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Homework 10b

# Homework 10b - *7 ‘5 = LC‘MMHDJQ 7.029K10 x 4 g Mano...

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Unformatted text preview: *7 ‘5 = LC‘MMHDJQ +7.029K10 x 4 g Mano ‘50 . 'i =:o.03‘l|3‘93 VI = 0x32 4/37 X‘l‘ = 0.533 qe‘ “ 'I =0 2 =ﬁ¢1q9xﬂo — 5.7?x/O I 9.85q6 II n =‘0 035317 :95 Kt: ow,— " la "F—:s~*——-—*— II ’0 = Cit/x0 ff :1 7 x/O 0.919” M I: “M Wk; I°lo&“ww_ “ a 5’9 3 la 9.33 The electric mobility of Mg2+(aq) and C1’(aq) at inﬁnite dilution and 298.15 K are 55.0 X 10‘5 cm2 V"1 s"1 and 79.1 X 10‘5 cm2 V"1 s‘], respectively. Compute the transference numbers and molar ionic conductances of Mg 2+(aq) and C1'(aq) at infinite dilution. What is the value of Aﬁgaz at 298.15 K? Solution The transference numbers are given by Eq. 9.124: u + -5 “‘48” = uMgz+Mi 140— = 55.0 x :32 : :21 x 10“5 = w' (1) Since transference numbers must add to unity, we have tcr = 1 — twp = 1 — 0.410 = m. (2) The ionic conductances are given by K, = fag-2,11,. (3) The molar ionic conductances are given by Eq. 9.125: 11,-: Ki/c, = .7-"2,u,-. (4) Therefore, at infinite dilution, we have 2%“ = (96,485 C mol’1)(2)(55.0 x 10*5 cm2 V" s“) = (2)(96,4_85 amp 5 moi—1)(55.0 X 10'5 cm2 V’1 511) = 106.1 amp cm2 V‘1mol“. But an ohm {0) is a volt amp“, so 1 amp V“ = 1 Q.“ = IS. Therefore, Aim» = 106.1 S cm2 mol’l. For the Cl' ion, we obtain A5— = (96,485 c moi—1)(1)(79.1 x 16556.2 V154) = 76.325 cm2 6661-1. The molar conductivity is given by Eq. 9.126: A3302 = C—1[CM32+A:182+ 'i" Gel-E]—] = AESZi' + 2 Aai = 106.1 + (2) 76.32 s cm2 6661'1 = 258.7 s cm2 r661". In Eq. 8, we have ca— = 2 cMgz+ = 2c. (5) (6) (7) (8) 9.45 Using the data given in Table 9.3, determine the voltage that would be obtained from the galvanic cell Pt|2n|2nso. (m : 0.005)||Pbso.(s)]1>b(s) at 298.15 K if junction potentials are ignored. You may use the Debye—Hiickel theory to estimate any activity coefficients needed. Solution The half-reaction occurring at the anode is Zn —> Zn2+ + 22‘. At the cath- ode, the half-reaction is PbSO4(s} + 29‘ -——> Pb(s) + 50420. The total cell reaction is the sum of these: Zn(s) + P0504“) —-> ZnSOﬂm = 0.005) + Pb(s). The cell potential is given by Eq. 9.159: 0.02569 [ “amine: ] in —— . c1: = ° — ll cell ce n (1) ‘1 211(st Pbso‘u) In this case, we have u = 2. If the activities of the solids are set to Lmity and that for 21150; is written in terms of the mean ionic activity coefficient and molali- ty, Eq. 1 becomes chm“ = gen _ bib/@112]. The difference in standard half-reaction potentials can be obtained from Eq. 9.161 and the data in Table 9.3: 2..” = 16D: — GDSI = Q)? — (I): = —0.3588 — (—0.7618) volts = 0.403 volt. (3) The mean ionic activity coefficient is given by log n = —0.509262+z_51/2, (4) where the ionic strength is obtained from Eq. 9.64: S = 0.5[0.005(2)2 + 0.005(2)“ = 0.02. (5) Combining Eqs. 4 and 5, we obtain 10g vi = -0.50926(2)(2)(0.02)”2 = —0.28808. (6) This gives 7i = 0.51513. Substituting into Eq. 2 then produces (1)09“ = 0.403 — 0.012845 ln[(0.51513)2(0.005)2] = 0.403 + 0.153 = 0.556 volts (7) The accuracy of this result is questionable, because of the use of Debye—Hiickel theory to compute the activity coefficient. A more accurate result could be obtained if a measured value were employed. If the Davies empirical equation, Eq. 9.68, is employed to compute the activity coefficient, the result is @ceu = 0.553 volt. 9.46 (A) Show a schematic diagram of a galvanic cell whose cell reaction is 5Cd + 2Mn0; +16H* ——> 50:12+ + 2Mn2+ + 81-120 if a salt bridge is ernployed. (B) Show a schematic diagram of a galvanic cell with transference and with the same cell reaction. Compute the standard electric potential for this cell at 298.15 K. Solution : i (A) We will make use