Electrochemistry Exercise # 5

Exercise # 5 1. Part # I [Previous Year Questions] Consider the following E0 values : E0 3  Fe / Fe 2 = + 0.77 V ; E0 Sn 2 / Sn = – 0.14 V Under standard conditions, the cell potential for the reaction given below is : Sn(s) + 2Fe3+(aq)  2Fe2+(aq) + Sn2+(aq) 2. 3. 4. 5. 6. (A) 1.68 V (B) 1.40 V (C) 0.91 V (D) 0.63 V The limiting molar conductivities º for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol–1 respectively. The value of º for NaBr is : (A) 128 S cm2 mol–1 (B) 176 S cm2 mol–1 (C) 278 S cm2 mol–1 (D) 302 S cm2 mol–1 In a cell that utilizes the reaction Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g), addition of H2SO4 to cathode compartment will : (A) lower the E and shift equilibrium to the left. (B) lower the E and shift the equilibrium to the right. (C) increase the E and shift the equilibrium to the right. (D) increase the E and shift the equilibrium to the left. 0 The E M 3+ /M 2+ values for Cr, Mn, Fe and Co are – 0.41, + 1.57, + 0.77 and + 1.97 V respectively. For which one of these metals, the change in oxidation state from +2 to +3 is easiest : (A) Cr (B) Mn (C) Fe (D) Co The highest electrical conductivity among the following aqueous solutions is of : (A) 0.1 M acetic acid (B) 0.1 M chloroacetic acid (C) 0.1 M fluoroacetic acid (D) 0.1 M difluoroacetic acid Aluminium oxide may be electrolysed at 1000ºC to furnish aluminium metal (At. Mass of Al = 27 amu ; 1 Faraday = 96,500 Coulombs). The cathode reaction is Al3+ + 3e–  Al0. To prepare 5.12 kg of aluminium metal by this method, one would require : (A) 5.49 × 107 C of electricity (B) 1.83 × 107 C of electricity (C) 5.49 × 104 C of electricity 7. (D) 5.49 × 1010 C of electricity The molar conductivities Λ 0NaOAc and Λ 0HCl at infinite dilution in water at 25°C are 91.0 and 426.2 Scm2/mol respectively. To calculate 0HOAc , the additional value required is : 0 (A) H2O 8. (B) 0KCl (D) 0NaCl Given data is at 25°C : Ag + – Ag + e– ; E° = 0.152 V Ag Ag+ + e– ; E° = – 0.800 V What is the value of log Ksp for AgI : 9. (C) 0NaOH (Take 0.474 = 8.065) 0.059 (A) – 8.12 (B) + 8.612 (C) – 37.83 (D) – 16.13 Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 . The conductivity of this solution is 1.29 Sm–1. Resistance of the same cell when filled with 0.02 M of the same solution 129 = 0.248) 520 (D) 12.4 × 10–4 Sm2mol–1 is 520 . The molar conductivity of 0.02 M solution of the electrolyte will be : (Take (A) 124 × 10–4 Sm2mol–1 (B) 1240 × 10–4 Sm2mol–1 (C) 1.24 Sm2mol–1 ELECTROCHEMISTRY 10. The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25°C are given below : Λ 0CH3COONa = 91.0 Scm2/equiv and Λ 0HCl = 426.2 Scm2/equiv What additional information/quantity one needs to calculate º of an aqueous solution of acetic acid : ° (A) The limiting equivalent conductance of H+ (λ H + ) 11. (B) º of chloroacetic acid (ClCH2COOH) (C) º of NaCl (D) º of CH3COOK The cell Zn | Zn2+(1M) || Cu2+(1M) | Cu : (E°cell = 1.10V) was allowed to completely