convention that is used in writing down the reaction corresponding to an electrode potential. State advantages in using this convention over writing the full reaction. 2. Write down the actual reactions corresponding to the three standard electrode potentials listed in SAQ 31.1. 3. Write down the reactions corresponding to the three standard electrode potentials listed in SAQ 31.1 using RULE 1. 4. Explain why the electrode potentials are considered as reduction potentials. 5. A book of constants indicates that the standard electrode potential of the +

−

reaction O 2 (g) + 4H (aq) + 4e → 2H 2 O(aq) to be 1.229 V at 298 K. What is the actual reaction for which the emf assigned at 298 K is 1.229 V.

64

Electrochemistry

Note that the reactions written according to RULE 1 does not represent what really happens. It merely defines a shorthand notation that can be used profitably, with the other rules specified in this session, in problem solving. Thus the rules developed in this session must not be used in serious derivations. In a book of constants you can find the half reactions and their standard electrode potentials. See Table 1. Table 1: Standard electrode potentials in water at 298 K (from Reference 2) E0 V

Reaction K + (aq) + e −

 → K(s)

Ca 2 + (aq ) + 2 e − +

−

Na (aq ) + e Mg

2+

(aq ) + 2 e

 → Ca (s)

− 2.84

 → Na (s)

− 2.714

−

− 2.356

Al 3 + (aq ) + 3 e − 2 H2O + 2 e

− 2.925

 → Mg (s)  → Al (s)

−

− 1.67

 → H 2 (g ) + 2 OH

2+

Zn (aq ) + 2 e

−

−

− 0.828

 → Zn (s)

− 0.763

Fe 2 + (aq ) + 2 e −

 → Fe(s)

− 0.44

Cd 2 + (aq ) + 2 e −

 → Cd (s)

PbI 2 (s) + 2 e

−

− 0.403 −

 → Pb (s) + 2 I (aq ) −

2− 4

− 0.365

 → Pb (s) + SO (aq )

− 0.351

 → Sn (s)(white)

− 0.136

 → Pb (s)

− 0.125

 → Fe(s)

− 0.04

2 D + (aq ) + 2 e −

 → D 2 (g )

− 0.01

2 H + (aq ) + 2 e −

 → H 2 (g )

PbSO 4 (s) + 2 e

Sn 2 + (aq ) + 2 e − 2+

Pb (aq ) + 2 e 3+

Fe (aq ) + 3 e

AgBr ( s) + e

−

AgCl ( s) + e

−

−

−

 → Ag ( s) + Br ( aq ) −

 → Ag ( s) + Cl ( aq )

Hg 2 Cl 2 ( s) + 2 e − 2+

Cu ( aq ) + 2 e +

−

Cu ( aq ) + e

0 −

I 2 ( s) + 2 e −

−

 → 2 Hg ( l) + 2 Cl − ( aq )  → Cu ( s)

 → Cu ( s)  → 2 I −

Hg 2 SO 4 ( s) + 2 e −

0.0709 0.2221 0.2680 0.340 0.520 0.536

 → 2 Hg ( l) + SO 42 − ( aq )

−

 → Fe ( aq )

0.771

+

−

 → Ag ( s)

0.7989

Fe ( aq ) + e Ag ( aq ) + e Br2 ( l) + 2 e −

2+

0.613

3+

 → 2 Br − +

O 2 ( g ) + 4 H ( aq ) + 4 e Cl 2 ( g ) + 2 e

−

−

 → 2 H 2 O

 → 2 Cl

−

1.065 1.229 1.358 65


5.3 Working with electrode potentials From the definition it is clear that electrode potentials are emfs assigned to cell reactions of a particular class; viz. cell reactions involving the oxidation of hydrogen gas into hydrogen ions as anode reaction. Moreover, hydrogen gas and hydrogen ions are in their standard states. Thus, all what you have learned so far in previous sessions can be applied to these emfs as well. However, as stated earlier, in this session you are interested in developing some shortcuts in working with electrode potentials. Already you have learned the following. •

How to obtain a cell emf using electrode potentials; Equation (43).

•

How to write down an abbreviated form of the reaction corresponding to an electrode potential using a convention.

5.3.1 An enlightening exercise In A49 you learned how to use arguments based on interfacial potential differences in calculating the cell emf using electrode potentials. Let us learn a more powerful but a simple method, based on thermodynamics, to obtain the same result. Q52

Find the emf of the cell Pb(s) PbSO 4 (s) SO 42 − (aq) Fe3 + (aq) , Fe 2 + (aq) Pt(s) in terms of E L and E R .

A52

Let us first write down the full reactions corresponding to the two electrode potentials, E L and E R . (Note that you write down full reactions since you are involved in a derivation.)

