CHAPTER 8 ELECTROCHEMICAL EQUILIBRIA



chapter 8 electrochemical equilibria

8.1 Introduction………..………………………………………………………………….…. 1

8.2 Redox Equilibria and Electron Activity……………………………………………….. 2

8.2.1 Redox Reactions………………………………………………………………….. 2

8.2.2 Electron Activity………………………………………………………………….. 2

8.3 Electrochemical Cells and Electrode Potentials……………………………………….. 7

8.3.1 Metal Electrodes………………………………………………………………….. 7

8.3.2 Electrochemical Cells and Redox Equilibria. ……………………………………. 8

8.3.3 Standard Electrode Potentials…...........................................……………………. 12

8.3.4 Electron Activity and Redox Potential…………………………………………... 15

8.3.5 Minimum Electrolyzing Voltage………………………………………………... 17

8.4 Electrochemical Selectivity…………………………………………………………………………. 23

8.4.1 Selectivity of the Anodic Process…………………………………………………………. 23

8.4.2 Selectivity of the Cathodic Process……………………………………………………….. 26

8.4.3 Electrodeposition Reactions………………………………………………………………… 27

8.4.4 Electrorefining Reactions…………………………………………………………………… 35

8.5 Semiconductor Electrodes……………………………………………………………... 38

8.5.1 Energy Levels of Redox Couples………………………………………………... 38

8.5.2 Energy Levels and Electron Transfer……………………………………………. 39

8.1 Introduction

The two broad categories of chemical reactions are atom transfer reactions and electron transfer reactions. In atom transfers, chemical reaction does not lead to changes in the oxidation states of the constituent atoms of the reactants. In Chapters 5-7 we limited our discussion to such reactions. In the present chapter, we turn our attention to the other major class of reactions, i.e., the case where chemical reaction introduces changes in oxidation states. While electron transfer reactions can occur in homogeneous solution, they oftentimes take place in heterogeneous systems. Such interfacial electron transfer reactions form the basis of important industrial processes such as electroplating, electrowinning, and electrorefining. The term electrochemical may be used strictly to refer only to electron transfer reactions at interfaces. However, a more flexible usage of the term is also widely accepted, where it refers to both homogeneous and heterogeneous processes and to the transfer of positive as well as negative charges. In this chapter we shall adopt this more general usage.

8.2 Redox Equilibria and Electron Activity

8.2.1 Redox Reactions

Consider the deposition of metallic copper via reaction of cupric ion with hydrogen gas:

Cu2+ + H2(g) = Cu + 2H+ (8.1)

This reaction can be viewed as consisting of two steps:

H2(g) = 2H+ + 2e- (8.2)

Cu2+ + 2e- = Cu (8.3)

That is, the copper deposition reaction involves the transfer of two electrons from a hydrogen gas molecule to a cupric ion. Reactions which involve such electron transfers are called oxidation-reduction or redox reactions.

The species which releases electrons, i.e., the electron donor is called the reductant, while the electron acceptor is called the oxidant. Another way of looking at this process is to think of a reductant as an electron complex. Thus an oxidation reaction is one in which an electron complex or reductant decomposes, i.e., releases its electron, while a reduction reaction is one in which an electron complex or a reductant is formed. Thus in reaction 8.1, Cu and H2 are reductants, since according to Equations 8.2 and 8.3, they are electron complexes. On the other hand, H+ and Cu2+ are oxidants. Furthermore, Equations 8.2 and 8.3 are respectively oxidation and reduction reactions.

8.2.2 Electron Activity

It is convenient to define a function pε which is a measure of electron activity (cf, pH, a measure of proton activity):

pε = -log{e-} (8.4)

When pε is large and positive, (i.e., low electron activity, {e}), we have strong oxidizing conditions. Since oxidants are electron acceptors, a system containing a high concentration of oxidant must necessarily contain a small amount of "free" electrons, i.e., the system will be characterized by a low level of electron activity. When pε is small or negative (i.e., high electron activity) we have strongly reducing conditions.

Consider the general redox reaction,

aA + bB + ne- = cC + dD (8.6)

The corresponding reaction quotient, Q, is given by:

[pic] (8.7)

where

Q' = (8.7a)

At equilibrium,

logQeq = logK (8.8)

Therefore,

[pic]

If each of the species A, B, C, and D is present at unit activity, it follows that

logK = (pε)eq, A, B, C, D at unit activity (8.10)

In order to provide a less clumsy way of writing Equation 8.10, the quantity pε° is defined, where

pε° = logK (8.11)

Therefore, Equation 8.9 can be rewritten as:

[pic]

where the subscript "eq" has been omitted for convenience.

Example 8.1 Redox Equilibria

Calculate pε values for the following equilibrium systems (assume that the solutions are infinitely dilute):

(a) An acidic solution 10-3 M in V3+ and 10-5 M in V2+

(b) A solution at pH 2 in equilibrium with hydrogen gas at 0.5 atm.

(c) A solution containing 10-5 mol/L RuO42- in equilibrium with RuO2⋅2H2O(s) at pH 14.

(d) A solution containing 10-4 mol/L Fe2+ in equilibrium with Fe3O4 at pH 6.

(e) An acidic solution containing 10-5 mol/L Cd2+ and at equilibrium with CdS(s) and elemental sulfur.

The following data are available:

V3+ + e- = V2+ logK = -4.31 (1)

2H+ + 2e- = H2(g) logK = 0 (2)

RuO42- + 4H+ + 2e- = RuO2⋅2H2O(s) logK = 67.33 (3)

Fe3O4(s) + 8H+ + 2e- = 3Fe2+ + 4H2O logK = 36.81 (4)

Cd2+ + S(s) + 2e- = CdS(s) logK = 12.16 (5)

Solution

(a) It follows from Equation 1 that

Q = [V2+]/[V3+] {e-}

At equilibrium log Q = log K and therefore

logK = log ([V2+]/[V3+]) - log {e-}

Thus,

pε = logK - log ([V2+]/[V3+])

= (-4.31) - log(10-5/10-3) = (-4.31) - (log 10-2)

= (-4.31) - (-2.0) = -2.31

Alternatively we can use Equation 8.12 directly:

pε = pεo - (1/n) log Q' (8.12)

Comparison of Equations 1 and 8.6 indicates that n= 1. Therefore from Equation 8.11,

pεo = (1/n) logK = -4.31 (6)

Also comparison of Equations 1 and 8.6 in the light of Equation 8.7a gives:

logQ' = log ([V2+]/[V3+]) = log (10-5/10-3) = - 2.0 (7)

It follows from Equations 6, 7, and 8.12 that

pε = -4.31 - (-2.0) = -2.31

(b) Referring to Equation 2,

Q' = PH2/{H+}2

Thus,

log Q' = log PH2 + 2pH = log (0.5) + 2(2) = (-0.30) + 4 = -3.70

Also n = 2, and log K = 0. Therefore from Equation 8.11, pεo = 0. Therefore by using Equation 8.12,

pε = (0) - (1/2) (-3.70) = 1.85

(c) From Equation 3,

logQ' = - log[RuO4 2-] + 4pH = -log(10-5) + 4(14) = 5 + 56 = 61

Also,

pεo = (1/n)logK = (1/2) (67.33) = 33.67

It follows from Equation 8.12 that:

pε = (33.67) - (1/2) (61) = 3.17

(d) From Equation 4,

logQ' = 3log[Fe2+] + 8pH = 3log(10-4) + 8 (6) = -12 + 48 = 36

Also,

pεo = (1/n) logK = (1/2) (36.81) = 18.4

Thus it follows from Equation 8.12 that

pε = (18.4) - (1/2) (36) = 0.4

(e) From Equation 5,

logQ' = -log[Cd2+] = - log(10-5) = 5

pεo = (1/2) (12.16) = 6.08

Therefore

pε = (6.08) - (1/2) (5) = 6.08 -2.5 = 3.58

example 8.2 Graphical Representation of Redox Equilibria

Consider the reaction

V3+ + e- = V2+ logK = -4.31 (1)

