CBSE Class 12 Chemistry Notes - Electrochemistry Concepts PDF

For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 17 UNIT 2 ELECTROCHEMISTRY CONCEPTS Points to Remember...

For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 17 UNIT 2 ELECTROCHEMISTRY CONCEPTS Points to Remember Electrochemistry may be defined as the branch of chemistry which deals with the quantitative study of inter-relationship between chemical energy and electrical energy and inter-conversion of one form into another relationships between electrical energy taking place in redox reactions. A cell is of two types : I. Galvanic cell II. Electrolytic cell In Galvanic cell, the chemical energy of a spontaneous redox reaction is converted into electrical work. In Electrolytic cell, electrical energy is used to carry out a non-spontaneous redox reaction. 1. Conductivity (k) : 1 1 li k= = × ρ R A where R is Resistance, l/A = cell constant (G*) and ρ is resistivity. 2. Relation between k and Λm 1000 × k Λm = C where Λm is molar conductivity, k is conductivity and C is molar concentration. Kohlrausch’s law : (a) In general, if an electrolyte on dissociation give y+ cations and γ− anions, then its limiting molar conductivity (Λºm) is given by Λ °m = v+ λ °+ + v− λ °− Here, lº+ and lº− are the limiting molar conductivities of cation and anion respectively and v+ and v− are the number of cations and anions furnished by one formula unit of the electrolyte. For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 18 | Chemistry-XII (b) Degree of dissociation (α) is given by : Λcm α= Λºm Here, Λcm = is molar conductivity at the concentration C and Λºm is limiting molar conductivity of the electrolyte. (c) Dissociation constant (K) of weak electrolyte : 2  Λcm  C   Cα 2  Λºm  =K = 1−α  Λm  1 −   Λºm  Dry cell : At anode (Oxidation) Zn → Zn2+ + 2e- At cathode (Reduction) 2NH4+ + 2MnO2 + 2e− → 2MnO (OH) + 2NH3 Overall Zn (s) + 2NH4+ + 2MnO2 → Zn2+ + 2MnO (OH) + 2NH3 Mercury cell : At anode (Oxidation) Zn (Hg) + 2OH− → ZnO (s) + H2O + 2e- At cathode (Reduction) HgO (s) + H2O + 2e− → Hg (l) + 2OH- Overall Zn (Hg) + HgO (s) → ZnO (s) + Hg (l) Lead storage cell At anode (Oxidation) Pb (s) → Pb2+ + 2e- Pb2+ + SO42− → PbSO4 At cathode (Reduction) PbO2 + 4H+ + 2e− → Pb2+ + 2H2O Pb2+ + SO42− → PbSO4 (s) For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 19 Overall Discharging  Pb(s) + PbO2 (s) + 2H2SO4 (aq)    2PbSO4(s) + 2H2O(l) Recharging 3. Nernst Equation for electrode reaction : Mn+ (aq) + ne− → M(s) 2.303RT 1 0.059 1 Eθ − E= log Eθ − = log nF Mn +  n Mn +      − The cell potential of electrochemical reaction : aA + bB → cC + dD is given ne by : 2.303RT 0.059 [C]c [D]d E θ cell − E cell = Eθ − log[Qc ] = log nF n [A]a [B]b 4. Relation between Eqcell and equilibrium constant (Kc) : 2.303RT 0.059 = E θ cell = log K c log K c nF n 5. ∆G0 = − nF E0cell where ∆G0 = standard Gibbs energy change and nF is the number of Faradays of charge passed. E0cell is standard cell potential. ∆G0 = − 2.303 RT log Kc Corrosion of metals is an electrochemical phenomenon. In corrosion, metal is oxidized by loss of electrons to oxygen and formation of oxides. At anode (Oxidation) : 2Fe (s) → 2Fe2+ + 4e- At cathode (Reduction) : O2 (g) + 4H+ (aq) + 4e− → 2H2O Atmospheric oxidation : 2Fe2+ (aq) + 2H2O (l) + ½O2 (g) → Fe2O3 (s) + 4H+ (aq) MULTIPLE CHOICE QUESTIONS (1 Mark) 1. When a lead storage battery is discharged: (a) SO2 is evolved (b) lead is formed (c) H2SO4 is consumed (d) PbSO4 is consumed For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 