Electrochemistry Exercise # 3

ELECTROCHEMISTRY Exercise # 3 1. Match Matrix Part # I [Matrix Match Type Questions] 0 ( E Ag+ /Ag = 0.8, KSP(AgCl) = 10–10). Column – I 2. 3. (A) Pt | H2 (0.1 bar) | H+ (0.1 M) || H+ (1 M) | H2 (0.01 bar) | Pt (p) Concentration cell (B) Ag | AgCl (KCl, 0.1M) || Ag+ (0.01M) | Ag (q) Ecell > 0 (C) Cu | Cu2+ (0.1 M) || Cu2+ (0.01 M) | Cu (r) Eºcell = 0 but cell is working. (D) Pt | Cl2 (1bar) | HCl (0.1 M) || NaCl (0.1M) | Cl2 | Pt (1 bar) (s) non working condition Match the column Column I (A) Zn / Zn+2 | | Mg2+ / Mg c1 c2 (c1 = c2) +2 (B) Zn / Zn | | Ag+ / Ag at. equilibrium (C) Ag / Ag+ | | Ag+ / Ag c1 c2 (c1 = c2) (D) Fe / Fe+2 | | Ag / Ag+ c1 c2 (c1 = c2) Column II (p) Ecell = 0 (q) E0cell 0 (r) E0cell = +ve (s) E0cell = –ve Column-I and Column-II contains four entries each. Entries of Column-I are to be matched with some entries of Column-II. One or more than one entries of Column-I may have the matching with the same entries of Column-II. Column-I Column-II (A) Very dilute solution of HCl (p) O2 evolved at anode (B) Very dilute solution of NaCl (q) H2 evolved at cathode (C) Concentrate solution of NaCl (r) Cl2 evolved at anode (D) Fairly concentrate solution of AgNO3 (s) Ag deposition at cathode 4. Column-I (A) (B) (C) (D) (E) 5. Column – II Cathode 1 Coulomb Dry cell Lead strong cell Zn|Zn2+(0.01M)||Zn2+(0.1M)|Zn Column-I (A) Cell constant (B) Anode (C) Conductance (D) Electrochemical equivalent (E) E0cell Column-II (p) Primary cell (q) Secondary cell (r) 6.24 × 1018 electrones (s) Concentration cell (t) Positive terminal of electrochemical cell Column-II (p) E0cathode + E0anode (q) /A (r) Mass of product deposited by 1 coulomb of electricity. (s) (Resistance)–1 (t) Involve oxidation 6. Column-I Column-II (A) Conductance (p) Cm–1 (B) Specific conductance (q) Ohm–1 cm2 mol–1 (C) Cell constant (r) Ohm–1 (D) Equivalent conductance (s) Ohm–1 cm–1 (E) (t) Ohm–1 cm2 equivalent–1 7. Molar conductance Column-I Column-II (A) Electrolytic cell (B) nFE0Cell (C) Ecell = (p) –G° (q) Concentration cell C 0.059 log cathode Canode n (r) 96500 Coulombs (D) Diffusion of ions (E) (s) Device converting electrical energy into chemical energy (t) Salt bridge 1 Faraday Part # II [Comprehension Type Questions] Comprehension # 1 G Zn electrode Cu electrode 1 L, 2 M ZnSO4 Given E ºZn +2 |Zn = – 0.76 V º E Cu +2 |Cu = 0.34 V 1 L, 0.2 M CuSO4 Kf [Cu(NH3)4]+2 = 4 × 1011 Answer the following. 1. The emf of cell at 200 k is [Given : (A) 1.7 V 2. 3. 2.303× R = 2  10-4 and assume that E0 values are independent on temperature.] F (B) 1.08 V (C) 1.09 V When 1 mole NH3 added to cathode compartment than emf of cell is (at 298K) (A) 0.81 V (B) 1.91 V (C) 1.1 V (D) 1.10 V (D) 0.72 V At what conc of Cu+2 emf of the cell will be zero (at 298K) and conc. of Zn+2 is remain same (A) 1.19 × 10–37 (B) 1. 19 × 10–20 (C) 3.78 × 10–4 (D) 0.0068 ELECTROCHEMISTRY Comprehension # 2 If an element can exist in several oxidation states, it is convenient to display the reduction potentials corresponding to the various half reactions in diagrammatic form, known as Latimer diagram. The Latimer diagram for chlorine in acid solution is ClO4– +1.20 V ClO3– +1.18 V HClO2 +1.60 V HClO +1.67 V Cl2 +1.36 V Cl– in basic solution. ClO4– 0.37 V ClO3– 0.30 V ClO2– 0.68V ClO– 0.42V Cl2 1.36V Cl– The standard potentials for two nonadjacent species can also be calculated by using the concept that G° as an additive property but potential is not an additive property and G° = – nFx0. If a given oxidation state is a stronger oxidising agent than in the next higher oxidation state, disproportionation can occur. The reverse of disproportionation is called comproportionation. The relative stabilities of the oxidation state can also be understood by drawing a graph of G°/F against oxidation state, known as Frost diagram, choosing the stability of zero oxidation state arbitrarily as zero. The most stable oxidation state of a species lies lowest in the diagram. Disproportionation is spontaneous if the species lies above a straight line joining its two product species. 