Under equilibrium conditions, we have (using concentrations in place of activities),

K = (2)

Also there is the following mass balance relation for the total dissolved vanadium, [V]T:

[V]T = [V2+] + [V3+] (3)

It follows from Equations 2 and 3 that:

[V3+] = (4)

and

[V2+] = (5)

Thus,

log[V3+] = log[V]T - logK - log[{e-} + K-1] (6)

log[V2+] = log[V]T - pε - log[{e-} + K-1 ] (7)

These results can be depicted graphically. We can make use of the following asymptotic conditions: When {e-}>>K-1, i.e., pε 0 (8.24)

Thus according to Equation 8.24, in a galvanic cell the more positive electrode is the cathode. On the other hand, for an electrolytic cell the overall reaction is not spontaneous, i.e., ΔG is positive and therefore Ecell = φc - φa < 0; that is, in this case the anode is the more positive electrode. Therefore, the terminals of electrochemical cells are classified as plus or minus, as shown in Figure 8.3. Note that electrons always leave a cell, whether galvanic or electrolytic, at the anode. On the other hand, electrons always enter a cell at the cathode.

[pic]

Figure 8.3 Sign convention for terminals of electrochemical cells.

According to Equations 8.23 and 8.24,

Ecell = φ c - φa = -ΔG/zF (8.25)

Thus recalling for the overall reaction that ΔG = ΔGo + RT ln Q, it follows that

Ecell = (-ΔGo/zF) - (2.303 RT/zF) log Q (8.26)

It should be noted that ΔG, ΔGo, and Q refer to the overall reaction, i.e., Equation 8.14. Recalling further, that ΔGo = - 2.303 RT log K, Equation 8.26 can be rewritten as:

Ecell = (2.303 RT/zF) logK - ) log Q (8.27)

or

Ecell = Eo - (2.303 RT/zF) log Q (8.28)

where Eo is the standard EMF of the cell and is given by:

Eo = -ΔGo/zF = (2.303 RT/zF) log K (8.29)

It can be seen by comparing Equations 8.28 and 8.29 that the standard EMF of the cell (Eo) represents the equilibrium cell potential under conditions where log Q has a value of unity. The definition of Eo is based on the specific case where all the aqueous species have unit activities and gases are at 1 atm pressure, and solid phases are in their most stable forms.

example 8.3 The Standard EMF of the Daniel Cell

The Daniel cell is a galvanic cell based on a zinc anode and a copper cathode. The cell diagram may be represented as:

Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)

(a) Write down the respective half-cell reactions

(b) Write down the overall cell reaction

(c) Determine the standard cell potential

The following data are available: ΔG= -35.18 kcal/mole, ΔG= 15.6 kcal/mole.

Solution

(a) Since the zinc electrode is the anode, it must undergo an oxidation reaction, i.e., it must release electrons:

Zn(s) = Zn2+(aq) + 2e- (1)

On the other hand, the copper electrode is the cathode and therefore it is the site of the corresponding reduction reaction, i.e., a reaction involving the consumption of electrons:

Cu2+(aq) + 2e- = Cu(s) (2)

(b) The overall reaction is obtained by combining Equation 1 and 2:

Zn(s) + Cu2+(aq) = Zn2+(aq) + Cu(s) (3)

(c) The standard cell potential is given by Equation 8.29:

Eo = -ΔGo/zF = (2.303RT/zF) log K (8.29)

It follows from Equation 3 that:

ΔGo = ΔG[Zn2+(aq)] + ΔG[Cu(s)] - ΔG[Zn(s)] - ΔG[Cu2+(aq)]

= (-35.18) + (0) - (0) - (15.6) = - 50.78 kcal/mole

Thus from Equation 8.29, with z = 2 (see Equations 1 and 2):

Eo = -(-50.78 kcal/mole)/2(23.0609 kcal V-1 mole-1) = 1.10 V.

8.3.3 Standard Electrode Potentials

The potential of an electrode does not have an absolute value, rather it is specified relative to the potential of a reference electrode. A common reference electrode is the standard hydrogen electrode (S.H.E.), which consists of a platinum electrode immersed in an aqueous solution of unit hydrogen ion activity and in contact with hydrogen gas at one atmosphere pressure. The standard hydrogen electrode is arbitrarily assigned a zero potential.

An electrochemical cell consisting of a metal electrode and the standard hydrogen electrode can be represented schematically as:

Pt, H2 (PH2 = 1 atm)| H+({H+} = 1) | | Mz+ ({Mz+})|M

The difference between the potential of the metal electrode and that of the standard hydrogen electrode is termed the electromotive force or EMF of the cell. The cell EMF is given by

Ecell = Eel - Eref = EM/Mz+ - EH2/H+ (8.30)

By definition, when PH2 = 1 atm and {H+} = 1, EH2/H+ = 0. Therefore, under these circumstances,

Ecell = EhM/Mz+ (8.31)

where the symbol "h" is used in Eh to signify that the potential is measured with reference to the standard hydrogen electrode. Thus it follows from Equation 8.31 that the electrode potential of a metal is the EMF of a cell formed between the metal electrode and the standard hydrogen electrode.

The half-cell reactions corresponding to the above electrochemical cell can be written for the metal and hydrogen electrodes respectively as:

Mz+ + ze- = M KM (8.32)

zH+ + ze- = (z/2)H2(g) KH (8.33)

The overall reaction is

Mz+ + (z/2) H2(g) = M + zH+ (8.34)

Thus invoking Equation 8.28, the corresponding cell potential is given by

Ecell = Eo - (2.303RT/zF) [zlog{H+} - log{Mz+} - (z/2) log PH2] (8.35)

Therefore, when {H+} = 1.0 and PH2 = 1 atm,

Ecell = EhM/Mz+ = Eh+ (2.303 RT/zF) log{Mz+} (8.36)

Furthermore, when {Mz+} = 1.0,

Ecell = Eh= -ΔGo/zF = (2.303 RT/zF) log K (8.37)

where EhoM/Mz+ is termed the standard electrode potential or the standard reduction potential for the M/Mz+ electrode. Table 8.1 presents values of standard electrode potentials for selected metals.

Metal electrodes with negative Eho values tend to behave as depicted in Figure 8.1b; that is, they will dissolve according to

M(electrode) = Mz+ (aq) + ze- (electrode) (8.38)

Thus, when electrical contact is made between the metal electrode and the Pt of the hydrogen electrode, electrons will flow to the platinum and result in the evolution of hydrogen:

2H+ + 2e- = H2(g) (8.39)

In Figure 8.4 electrons leave the cell through the M electrode and therefore this electrode represents the anode of the cell, while the hydrogen electrode is the cathode.