of a Cd electrode for the anode and an inert Pt rod for the i _ cathode. The salt bridge will contain concentrated K2504. A chloride should be i avoided, to prevent the reaction of MnO; with C1" to form C12. The required hydroniurn ion can be obtained from H1504. Group VIIA acids should be avoided, since the anion will be oxidized by the permanganate. A schematic of a possible setup is shown in the diagram at the top of the next page. (B) The same cell with transference is obtained by eliminating the salt bridge, as in the second diagram on the next page. For both cells, the standard potential (133?“ is given by Eq. 9.161: Ea: = I‘DZ — ¢3| = ‘1)?— ‘1’3- (1) The data in Table 9.3 give the following result: (1);“ = 1.507 — (-0.4030) volts = 1.910 volts. (2) I that I _ Negative charge flow alali- ——i— (2) Pt rod P.161 (3) Anode * Cathode 0x1dation ' ' ‘ " ' ; reduction I i (4) M110; + 3H+ —» Mn2+ + 41120 Salt bridge: K2804 (5} Negative charge flow (6] I - Pt rod (’7) f Anode Cathode o oxrdation reduction ate es I 1t, MnO; + 8H‘ -> Mn1+ + 4H20 Porous plu'g Chapter 9 Thermodynamics of Electrolytic Solutions 365 is measured at 298.15 K using a bridge circuit so that no current flows during the activ- wr the measurement. The voltage is fotmd to be 0.2411 volt. 2’ tht‘ (A) Determine the ionic activity coefficient for HQ with m = 1.000 mol kg" if the contact potential between the Pt and Ag electrodes is ignored. (B) If the contact potential is _8 millivolts, what percent error does ignoring the (2} contact potential introduce into the value of mm? Solution 3 (A) The reaction at the anode i505 HzigXP = 1 bar) —-> H‘ (m I 1.000) t 1e". ( ) At the cathode, the reaction is AgCl(s‘_) + 10" ——> Ag(s) + Cl’(m = 1.000). The overall cell reaction is, therefore, 0.5 H2(g)(P = 1 bar) + Agles) —> HCle 2 1.000 molal) + Ag(s). (4) From Eq. 9.159, the cell voltage is 0.02569 ‘iiitilam (Dcull : Sell ’ *“——ln[ 1m 0 (1) ’7 firlgaAgCI For this cell reaction, it = 1, and the fugacity of the 1-12 gas can be equated to its pressure at this low value of the pressure. If we take the pure-component active (5) ities of Ag(s) and Agles) to be unity, Eq. 1 becomes ¢cell : 3211 _ 0-025691T‘lﬂHcr: [6’ = Nigger“ — CD‘fﬁl — 0.025691n[y1im:] : 0.2411 volt. (2) The standard half-reaction potentials can be obtained from Table 9.3. These data show that <Dh2 = 0.0000 volt and (DSWUHF. = 0.22233 volt. Thus, Eq. 2 becomes 7) t . 0.22231 — 0.025691n y: = 0.2411, (3) ;- since m = 1,000. Solving for vi, we obtain l I’ 0.2411 — 0.22231 2 2 I"— : 0,4812, 3) I Yi exp: 0.02569 (4) which gives 7/1 = 0.99:1. (5) ) .3 (B) If the effect of the contact potential, (D1, is taken into account, Eq. 2 becomes t' E. a)...“ = [origcw — (13%| + <1), — 0.025691n[ﬁm2] : 0.2411 volt. (e) g. Rearranging terms in Eq. 6 gives l imam, e v.2 — 0.02569 mmmi] = 0.2411 - «1),. (7) If (bi = 0.008 volt, then Eq. 4 becomes 2 — ,X 20:2ililj.w2.2§l _ 0 3132; 8 7'1 t p 0.02569 _ “’ H which gives y, : 0.594. (9) The percent error produced by the B—millivolt contact potential is 04 7 100(0694 7 0.594) 16 8“ to error 7 0-594 7 . .20. (10) Obviously, it is important to account properly for iunction potentials. Chapter 9 Thermodynamics of Electrolytic Solutions 369 Equation 17 is the desired expression. The plot shown below demonstrates k- graphically the variation of the cell potential with the volume of NaOH added. 'ed Equation 1? predicts that, as V approaches 1 L, the cell potential approaches F infinity. This result is an artifact of our neglect of the contribution of the self- ionization of H20 to the H‘ concentration. Nevertheless, the result makes it (6) very clear that the endpoint can easily be identified by monitoring the point at ; which Chm, increases dramatically. 1.. 