20 | Chemistry-XII Electrochemistry Concepts 2. How many coulomb are required for the oxidation of 1 mol of H2O2 to O2? (a) 9.65 × 104 C (b) 93000 C 5 (c) 1.93 × 10 C (d) 19.3 × 102C 3. KCl is used in salt bridge because: (a) It forms a good jelly with agar-agar (b) It is a strong electrolyte (c) It is a good conductor of electricity (d) Migration factor of K+ and Cl– ions are almost equal 4. The nature of curve of E° cell against log KC is: (a) a straight line (b) parabola (c) a hyperbola (d) an elliptical curve 5. For a spontaneous reaction the ∆G, equilibrium constant (K) and E°cell will be respectively. (a) – ve, < 1, – ve (b) – ve, > 1, – ve (c) – ve, > 1, + ve (d) + ve, > 1, – ve 6. Determine the value of E°cell for the following reaction, cu2+ + Sn+2 → Cu + Sn+4, equilibrium constant is 106. (a) 0.1773 (b).01773 (c) 0.2153 (d) 1.773 7. Which is the best reducing agent? (a) F– (b) Cl– (c) Br– (d) I– 8. If a salt bridge is removed between the half cells, the voltage: (a) drops to zero (b) does not change (c) increase gradually (d) increases rapidly 9. Faraday's law of electrolysis are related to the: (a) Atomic number of the cation (b) atomic number of the anion (c) equivalent weight of the electrolyte (d) speed of the cation 10. The process in which chemical change occurs on passing electricity is termed: (a) Ionisation (b) neutralisation (c) electrolysis (d) hydrolysis For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 21 11. The charge required for the reduction of 1 mol of MnO4– to MnO2 is: (a) 1 F (b) 3 F (c) 5 F (d) 4 F 12. The value of Λºm for NH4Cl, NaOH and NaCl are 129.8, 248.1 and 126.4 Ohm–1 cm2 mol–1 respectively. Calculate Λºm for NH4OH solution. (a) 215.5 (b) 251.5 (c) 244.7 (d) 351.5 13. A current of 9.65 amp flowing for 10 minutes deposits 3.0 g of a metal. The equivalent wt. of the metal is: (a) 10 g (b) 30 g (c) 50 g (d) 96.5 g 14. In a Golvenic cell the electrical work done is equal to: (a) free energy change (b) mechanical work done (c) thermodynamic work done (d) all of the above 15. When lead storage battery is charged it is: (a) an electrolyte cell (b) a galvenic cell (c) a daniel cell (d) a and b both 16. In a galvenic cell the direction of current is: (a) anode to cathode (b) cathode to anode (c) Zn rod to Cu rod (d) Depend on concentration of ZnSO4 and CuSO4 17. Which metal does not give the following reaction M + water → oxide or hydroxide + H2 (a) Fe (b) Na (c) Hg (d) Ag 18. Electrolysis of aq. CuSO4 produces: (a) an increase in pH (b) a decrease in pH (c) either decrease or increase (d) H2SO4 in the solution 19. Zn cannot displace following ions from their aquous solution: (a) Al+3 (b) Cu2+ (c) Fe2+ (d) Na+ 20. Which one is not a secondary battery? (a) laclanche cell (b) Ni-Cd cell (c) Mercury cell (d) Daniel cell For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 22 | Chemistry-XII Electrochemistry Concepts 21. Which of the following decrease with increase in concentration? (a) conductance (b) specifci conductance (c) Molar conductance (d) Conductivity Fill in the blanks type question: 22. To deposite 2 mol of Ca from CaCl2............... electricity is required. 