Research: Science and Education
A Graphical Presentation of the Born–Haber Cycle for Estimating the Electrode Potentials of Metals Michael Laing School of Pure and Applied Chemistry, University of Natal, Durban 4041, Republic of South Africa;
[email protected] All modern textbooks on inorganic chemistry discuss redox reactions, electrochemistry, and standard electrode potentials (1). Data pertaining to electrode potentials are presented in several different ways: tabulated half-reactions and values for E ° (2, 3), Born–Haber cycles (4), Latimer potential diagrams (5), Frost volt-equivalent diagrams (6), and Pourbaix predominance diagrams (7). Each method has different advantages and each plays an important role. Which format one chooses depends upon convenience and upon what is under discussion.
sented in the recent text by Wulfsberg (9) and reproduced below: ⫹ M(s) + n H (hyd)
∆ HBH
− n ∆H hyd ∆H at
n H⫹(g)
−n2D ∆H hyd
− n IE
The Born–Haber Cycle Approach
n H(g)
n
∑ IEi
To get an estimate of the ease of oxidation of a metal, one can set up the well-known Born–Haber cycle for the reaction: Mn⫹(hyd)
M(s)
n H (g) + Mn⫹(hyd) 2 2
(1)
i =1
M(g)
n⫹
M
The overall enthalpy for the reaction, ∆HBH is made of two parts: the enthalpy for the half-reaction, ⫹
Mn⫹(hyd)
M(s) ∆H at
i =1
∆H rxn = ∆H at +
M(s) → Mn+(hyd) which will differ for each metal. The value of ∆H for the reaction 5 is:
n⫹
M (g) n
∑ IEi
i =1
+ ∆H hyd
(2)
∆Hrxn is positive and characteristic of each metal. The larger the value of ∆Hrxn, the more costly it is to oxidize the metal to form the hydrated cation. However, this hypothetical “reaction” is not directly measurable. More realistically, it is the standard electrode potential that is of interest, which is equivalent to the case of a metal “dissolving” in acid (8) ⫹ M(s) + nH (hyd)
n n⫹ M (hyd) + 2 H2(g) (3)
The corresponding electrochemical cell is represented as M(s) Mn + (hyd) H +(hyd) H2 (g) and will have a cell emf of E °cell V corresponding to the Gibbs free energy, ∆G °cell, for eq 3. E ° and ∆G° are related by the well-known equation: ∆G°cell = ᎑nFE °cell
(5)
which will be constant and, by definition, the value of its E ° will be zero volts; and the enthalpy for the reaction,
∆H hyd
n
∑ IEi
M(g)
n H (g) 2 2
nH (hyd)
∆H rxn
(g)
(4)
The enthalpies involved in this reaction may be conveniently displayed in a Born–Haber cycle as was nicely pre-
−n
D + IE + ∆H hyd 2
= −n
436 + 1312 − 1091 2
= −n × 439 kJ This value of 439 kJ corresponds to an E ° = 0 V. For each metal one can calculate the enthalpy sum, ∑∆H, n
∑ ∆H = ∆Hat + ∑ IEi + ∆Hhyd i =1
(6)
This sum will differ for each metal and will give an indication of the value of ∆G° in eq 4. Pertinent enthalpy data for metals representative of oxidation numbers +1, +2, and +3, negative values of E °, and positive values of E ° are given in Table 1 (8–11). One can plot these data step-by-step (as in Figure 1) and observe which energy factors dominate the activity of the metal. In all cases the enthalpy sum, ∑∆H, is a positive number or endothermic; the major difference is that the sum for metals with a negative value of E ° lies below the dotted line that corresponds to energies of n × 439 kJ兾mole (left hand side, Figure 1A). Conversely, the metals with positive values of E ° have sums that lie above the dotted line (right hand
JChemEd.chem.wisc.edu • Vol. 80 No. 9 September 2003 • Journal of Chemical Education
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IE3
5000 4500 4000
Σ∆H / kJ
This value is half the dissociation energy in units of kJ mol᎑1.
IE3
5500
∆Hhyd
3500
∆Hhyd
3000
IE2
IE2 2500 2000 1500
Co
IE1 1000
f
Values of Eesto are calculated from: [(∑∆H/n) − 439]/96.5 = Eest o. c
% Error = |[(∑∆H/n) − ∆Gm]/∆Gm|. e
∑∆H is the enthalpy sum given by eq 6 in the text.