Table 8.1. Standard Reduction Potentials (Eho) in Acid Solution*

Couple Eo Volts

Li+ + e- = Li -3.045

K+ + e- = K -2.925

Rb+ + e- = Rb -2.925

Ba2+ + 2e- = Ba -2.90

Sr2+ + 2e- = Sr -2.89

Ca2+ + 2e- = Ca -2.87

Couple Eo Volts

Na+ + e- = Na -2.174

La3+ + 3e- = La -2.52

Mg2+ + 2e- = Mg -2.37

Sc3+ + 3e- = Sc -2.08

Th4+ + 4e- = Th -1.90

Be2+ + 2e- = Be -1.85

Hf4+ + 4e- = Hf -1.70

Al3+ + 3e- = Al -1.66

Ti2+ + 2e- = Ti -1.63

Zr4+ + 4e- = Zr -1.53

U4+ + 4e- = U -1.50

______________________________________________________________________________

Mn2+ + 2e- = Mn -1.18

Nb3+ + 3e- = Nb -1.1

Zn2+ + 2e- = Zn -0.763

Cr3+ + 3e- = Cr -0.74

Ga3+ + 3e- = Ga -0.53

Fe2+ + 2e- = Fe -0.44

Cd2+ + 2e- Cd -0.403

In3+ + 3e- = In -0.342

Co2+ + 2e- = Co -0.277

Ni2+ + 2e- = Ni -0.250

Mo3+ + 3e- = Mo ~-0.20

Sn2+ + 2e- = Sn -0.136

Pb2+ + 2e- = Pb -0.126

2H+ + 2e- = H2 -0.00

Cu2+ + 2e- = Cu +0.337

Hg+ 2e- = 2Hg +0.789

Ag+ + e- = Ag +0.7991

Rh3+ + 3e- = Rh ~ + 0.8

Pd2+ + 2e- = Pd +0.987

Au3+ + 3e- = Au +1.50

*Source: W. M. Latimer, Oxidation Potentials

[pic]

Figure 8.4 Behavior of metal electrodes with negative standard

electrode potentials (Eho): evolution of hydrogen gas.

8.3.4 Electron Activity and Redox Potential

According to convention the equilibrium constant for the hydrogen reaction (Equation 8.33) is unity, i.e., KH = 1. Thus, the equilibrium constant for the overall reaction (Equation 8.34) is given by

logK = logKM - logKH = logKM (8.40)

It follows therefore from Equations 8.36 to 8.40 that

[pic]

In view of the previous discussion in Section 8.2, Equation 8.32 can also be considered in terms of electron activity, i.e.,

pε = (1/z) logKM + (1/z) log {Mz+} (8.42)

where pεo = (1/z) logKM. A comparison of Equations 8.41 and 8.42 reveals the following relationships between Eh and pε:

Eh = (2.303 RT/F) pε (8.43)

Eho = (2.303 RT/F) pεo (8.44)

example 8.4 Standard Reduction Potentials

For each of the equilibrium systems listed in Ex. 8.1, determine the standard reduction potential as well as the Eh for the indicated solution conditions.

Solution

According to Equations 8.43, and 8.44,

Eh = (2.303RT/F) pε (8.43)

Eho = (2.303RT/F) pεo (8.44)

(a) V3+ + e- = V2+ logK = -4.31

(i) From Ex. 8.1, for logK = -4.31, pε° = -4.31. Therefore Eh° = 0.059 pε°,

volt = (0.059)(-4.31) V = -0.25 V

(ii) For 10-3M V3+, 10-5 M V2+, pε = -2.31 (Ex.8.1). Thus Eh = 0.059 pε, volt

= (0.059)(-2.31)V = -0.136 V.

(b) 2H+ + 2e- = H2(g) logK = 0

(i) From Ex. 8.1, for logK= 0, pε° = 0. Therefore Eh° = 0V

(ii) For pH = 2, PH2 = 0.5 atm, pε = 1.85. Thus Eh = 0.059pε, volt =

(0.059)(1.85)V = 0.109V.

(c) RuO42- + 4H+ + 2e- = RuO2.2H2O(s) logK=67.33

(i) From Ex. 8.1, for logK = 67.33, pε° = 33.67. Thus Eh° = 0.059 pε° =

(0.059)(33.67)V = 1.99V.

(ii) For [RuO42-] = 10-5 mol/L and pH 14, pε = 3.17 (Ex. 8.1). Thus Eh - 0.059

pε = (0.059)(3.17) V = 0.187V.

(d) Fe3O4(s) + 8H+ + 2e- = 3Fe2+ + 4H2O logK = 36.81

(i) From Ex. 8.1, pε° = 18.4. Thus Eh° = 0.059 pε° = (0.059)(18.4)V = 1.09V.

(ii) For [Fe2+] = 10-4 mol/L and pH 6, pε = 0.4 (Ex.8.1). Thus Eh = 0.059 pε

= (0.059)(0.4)V = 0.024V.

(e) Cd2+ + S(s) + 2e- = CdS(s) logK = 12.16

(i) From Ex.8.1, pε° = 6.08. Thus Eh° = 0.059 pε° = (0.059)(6.08)V = 0.36V.

(ii) For [Cd2+] = 10-5 mol/L, pε = 3.58 (Ex.8.1). Thus Eh = 0.059 pε =

(0.059)(3.58)V = 0.211V.

8.3.5 Minimum Electrolyzing Voltage

Consider the reaction

Mz+ + (z/2)H2O = z/4O2 + zH+ + M (8.45)

where

ΔG1 = z/4μO2 + zμH+ + μM - μMz+ -(z/2)μH2O (8.46)

Table 8.2 presents the ΔGo values of this reaction for several metals. It can be seen that in all cases, the free energy change is positive and therefore unfavorable for spontaneous reaction in the forward direction. Equation 8.45 can be re-expressed in terms of two half-cell reactions, i.e., a cathodic reaction yielding a metallic phase (Equation 8.47) and an anodic reaction giving rise to oxygen evolution (Equation 8.48).

Mz+ + ze- = M (8.47

(z/2)H2O = z/4 O2 + zH+ + ze- (8.48)

Table 8.2 Values of ΔG°r for the reactions Mz+ + (z/2) H2O = (z/4)O2 + zH+ + M

|___Mz+___ |ΔG°r (kcal/mol) |___Mz+___ |ΔG°r (kcal/mol) |

|Ag+ |9.9 |Ni2+ |67.5 |

|Au3+ |20.2 |Pb2+ |62.4 |

|Cr3+ |36.2 |Pd2+ |25.3 |

|Cu2+ |41.0 |Sn2+ |63.1 |

|Fe2+ |78.7 |Zn2+ |91.8 |

The ability to achieve metal deposition via electrolysis is based on the fact that the chemical potentials of the respective electrons associated with the cathodic and anodic reactions can be altered sufficiently to reverse the sign of the free energy change corresponding to the overall reaction (Equation 8.45).

Imagine a (divided) cell within which a metal deposition process is occurring. When an external source of electrical energy is applied, the electrons at the cathode and anode acquire different chemical potentials. Thus the overall reaction (obtained by combining Equations 8.47 and 8.48) can be written as

Mz+ + (z/2) H2O + ze= z/4 O2 + zH+ + M + ze (8.49)

where subscripts c and a denote the cathode and anode respectively. Assuming that aside from the electrons, the application of the electrical potential does not change the chemical potentials of the reactants and products, it follows that the free energy change associated with Equation 8.49 can be represented as:

ΔG2 = z/4 μO2 + zμH+ + μM - μMz+ - (z/2)μH2O + zμea - zμec (8.50)

Equations 8.46 and 8.50 can be combined to give

ΔG2 = ΔG1 + zμea - zμec (8.51)

In the absence of an externally applied voltage, the cathode and anode electrons have the same chemical potential, i.e.,

μea = μec (8.52)

Therefore under these circumstances,

ΔG2 = ΔG1 (8.53)

On the other hand, at equilibrium in the presence of applied potential, ΔG2 = 0 and therefore Equation 8.51 becomes

ΔG1 = - z(μea - μec)eq (8.54)

But at equilibrium, the chemical potential of an electron in an electrode is the same as that of an electron in the corresponding terminal. It follows therefore that Equation 8.54 can be written as

ΔG1 = -z(μeat - μect)eq (8.55)

where the subscripts at and ct respectively represent the anode and cathode terminals.

Recalling the relationship between electrical potential and the chemical potential of an electron (Equation 8.22), we can rewrite Equation 8.55 as:

ΔG1 = zF (φat - φct)eq (8.56a)

= zFEmin (8.56b)

where Emin = (φat - φct)eq is the minimum electrolyzing voltage (MEV).