1. Which of the following couple have same value of potential at pH = 0 and pH = 14? (A) 2. ClO 4– ClO3– What is the potential of couple (A) 1.78 V 3. 4. (B) ClO 2– Cl 2 (C) ClO – Cl 2 Cl2 Cl – ClO – at pH = 14 ? Cl – (B) – 0.94 V (C) 0.89 V (D) – 0.89 V Which of the following statement is correct ? (A) Cl2 undergoes disproportionation into Cl– and ClO– both at pH = 0 and pH = 14. (B) Cl2 undergoes disproportionation into Cl– and ClO– at pH = 14 but not at pH = 0. (C) Cl2 undergoes disproportionation into Cl– and ClO– at pH = 0 but not at pH = 14. (D) None of these For a hypothetical element, the Frost diagram is shown in figure? Which of the following oxidation state is least stable? (A) – 1 (B) 0 (C) + 2 5. (D) Which of the following statement is correct ? (A) A+1 undergoes disproportionation into A and A2+. (B) A2+ undergoes disproportionation in A and A3+. (C) A undergoes comproportionation in A+1 and A–1. (D) All of the above. (D) + 3 Comprehension # 3 Strong Acid Versus Strong Base The principle of conductometric titrations is based on the fact that during the titration, one of the ions is replaced by the other and invariably these two ions differ in the ionic conductivity with the result that the conductivity of the solution varies during the course of the titration. Take, for example, the titration between a strong acid, say HCl, and a strong base, say NaOH. Before NaOH is added, the conductance of HCl solution has a high value due to the presence of highly mobile hydrogen ions. As NaOH is added, H+ ions are replaced by relatively slower moving Na+ ions. Consequently, the conductance of the solution decreases and this continues right upto the equivalence point where the solution contains only NaCl. Beyond the equivalence point, if more of NaOH is added, then the solution contains an excess of the fast moving OH– ions with the result that its conductance is increased and it continues to increase as more and more of NaOH is added. If we plot the conductance value versus the amount of NaOH added, we get a curve of the type shown in Fig. The descending portion AB represents the conductances before A the equivalence point (solution contains a mixture of acid HCl and the salt NaCl) and the ascending portion CD represents the conductances D after the equivalence point (solution contains the salt NaCl and the excess of NaOH). The point E which represents the minium conductance is due to the solution containing only NaCl with no free B E C acid or alkali and thus represents the equivalence point. This point can, however, be obtained by the extrapolation of the lines AB and DC, and therefore, one is not very particular in locating this point volume of NaOH experimentally as it is in the case of ordinary acid-base titrations involving the acid-base indicators. Weak Acid versus Strong Base Let us take the specific example of acetic acid being titrated against NaOH. Before the addition of alkali, the solution shows poor conductance due to feeble ionization of acetic acid. Initially the addition of alkali cause not only the replacement of H+ by Na+ but also suppresses the dissociation of acetic acid due to the common ion Ac– and thus the conductance of the solution decreases in the beginning. But very soon the conductance starts increasing as addition of NaOH neutralizes the undissociated HAc to Na+Ac– thus causing the replacement of non-conducting HAc with strong-conducting electrolyte Na+ Ac–. The increase in conductance continues right up to the equivalence point. Beyond this point conductance increases more rapidly with the addition of NaOH due to the highly conducting OH– ions. The graph near the equivalence point is curved due to the hydrolysis of the salt NaAc. The actual equivalence point can, as usual, be obtained by the extrapolation method. In all these graphs it has been assumed that the volume change due addition of solution from burette is negligible, hence volume change of the solution in beaker the conductance of which is measured is almost constant throughout the measurement. ELECTROCHEMISTRY (A) (B) Conductance The nature of curve obtained for the titration between weak acid versus