0.18 l - 'i. {7) 0.16 :- ‘J E 0.14 r 7 E * ‘ — 0.12 i (8) 7‘ i ‘; *3 0.10e - .' E. J _ 8 0,08 l— —' (9) L _ . 0.06 E— — v 4 in 0.0 0.2 0.4 0.0 (1.8 1.0 rot Volume of NuOH added (L) W 1) 9.51 (been for the cell f Ag(s)ll—l2(g) (P : 1bar), HCl (m : 1.00 molal). AgCl(s}|Ag(s) V‘ is measured at different temperatures. The data are fitted to a Taylor series 2) ' expansion. The result of the fitting is (1);.“ = 0.22233 - 0.0006477 (T — 298.15) — 3.241 x 10‘“(T — 298.15)2 volts. 3} ;. Determine the equilibrium constant, A3", All", and AG; for the cell reaction 3 at 280 K. a) Solution The cell reaction is 0.5 H2(g)(P = 1 bar] + AgCl(s) = I-lCl(m = 1 molal) +Agis). 9_ For this reaction, ’e Au” = —n}7(l>ﬂeu 5’ : —nﬂ022213 — 0.0006477(T 7 29815] r 3.241 x 10'“{T — 298.132]. (1) At 280 K, we obtain i, Ap.“(T = 230 K) = -(1)(96,485C mol'1)[0.22213 e 0.0006477(~1s.15) 7 3.241 >< 10'“(—18.15)2} volts = 722,463Jmol'”. (2) The corresponding equilibrium constant is ’l 3,0." F 22 463 I mol’ ‘ K=ex[- ]:ex{ I i=15>< 4 P RT l (8314] mor' K '1)(280 K) 5 ﬂ (3) :3 3 2. ,4 5, .3? 370 Section 2 Solutions and Comments on Problems 1 The change in the standard partial molar entropy for the reaction is given by Eq. 9.170: adjiiell as“ = n?( ) = r:f[—0.0005477 — 6,482 x 10m" 7 2985)] volts K“. (4) F (J At T : 280 K, we obtain as“ : (1)(96,485 c mol")[70.OOOb-177 — 6.482 x 10"“(—18.15)]volts K ‘ = —51.1-4_1mol'] KL}. (5) The change in the standard partial molar enthalpy for the reaction is given by Eq. 9.17]: AH" = a...” + T33“ = —22,463jmol '1 + (_280K](—51.14]mol'1K'l) = —_3.678 X 104Jmol". (6) AC; is given by Eq. 9.172: .2 n i. if") = r-szl—6482 >< iii—“voitK 2] at“ r ac; = nrr( = (1)(96,485C mol")(280 K)(-6.482 >< 10"“voltK 2) : 475.1}:1101’ 1 15:. (7) 9.52 An investigator measures the standard potential \$23“ for an electrochemical cell as a function of temperature over the range 280 K E T S 310 K and obtains the data shown in the following table. ‘ mo , mpg... voila " roof E)" “L i ‘5ng ‘js ': 280 023302 296 0.22371 282 0.23195 298 022243 284 0.23085 300 0.22112 286 0.22972 302 0.21979 288 0.22857 304 0.21843 290 0.22739 306 0.21705 292 0.22619 308 0.21564 294 0.22496 310 0.21420 ; _ i (A) Using these data, determine A5" for the cell reaction at 290 K, 292 K, 294 K, “r 298 K, 300 K, and 302 K. Assume that n = 1 for the cell reaction. i (B) Using the results obtained in (A), determine the value of AC; at 296 K. Solution (A) as“ is given by Eq. 9.170: A5" = HIF( fume“) . {1) 87' P -. Therefore, we need the derivative (61¢Seilfﬁij at each of the six temperatures at which we wish to compute 3.5“. These data are most conveniently obtained by using Eqs, 9.173 and 9.174. We illustrate the procedure for the derivative at T : 290 K. The value of S1, at this point is -. -~ . :.- [clam — chm] _] [0.22619 — 0.22357] _ —mm< : —— = —0.00119volt1<"1, (2) 5 : 1 2 2 E g Chqﬁer9 ThennodynanucsofEchoWﬁcSoMﬁons 375 Therefore, the final result is (26) (14) which is exactly Eq. 9.173. (15 ) 9.54 An acidic solution X is placed in the anode compartment of the electrochem- ical cell (15) Pt(Pd)(Sl|H2(g) (P : 1barLH‘lmtllKCl (satllAgCllsllAglsl, (17) and the voltage of the cell is measured at 298.15 K. This potential is found to be 0.16067 volt higher than that obtained using a 0.05-molal potassium hydrogen phthalate solution in place of solution X. What is the pH of solution X? (18) ‘ Solution that The pH is given by Eq. 9‘178: 17, E “Dix-n ’ (Weill-F r HI : H. + 1 p P lnumRT ( l (19) g Substituting values given in the statement of the problem and in the text yields H 4 005 + 0.16067 volts(96,485 C mol") 4 005 2 7160 6 721 (2 y = ' - e " . + . : . . P 1n(10)(8.314 J mor‘ K 1)(29815 K) ~— } (20) 121) E [22) . 16 :23) ‘ 24) 25) ...
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