23................... gives a constant voltage throughout its life. 24. Match the column and choose correct option: (A) Conductance P. m–1 (B) Conductivity Q. 5 cm–1 (C) Molar conductance R. Siemen (D) Cell constant S. 5 cm2 mol–1 (a) A–R, B–Q, C–S, D–P (b) A–R, B–S, C–Q, D–P (c) A–R, B–Q, C–P, D–S (d) A–R, B–P, C–Q, D–S 25. (A) MnO4– → Mn+2 (1 mol)    P. Required 1F (B) CuSO4 → Cu (1 mol) Q. Required 5 F (C) Al2O3 → Al (1 mol) R. Required 3 F (D) NaCl → Na (1 mol) S. Required 2 F (a) A–Q, B–P, C–S, D–R (b) A–P, B–Q, C–S, D–R (c) A–Q, B–S, C–P, D–R (d) A–Q, B–S, C–R, D–P Assertion-Reason type 26. Statement 1 : Galvanised iron does not rust. Statement 2 : Zn has more (–) ve electrode potential than Fe. 27. Statement 1 : Conductivity decreases with dilution. Statement 2 : Number of ions per unit volume decreases on dilution. ANSWERS 1. (c) 2. (c) 3. (d) 4. (a) 5. (c) 6. (a) 7. (d) 8. (a) 9. (c) 10. (c) 11. (b) 12. (b) 13. (c) 14. (a) 15. (a) 16. (b) 17. (c, d) 18. (b, d) 19. (a, d) 20. (a, c, d) 21. (a, c) 22. 4 F 23. Mercury cell 24. (a) 25. (d) 26. (a) 27. (a) For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 23 VERY SHORT ANSWER TYPE QUESTIONS (1 Mark) Q. 1. Why is it not possible to measure single electrode potential ? Ans. Because the half cell containing single electrode cannot exist independently, as charge cannot flow on its own in a single electrode. Q. 2. Name the factor on which emf of a cell depends. Ans. Emf of a cell depends on following factors : (a) Nature of reactants (b) Concentration of solution in two half cells (c) Temperature Q. 3. What is the effect of temperature on the electrical conductance of metal ? Ans. Temperature increases, electrical conductance decreases. Q. 4. What is the effect of temperature on the electrical conductance of electrolyte ? Ans. Temperature increases, electrical conductance increases. Q. 5. What is the relation between conductance and conductivity ? k Ans. Λ cm = C Q. 6. Reduction potentials of 4 metals A, B, C and D are – 1.66 V, + 0.34 V, + 0.80 V and – 0.76 V. What is the order of their reducing power and reactivity ? Ans. A > D > B > C Q. 7. Why does a dry cell become dead even if it has not been used for a long time ? Ans. NH4Cl is acidic in nature. It corrodes zinc container. Q.8. Why Na cannot be obtained by the electrolysis of aqueous NaCl solution ? Ans. Due to low reduction potential, Na+ ions are not reduced at cathode. Instead, H+ are reduced and H2 is obtained. Q.9. What is the use of platinum foil in the hydrogen electrode ? Ans. It is used for the in and out flow of electrons. Q.10. Why Λmº for CH3COOH cannot be determined experimentally ? Ans. Molar conductivity of weak electrolytes keeps on increasing with dilution and does not become constant even at very large dilution. Q.11. Why is it necessary to use a salt bridge in a galvanic cell ? Ans. To complete the inner circuit and to maintain electrical neutrality of the electrolytic For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 24 | Chemistry-XII solutions of the half cells. Q.12. Why does mercury cell gives a constant voltage throughout its life ? Ans. This is because the overall cell reaction does not have any ionic concentration in it. Q.13. What is the role of ZnCl2 in a dry cell ? Ans. ZnCl2 combines with the NH3 produced to form a complex salt [Zn(NH3)2]Cl2. Q.14. Why does the conductivity of a solution decrease with dilution ? Ans. Conductivity of a solution is dependent on the number of ions per unit volume. On dilution, the number of ions per unit volume decreases, hence the conductivity decreases. Q.15. Suggest two materials other than hydrogen that can be used as fuels in fuel cells. Ans. Methane and methanol. Q.16. How does the pH of Al-NaCl solution be affected when it is electrolysed ? Ans. When Al-NaCl solution is electrolysed, H2 is liberated at cathode, Cl2 at anode and