discharge at 298 K. The relative   Zn 2+   1.1   concentration of Zn to Cu  = 18.65) 2+  is : (Take Cu  0.059   2+ 2+ (A) 1037.3 12. Given : (B) 9.65 × 104 E0 Cr3+ /Cr (C) antilog (24.08) 0 = – 0.72, E Fe2+ /Fe = – 0.42 V The potential for the cell Cr | Cr3+(0.1 M) || Fe2+(0.01 M) | Fe at 298 K is : (Take (A) 0.339 V 13. Given : (D) 37.3 (B) – 0.339 V (C) – 0.26 V 2.303R (298) = 0.06) F (D) 0.26 V E 0Fe3+ /Fe = – 0.036 V, E 0Fe2+ /Fe = – 0.439 V 2+ – The value of standard electrode potential for the change, Fe3+ (aq) + e  Fe (aq) will be : 14. (A) 0.385V (B) 0.770V (C) –0.270V The Gibbs energy for the decomposition of Al2O3 at 500ºC is as follows : (D) –0.072V 2 4 Al2O3  Al + O2 ; rG = + 966 kJmol–1. The potential difference needed for electrolytic reduction of Al2O3 at 3 3 500ºC is at least : (A) 4.5 V 15. (B) 3.0 V (B) p(H2) = 1 atm and [H+] = 1.0 M (D) p(H2) = 2 atm and [H+] = 2.0 M The standard reduction potentials for Zn2+/Zn, Ni2+/Ni and Fe2+/Fe are –0.76, – 0.23 and –0.44 V respectively. The reaction X + Y2+  X2+ + Y will be spontaneous, when : (A) X = Ni, Y = Fe 17. (D) 5.0 V The reduction potential of hydrogen half-cell will be negative, if : (A) p(H2) = 1 atm and [H+] = 2.0 M (C) p(H2) = 2 atm and [H+] = 1.0 M 16. (C) 2.5 V (B) X = Ni, Y = Zn (C) X= Fe, Y = Zn (D) X= Zn, Y = Ni Given : 0 E 0Cr 3+ /Cr = –0.74 V ; E MnO– /Mn 2+ = 1.51 V 4 E 0Cr O2– /Cr 3+ = 1.33 V ; E 0Cl/Cl – = 1.36 V 2 7 Based on the data given above, strongest oxidising agent will be : (A) Cl (B) Cr3+ (C) Mn2+ (D) MnO4– 18. Given E 0Cl 2 /Cl  0 = 1.36 V, E Cr3 /Cr = –0.74 V E 0Cr O 2 /Cr3 = 1.33V, E 0MnO / Mn 2 = 1.51 V . 2 7 4 Among the following, the strongest reducing agent is : (A) Cr 19. (B) Mn2+ (D) Cl– How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane ? (Atomic weight of B = 10.8 u) (A) 0.8 hours Part # II 1. (C) Cr3+ (B) 3.2 hours (C) 1.6 hours (D) 6.4 hours [Previous Year Questions] The emf of the cell, Zn | Zn2+(0.01 M) || Fe2+(0.001 M) | Fe at 298 K is 0.2905 V. Then the value of equilibrium constant for the cell reaction is : 0.32 (A) e 0.0295 2. 0.26 0.32 (C) 10 0.0295 (B) 10 0.0295 0.32 (D) 10 0.059 Find the equilibrium constant at 298 K for the reaction : Cu2+(aq) + In2+(aq) Cu+(aq) + In3+(aq) Given that EºCu2+ / Cu+ = 0.15 V, EºIn3+ / In+ = – 0.42 V, EºIn2+ / In+ = – 0.40 V. 3. The half cell reactions for rusting of iron are : 2H+ + 1 O + 2e–  H2O; Eº = +1.23 V & Fe2+ + 2e–  Fe ; Eº = – 0.44 V 2 2 Gº (in kJ/mol) for the overall reaction is : (A) – 76 4. (B) – 322 (C) – 122 (D) – 176 (a) Calculate Gºƒ of the following reaction : Ag+(aq) + Cl–(aq)  AgCl(s) Given : Gºƒ(AgCl) = – 109 kJ/mole, Gºƒ(Cl–) = – 129 kJ/mole, Gºƒ(Ag+) = 77 kJ/mole. Represent the above reaction in form of a cell. Calculate Eº of the cell. Find log10Ksp of AgCl at 25ºC. (b) 6.539 × 10–2 g of metallic Zn (atomic mass = 65.39 amu) was added to 100 mL of saturated solution of AgCl. Calculate log10 [Zn 2+ ] at equilibrium at 25ºC, given that : [Ag + ]2 Ag+ + e–  Ag Eº = 0.80 V