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150, CHEMISTRY FOR DEGREE STUDENTS SEMESTER-II (PHYSICAL CHEMISTHY), Appl, PROBLEMS FOR PRACTICE, 1. The AE for the reaction, In, Fe (s) + Zn Zn (s) + Fe, I8 - 0.0032 volt. What is the concentration of Fe2 reached when a piece of iron 15 placed, in a 1 M Zn 2 solution?, (M), th, The, 2.303 RT, The value of, at 25°C is 0.059, [Ans. 1.283 M], 2. What will be the EMF of the cell, aisto, Zn, Zn2* (a = 0.1) || Fe2 (a = 0.005), Fe at 25°C?, Put, 2.303RT, 273K (as, Given:, = 0.059, [Ans. 0.314], E = 0.323 volts,, 3. Calculate the EMF of the cell, Pb Pb2+, Given that E for Sn, Sn2+, (Sn, Thi, (b), (a 0.200) || Sn2+ (a = 0.800) | Sn, Sn2+ + 2e ) = + 0.140 volt and, Pb2+ +2e) = + 0.126 volt and log 4 = 0.60., E° for Pb, Pb2+ (Pb, The elect, [Ans. 0.00373 volt], and the reaction of the cell will be Pb+ Sn2+ Pb* + Sn.], Ap, 4.18 NERNST EQUATION FOR SINGLE ELECTRODE POTENTIAL, (a) Metal-metal ion electrode. Take the example of zinc electrode in which zinc metal is in, equilibrium with zinc ions. The electrode reaction, written as reduction reaction will be, If, Zn2+ + 2e, Zn (n = 2), Applying Nernst equation, the electrode potential of the zinc electrode is given by, Eq, E, 'Zn2+,Zn, RT, az, In, ... (), %3!, Zn,Zn, 2F, of t.0, RT, a, ad lo 3, In Zn, 2F azna, E°, Zn,Zn, %3D, In og, where E, 7, represents the standard electrode potential of the zinc electrode. In the symbols, E, Ac, Zn and E, the ions have been written first because these represent reduction potentials., 'Zn,Zn, Putting am, = 1 (because the activity of a pure solid is unity), equation (i) becomes, Sir, RT, In az, 2F, Ez Zn, Zn,Zn, He, Assuming that the solution is dilute, the activity of Zn+2 ions may be replaced by mo, concentration., (c), E, E Zn, RT, In [Zn*], 2F, %3D, Zn,Zn, By, Similarly, the expression for the silver electrode may be derived. The electrode reaction., written as reduction reaction, is, Ag +e, Ag (n = 1), %3D, 1

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A the electrode reaction, written as reduction reaction, is, The olectrode reaction, written as reduction reaction, is, The electrode potential will then be given by the expression, Putting R-8314 joules, F 96500 coulombs, In[M*"] 2.303 log [M"] and T 25 +, In general, for any metal electrode, consisting of the metal (M) in equilibrium with its ions,, Applying Nemst equation and simplifying the result obtained, we have, 23K (assuming that the measurements are carried out at 25°C), the above equation takes the form, ELECTROCHEMISTRY 151, RT, In [Ag'], M +ne, M., ENM, RT, In [M*t1, nF, M*. M, 0.0591, log [M"], M, This is Nernst equation for the calculation of electrode potential of metal-metal ion electrode., ) Hydrogen electrode. Hydrogen electrode, consists of H, gas in equilibrium with H ions., H* +e, H, (n = 1), Applying Nernst equation, the electrode potential of the hydrogen electrode is given by, RT (4), Er M, In, H,H, F, ... (1), %3D, H, If H, gas is at 1 atmospheric pressure, a = 1, Equation (i) above takes the form, RT, In, a, H*.H, "H,H, RT, Er M, In a, F, H,H, Activity of H* ions may be replaced by its molar concentration, Thus, we have, RT, - In[H*], F, E, H".H, Since the standard electrode potential of hydrogen electrode is taken as zero, = 0, H.H, RT, In (H'|=0.0591 log[H'| at 25°C, %3D, Hence, E H2, olr, ) Calomel electrode. The electrode reaction, written as reduction reaction, is, 2Hg(/) +2C1 (n = 2), Hg,Cl,(s) + 2e, *y Nernst equation, the electrode potential of this electrode is given by, ax a, ... (7), RT, In, 2F, cal, cal, 2

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152 CHEMISTRY FOR DEGREE STUDENTS SEMESTER-M (PHYSICAL CHEMISTRY), As Hg is a pure iquid and Hg Cl, isa pure solid, therefore 1 and . Hence, takes the Rwm, RT, Taking molar conentration in place of activit, we have, In (Cr), 4.19 ELECTRODE POTENTIAL OF REDUCTION-OXIDATION, ELECTRODE, Consider the clectrode prepared by putting a platinum wire in a solution of a mixture of a, ferrous salt and a ferric salt, The electrode reaction, expressed as reduction reaction, is, Fe, Fe e, (*1), Applying Nemst equation, the electrode potential of this electrode is given by, salpha, RT, In, 2F, copper, be com, Using molar concentrations in place of activities, we have, RT Fe1, In, RT, General expression for any reversible electrode, The electrode reaction can be written in a general form as, Onidized state n electrons - Rechced state, Substi, Applying Nemst equation, the expression for the electrode potential will be, E-E, RT, In, IReduced statel, lOxidized statel, E-E+, RT, Oxidized state], or, In, (Reduced state], Here E represents the electrode potential at the given concentrations of the ions and, represents the standard electrode potential., degre, NUMERICAL PROBLEMS ON NERNST EQUATION, poten, FOR ELECTRODE POTENTIAL, Example 1. Calculate the electrode potentials (reduction potentials) of each of, following single electrodes at 25°C., () Sn | Sn (a-0.01) E",, --0.14V, S" S, (ii) Ag AgCl(s), CF (a-0.0001) E, 0.22V, ARCLCT, Solution. () The electrode reaction expressed as reduction reaction is, 3