NaOH is formed in the solution. Hence pH of solution increases. Q.17. Which reference electrode is used to measure the electrode potential of other electrodes. Ans. SHE, whose electrode potential is taken as zero. Q.18. Out of zinc and tin, which one protects iron better even after cracks and why ? Ans. Zinc protects better because oxidation of zinc is greater but that of tin is less than that of iron. Q.19. Define corrosion. What is the chemical formula of rust ? Ans. Corrosion is the slow eating away of the surface of the metal due to attack of atmospheric gases. Fe2O3.xH2O. Q.20. What is the electrolyte used in a dry cell ? Ans. A paste of NH4Cl. For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 25 SHORT ANSWER-I TYPE QUESTIONS (2 Marks) Q. 1. How can you increase the reduction potential of an electrode for the reaction : Mn+ (aq) + ne– → M (s) Ans. Nernst equation is : E M n+ /M = E M n+ /M – E M n+ /M can be increased by (a) Increase in concentration of Mn+ ions in solution. (b) By increasing the temperature. Q. 2. Calculate emf of the following cell at 298 K : Mg (s) + 2Ag+ (0.0001M) → Mg2+ (0.130 M) + 2Ag (s) [Given : Eθcell = 3.17 V] Ans. n = 2 The Nernst equation for the cell is : θ 0.059  Mg 2+  = E E − log 2 2  Ag +  0.059.130 = 3.17 − log (.0001) 2 2 = 3.17 – 0.21 = 2.96V Q. 3. Suggest a way to determine the Λmº value of water. Ans. Λ ºm ( H 2 O ) = Λ ºm H+ + Λ ºm OH− It can be determine from the value of Λmº (HCl), Λmº (NaOH) and Λmº (NaCl). Then, Λmº (H2O) = Λmº (HCl) + Λmº (NaOH) − Λmº (NaCl) Q. 4. How much electricity in term of Faraday is required to produce 40 gram of Al from Al2O3 ? (Atomic mass of Al = 27 g/mol) Ans. Al3+ + 3e– → Al 27 gram of Al require electricity = 3F 3F 40 gram of Al require electricity = × 40 = 4.44 F 27 For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 26 | Chemistry-XII Q. 5. Predict the product of electrolysis of an aqueous solution of CuCl2 with an inert electrode. Ans. CuCl2 (s) + Aq → Cu2+ + 2Cl– H2O → H+ + OH– At cathode (Reduction) : Cu2+ will be reduced in preference to H+ ions. Cu2+ + 2e– → Cu(s) At anode (Oxidation) : Cl– ions will be oxidized in preference to OH– ions. Cl– → ½Cl2 + 1e– Thus, Cu will be deposited at cathode and Cl2 will be liberated at anode. Q. 6. Calculate Λ m for CaCl2 and MgSO4 from the following data : º Λ ºm ( Ca ) = 119.0, Mg = 106.0, Cl = 76.3 and SO4 = 160.05 cm mol 2+ 2+ – 2– 2 –1 Λ ºm ( CaCl2 ) =Λ ºm Ca 2+ + 2Λ ºm Cl− Ans. ( ) ( ) = 119 + (2 × 76.3) = 271.6 S cm2 mol–1 Λ ºm ( MgSO4 ) =Λ ºm Mg 2 + + 2Λ ºm SO 2 − ( ) ( 4 ) = 106 + 160 = 266 S cm2 mol–1 Q. 7. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. Ans. H+ + e– → ½H2 n = 1 0.0591 1 E = E ”− log + n  H  0.0591 E= 0− × pH 1 E = − 0.0591 × 10 v E = – 0.591/V Q. 8. If a current of 0.5 amp flows through a metallic wire for 2 hours, how many electrons would flow through the wire ? Ans. q = i × t = 0.5 × 2 × 60 × 60 = 3600 C 96500 Coulombs are equal to 6.022 × 1023e- 6.022 × 1023 So, 3600 Coulombs = × 3600 = 2.246 × 1022 electrons 96500 For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 27 Q.9. How much electricity is required in Coulomb for the oxidation of 1 mole of FeO to Fe2O3 ? Ans. Fe2+ → Fe3+ + 1e- So, 1F = 1 × 96500 C = 96500 C Q.10. The conductivity of a 0.20M solution of KCl at 298K is 0.0248 S cm–1. Calculate molar conductivity. k × 1000 0.0248 S cm −1 × 1000 cm3 L−1 Ans. Molar conductivity = = M 0.2 mol L−1 = 124.0 S cm2 mol–1 Q.11. Define conductivity and molar conductivity for a solution of an