Zn2+ + 2e–  Zn Eº = – 0.76 V Also find how many moles of Ag will be formed. (Take 114 1.56 = 0.59, = 26.44) 193 0.059 ELECTROCHEMISTRY Comprehension # 1 Tollen’s reagent is used for the detection of aldehyde. When a solution of AgNO3 is added to glucose with NH4OH, then gluconic acid is formed. Ag+ + e–  Ag ; Eºred = 0.8 V C6H12O6 + H2O  C6H12O7 (Gluconic acid) + 2H+ + 2e– ; Eºoxd = – 0.05 V Ag(NH3)2+ + e–  Ag(s) + 2NH3 ; Eº = – 0.337 V [Use 2.303 × RT F = 0.0592 and = 38.92 at 298 K] F RT Now answer the following three questions : 5. 2Ag+ + C6H12O6 + H2O  2Ag(s) + C6H12O7 + 2H+ Find ln K of this reaction : (A) 66.13 6. 7. (B) 58.38 (C) 28.30 (D) 46.29 When ammonia is added to the solution, pH is raised to 11. Which half-cell reaction is affected by pH and by how much : (A) Eoxd will increase by a factor of 0.65 for Eºoxd (B) Eoxd will decrease by a factor of 0.65 for Eºoxd (C) Ered will increase by a factor of 0.65 for Eºred (D) Ered will decrease by a factor of 0.65 for Eºred Ammonia is always is added in this reaction. Which of the following must be INCORRECT : (A) NH3 combines with Ag+ to form a complex. (B) Ag(NH3)2+ is a weaker oxidising reagent than Ag+. (C) In absence of NH3, silver salt of gluconic acid is formed. (D) NH3 has affected the standard reduction potential of glucose/gluconic acid electrode. 8. We have taken a saturated solution of AgBr. Ksp of AgBr is 12 × 10–14. If 10–7 mole of AgNO3 are added to 1 litre of this solution, find conductivity (specific conductance) of this solution in terms of 10 –7 Sm–1. Given : º(Ag+) = 6 × 10–3 Sm2mol–1, º(Br–) = 8 × 10–3 Sm2mol–1, º(NO – = 7 × 10–3 Sm2mol–1. 3 ) Comprehension # 2 Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with its atomic/molecular masses. To handle such large number conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes. (Given : Atomic masses : Na = 23, Hg = 200 ; 1 Faraday = 96500 coulombs) Now answer the following three questions : 9. The total number of moles of chlorine gas evolved is : (A) 0.5 (B) 1.0 (C) 2.0 (D) 3.0 10. If the cathode is Hg electrode, the maximum weight (in g) of amalgam formed from this solution is : (A) 200 11. (B) 225 (C) 400 (D) 446 The total charge (in coulombs) required for complete electrolysis is : (A) 24125 (B) 48250 (C) 96500 (D) 193000 Comprehension # 3 Redox reactions play a pivoted role in chemistry and biology. The values of standard redox potential (Eº) of two halfcell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below is a set of half-cell reactions (acidic medium) along with their Eº values with respect to normal hydrogen electrode. Using this data, obtain the correct explanations to questions 15 - 16. 12. 13. 