PbSO 4 (s) + H 2 (g) 2 Fe 3 + ( aq ) + H 2 ( g)

 → Pb(s) + 2 H + (aq ) + SO 24 − (aq )  → 2 Fe 2 + ( aq ) + 2 H + (aq )

EL (46a) ER (46b)

The cell reaction of the above given cell is Pb( s) + 2 Fe 3 + (aq ) + SO 24 − ( aq ) → PbSO 4 ( s) + 2 Fe 2 + ( aq ) E (46c) You observe that reaction (46c) can be obtained by subtracting reaction (46a) from (46b). However at this point you do not know whether you could do the same with the corresponding electrode potentials. But you know that you could do it with Gibbs free energy changes of reactions. Thus, if you denote the Gibbs free energy change of reactions (46a), (46b) and (46c) by ∆G L , ∆G R and ∆G , then you have (47) ∆G = ∆G R − ∆G L Charge number of all three reactions is 2. Thus, substituting ∆G = −n F E (Equation 25) in (47) you obtain 66

Electrochemistry

− 2 F E = − 2 F ER − ( − 2 F EL ) Cancellation of − 2 F gives the familiar equation E = E R − E L Q53

Do the problem in Q52 using abbreviated notation

A53

PbSO 4 (s) + 2 e − Fe 3 + (aq ) + e −

 → Pb(s) + SO 24 − (aq )  →

EL

Fe 2 + (aq )

ER

(48a) (48b)

The cell reaction of the above given cell is Pb( s) + 2 Fe 3 + ( aq ) + SO 24 − ( aq ) → PbSO 4 ( s) + 2 Fe 2 + ( aq )

E (48c)

You can obtain the cell reaction, (48c), by multiplying (48b) by 2 and subtracting (48a) from it. So you can write down the relationship between Gibbs free energy changes of these reactions as ∆G = 2 × ∆G R − ∆G L

(49)

Now you agree to apply ∆G = −n F E even for half cell reactions with the understanding that n is the stoichiometric coefficient of electrons in the half cell reaction.

− 2 F E = 2 × (− F ER ) − (− 2 F EL ) Which gives the familiar equation E = E R − E L . Problem in Q52 and Q53 may suggest that you could subtract electrode potentials as you subtract reactions corresponding to them. However, this is deceiving. Simple subtraction can be done only when the final reaction (written in our shorthand notation) does not have electrons. Q54

Given that the standard electrode potentials, at 298K, of reactions

Cr 3 + (aq ) + e −

 → Cr 2 + (aq )

(50a)

Cr 3 + ( aq ) + 3 e −

 → Cr (s)

(50b)

as − 0.424 V and − 0.90 V , calculate that of Cr 2 + ( aq ) + 2 e − A54

 → Cr ( s) .

(50c)

Denote the standard Gibbs free energy change of equation (50a), (50b) and (50c) by ∆G 10 , ∆G 02 and ∆G 0 . You observe that equation (50c) can be obtained by subtracting (50a) from (50b). The relationship among standard Gibbs free energy changes is ∆G 0 = ∆G 02 − ∆G 10 Now substitution from ∆G = −n F E as before leads to

(

− 2 F E 0 = − 3 F × ( − 0.90 V ) − − F × ( − 0.424 V )

) 67


∴ E0 =

1 × [ 3 × 0.90 − 0.424] V = − 1138 . V 2

In the problem in Q54, you see that the electrode potential of the resultant reaction is not equal to the difference of those of reactions (50b) and (50a). This is always true when the final reaction has electrons (i.e. when it is a half reaction itself). In summary you have the following rules when working with electrode potentials. RULE 2

When two half reactions are subtracted to obtain a cell reaction, i.e. a reaction that does not contain electrons, then the emf corresponding to it can be obtained by subtraction of the electrode potentials corresponding to the half reactions. RULE 3

When two half reactions are subtracted to obtain another half reaction, i.e. a reaction having electrons, then one must use the relationship among Gibbs free energy of the reactions to obtain the electrode potential of the final reaction. The Gibbs free energy change of a reaction in abbreviated form, is calculated using ∆G = − n F E where n is the stoichiometric coefficient of the electrons.

Activity 33 1. Using Table 1, find out the E 0 values for the following reactions at 298K. (a) Cl 2 (g) + D 2 (g)  → 2 Cl − (aq) + 2 D + (aq)

Cu + (aq)

(b)

 → Cu 2 + (aq) + e −

5.3.2 Electrode potentials and Nernst equation Q55

What is the relationship between the activity of silver ion and the electrode potential of a silver/silver ion electrode?

A55

The (full) reaction corresponding to the electrode potential of Ag (s) Ag (aq ) is H 2 ( g) + 2 Ag + ( aq )

 → 2 H + ( aq ) + 2 Ag( s)

If you denote the electrode potential by E and standard electrode potential for this reaction by E 0 , then application of the Nernst equation (as in session 4) you obtain 2 2 RT  a H+ (aq) a Ag(s)  E=E − ln 2 F  a H2 (g) a 2 +  Ag (aq)   0

68

Electrochemistry

Remember that electrode potentials are measured relative to a standard hydrogen electrode. Thus you have a H 2 ( g ) = a H + ( aq ) = 1. Then the above equation becomes 2 RT  a Ag ( s)   E=E − ln  2 2 F  a Ag + ( aq )  0

Which is reduced to

E = E0 −

 a  RT Ag ( s )  ln   F a  Ag + ( aq ) 

(51)

Note only the activities of species involved in the half reaction of the electrode of interest appears in the Nernst equation. So, you may think of finding a simpler way of applying Nernst equation to relate electrode potentials to activities of species. Q56

How do you apply the Nernst equation to a reaction written in abbreviated form?

