32
Chemical equilibrium: electrochemistry 자자자자자자 자자자 자자자 자자

Chemical equilibrium: electrochemistry 자연과학대학 화학과 박영동 교수

Embed Size (px)

Citation preview

Chemical equilibrium: electrochemistry

자연과학대학 화학과박영동 교수

ExamplesIn distilled water

Hg2(IO3)2 ⇄ Hg22+ + 2IO3

Ksp = 1.3 × 10-18

Ksp = [Hg22+][IO3

]2 = x(2x)2 → x = [Hg22+] = 6.9 × 107

In 0.050 M KNO3 [Hg22+] = 1.0 × 106

→ 이온세기가 커지면 Hg22+ 와 IO3- 간의

인력은 순수한 물에서보다 줄어듦 → 서로 합치려는 경향이 줌 → Hg2(IO3)2 의 용해도가 !

Typical aqueous solution

q1q2

ϕ=q

4 𝜋𝜀 r

Examples

-

--

++

+

-

--

++

+

The activities of ions in solution

Fig. 5.33 The picture underlying the Debye-Hückel theory is of a tendency for anions to be found around cations, and of cations to be found around anions (onesuch local clustering region is shown by the circle). The ions are in ceaseless motion, and the diagram represents a snapshot of their motion. The solu-tions to which the theory applies are far less con-centrated than shown here.

where A = 0.509 for an aqueous solu-tion at 25°C and I is the dimension-less ionicstrength of the solution:

Mean activity coefficients

Debye-Hückel limiting law.

extended Debye-Hückel law

ionic strength of the solution:

Fig. 5.34 An experimental test of the Debye-Hückel lim-iting law.

Fig.5.35 The extended Debye-Hückel law gives agreement with experiment over a wider range of molalities.

The variation of the activity coefficient with ionic strength according to the extended Debye–Hückel theory. (a) The limiting law for a 1,1-electrolyte. (b) The extended law with B = 0.5. (c) The extended law, extended further by the addi-tion of a term CI; in this case with C = 0.2. The last form of the law reproduces the observed behaviour reasonably well.

Calculate the ionic strength and the mean ac-tivity coefficient of 5.0×10-3 mol kg-1 KCl(aq) at 25°C.

Self-test 5.8 Calculate the ionic strength and the mean activity coefficient of 1.00 mmol kg-1 CaCl2 (aq) at 25°C.

where A = 0.509 for an aqueous solu-tion at 25°C and I is the dimension-less ionicstrength of the solution:

Answer: [3.00 mmol kg-1, 0.880]

I =½(b+ + b_)/bo= b/ bo

where b is the molality of the solution (and b+ = b_ = b).

log γ±= -0.509 × (5.0 × 10-3)1/2 = -0.036Hence, γ± = 0.92. The experimental value is 0.927.

migration of ions

V = IR

R = ρL/A

κ = 1/ρ : conductivity

Λm = κ/c : molar conductivity

Λm = Λmo

– Kc1/2

Λmo : limiting molar

conductivityΛm

o : = λ+ + λ−

Ionic conductivitiesTable 9.1 Ionic conductivities, λ/(mS m2 mol−1)

Ionic mobilities in waterIonic mobilities in water at 298 K, u/(10−8 m2 s−1 V−1)

Grotthus mechanism

Electrochemical cells

two electrodes share a common electrolyte.

Figure 9.6 When the electrolytes in the electrode compartments of a cell are different, they need to be joined so that ions can travel from one com-partment to another. One device for joining the two compartments is a salt bridge.

a galvanic cell

Figure 9.7  The flow of elec-trons in the external circuit is from the anode of a galvanic cell, where they have been lost in the oxidation reaction, to the cathode, where they are used in the reduction re-action. Electrical neutrality is preserved in the electrolytes by the flow of cations and an-ions in opposite directions through the salt bridge.

an electrolytic cell

Figure 9.8  The flow of elec-trons and ions in an elec-trolytic cell. An external sup-ply forces electrons into the cathode, where they are used to bring about a reduction, and withdraws them from the an-ode, which results in an oxida-tion reaction at that electrode. Cations migrate towards the negatively charged cathode and anions migrate towards the positively charged anode. An electrolytic cell usually consists of a single compart-ment, but a number of indus-trial versions have two com-partments.

a (H+) = 1.00

H2 (g)

e-

Pt gauze

The Standard Hydrogen Electrode

• Eo (H+/H2) half-cell = 0.000 V

f{H2(g)} = 1.00

a silver–silver-chloride electrode

Figure 9.10  The schematic structure of a silver–silver-chloride elec-trode (as an example of an insoluble-salt electrode). The electrode consists of metallic silver coated with a layer of silver chloride in contact with a solution containing Cl− ions.

a redox electrode

Figure 9.11  The schematic structure of a redox electrode. The plat-inum metal acts as a source or sink for elec-trons required for the in-terconversion of (in this case) Fe2+ and Fe3+ ions in the surrounding solution.