Example 8.5 Minimum Electrolyzing Voltage for Copper Electrowinning

The higher the minimum electrolyzing voltage (MEV), the greater the energy requirement for a given electrolytic process. The MEV of a given electrochemical reaction cannot be changed; however, by changing the reaction itself, e.g., the anodic half-cell reaction, it is possible to substantially decrease the energy needed to obtain a given product.

Currently, oxygen evolution is the main anode reaction utilized in copper electrowinning. However, a number of novel reactions are receiving increasing attention. For each anode reaction listed below, determine the corresponding MEV for copper electrowinning, assuming standard conditions:

(a) H2O = 1/2O2(g) + 2H+ + 2e- ΔGo = 237.13 kJ mol-1 (1)

(b) 2Cl- = Cl2(g) + 2e- ΔGo = 262.46 kJ mol-1 (2)

(c) SO2(g) + 2H2O = SO+ 4H+ + 2e- ΔGo = 29.92 kJ mol-1 (3)

The cathodic reaction during the electrowinning process is:

Cu2+ + 2e- = Cu(s) ΔGo = -65.49 kJ mol-1 (4)

Solution

Assuming standard conditions means that PO2 = PCl2 = PSO2 = 1 atm, and {H+} = {Cl-} = {SO} = {Cu2+} = 1.0

(a) In this case the anodic reaction involves oxygen evolution. The corresponding overall reaction is obtained by combining Equations 1 and 4:

Cu2+ + H2O = Cu(s) + 1/2O2(g) + 2H+ ΔGo = 171.64 kJ mol-1

Recalling Equations 8.45, 8.46, and 8.56, and the fact that z = 2 (Eq. 4) it follows that

Emin = ΔG/zF = (171.64 kJ mol-1)/(2)(96.487 kJ V-1 mol-1) = 0.89 V

(b) The anodic reaction is chlorine evolution. Combining Equations 2 and 4 gives:

Cu2+ + 2Cl- = Cu(s) + Cl2(g) ΔGo = 196.97 kJ mol-1

Thus,

Emin = ΔG/zF = 1.02 V

(c) The anodic reaction is SO2 reduction. Combining Equations 3 and 4 gives the following overall reaction:

Cu2+ + SO2(g) + 2H2O = Cu(s) + SO+ 4H+ ΔGo = -35.57 kJ mol-1

Thus,

Emin = ΔG/zF = -0.184 V

EXAMPLE 8.6 Free energy of formation of barium titanate via electrochemical synthesis

Barium titanate films have been prepared by anodizing titanium in barium hydroxide solution. The anodic reaction can be written as:

Ti + Ba2+ + 3H2O = BaTiO3 (s) + 6H+ + 4e- (1)

For each of these two cases, determine whether or not the overall reaction is spontaneous: (a) The cathodic reaction is the hydrogen evolution reaction. (b) The cathodic reaction is the oxygen reduction reaction. Relevant thermodynamic data are provided below.

Species ΔGof (kcal/mole) Species ΔGof (kcal/mole)

Ba2+ -130.86 OH- -37.594

BaOH+ -171.48 H2O -56.687

BaTiO3 (s) -502.6

Solution

(a) With hydrogen evolution (Equation 2) as the cathodic reaction, the overall reaction is obtaianed by combining Equations 1 and 2:

Ti + Ba2+ + 3H2O = BaTiO3 (s) + 6H+ + 4e- (1)

4H+ + 4e- = 2H2 (2)

_________________________________________________________________________________

Ti + Ba2+ + 3H2O = BaTiO3 (s) + 2H+ + 2H2 (3)

The corresponding standard free energy change is given by:

ΔGor = ΔGof (BaTiO3 (s)) + 2ΔGof (H+) + 2ΔGof (H2) - ΔGof (Ti) - ΔGof (Ba2+)

- 3ΔGof (H2O)

= (-502.6) + 2(0) + 2(0) - (0) - (-130.86) - 3(-56.687) = -201.679 kcal/mole.

(b) With oxygen reduction as the cathodic reaction, the following half-reactions must be combined to get the overall reaction:

Ti + Ba2+ + 3H2O = BaTiO3 (s) + 6H+ + 4e- (1)

O2 + 4e- + 4H+ = 2H2O (4)

__________________________________________________________________________________

Ti + Ba2+ + H2O + O2 = BaTiO3 (s) + 2H+ (5)

ΔGor = ΔGof (BaTiO3 (s) ) + 2ΔGof (H+ ) - ΔGof (Ti) - ΔGof (Ba2+)

- ΔGof (H2O) - ΔGof (O2)

= (-502.6) + 2(0) - (0) - (-130.86) - (-56.687) - (0)

= -315.053 kcal/mole.

It can be seen from the above calculations that irrespective of the type of cathodic reaction, the free energy of reaction is negative. It may be concluded that the need for applied potential is a kinetic rather than a thermodynamic requirement.

Example 8.7 Electroleaching of Copper Sulfides in Chloride Solution

The Cymet Process for the electrodissolution and production of copper in chloride media is based on the anodic and cathodic reactions given below:

CuFeS2 + 5Cl- = CuCl+ FeCl2(aq) + 2S + 3e- logK = -31.55 (1)

CuCl+ e- = Cu + 3Cl- logK = 3.14 (2)

These reactions occur respectively in the anode and cathode chambers of a divided electrolytic cell. Assuming 100% current efficiency:

(a) Demonstrate that the free energy change for the overall reaction does not favor spontaneous chemical reaction.

(b) Determine the minimum electrolyzing voltage.

Solution

(a) The overall reaction is obtained by combining Equation 1 and 2:

(1): CuFeS2  + 5Cl- = CuCl+ FeCl2(aq) + 2 S + 3e- logK = -31.55

3x (2): 3CuCl+ 3e- = 3Cu + 9Cl- logK = 9.42

CuFeS2 + 2CuCl= 3Cu + FeCl2(aq) + 2S + 4Cl- logK = -22.13

The negative value of logK for the overall reaction indicates that the corresponding free energy change is positive and therefore unfavorable for spontaneous reaction. Hence the need for electrolytic processing.

(b) Based on the information provided in Equations 1 and 2,

Eh= (0.059) (1/3) (31.55)V = 0.62V

Eh= (0.059) (1) (3.14)V = 0.19V

Thus recalling Equation 15.7,

V˚min = Eh- Eh= (0.62 - 0.19)V = 0.43V

Example 8.8 Electroleaching of Copper Sulfides in Chloride Solution

The Dextec Process differs from the Cymet Process in that air is bubbled through the anode compartment to give an insoluble iron oxide reaction product. The resulting anodic reaction may be considered as:

CuFeS2 + 1/4 O2 + H2O + 3Cl- = CuCl+ 1/2 Fe2O3 + 2S + 2H+ + 3e- logK = -10.95

Repeat Example 8.7, assuming the same cathodic reaction as that given before.

Solution

a) The overall reaction is obtained by combining the above anodic reaction with the cathodic reaction given in Example 8.7:

CuFeS2 + 2CuCl  + 1/4 O2 + H2O = 3 Cu + 1/2 Fe2O3 + 2S + 2H+ + 6Cl- logK = -1.53

The free energy change in negative, which means that electricity need to be applied to drive the reaction in the forward direction. Note, however, that the equilibrium constant obtained here has a much lower absolute value than that found for the Cymet Process.

b) Eh= (0.059/3) (10.95) = 0.22 V

Eh= 0.19V (from Ex. 8.7)

Therefore,

V= Eh- Eh= 0.22 - 0.19 = 0.03V

8.4 Electrochemical Selectivity

8.4.1 Selectivity of the Anodic Process

As discussed above, in an electrolytic cell, one alters the chemical potentials of the cathodic and anodic electrons in order to obtain a negative free energy change for the desired overall reaction. Figure 8.5 presents a number of electrochemical processing systems exploited in industry. Electrodissolution is used as a leaching method in hydrometallurgical extraction and in the stripping of metallic coatings from defective electroplated articles; it is also an important step in the electrorefining of metals. Electrodeposition techniques are used in electrorefining and electrowinning operations as well as in electroplating. The term electrosynthesis is generally reserved for the processes in which non-elemental compounds are produced by electrolysis. The compound thus produced can be a final product which is packaged for sale, or it can be an intermediate product which is consumed by subsequent reaction. As indicated in Figure 8.5, electroleaching processes can involve electrosynthesis, e.g., an electrogenerated reagent (produced via electrosynthesis) can serve as a lixiviant.