Enthalpy data are taken from Tables 2.2, 2.3, 5.4 of ref 8. a
2957
-
1561
218f H+/H
759 415 Fe3+/Fe
1312
-
1450
1412 709
716
302
195
Sn /Sn
2+
Pb2+/Pb
-
1850 1067
1958 746
538
338
423
Cu2+/Cu
La3+/La
1723 131 Zn2+/Zn
906
3232
1810
1646 758 Co3+/Co
1007 61
425
Hg2+/Hg
284 Ag+/Ag
731
2745 1817 578 326 Al3+/Al
Values of ∆Gm for the half reaction M(s) → Mn+(hyd) are equal to (96.5 × Eo) + 439 kJ.
0 439
439 439
᎑4376
᎑1091
1316
1
᎑4 435
440
869
881
᎑1554
᎑1480
33
198
᎑241
472
595 ᎑3283
᎑2099
943
᎑77 362 ᎑2047
723
30 469
86 525
1408 ᎑4653
᎑1829
1049
108 547 ᎑468
547
᎑177 262 ᎑4680
786
᎑233 206
᎑240 199
411
᎑406
199
-
-
1451 738 148 Mg2+/Mg
496 109 Na+/Na
d
7 7 0.00
᎑0.037
43 2
25
22 413
415
26
24
᎑0.137
᎑0.125
0.00
0.00
0
9 2 09
472 33
᎑230 ᎑2.38
0.34 0.34
᎑2.49
4 3 66 ᎑73 ᎑0.76 ᎑0.80
13
4 521
482 43
82 0.85
0.45 0.31
0.89
31 5 16 77 0.80 1.12
15 2 77 ᎑162 ᎑1.68 ᎑1.83
5 211
1 78 ᎑261
᎑228 ᎑2.36
᎑2.71 ᎑2.49
᎑2.41
21
6000
b
1058
B
2
5
6
4
0
3
1
1
5
6
2
12
A
᎑1926
∆Gmd/ |(∑∆H/n) −∆Gm|/ % Errore kJ kJ ∆Gcell°/ kJ Eo/ V Eesto/ Vc (∑∆H/n)/ (∑∆H/n) − 439/ kJ kJ ∑∆H b/ (kJ mol᎑1) ∆Hhyd/ (kJ mol᎑1) IE3/ (kJ mol᎑1) IE2/ (kJ mol᎑1) Metal
∆Hat/ (kJ mol᎑1)
IE1/ (kJ mol᎑1)
Table 1. Pertinent Enthalpy Dataa for Metals Representative of Oxidation Numbers +1, +2, and +3
Research: Science and Education
Hg
IE1
∆Hat
Al
500
Mg
∆Hat
Ag
Na 0 0
ⴙ1
ⴙ2
ⴙ3
0
ⴙ1
ⴙ2
ⴙ3
Oxidation State Figure 1. Step-by-step plots of ∆Hat, IEi, ∆Hhyd for metals with common oxidation states +1, +2, and +3; the more active metals on the left (A), the less active on the right (B). The dashed line corresponds to the enthalpy for the reverse process shown in eq 5, i.e., n x 439 kJ. It is clear that the sums of enthalpies for Na+, Mg2+, and Al3+ fall below this line, while the sums for Ag+, Hg2+, and Co3+ fall above it.
side, Figure 1B). What is striking is how small the enthalpies of atomization of the metal are compared to the ionization energies and the hydration energies. A plot of the enthalpy sums for metals in oxidation states +1, +2, and +3 is shown in Figure 2. The energy differences between “reactive” and “unreactive” metals is relatively small. For example, the range for the +2 series, Mg to Hg, is only about 600 kJ; the gap between Na and Ag is less than 400 kJ; and the difference between Zn and Cu is only 220 kJ. If we plot ∑∆H兾n against oxidation state n (Figure 3), nH+兾(nH2兾2), from eq 5, is a horizontal line with a value of 439 kJ (corresponding to E ° = 0 V). The metals are clearly separated into two classes depending on whether ∑∆H兾n is less or more than 439 kJ: E ° is negative below the line and E ° is positive above the line; that is, the sums for the more active metals lie below this line. For each metal one can now plot ∑∆H兾n versus E ° for Mn+(hyd)兾M (Figure 4). The remarkable linearity over a range of 3 V and 300 kJ is immediately obvious, which shows that the sum of the enthalpies gives a good indication of the value of E °. If the line is extrapolated back, it intersects the horizontal axis at ᎑439兾96.5 ≈ ᎑4.5 V (the figure given by Sanderson in his eq 3, p 157, ref 10, and in a different form by Wulfsberg in his eq 7.8, p 325, ref 9).