electrolyte. Ans. Conductivity is defined as ease with which current flows through electrolyte. It is reciprocal of specific resistance. Molar conductivity is conductance of all the ions produced by one mole of electrolyte when electrodes are at unit distance apart and have sufficient area of cross-section to hold electrolyte. Q.12. The resistance of conductivity cell containing 0.001M KCl solution at 298K is 1500Ω. What is the cell constant if the conductivity of 0.001M KCl solution at 298K is 0.146 × 10–3 S cm–1. Ans. Cell constant = Conductivity × Resistance = 0.146 × 10–3 S cm–1 × 1500Ω = 0.219 cm–1 Q.13. Indicate the reactions which take place at cathode and anode in fuel cell. Ans. At cathode : O2 (g) + 2H2O + 4e− → 4OH− (aq) At anode : 2H2 (g) + 4OH− (aq) → 4H2O + 4e- The overall reaction is : 2H2 (g) + O2 (g) → 2H2O (l) Q.14. Explain Kohlrausch’s law of independent migration of ions. Ans. It states that at infinite dilution, molar conductivity of an electrolyte is equal to sum of contributions due to cation as well as anion. Λ ∞m ( Na 2SO4 ) = 2Λ ºm + Λ ∞m SO 2 − ( Na ) + ( 4 ) Q.15. The standard reduction potential for the Zn2+ (aq)/Zn (s) half cell is – 0.76V. Write the reactions occurring at the electrodes when coupled with standard hydrogen electrode (SHE). Ans. At anode : Zn (s) → Zn2+ (aq) + 2e- At cathode : 2H+ + 2e− → H2 (g) Zn (s) + 2H+ (al) → Zn2+ (aq) + H2 (g) For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 28 | Chemistry-XII Q.16. Calculate the electrode potential of a copper wire dipped in 0.1M CuSO4 solution at 25ºC. The standard electrode potential of copper is 0.34 Volt. Ans. The electrode reaction written as reduction potential is Cu2+ + 2e− → Cu n=2 0.0591 1 0.0591 1 E Cu 2+ /Cu =− E 0 Cu 2+ /Cu log = 0.34 − log = 0.3104 V 2 [ Cu ] 2 0.1 Q.17. Two metals A and B have reduction potential values – 0.76 V and + 0.34 V respectively. Which of these will liberate H2 from dil. H2SO4 ? Ans. Metal having higher oxidation potential will liberate H2 from H2SO4. Thus, A will liberate H2 from H2SO4. Q.18. How does conc. of sulphuric acid change in lead storage battery when current is drawn from it ? Ans. Concentration of sulphuric acid decreases. Q.19. What type of a battery is lead storage cell ? Write the anode and cathode reaction and overall reaction occurring in a lead storage battery during discharging and recharging of cell. Ans. It is a secondary cell. Anode reaction : Pb + SO42− → PbSO4 + 2e− Cathode reaction : PbO2 + 4H+ + SO42− + 2e− → PbSO4 + 2H2O  Discharging Pb (s) + PbO2 (s) + 2H2SO4    2PbSO4(s) + 2H2O (l) Recharging Q.20. Why is alternating current used for measuring resistance of an electrolytic solution ? Ans. The alternating current is used to prevent electrolysis so that the concentration of ions in the solution remains constant. Q.21. Eθ values of MnO4−, Ce4+ and Cl2 are 1.507, 1.61 and 1.358 V respectively. Arrange these in order of increasing strength as oxidizing agent. Ans. Cl2 < MnO4− < Ce4+ Q.22. Draw a graph between Λºm and C for strong and weak electrolyte. Ans. For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 29 Q.23. The conductivity of 0.02M solution of NaCl is 2.6 × 10-2 S cm-1. What is its molar conductivity ? Ans. k = 2.6 × 10-2 S cm-1 C = 0.02M k × 1000 Λm = C(M) 2.6 × 10−2 × 1000 = 0.02 26 × 100 26 × 102 = = 0.02 × 100 2 = 13 × 102 S cm mol-1 Q.24. Give products of electrolysis of an aqueous solution of AgNO3 with silver electrode. Ans. At anode : Ag (s) → Ag+ + e- At cathode : Ag+ + e− → Ag (s) SHORT ANSWER-II TYPE