Electrolytic Processing

Electrodeposition Electrodissolution Electrosynthesis

Electrowinning Electroplating Electrorefining Electroleaching

Figure 8.5 Types of Electrolytic Processing Systems

Consider an impure metal electrode undergoing electrolytic dissolution. The anodic reaction of one of the metallic components, M, can be described as

[pic]

In order for this dissolution process to be selective, impurities initially present in the anode must be prevented from dissolving.

Let Ehbe the standard electrode potential of M, then it follows that

Eh1 = Eh+ (0.059/z) log {Mz+} (8.58)

According to our previous discussion on electrolytic cells, Eh1 corresponds to the EMF of an electrochemical cell, as shown below:

M|Mz+||H+(aH = 1)| H2(Pt),(PH2 = 1atm)

Thus the reactions of interest during the anodic dissolution of M1 are:

M = Mz+ + ze- (anode) (8.59)

ze- + zH+ = (z/2)H2 (cathode) (8.60)

M + zH+ = Mz+ + (z/2)H2 (overall) (8.61)

The corresponding cell potential is given by

Ecell = Ehanode - Ehcathode = Eh1 (8.62)

since Ehcathode refers to the standard hydrogen electrode whose potential is zero by convention. Similarly, the cell potential needed to dissolve N electrolytically is Eh2, where

Eh2 = Eh+ (0.059/x) log {Nx+} (8.63)

Suppose N is an impurity in the anode. If the anodic potential is fixed at Eha = Eh1, then it follows that the available cell voltage would be insufficient to drive the dissolution reaction of N if

Eha = Eh1 < Eh2 (8.64)

Thus under these circumstances, M would be expected to dissolve selectively. In general then, during electrodissolution at Ea = Eh1, the more electropositive metals relative to M (i.e., Ehj > Eh1) do not dissolve while the less electropositive metals (or the more electronegative metal; Ehj < Eh1) dissolve. This situation is illustrated schematically in Figure 8.6a.

[pic]

Figure 8.6 (a) Selective electrodissolution, (b) Selective electrodeposition

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example 8.9 Selectivity of the Anodic Process in Electrorefining

Ag and Fe are metallic impurities in copper anodes to be used in an electrorefining process. Determine on the basis of thermodynamic considerations whether a given impurity will report to the aqueous electrolyte or will remain in the anode slime. The electrolyte contains 200 g/L H2SO4. The relevant chemical equations are:

Ag = Ag+ +e- logK = -13.50

Cu = Cu2+ +2e- logK = -11.44

Fe = Fe2+ +2e- logK = 16.17

Pb = Pb2+ +2e- logK = 4.27

Solution

Based on the available data, and the fact that Eho = (1/n) 0.059 log K, where K is the equilibrium constant for the reduction reaction, it follows that the expected results can be summarized as shown below:

Comparison

Reaction Eho(V) of Ehand Eh Conclusion

Ag = Ag+ + e- 0.797 Eh> Eh Slime

Cu = Cu2+ + 2e- 0.337 Eh Electrolyte

Fe = Fe2+ + 2e- -0.477 Eh< Eh Electrolyte

Pb = Pb2+ + 2e- -0.126 Eh< Eh Electrolyte

8.4.2 Selectivity of the Cathodic Process

Following our consideration of the electrode potential in terms of the EMF of an electrochemical cell, the pertinent reactions describing the cathodic deposition of M are given by the reverse of the reactions given in Equations 8.57, 8.59 and 8.60. The cell potential is now given by

Ecell,1 = Eha - Ehc = -Eh1 (8.65)

since in this case the anode is the standard hydrogen electrode. Considering the deposition of a metal N, the corresponding electrolyzing voltage is

Ecell,2 = -Eh2 (8.66)

Thus if Ecell,1 is less than Ecell,2 and the cathodic potential is fixed at Ehc = Eh1, Nx+ will remain in the electrolyte and will not be deposited. That is, selective deposition of M occurs if

Ehc = Eh1 > Eh2 (8.67)

This situation is illustrated schematically in Figure 8.6b

example 8.10 Selectivity of the Cathodic Process in Electrorefining

(a) A copper refining electrolyte contains 40 g/L Cu, 20 g/L Ni, 1 g/L As and 0.1 g/L Bi

(b) A silver refining electrolyte contains 1 g/L Cu and 1 g/L Pd

(c) A zinc refining electrolyte contains 1 g/L Mn

For each of these electrorefining systems, determine the feasibility of selectively depositing the valued metal.

Example 8.11 Minimum Fe3+ Concentration Needed to Prevent Interference with Cu Deposition

In a certain copper electrodeposition process, the cathode potential is maintained at 0.33 volts. A possible side reaction is the reduction of ferric to ferrous ions. For an electrolyte that contains 0.01 mol/L Fe2+, determine the highest Fe3+ concentration that can be tolerated without adverse effects on copper deposition.

Solution

The reactions of interest are:

Cu2+ + 2e- = Cu Eh1 = 0.33 volt

Fe3+ +e- = Fe2+ Eh= 0.771 volt

Therefore

Eh2 = 0.771 - 0.059 {log[Fe2+] - log [Fe3+]}

According to Equation 8.67 the copper deposition reaction proceeds selectively if

Eh1 > Eh2

That is, if

0.33 > 0.771 - 0.059 {log [Fe2+] - log [Fe3+]}

or if

log[Fe3+] < ((0.33 - 0.771)/0.059) + log [Fe2+]

i.e., if

[Fe3+] < 3.4 x 10-10 mol/L

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8.4.3 Electrodeposition Reactions

Electrolytic Reactions. The simplest electrodeposition electrolyte contains a dissolved metal salt, e.g., Mz+, A-, as well as the molecular and ionic constituents of water, i.e., H+, OH-, H2O. When electric current is passed through such an electrolyte, the possible anodic and cathodic reactions are:

Anode

Mz+ = M(z+1)+ + e- (8.68)

2Mz+ + (z + 1) H2O = M2Ox(s) + 2(z + 1)H+ + 2e- (x = z + 1) (8.69)

A- = A + e- (8.70)

H2O = 2H+ + 1/2 O2 + 2e- (8.71)

Cathode

Mz+ + ze- = M (8.73)

A- + e- = A2- (8.74)

2H+ +e- = H2 (8.75)

Table 8.3 presents the main electrode reactions for selected electrodeposition systems. For efficient electrodeposition operations, it is desired that the predominant cathodic reaction be the metal deposition reaction, i.e., Equation 8.73. Whether this requirement is achieved or not depends on the chemical behavior of Mz+, A-, and H+ as well as on the respective concentrations of these ions. The metal deposition reaction will be thermodynamically favored if the reduction potentials fulfill the conditions,