Journal of Chemical Education • Vol. 80 No. 9 September 2003 • JChemEd.chem.wisc.edu
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ⴙ1
unreactive
1500 600
Ag
/ kJ Al
Zn Ag H
500
Mg
ⴙ1
0
Zn
ⴚ1
300
Al 200
negative E ¡
ⴚ2
Mg
Na
ⴚ3
reactive
100
0
ⴙ3
ⴙ2
Co
400
Na 0
Cu
H
n
positive E ¡
Σ∆H
Σ∆H / kJ
Hg Cu
1000
ⴙ1
Hg
500
E¡ / V
Co
ⴙ1
0
Oxidation State
ⴙ3
ⴙ2
Oxidation State
Figure 2. A plot of the enthalpy sum, eq 6, versus the oxidation state for the metals displayed in Figure 1 together with Zn and Cu. The diagonal line corresponds to n x 439 kJ, the H+/(H2/2) reaction, and clearly separates active metals (negative value for E ⬚) from inactive metals (positive value for E ⬚).
Figure 3. A plot of ∑∆H/n versus the oxidation number. The horizontal line corresponds to 439 kJ and 0 V. The units of the right hand axis are volts.
One can estimate the value of E ° for the cell from the equation:
The above comments emphasize again that for each metal it is the sum, ∆Hat + ∑IEi + ∆Hhyd, that dictates the value of E °, and that no one part of the Born–Haber energy cycle is proportional to E °. It is however clear that, depending on the metal, one or some combination of the energies may be the major influence. As might be expected, generation of the cation in the gas phase is the costly process and whether ∆Hhyd can overcome this determines whether the metal will be active, with a negative E °, that is (∆Hat + ∑IEi + ∆Hhyd)兾n < 439 kJ. The size of the influence is seen in Figure 1 for the case of Al3+ and Co3+ where the maxima are marked with a star. Figure 2 shows that the energy difference between active and inactive metal is quite small: about 350 kJ for Na versus Ag and less than 250 kJ for Zn versus Cu.
n
E °est =
∆H H at + ∑ IEi + ∆H hyd i =1
n
− 439
1 V 96.5
(7)
The value of E °est obtained from the enthalpy sum is close to, but not equal to E °. The lack of ideality is due to ignoring entropy effects, that is, the assumption that ∆H = ∆Gm for the reaction M(s) → Mn+(hyd) is not correct (8). The difference between the values of ∆H and ∆Gm can be up to about 10% of ∆Gm. The implications of this error are considered later. Which Energy Dominates?
600
Ag Cu
n
/ kJ
500
Σ∆H
One may now ask, in each case does any one of the three enthalpies dominate the value of E °? Figure 1 shows quite well that for Na versus Ag the combination of the large values of both ∆Hat and IE1 account for the low activity of Ag. Comparing Hg and Mg, it is the large value for the sum of IE 1 + IE 2 that dictates the positive value of E ° for Hg2+兾Hg. The case of Al3+兾Co3+ is not so obvious: the values of ∆Hhyd are similar. It is the sum of IE1 + IE2 + IE3 that causes the difference. (It is the large value of IE3 that makes Co3+ unstable relative to Co2+.) If one examines the lanthanides, for example La, it is evident that the values of IE2 and IE3 are significantly smaller than the corresponding values in Al and Co (as a result of the much larger atomic radius; see Table 1). While ∆Hhyd is much smaller than, for example, Al3+, it more than overcomes ∑IEi and thus makes La3+(hyd) stable and causes the metal to be a strong reducing agent, E ° = ᎑2.38 V (Figure 5).