strong base as described in the above passage will be : Conductance 1. volume of NaOH (D) Conductance (C) Conductance volume of NaOH volume of NaOH (A) (B) Conductance The most appropriate titration curve obtained when a mixture of a strong acid (say HCl) and a weak acid (say CH3COOH) is titrated with a strong base(say NaOH) will be Conductance 2. volume of NaOH V1 + V2 V1 volume of NaOH (D) V1 Conductance Conductance (C) V2 V1 V2 volume of NaOH volume of NaOH 3. V2 volume of NaOH If a 100 mL solution of 0.1M HBr is titrated using a very concentrated solution of NaOH, then the conductivity (specific conductance) of this solution at the equivalence point will be (assume volume change is negligible due to addition of NaOH) . Report your answer after multiplying it with 10 in Sm–1. º º –3 [Given λ (Na Sm2 mol–1 , λ (Br = 4 × 10–3 S m2 mol–1 ] + = 8 × 10 – ) ) (A) 6 (B) 12 (C) 15 (D) 24 Comprehension # 4 The molar conductance of NaCl varies with the concentration as shown in the following table . and all values follows the equation Cm = m – b C Where Cm = molar specific conductance m = molar specific conductance at infinite dilution C = molar concentration Molar Conductance Molar Concentration of NaCl -1 2 in ohm cm mole 4 x 10 -4 107 9 x 10 -4 97 16 x 10 -4 -1 87 When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl– and SO4–2 are 80 ohm–1 cm2 mole–1 and 160 ohm–1 cm2 mole–1 respectively. 1. What is the molar conductance of NaCl at infinite dilution ? (A) 147 ohm–1 cm2 mole–1 (B) 107 ohm–1 cm2s mole–1 –1 2 –1 (C) 127 ohm cm mole (D) 157 ohm–1 cm2 mole–1 2. What is the cell constant of the conductivity cell (C) (A) 0.385 cm–1 (B) 3.85 cm–1 (C) 38.5 cm–1 3. (D) 0.1925 cm–1 If the cell (C) is filled with 5 x 10–3 (N) Na2SO4 the observed resistance was 400 ohm. What is the molar conductance of Na2SO4. (A) 19.25 ohm–1 cm2 mole–1 (B) 96.25 ohm–1 cm2 mole–1 –1 2 –1 (C) 385 ohm cm mole (D) 192.5 ohm–1 cm2s mole–1 Comprehension # 5 Copper reduces NO 3– into NO and NO 2 depending upon conc. of HNO 3 in solution. Assuming [Cu2+] = 0.1 M, and PNO = PNO = 10–3 atm and using given data answer the following questions : 2 E°Cu2+/Cu = +0.34 volt ; E °NO – / NO = +0.96 volt ENO  / NO = +0.79 volt ; at 298 K 3 1. 2 RT (2.303) = 0.06 volt F ECell for reduction of NO3– NO by Cu(s), when [HNO3] = 1 M is [At T = 298] (A) ~0.61 2. 3 (B) ~0.71 (C) ~0.51 (D) ~0.81 At what HNO3 concentration thermodynamic tendency for reduction of NO3– into NO and NO2 by copper is same? (A) 101.23 M (B) 100.56 M (C) 100.66 M (D) 100.12 M ELECTROCHEMISTRY Comprehension # 6 Accidentally chewing on a stray fragment of aluminium foil can causes a sharp tooth pain if the aluminium comes in contact with an amalgam filling. The filling, an alloy of silver, tin and mercury, acts as the cathode of a tiny galvanic cell, the aluminium behaves as the anode, and saliva serves as the electrolyte. When the aluminium and the filling come in contact, an electric current passage from the aluminium to the filling which is sensed by a nerve in the tooth. Aluminium is oxidized at the anode, and O2 gas is reduced to water at the cathode. E°Al3+ / Al = –1.66 V, E°O2 ,H+ /H2O = 1.23 V 1. Net reaction taking place when amalgam is in contact with aluminium foil : (A) Al + O2 + OH–  Al(OH–) + H2O (B) 4Al + 3O2 + 12H+  4Al3+ + 6H2O (C) 4Al + 3O2  4Al2O3 (D) 2H2 + O2  2H2O 2. Standard E.M.F. experienced by the person with dental filling is : (A) +2.89 V 3. (B) –2.89 V (C) –0.93 V (D) +0.43 V The standard reduction potential of the reaction, H 2O + e –  1 H + OH– at 298 K is : 2 2 (A) E° = RT lnK w 2F (C) E° = RT  PH2  ln F [H + ] (B) E° = RT 1/2 ln  PH2  [OH – ] F (D) E° = RT lnK w F 1/2 EXERCISE - 3 : PART # I 1. A p, q, r, B q, C p, s, D p, s 2. A s, B p, r, C p, q, D r 3. A p, q, B p, q, C q, r, D p, s 4. A u, B r, C p, D q, E s 5. A q, B u, C s, D r, E p 6. A r, B s, C p, D u, E r 7. A s, B p, C q, D u, E r PART # II Comprehension # 1 : 1. B 2. A 3. A Comprehension # 2 : 1. D 2. C 3. B Comprehension # 3 : 1. A 2. C 3. B Comprehension # 4 : 1. C 2. D 3. D Comprehension # 5 : 1. B 2. C Comprehension # 6 : 1. A 2. C 3. A 4. D 5. A