Table 8.3 Electrode Reactions in Electrodeposition

Anion Metals Main Anode Reaction Desired Cathode Reaction

SO Cu,Zn,Ni,Co H2O = 1/2 O2 + 2H+ + 2e- Mz+ + ze- = M

Cd,Cr,Mn,Fe

Cl- Cu Cl- = 1/2 Cl2 + e- CuCl+ e- = Cu + 3Cl-

S2O Ag 2S2O= S4O+ 2e- AgS2O+ e- = Ag + S2O

S2- Sb 2S2- = S+ 2e- SbSx+ me- = Sb + nS2-

CN- Au,Ag 4OH- = O2 + 2H2O + 4e- Au(CN)+ e- = Au + 2CN-

OH- Ga,Te 4OH- = O2 + 2H2O + 4e- Ga(OH)+ 3e- = Ga + 4OH-

EM > EA and EM > EH (8.76)

where the subscripts M, A and H refer to the metal, anion, and hydrogen species in the cathodic reactions of Equations 8.73, 8.74, and 8.75 respectively. Slow rate processes can however influence the observed selectivities. Thus zinc deposition is feasible primarily because of the extremely slow rate of the hydrogen evolution reaction on the zinc electrode surface. A similar effect permits metals such as Mn, Cr to be electrodeposited in spite of the fact that their standard electrode potentials are below that of the standard hydrogen electrode. Similarly if selective reduction of the anion is thermodynamically favored (i.e., EA > EM) but proceeds very slowly, it is possible to obtain selective metal deposition instead.

______________________________________________________________________________

example 8.12 Copper Electrodeposition in Sulfate Media

The following are possible cathodic reactions during copper electrodeposition in acidic sulfate solution:

Cu2+ + 2e- = Cu logK = 11.44

HSO+ 7H+ + 6e- = S + 4H2O logK = 33.92

2H+ + 2e- = H2 logK = 0

Determine the thermodynamic feasibility of selective copper deposition. Assume pH = 0, 10-2 < {Cu2+} < 1.0, {HSO} = 1.0.

Solution

Let the subscripts 1,2, and 3 refer to the copper, sulfate and hydrogen reactions respectively. Then

Eh= (0.059/2) (11.44) = 0.337V for Cu

Eh= (0.059/6) (33.92) = 0.334V for HSO

Eh= 0.0V for H+

and

Eh1 = 0.337 + (0.059/2) log {Cu2+}

Eh2 = 0.334 + (0.059/6) log {HSO} - (0.059/6) 7pH

Eh3 = - (0.059/2)pH

Using the given conditions, i.e., 10-2 < {Cu2+} < 1.0, {HSO} = 1.0

and pH = 0,

Eh1 = 0.337V for {Cu2+} = 1.0

= 0.278V for {Cu2+} = 10-2

Eh2 = 0.334V

Eh3 = 0.0V

Thus invoking Equation 8.67, the cathodic reaction is selective towards the copper deposition reaction if

Eh1 > Eh2 and Eh1 > Eh3

The second condition is clearly fulfilled for both {Cu2+} = 1.0 and 10-2. However, the first condition is barely fulfilled when {Cu2+} = 1.0 and not at all when {Cu2+} = 10-2. Thus in principle, during cathodic deposition of copper in sulfate solution, simultaneous deposition of sulfur should occur. The fact that the sulfur reaction is a result of the slow kinetics (i.e., high overpotential) of the sulfate reduction reaction.

Example 8.13 pH Constraints on Ni Electrodeposition

In what pH range should Ni electrodeposition be conducted in order to avoid H2 evolution and hydrolysis to Ni(OH)2 (s)? The anticipated nickel activity in solution would range from 1.0 to 10-2.

Solution

The competing cathodic reactions are:

2H+ + 2e- = H2 logK = 0

Ni2+ + 2e- = Ni logK = -7.99

Thus using the subscripts "1" and "2" for nickel and hydrogen respectively,

Eh1 = (0.059/2) (-7.99) + (0.059/2) log{Ni2+}

= -0.236 + 0.0295 log{Ni2+}

Eh2 = 0 - (0.059/2)2pH + (0.059/2) log PH2

= -0.059 pH, taking PH2 = 1 atm.

To avoid hydrogen evolution, we must have

Eh1 > Eh2

i.e.,

- 0.236 + 0.0295 log {Ni2+} > - 0.059 pH

Thus,

pH > (0.236/0.059) - (0.0295)/(0.059) log {Ni2+}

> 4.0 - 0.5 log{Ni2+}

i.e.,

pH > 4.0 for {Ni2+} = 1.0

> 5.0 for {Ni2+} = 10-2

Concerning hydroxide precipitation the relevant reaction is

Ni2+ + 2H2O = Ni(OH)2(s) + 2H+ logK = -12.73

This reaction proceeds from left to right if log Q < logK

i.e., if

- 2pH - log{Ni2+} < -12.73

pH > (12.73/2) - (1/2) log{Ni2+}

pH > 6.37 -0.5log{Ni2+}

Thus nickel hydroxide precipitation occurs when

pH > 6.37, for {Ni2+} = 1.0

> 7.37, for {Ni2+} = 10-2

Combining the results obtained for hydrogen evolution and hydroxide precipitation, we see that at the beginning of the process (i.e., {Ni2+} = 1.0) the constraint on pH is:

4.0 < pH < 6.37

On the other hand, at the end, where {Ni2+} = 10-2, the corresponding constraint on pH is:

5.0 < pH < 7.37

Example 8.11, involving nickel electrodeposition, illustrates the fact that the simultaneous need to minimize hydrogen evolution and avoid hydrolytic precipitation of metal may lead to a situation where electrodeposition is feasible only within a narrow pH range. In nickel electrowinning boric acid is used as a buffer to provide the necessary pH regulation.

In the case of the anodic reactions, Equations 8.68-8.71 give four different possibilities. Depending on the particular M-A-H2O system, anyone of these four may be exploited to advantage. Thus in sulfate solution, oxygen evolution is the predominant anodic reaction while in chloride solution, chlorine evolution may be favored. Where Equation 8.70 or 8.71 is the preferred anodic reaction, it is necessary to suppress the anodic reaction of the metal ion (Equations 8.68 and 8.69) otherwise the anodic current efficiency would be excessively low. The tendency for Equations 8.68 and/or 8.69 to interfere with the electrowinning process is especially serious when solutions containing multivalent metal ions (e.g. Mn, Cr, Fe systems) are electrolyzed. However, there may be specific instances when the preferred anodic reaction is that involving the metal ion. This is the case in certain copper-chloride process systems. Also, the anodic reaction described by Equation 8.69 can be exploited in the synthesis of oxide films and powders.

Example 8.14 Chlorine Evolution in Chloride-containing Electrodeposition Systems

Under what conditions will a chloride electrowinning electrolyte evolve Cl2 rather than O2?

Solution

The pertinent reactions are

2Cl- = Cl2(g) + 2e- logK = -46.0 (1)

H2O = 1/2 O2(g) + 2H+ + 2e- logK = -41.56 (2)

Using the subscripts 1 and 2 to refer to the chlorine and oxygen reactions respectively, and recalling that the standard electrode potential (Eh˚) is defined for a cathodic reaction (i.e. the reverse of Equations 1 and 2), it follows that:

Eh= (0.059/2) (46.0) = 1.36 volt

Eh= (0.059/2) (41.56) = 1.23 volt

Therefore,

Eh1 = 1.36 - (0.059/2) [2 log{Cl-}-logPCl2]

= 1.36 - 0.059 log {Cl-}, assuming PCl2 = 1atm.

Eh2 = 1.23 - (0.059/2) [2pH - 1/2 log PO2]

= 1.23 - 0.059 pH, assuming PO2 = 1atm.