H 400
Zn
300 200
Hg Co
Al Na Mg
100 0
ⴚ2
ⴚ1
0
ⴙ1
E° / V Figure 4. A plot of ∑∆H/n versus known E ⬚ for Mn+/M for the metals presented in Figures 1, 2, and 3. The good linearity indicates that the assumption that ∑∆H ≈ ∆Gm is remarkably good.
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Exercises for Students E°’s and Enthalpy Sums If students are to develop a feel for the activity of metals, they should understand which factors are involved and how each factor influences the value of E °. To do this the students should collect enthalpy data, ∆Hat, IEi, and ∆Hhyd, for some representative metals (other than those in Table 1 and Figures 1–5) They should plot the data step-by-step as in Figure 1, calculate the sums and plot these as in Figure 2, followed by plotting values of ∑∆H兾n, firstly versus oxidation state, and then versus known values of E ° for Mn+兾M to see how good the approximation is. The students should plot the values for Sn, Pb, and Fe on Figure 4 to see the effect of ignoring the contribution of the entropy factor. The visual presentation of the data should help the student see the relative importance of each energy factor, and understand that the Born–Haber cycle is not solely an exercise in numbers. They should then calculate E °est from eq 7 and consider what aspect of the energy terms gives rise to the difference between E ° and E °est. (See ref 8, p 144, Table 5.1)
6000 5500 5000 4500
Σ∆H / kJ
4000
IE3
3500 3000 2500 2000
∆Hhyd
IE2
1500 1000
Co3ⴙ
IE1 Al 3ⴙ
La3ⴙ
500
Co2ⴙ
∆Hat
0 0
ⴙ3
0
ⴙ3
0
ⴙ3
0
ⴙ2
Oxidation State Figure 5. Plots of the comparative sums of ∑∆H/n for La3+/La, Al3+/ Al, Co3+/Co, and Co2+/Co. The plimsol symbols ( ) mark the equivalent enthalpies for n H+兾(n H2兾2), and show clearly that E⬚ for Co3+/Co will be positive while all other E⬚’s will be negative. For the case of Co3+, it is evident that despite its large negative value, the value of ∆Hhyd is not enough to overcome the large value of IE3 (bigger by far than the values of IE3 for both Al and La).
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Relative Stabilities of Different Oxidation States The study can then be extended for metals with two easily attained oxidation states, for example Fe2+ and Fe3+; Cr2+ and Cr3+; Eu2+ and Eu3+; and Mn2+ and Mn3+. It should quickly become evident that it is the sizes of the enthalpies of hydration of the M2+ and M3+ cations that will determine the relative stability of the two oxidation states (a possibly unexpected result). The reasons for irregularities in the energy terms can be explained in terms of the electron configurations of the atoms (for ∆Hat and IE1) and of the ions (for IE2 and IE3). By comparing the different plots, the students can see how the different values of nE °, volt-equivalents, can be used in Frost diagrams to compare relative stabilities of adjacent species, for example M2+, M3+, M4+, while the values of (∆Hat + ∑IEi + ∆Hhyd)兾n can be used to estimate a value of E ° for the half-cell potential corresponding to Mn+兾M. The Difference between E° and E °est: Entropy Effects The values of E ° and E °est for the various metals are given in Table 1. The magnitude of the difference between Gibbs free energy change, ∆Gm, and the enthalpy sum, ∑∆H, for the half reaction M(s) → Mn+(hyd) ranges from 0 up to 31 kJ (Table 1); that is, the % error in assuming that ∆H = ∆Gm is between 0 and 12%. The values of E° for Na and Ag are overestimated by 21 and 31 kJ, respectively, while the values for both Al and La are underestimated, by 15 and 9 kJ, respectively. The difference between E °est and E° for each metal (Table 1) shows that the errors are between 0 and 0.3 volts, with the largest error being for Na+兾Na. (It is immediately obvious that one cannot calculate percentage error in E °est by the function |(E °− E °est)兾E°| because it would become infinity for E ° = 0 V, which clearly is nonsense. The same applies to using ∆G°cell as this too can be zero. One must therefore use the energies ∆H and ∆Gm for M(s) → Mn+(hyd) for estimating percentage errors.) The size of the error that results from