14. I2 + 2e–  2I– Eº = 0.54 V Cl2 + 2e–  2Cl– Eº = 1.36 V Mn3+ + e–  Mn2+ Eº = 1.50 V Fe3+ + e–  Fe2+ Eº = 0.77 V O2 + 4H+ + 4e–  2H2O Eº = 1.23 V Among the following, identify the correct statement : (A) Chloride ion is oxidised by O2 (B) Fe2+ is oxidised by iodine (C) Iodine ion is oxidised by chlorine (D) Mn2+ is oxidised by chlorine While Fe3+ is stable, Mn3+ is not stable in acid solution, because : (A) O2 oxidises Mn2+ to Mn3+ (B) O2 oxidises both Mn2+ to Mn3+ and Fe2+ to Fe3+ (C) Fe3+ oxidises H2O to O2 (D) Mn3+ oxidises H2O to O2 Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mole of H2 gas at the cathode is : (1 Faraday = 96500 C) (A) 9.65 × 104 sec 15. (B) 19.3 × 104 sec (C) 28.95 × 104 sec (D) 38.6 × 104 sec For the reduction of NO3– ion in an aqueous solution, Eº is +0.96 V. Values of Eº for some metal ions are given below V2+(aq) + 2e–  V Eº = –1.19 V Fe (aq) + 3e  Fe Eº = –0.04 V Au3+(aq) + 3e–  Au Eº = +1.40 V 3+ – Hg2+(aq) + 2e–  Hg Eº = +0.86 V The pair(s) of metals that is(are) oxidized by NO3 in aqueous solution is(are) : – (A) V and Hg (B) Hg and Fe (C) Fe and Au (D) Fe and V ELECTROCHEMISTRY Comprehension # 4 The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is : M(s) | M+(aq ; 0.05 molar) || M+(aq ; 1 molar) | M(s) For the above electrolytic cell, the magnitude of the cell potential is | Ecell | = 70 mV. Now answer the following two questions : 16. For the above cell : (A) Ecell < 0 ; G > 0 17. (C) Ecell < 0 ; Gº > 0 (D) Ecell > 0 ; Gº < 0 If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be : (A) 35 mV 18. (B) Ecell > 0 ; G < 0 (B) 70 mV (C) 140 mV (D) 700 mV AgNO3(aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. The plot of conductance () versus the volume of AgNO3 is : x volume (Q) volume (P) (A) (P) 19. x xx (B) Q x x x x x x x x x x x x x x x x x x x x x x x volume (S) volume (R) (C) (R) (D) (S) Consider the following cell reaction : 2Fe (s) + O2 (g) + 4H+ (aq)  2Fe2+ (aq) + 2H2O (l) Eº = 1.67 V At [Fe2+] = 10–3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25ºC is : (Take 2.303R (298) = 0.06) F (A) 1.47 V (D) 1.57 V (B) 1.77 V (C) 1.87 V Comprehension # 5 The electrochemical cell shown below is a concentration cell. M|M2+ (saturated solution of a sparingly soluble salt, MX2)|| M2+ (0.001 mol dm–3) |M The emf of the cell depends on the difference in concentration of M2+ ions at the two electrodes. The emf of the cell at 298 is 0.059 V 20. The solubility product (Ksp ; in mol3 dm–9) of MX2 at 298 K based on the information available in the given concentration cell is : (Take 2.303× R × 298/F = 0.059 V) (A) 1 × 10–15 21. (B) 4 × 10–15 (B) 5.7 (C) 11.4 (D) –11.4 An aqueous solution ox X is added slowly to an aqueous solution of Y as shown in list I. The variation in conductivity of these reactions is given in List II. Match List I with List II and select the correct answer using the code given below the lists : P. Q. R. S. Codes : 23. (D) 4 × 10–12 The value of G (in kJ mol–1) for the given cell is : (Take 1F = 96500 C mol–1 ) (A) – 5.7 22. (C) 1 × 10– 12 List I (C2H5)3N + CH3COOH X Y KI (0.1M) + AgNO3(0.01M) X Y CH3COOH + KOH X Y NaOH + HI 1. 2. 3. 