According to Equation 8.64, in order for chlorine evolution to be the preferred anodic reaction Eh1 must satisfy the condition,

Eh1 < Eh2

That is,

1.36 - 0.059 log {Cl-} < 1.23 - 0.059 pH

pH < log {Cl-} - 0.13/0.059

pH < log {Cl-} - 2.20

Thus when {Cl-} = 1.0, the chlorine evolution becomes the thermodynamically selective anodic reaction provided the pH is below -2.2. Such a pH cannot be practically determined, however this result emphasizes the fact that extremely acidic solutions are required to ensure chlorine evolution.

example 8.15 Electrodeposition of Multivalent Metals: MnO2 Precipitation During Mn Electrowinning

What should be the maximum tolerable Mn2+ concentration in order to avoid MnO2 precipitation during O2 evolution at the Mn electrodeposition anode? Assume an anolyte pH of 1.0.

Solution

The MnO2/Mn2+ reaction is:

MnO2 + 2e- + 4H+ = Mn2+ + 2H2O logK = 41.90

This reaction proceeds from left to right if

log{Mn2+} + 2pε + 4 pH < 41.90

i.e., if

log{Mn2+} < 41.90 - (2/0.059)Eh - 4pH

The oxygen evolution reaction is given by

H2O = 2H+ + 1/2O2 + 2e- logK = -41.56

Thus

Eh = (0.059/2) (41.56) - (0.059/2) [2pH + 1/2 log PO2]

= 1.23 - 0.059 pH, assuming PO2 = 1atm.

= 1.7V for pH = 1

Using Eh = 1.17V and pH = 1,

log {Mn2+} < 41.90 - {2/0.059} (1.17) - 4(1)

< 41.90 - 39.66 - 4 = -1.79

Diaphragms in Electrowinning. If the predominant anodic reaction is that involving oxygen evolution, then as can be seen from Equation 8.71, increasing amounts of hydrogen ions are released into solution as the reaction progresses. Thus where the thermodynamic feasibility of the cathodic reaction requires a relatively high pH, the hydrogen ions produced by the anodic reaction must be prevented from reaching the cathode region, otherwise, the pH of the catholyte would fall below the necessary lower limit. In such cases then a divided cell is employed in which a porous cell divider or diaphragm is used to separate the catholyte and anolyte. Separation of anolyte from catholyte is also critical when multivalent metal ions are processed. For example, in manganese electrowinning, Mn2+ must be prevented from reaching the anode where it might be lost from solution as MnO2 (see Ex. 8.13). In chromium electrowinning, Cr6+ is produced at the anode and if it is allowed to enter the cathode compartment, it will decrease the efficiency of the cathodic reaction by oxidizing Cr2+ produced at the cathode from Cr3+.

example 8.16 Copper Electrowinning in Chloride Solution

In the Clear process, the electrodeposition of copper is conducted in chloride solution. Process conditions are such that both the cathodic and anodic reactions involve cuprous chloride complexes:

CuCl= CuCl+ + 2Cl- + e- logK = -7.91 (anode)

CuCl+ e- = Cu + 3Cl- logK = 3.14 (cathode)

a) What is the minimum electrolyzing voltage for this process? Assume unit activities of dissolved copper and chloride species.

b) Would this process require a divided cell?

Solution

a) Let the subscripts "a" and "c" refer to the anodic and cathodic reactions respectively. Then

Eh= (0.059) (7.91) = 0.47V

Eh= (0.059) (3.14) = 0.19V

Thus

Eha = 0.47 - 0.059 [log {CuCl} -2log {CuCl+} -2log {Cl-}]

If {CuCl} = {CuCl+} and {Cl-} = 1.0, then

Eha = 0.47V

Similarly,

Ehc = 0.19 - 0.059 [3log {Cl-} - log {CuCl}]

If {CuCl} = {Cl- = 1.0, then

Ehc = 0.19V

Therefore using Equation 8.56,

Vmin = Eha - Ehc = 0.47 - 0.19 = 0.28V.

b) The overall reaction is obtained by combining the cathodic and anodic half-cell reactions:

2CuCl= Cu + CuCl+ + 5Cl- logK = -4.77

The negative value of logK demonstrates that the forward direction does not proceed spontaneously, hence, the need for electrolysis. On the other hand, the reverse reaction has a favorable free energy change and should therefore proceed spontaneously. That is, in the presence of chloride ions, the cupric ion dissolves metallic copper to give cuprous chloride ions. Thus if the cupric ions generated at the anode are allowed to reach the cathode, the deposited copper will redissolve. Therefore a divided cell is needed to prevent this.

It must be noted however, that there is a net increase in chloride ion concentration in the cathode chamber since the cathodic reaction of one mole of CuClgenerates 3 moles of Cl- compared with 2 moles of Cl- for the anodic reaction. Consequently there is a chloride concentration gradient across the cell. Use of an anion-selective membrane as the cell divider permits chloride ions to be transported from the cathode to the anode chamber. This movement of chloride ions also serves to carry the current through the electrolyte.

8.4.4 Electrorefining Reactions

In electrorefining, an impure metal cast into an electrode is electrolytically dissolved by means of the anodic reaction,

M → Mz+ + ze- (8.87a)

Simultaneously, the pure metal is deposited electrolytically at the cathode via the reverse reaction:

Mz+ + ze- → Μ (8.87b)

In order for this refining process to be effective, impurities initially present in the anode must be prevented from dissolving; or failing this, they must be prevented from codepositing with the desired metal at the cathode. In other words, conditions must be created that promote selective electrodissolution as well as selective electrodeposition.

Thermodynamic data for a number of electrorefining reactions have been collected in Table 8.4. It will be recalled from Section 8.4.1 that for an M electrode containing N as an impurity, the anodic reaction is selective towards M so long as

Eh2 > Eh1 (8.78)

where Eh1 and Eh2 refer to the electrode potentials of M and N respectively. Thus any metals located above a given metal M in Table 8.4 will remain in the slime whereas any metals below M in the table would be expected to dissolve with M. On the other hand, in the case of the cathodic reaction, M will be deposited selectively if

Eh1 > Eh2 (8.79)

Accordingly, from a thermodynamic basis, metals below M in Table 8.4 dissolve with M but are not codeposited.

Table 8.4 Electrorefining Reactions

Reaction logK Eho(V)

Au = Au3+ + 3e- -76.03 2.24

Pt = Pt2+ + 2e- -32.55 0.96

Pd = Pd2+ + 2e- -30.94 0.91

As + 3H2O = AsO+ 6H+ + 3e- -46.23 0.91

Ag = Ag+ + e- -13.50 0.80

Cu = Cu2+ + 2e- -11.44 0.336

Bi = Bi3+ + 3e- -16.09 0.316

As + H2O = AsO+ + 2H+ + 3e- -14.13 0.28

As + 3/2 H2O = 1/2 As2O3 + 3e- + 3H+ -10.80 0.21

Sb + 2H2O = Sb(OH)+ 2H+ + 3e-  -10.1 0.20

H2 = 2H+  + 2e- 0 0

Pb = Pb2+ + 2e- 4.27 -0.125

Sn = Sn2+ + 2e- 4.75 -0.14

Ni = Ni2+ + 2e- 7.99 -0.236

Co = Co2+ + 2e- 9.53 -0.281

In = In3+ + 3e- 17.16 -0.337

Fe = Fe2+ + 2e- 16.17 -0.477

Zn = Zn2+ + 2e- 25.79 -0.761

Table 8.5 Electrorefining Anodes and Electrolytes

Anode Anode Composition Electrolyte

Sn 96%Sn, 1.0%Bi, 0.25%Sb, 0.15%As, 0.25Cu, 1.0%Pb H2SO4

In 99%In NaCl?

Bi 94%Bi, 2.2%Pb, 3.1%Ag, 0.5%Cu, 0.1%Sb, 0.1%Au HCl; SiF

Sb 87-94%Sb, 2-6%Pb, 3-6%Sn, 0.5%Cu, 0.5%Fe, 0.005%As NaCl; HF/H2SO4; NaOH?