assuming ∆H = ∆Gm for the half-reaction M(s) → Mn+(hyd) is nicely seen for metals whose standard electrode potentials are close to zero; that is, whose enthalpy sums are close to n × 439 kJ. Tin, lead, and iron are examples in which E ° values are: Sn2+兾Sn = ᎑0.137 V, Pb2+兾Pb = ᎑0.125 V, Fe3+兾Fe = ᎑0.037 V. Their enthalpy data are given in Table 1. The enthalpy sums, ∑∆H兾n, are close to 439 kJ; that is, the values of E °est are zero volts. The values of ∆Gm, deduced from the standard electrode potential E °, differ by less than 10% from the enthalpy sum ∆H—an encouraging result. For the student it is valuable to see that the enthalpy sum gives a value of E °est that is satisfactorily close to the known standard value of E °, and hence this value of E °est is a fair indicator of the activity of a metal. The magnitude and direction of the difference between E ° and E °est is difficult to explain. The entropy contribution, T∆S, is small, about 10% of ∆Gm for group 1 metals. What we do observe is an underestimate of E ° for M+ species and an overestimation for M3+ species. The entropy effect is related to the extent of structure-making associated with the [Mn+(H2O)x] species in aqueous solution. One gets the feeling that ordering of H2O molecules around the M3+ species will be significantly greater
Journal of Chemical Education • Vol. 80 No. 9 September 2003 • JChemEd.chem.wisc.edu
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than around the M+ species. The effect will also be related to the radius of the Mn+ cation because this will dictate the number of H2O molecules that can comfortably coordinate in the hydration sphere. A more detailed consideration is beyond the scope of this article, but the problem is discussed in detail by Dasent (8) in Chapter 5. Conclusion All that has been said in this paper has been in terms of thermodynamics—enthalpies of idealized reactions. Although these enthalpies indicate whether a reaction is possible, none of this can tell us whether a reaction will occur under ambient conditions, nor how fast it will go. This is the realm of kinetics and the properties of the surface of the metal in question. Nevertheless, displaying the energies step-by-step, as in Figure 1, and as sums in Figures 2 and 3, gives a visual indication of which energy factors are most influential and which Mn+(hyd)兾M(s) half reactions will have negative values of E ° and hence which metals should react with dilute acids. For the student it is important to see that the sum of three welldefined enthalpies (plotting the enthalpy data ∆Hat, IEi, and ∆Hhyd for the metals Sn, Pb, and Fe is a suggested exercise to learn this) can give a good indication of E ° for a metal, and hence that the Born–Haber cycle approach will give a good estimate of the chemical reactivity of the metal.
Literature Cited 1. Moeller, T. Inorganic Chemistry; Wiley: New York, 1982; pp 565–583. 2. Huheey, J. E.; Keiter, E. A.; Keiter, R. L. Inorganic Chemistry, 4th ed.; Harper Collins: New York, 1993; pp 378–381, 587– 599, A35–A37. 3. Porterfield, W. W. Inorganic Chemistry, 2nd ed.; Academic: San Diego, 1993; pp 411–420. 4. Parish, R.V. The Metallic Elements; Longman: London, 1977; pp 22–32, 244. 5. Bowser, J. R. Inorganic Chemistry; Brooks/Cole: Pacific Grove CA, 1993; pp 270–293, 484. 6. Phillips, C. S. G.; Williams, R. J. P. Inorganic Chemistry; Oxford University Press: Oxford, 1965; Vol. I, pp 311–325; Vol II, p 170–172. 7. Douglas, B. E.; McDaniel, D. E.; Alexander, J. J. Concepts and Models of Inorganic Chemistry, 3rd ed.; Wiley: New York, 1994; pp 354–361. 8. Dasent, W. E. Inorganic Energetics, 2nd ed.; CUP: Cambridge, England, 1982; pp 35–41, 44–45, 144, 152, 158–163. 9. Wulfsberg, G. Inorganic Chemistry; University Science: Sausalito CA, 2000; pp 287–297, 315–326. 10. Sanderson, R. T. Inorganic Chemistry; Reinhold: New York, 1967; p 157. 11. Johnson, D. A. Some Thermodynamic Aspects of Inorganic Chemistry, 2nd ed.; Cambridge University Press: Cambridge, UK, 1982; pp 89–95, 106–109, 144–147, 150–156, 179–182.
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