A Daniell cell

Figure 9.12  A Daniell cell consists of copper in contact with copper(II) sulfate solu-tion and zinc in contact with zinc sulfate solu-tion; the two compart-ments are in contact through the porous pot that contains the zinc sulfate solution. The copper electrode is the cathode and the zinc electrode is the anode.

measurement of cell poten-tial

Figure 9.13  The po-tential of a cell is measured by bal-ancing the cell against an external potential that op-poses the reaction in the cell. When there is no current flow, the external potential difference is equal to the cell potential.

The Nernst Equation

At 25.00°C,

Calculating an equilibrium constantExample 9.5                               Calculate the equilibrium constant for the disproportionation reaction 2 Cu+(aq)   Cu(s) + Cu2+(aq) at 298 K.Strategy The aim is to find the values of  corresponding to the reaction, for then we can use eqn 9.16. To do so, we express the equation as the difference of two reduction half-reactions. The stoichiometric number of the electron in these matching halfreactions is the value of V we require. We then look up the standard potentials for the couples corresponding to the half-reactions and calculate their difference to find  . Use RT/F = 25.69 mV (written as 2.569 × 10−2 V).Solution The two half-reactions are                                                                                                                                                                                                                                                                                   The difference is                                                                                                                                                                                                                                        It then follows from eqn 9.16 with ν= 1, that                                                                                                                                                      Therefore, because K = elnKK,K = e37/2.569 = 1.8 × 106

Because the value of K is so large, the equilibrium lies strongly in favour of products, and Cu+ disproportionates almost totally in aqueous solution

A glass electrode

Figure 9.14  A glass elec-trode has a potential that varies with the hydrogen ion concentration in the medium in which it is immersed. It consists of a thin glass mem-brane containing an elec-trolyte and a silver chloride electrode. The electrode is used in conjunction with a calomel (Hg2Cl2) electrode that makes contact with the test solution through a salt bridge; the electrodes are normally combined into a single unit.

Silver/Silver Chloride Electrode

m E m1/2 E+C

0.00100 0.5788 0.03162 0.22404

0.00322 0.5205 0.05670 0.22575

0.00450 0.5037 0.06708 0.22619

0.00562 0.4926 0.07496 0.22646

0.00600 0.4894 0.07746 0.22666

0.00750 0.4784 0.08660 0.22712

0.00914 0.4686 0.09559 0.22747

0.02563 0.4182 0.16009 0.23007

0.04000 0.3965 0.20000 0.23119

0.06000 0.3770 0.24495 0.23251

0.08000 0.3630 0.28284 0.23329

0.10000 0.3525 0.31623 0.23425

0.12380 0.3420 0.35185 0.23470

Temperature dependence of the standard potential of a cell

Figure 9.15  The variation of the standard potential of a cell with temperature depends on the standard entropy of the cell reaction.

Silver/Silver Chloride Electrode

Ag/AgCl Electrode

T (°C) E0(V vs SHE)

25 0.22233

60 0.1968

125 0.1330

150 0.1032

175 0.0708

200 0.0348

225 -0.0051

250 -0.054

275 -0.090

E0(V) = 0.23735 - 5.3783x10-4t - 2.3728x10-6t2 - 2.2671x10-9(t+273) for 25 < t < 275 °C

0 50 100 150 200 250 300-0.10

-0.05

0.00

0.05

0.10

0.15

0.20

0.25

E0(T) Ag/AgCl

T/( oC)

Eo/V

Thermodynamic properties

Δr G0=−𝜈 F E0 and 𝜕G0/𝜕T=− S0

Δr G0=−𝜈 F E0

d E0

dT=

Δr S0

𝜈 For Δr S0=𝜈 F

d E0

dT

)

Thermodynamic properties

E0 = 0.22232 VF = 96486 C mol-1