Pb 93-98%Pb, 1.2%Sb, 0.17%As, 0.06%Bi, 0.05%Cu, SiF

0.004%Sn

Co 95%Co, 1.5%Ni, 0.8%Fe, 0.4%Cu, 0.6%S

Ni 95%Ni, 2.5%Cu, 0.75%Fe, 0.75%S, ?%Co H2SO4/HBO4; H2SO4/NaCl

Au 98%Au, 1%Ag, 0.2-0.5%Pt, 0.5%Pd, trCu, trPb HCl

Ag 95%Ag, 1-3%Au, 2%(Cu,Bi,Pb,Te,Fe,Ni,Pt) HNO3 

Cu 99%Cu H2SO4

Zn

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Table 8.5 presents a summary of typical compositions of electrorefining anodes, as well as the types of electrolytes that are used. Tables 8.4 and 8.5 may be used to speculate about the feasibility of effecting selective dissolution or selective deposition. For example, Table 8.5 indicates that the main impurities..........

In practice, the predicted thermodynamic selectivities are only observed in the special cases where overpotentials are low for the dissolution and cathodic reactions (e.g., Ag, Cu, Pb) and where the Eho values of impurities, differ substantially from Ehof the refined metals. In cases where there is partial dissolution of noble metals, such metals would be expected to precipitate on the cathodes and to avoid this, it is necessary to use diaphragms e.g., in Ni electrorefining.

The interaction of a metal with an anion present in the electrolyte can also influence the disposition of a given metallic impurity. Thus even though the analysis in Ex. 8.7 predicts that Pb will report to the electrolyte, in practice, the presence of sulfate ions in the electrolyte causes the potentially soluble lead ions to precipitate out of solution as PbSO4. Insoluble products may also result when anions generated by anodic dissolution react with other metal ions also released from the anode. Thus Sb dissolves from a copper anode initially as Sb(OH)3(aq) but subsequently reacts with dissolved arsenic to give an insoluble product of antimonous arsenate. In some cases, an anion is intentionally added to the electrolyte to ensure the precipitation of a specific metal. It was found in Ex. 8.7 that silver reports to the anode slime. In practice, however, a small amount of silver dissolution occurs which later results in Ag precipitation and therefore contamination of the Cu cathodes. Thus in order to further suppress Ag dissolution, the electrolyte is spiked with a small amount of chloride ions (~30 mg/L) which causes the precipitation of insoluble silver chloride. Another factor that can influence the selectivity of the anodic reaction is the chemical nature of the impurities in the solid phase of the anode. Thus the presence of an impurity as a relatively unreactive compound will enhance the selectivity of the anodic reaction.

Example 8.17 Effect of Anions on the Selectivity of the Anodic Process in Electrorefining

Re-examine Example 8.7, taking into consideration the possibility of forming sulfate complexes and precipitates.

example 8.18 Effect of Mineralogical Characteristics on the Anodic dissolution of Ag and Se.

Show that Ag and Se present in copper anodes as Ag and Cu selenides are likely to report to the anode slime.

8.5 Semiconductor Electrodes

8.5.1 Energy Levels of Redox Couples

In solid state physics, energy levels are usually expressed in units of electron volts (eV);

the reference state is taken as the energy of the free electron at infinity (vacuum), which is arbitrarily set at a zero value. The vacuum (eV) scale of energy levels may be related to the conventional electrochemical potential, Eh(volt), scale as:

E(eV) = -eEh (volt) - 4.5 (8.80)

where e represents the electronic charge. Thus relative to the vacuum scale, the standard hydrogen electrode potential represents an energy level at -4.5eV. Using Equation 8.80 it is possible to express redox and electrode potentials in terms of energy levels as illustrated in Figure 8.7.

In Figure 8.7a each redox couple is depicted in terms of a single energy level, i.e. Eredox. However, in fact, this energy level lies midway between the energy level of the oxidant (Eox) and that of the reductant (Ered) which constitute the redox couple (O>R). The energy levels of O and R occur at different energies because these species carry different ionic charges and therefore interact differently with the water dipoles. The two-energy level representation of the redox couple O/R is illustrated in Figure 8.7b. This figure also illustrates an additional effect, i.e, the Guassian distribution of energy levels in the aqueous phase; the thermal motions of the water molecules results in the time-dependent fluctuation of the energy levels.

[pic]Figure 8.7 Correlation between energy levels (E, eV, vacuum scale) and electrochemical potentials (Eh, V).

8.5.2 Energy Levels and Electron Transfer

An electrochemical reaction at the solid/aqueous interface involves interfacial charge transfer. In order for charge transfer to occur, the energy level of the aqueous species must fluctuate to the same energy as the conduction or valence band in the solid. The resulting alignment of the energy levels in the solid and aqueous species permits charge transfer to occur without an energy change. Thus the probability of charge transfer can be ascertained by comparing the energy levels of the redox couple (Ered, Eox) with the energy bands of the semiconductor (Ec, Ev). Depending on the specific energy correlations, charge transfer may involve the conduction band or the valence band. As illustrated in Figure 8.8a, if Eox is relatively near Ec, electron transfer with the conduction band is the more favored pathway for charge transfer. On the other hand, as can be seen in the Figure 8.8b, if Eox is closer to Ev than to Ec, then hole exchange with the valence band is the preferred pathway.

[pic] (a) (b)

Figure 8.8 Charge transfer at the semiconductor/aqueous interface: (a) Electron transfer with the conduction band, (b) Hole exchange with the valence band.

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FURTHER READING

1. J. N. Butler, Ionic Equilibria: A Mathematical Approach, Addison-Wesley, Reading, MA, 1964.

2. W. Stumm and J. J. Morgan, Aquatic Chemistry, Wiley, New York, 1970.

3. A. J. Bard and L. R. Faulkner, Electrochemical Methods, Wiley, New York, NY, 1980, pp. 630-636.

4. J. O'M. Bockris and A. K. N. Reddy, Modern Electrochemistry, Vols. 1 and 2, Pergamon, New York, NY, 1970.

5. C. L. Mantell, Electrochemical Engineering, 4th ed., McGraw-Hill, New York, 1960.

6. A. T. Kuhn, ed., Industrial Electrochemical Processes, Elsevier, New York, 1971.

7. F. A. Lowenheim, ed., Modern Electroplating 3rd. ed., Wiley, New York, 1974.

8. A. K. Graham, ed., Electroplating Engineering Handbook, 3rd. ed., Van Nostrand Reinhold, New York, 1971.

9. D. Pletcher, Industrial Electrochemistry, Chapman and Hall, New York, 1982.

10. R. G. Bautista and R. J. Wesely, Energy Reduction Techniques in Metal Electrochemical Processes,

TMS, Warrendale, PA 1985.

11. P. D. Parker, ed., Chloride Electrometallurgy, TMS, Warrendale, PA 1982.

12. D. G. Robinson and S. E. James, Anodes for Electrowinning, TMS, Warrendale, PA, 1984.

13. J. E. Hoffmann, R. G. Bautista, V. A. Ettel, V. Kudryk, and R. J. Wesely, The Electrorefining and Winning

of Copper, TMS, Warrendale, PA, 1987.

14. Energy Considerations in Electrolytic Processes, Society of Chem. Ind., London, 1980.

15. C. A. Vincent, Modern Batteries, Edward Arnold, Baltimore, MD, 1984.

16. C. L. Mantell, Batteries and Energy Systems, 2nd ed., McGraw-Hill, New York, 1983.

17. S.R. Morrison, Electrochemistry at Semiconductor and Oxidized Metal Electrodes, Plenum, New York, 1980, Chapters 1 and 2, pp. 1-78.

18. A.K. Vijh, Electrochemistry of Metals and Semiconductors, Marcel Dekker, New York, 1973, pp. 1-34.

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