4. List II Conductivity decreases and then increases Conductivity decreases and then does not change much Conductivity increases and then does not change much Conductivity does not change much P Q R S (A) 3 4 2 1 (B) 4 3 2 1 (C) 2 3 4 1 (D) 1 4 3 2 The standard reduction potential data at 25ºC is given below. Eº (Fe3+.Fe2+) = + 0.77 V ; Eº (Fe2+.Fe) = – 0.44 V ; Eº (Cu2+.Cu) = + 0.34 V ; Eº (Cu+.Cu) = + 0.52 V ; Eº (O2(g) + 4H+ + 4e–  2H2O) = + 1.23 V ; Eº (O2(g) + 2H2O + 4e–  4OH) = + 0.40 V Eº (Cr3+.Cr) = – 0.74 V ; Eº (Cr2+.Cr) = – 0.91 V Match Eº of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists : P. Q. R. S. Codes : (A) (B) (C) (D) List I / I Eº (Fe3+, Fe) Eº (4H2O 4H+ + 4OH–) 2+ Eº (Cu + Cu  2Cu+) Eº(Cr3+, Cr+2) P 4 2 1 3 Q 1 3 2 4 R 2 4 3 1 1. 2. 3. 4. S 3 1 4 2 List II / II – 0.18 V –0.4 V –0.04 V –0.83 V ELECTROCHEMISTRY 24. The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm2. The conductance of this solution was found to be 5 × 10–7 S. The pH of the solution is 4. The value of limiting molar conductivity (  om ) of this weak monobasic acid in aqueous solution is Z × 102 S cm–1. The value of Z is For the following cell, 25. Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s) when the concentration of Zn2+ is 10 times the concentration fo Cu2+, the expression for G (in J mol–1) is [F is Faraday constant ; R is gas constant ; T is temperature ; E° (cell) = 1.1 V] (A) 2.303 RT – 2.2 F (B) –2.2 F (C) 2.303 RT + 1.1 F 26. (D) 1.1 F For the electrochemical cell. Mg(s) | Mg2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu(s) the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg+2 is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is _____. (given, 27. F = 11500 K V–1, where F is the Faraday constant and R is the gas constant, ln(10) = 2.30) R The surface of copper gets tamished by the formation of copper oxide. N2 gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below : 2Cu(s) + H2O (g)  Cu2O(s) + H2(g) pH2 is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln(p H 2 ) is _______. (Given : total pressure = 1 bar, R (universal gas constant) = 8 J K–1 mol–1, ln(10) = 2.3, Cu(s) and Cu2O(s) atr mutually immiscible. 1  Cu 2 O(s) ; G° = –78,000 J mol–1 At 1250 K : 2Cu(s)  O 2 (g)  2 1 H 2 (g)  O 2 (g)   H 2 O(g) 2 ; G° = –1,78,000 J mol–1 ; G is the Gibbs energy Consider an electrochemical cell : A(s) | An+ (aq, 2M) || B2n+ (aq, 1 M) | B(s). The value of H° for the cell reaction is twice that of G° at 300K. If the emf of the cell is zero, S° (in J K–1 mol –1) of the cell reaction per mole of B formed at 300 K is_____. (Given : ln (2) = 0.7, R (universal gas constant) = 8.3 J K–1 mol–1. H, S and G are enthalpy, entropy and Gibbs energy, respectively.) 28. EXERCISE - 5 : PART # I 1. C 14. C 2. A 15. C 3. C 16. D 4. A 17. D 5. D 12. D 13. B (A) Eº = 0.59 V, log10Ksp = – 10 ; (B) 52.88, 10–6 mole. 5. B A 12. C 18. D 18. A 6. A 7. D 8. D 9. D 10. C 11. A 19. B PART # II B 8. 55 S m–1 9. 20. B 2. KC = 1010 1. 21. D B 3. 10. D 22. A B 4. 11. D 23. D 24. 6 13. D 25. A 14. B 15. A, D 26. 10 27. –14.6 16. B 17. C 28. –11.62 JK–1 mol–1 6. 7. 19. D D