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ELECTROCHEMISTRY, 153, Hence, Sn2 + 2e, Sn (n= 2), E Sn, RT, In a, 2F, /Sn, Substituting the values, we have, 0.0591, log 0.01, Es Sn, -0.14+, 0.14-0.0591=-0.1991 V, () The electrode reaction is, AgCl (s) + e, Ag(s) + CF (n = 1), RT, In acr, ABCI/CI, are of a, Substituting the values, EASCUCI, = 0.22 - 0.0591 log 0.0001, = 0.22 + 0.2364 0.4564 V, Example 2. Calculate the electrode potential of a copper wire dipped in 0.1 molar copper, sulphate solution at 25°C. At this temperature, the standard electrode potential of, copper is 0.34 volt. (F = 96500 coulombs, R = 5.314 joules/degree/mol). Assume CuSo, to, be completely ionized and take the activity of copper ions equal to the molar concentration, Solution. The electrode reaction as reduction reaction may be written as, Cu2+ + 2e, Cu, Here, n = 2, By Nernst equation, the electrode potential is given by, [Cu ], RT, In, E° +, nF, E =, [u], The concentration of solid is taken as unity, thus [Cu] 1, Substituting for [Cu²*] and [Cu], we have, 2.303RT, E = E' +, log [Cu ], nF, 2.303x8.314x 298, E = 0.34 +, log [0.1], or, 2x96500, 1014, E = 0.34 + (- 0.0296) = 0.3104 volt, or, Example 3. A zinc electrode is placed in 0.1 M solution of zine sulphate at 25°C. If the, egree of dissociation of salt at this concentration is found to be 0.95, calculate the electrode, potential of the electrode at 25°C. Given that E, and E, = -0.76volt., Zn*/ Zn, 2+, Solution, As a =0.95, the salt is 95% dissociated. Concentration of Zn2* ions is given by, 95, X0.130.095, 100, of the, [Zn"2], %3D, 0.0591, 0.0591, E +, log 0.095 =-0.76 +, 2, (-1.022), E =, 0.76 0.03 =-0.79 volt, 4

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154, CHEMISTRY FOR DEGREE STUDENTS SEMESTER-III (PHYSICAL CHEMISTRY, PROBLEMS FOR PRACTICE, 1. A zinc electrode is placed in 0.1 M solution of zinc sulphate at 25°C. If the degree of, dissociation of the salt at this concentration is found to be 0,95, calculate the electrode, potential of the electrode at 25°C. Given that E, =-0.76 volt., 95, x0.1 0.095 M], 100, [Hint As a = 0.95 i.e., the salt is 95% dissociated, [Zn]=, (Ans,-0.79 volt), 2. Calculate the electrode potential of the following electrode at 25° C., Zn | Zn* (a = 0.1) Given that E = 0.7618 volt., =-0.7618 volt], [Ans. E -0.7914 volt], [Hint. E., Zn,Zn, 4.20 CALCULATION OF EQUILIBRIUM CONSTANT (K) FROM, NERNST EQUATION, Nernst equation for a reversible cell in a state of equilibrium can be represented as:, 2.303 RT, -log K, nF, Ecell, E, rel, K denotes the equilibrium constant. Consider the Zn-Cu cell as represented below:, Zn(s) | ZnSO, (aq) CuSO, (aq) | Cu(s), Zinc acts as anode and copper acts as cathode in the above cell. The cell reaction can be, written as :, (a), Zn +2e, Cu (s), Zn (s), Cu* + 2e, (), As the reaction takes place, the concentration of ZnSO, increases while that of CuSO,, decreases with time. Thus the potential of zine and copper electrodes keep changing in accordance, with Nernst equation. At the point of equilibrium, the two electrode potentials become equal. At, this stage, the emf of the cell becomes zero (Fig 4.16). We observe no flow of electrons from the, anode to the cathode in the outer circuit. The cell reaction in a state of equilibrium can be written, by adding equations (ii) and (iii), Zn (s) + Cu+, 1 Zn (aq) + Cu (s), The equilibrium constant for the above reaction can be written as:, [Zn*] [Cu(s)], Reduction half reaction, K =, [Cu][Zn (s)], Taking the values of Cu (s) and Zn (s), as unity each, we have, Equilibrium, [Zn], [Cu"], state, %3D, Oxidation half reaction, Substituting the value of K in (i) above, 2.303 RT, [Zn], -log, (Cu*], Eell= Ell, ..(iv), Progress of reaction, nF, Fig. 4.16. The cell stops working after sometime when, the equilibrium is reached, 5, Reduction potential