A56

Write down the half reaction corresponding to the electrode potential.

Ag + (aq ) + e −

 → Ag(s)

Taking n to be equal to the stoichiometric coefficient of electrons (and disregarding electron activities) apply Nernst equation to this reaction to obtain

E = E0 −

 a  RT Ag ( s )  ln   F  a Ag + ( aq ) 

which is the same as equation (51). RULE 4

In applying the Nernst equation to relate the electrode potential to activities of chemical species involved in a half reaction, you take charge number to be the same as the stoichiometric coefficient of electrons and considers only the activities of chemical species in the half reaction.

Activity 34 1. Using Table 1, calculate the electrode potentials of the following, at 298K. −

(a) Ag ( s) AgBr ( s) Br ( aq ) (b) Pt ( s) Fe

3+

( aq ) , Fe 2 + ( aq )

when

a Br − ( aq ) = 0.5

when

a Fe 3+ ( aq ) = 0.7 & a Fe 2 + ( aq ) = 0.9 .

69


5.3.3 How do you deal with oxidation potentials? In the beginning of this session, you learned that, conventionally, electrode potential is defined so that reduction takes place at the electrode of interest. So they are reduction potentials. So if the electrode potential of an electrode is given, (without any other information) you always assign that to the reduction half cell reaction. However, one may want to find out the emf associated with the oxidation half cell reaction. (i.e. reverse of the reduction reaction.) You have the following rule to deal with the situation. RULE 5

If emf assigned to a half reaction (abbreviated form) is E then the emf assigned to the reverse half reaction is − E. Q57

Give an example of application of Rule 5.

A57

According to Table 1, the standard emf assigned to the half reaction Fe 3 + ( aq ) + e − → Fe 2 + ( aq ) is 0.771 V at 298 K. Then the standard emf assigned to the reaction Fe 2 + ( aq )  → Fe 3 + ( aq ) + e − , at 298 K is − 0.771 V.

Q58

Using Table 1, calculate the emf assigned to 2 I − ( aq ) + Pb( s) → PbI 2 ( s) + 2 e − at 298 K when a I − ( aq ) = 0.75

A58

Using Rule 4 you apply Nernst equation to obtain   RT  a PbI 2 ( s)  RT 1 0   = E − E=E − ln  ln   2 F  a Pb ( s) a 2I − ( aq )  F a  I − ( aq )  0

= E0 +

(

RT ln a I − ( aq ) F

)

Now using Rule 5 and Table 1, you find that E 0 = 0.365 V . Substitution of values in the above equation gives 8.314 × 298   E =  0.365 + ln ( 0.75)  V = 0.358 V   96500

Activity 35 1. Construct a justification for RULE 5.

70

Electrochemistry

Summary

An electrode potential is an emf of a particular class of cells: viz. standard hydrogen electrode appears as one of the electrodes. Thus the reaction corresponding to an electrode potential always involves hydrogen gas and aqueous hydrogen ions in their standard states. From a theoretical point of view, there is nothing special about these emf values than those discussed in previous sessions. However they are of great practical importance. Standard electrode potential is a characteristic of a particular electrode (at a given temperature). You could find tables of them in a book of constants. These values are indispensable in applying Nernst equation.

In the absence of any other information, standard electrode potential refers to the reduction reaction of the electrode of interest. Thus quite often, the reduction half reaction and its standard electrode potential are listed in tables. In this session you learned a number of shortcuts that could be used when working with electrode potentials. As the first rule you learned to disregard the hydrogen gas and hydrogen ions in writing reactions corresponding to electrode potentials. The other rules indicated how to use electrode potentials and the reactions written in abbreviated form in problem solving. Objectives

At the end of a successful study of this session, you should be able to do the following. • State the significance of standard emf of a reaction. •

Describe an advantage of using a single reference electrode in measuring the interfacial potential difference of an electrode under standard conditions.

•

Define the electrode potential and standard electrode potential of an electrode.

•

Deduce E cell = E R − E L .

•

Write down the actual chemical reaction corresponding to an electrode potential.

•

State the convention used in writing down the reaction corresponding to an electrode potential and describe its advantages.

•

Write down the chemical reaction corresponding to an electrode potential using the convention.

•

Calculate the emf assigned to a cell reaction using the electrode potentials of the half reactions.

•

Calculate the electrode potential of a half reaction using the electrode potentials of constituent half reactions.

•

Write down the Nernst equation corresponding to an electrode potential using the reaction written in abbreviated form.

•

Calculate the electrode potential of an electrode using its standard electrode potential.

71


Reference 2 “Physical Chemistry”, 3rd edition, Ira N. Levine, McGrawHill International Editions, 1988.

72