QUESTIONS Q. 1. A solution of CuSO4 is electrolysed for 10 mins. with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ? Ans. I = 1.5 Ampere Time = 10 × 60s = 600s Q =I×t = 1.5 × 600 = 900 C Cu2+ + 2e− → Cu (s) 2F amount of electricity deposit copper = 63.5 g 63.5 × 900 900 C amount of electricity deposit copper = 2 × 96500 = 0.296 g Q. 2. Depict the galvanic cell in which the reaction Zn (s) + 2Ag+ → Zn2+ + 2Ag (s) takes place. Further show : (a) Which of the electrode is negatively charged ? For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 30 | Chemistry-XII Electrochemistry Concepts (b) The carriers of the current in the cell. (c) Individual reaction at each electrode. Ans. Zn (s)|Zn2+ (aq) || Ag+ (aq)|Ag (s) (a) Zn electrode (anode) (b) Ions are carriers of the current in the cell. (c) At anode : Zn (s) → Zn2+ + 2e- At cathode : Ag+ + e− → Ag (s) Q. 3. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10-3 S cm-1 ? Ans. Cell constant = k × R = 0.146 × 10-3 × 1500 = 0.219 cm-1 Q. 4. Predict the products of electrolysis in each of the following : (a) An aqueous solution of AgNO3 with platinum electrodes. (b) An aqueous solution of CuCl2 with Pt electrodes. Ans. (a) At anode (Oxidation) 4OH− − 4e− → 2H2O + O2 At cathode (Reduction) Ag+ + e− → Ag (s) (b) At anode (Oxidation) Cl− − e− → Cl (g) Cl + Cl → Cl2 At cathode (Reduction) Cu2+ + 2e− → Cu (s) Q. 5. Determine the values of equilibrium constant Kc and ∆Gθ for the following reaction : Ni (s) + 2Ag+ (aq) → Ni2+ (aq) + 2Ag (s) Eθ = 1.05 V Ans. ∆Gθ = − nFEθcell n = 2, Eθcell = 1.05 V For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 31 F = 96500 C mol-1 ∆Gθ = − 2 × 1.05 × 96500 = – 202.650 kJ ∆G θ = − RT ln Kc ∆Gθ −202.650 × 103 ln Kc = − = RT 8.314 × 298 Kc = 3.32 × 1035 Q. 6. The Ksp for AgCl at 298 K is 1.0 × 10-10. Calculate the electrode potential for Eθ + Ag+/Ag electrode immersed in 1.0M KCl solution. Given Ag / Ag = 0.80 V. Ans. AgCl (s)  Ag+ + Cl- Ksp = [Ag+][Cl-] [Cl-] = 1.0 M k sp 1 × 10−10 = [Ag ] =+ = 1 × 10-10 M Cl−  1 Now, Ag+ + e− → Ag (s) θ 0.059 1 E = E − log 1  Ag +  0.059 1 = 0.80 − log 1 10−10 = 0.80 – 0.059 × 10 = 0.21 V Q. 7. Estimate the minimum potential difference needed to reduce Al2O3 at 500ºC. The free energy change for the decomposition reaction : 2 4 Al 2 O 3 → Al + O 2 is ∆G = + 960 kJ, F = 96500 C mol-1. 3 3 2 4 Ans. Al 2 O3 → Al + O 2 3 3 6× 2 n = = 4e- 3 ∆G = − nFE For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 32 | Chemistry-XII Electrochemistry Concepts ∆G = 960 × 103 J, n = 4, F = 96500 C mol-1 960 × 103 = − 4 × 96500 × E E = − 2.487 V Minimum potential difference needed to reduce Al2O3 = – 2.487 V. Q. 8. Two electrolytic cells containing silver nitrate solution and copper sulphate solution are connected in series. A steady current of 2.5 amp was passed through them till 1.078 g of Ag were deposited. How long did the current flow ? What weight of copper will be deposited ? (Ag = 107.8 u, Cu = 63.5 u) Ans. w=z×i×t w t = z×i 1.078 × 1 × 96500 t = = 386 seconds 107.8 × 2.5 Cu2+ + 2e− → Cu 63.5 w = × 2.5 × 386 = 0.3175 gram 2 × 96500 Q.9. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5.0 amp for 20 minutes. What mass of the nickel will be deposited at the cathode ? (Ni = 58.7 u) Ans. w =z×i×t 58.7 z = 2 × 96500 w = 1.825 gram Q.10. The cell in which the following reaction occurs : 2Fe3+ (aq) + 2I− (aq) → 2Fe2+ (aq) + I2 (s) has E0cell = 0.236 V. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. Ans. n = 2 ∆Gº = − nFE0cell = − 2 × 96500 × 0.236 J = − 45.55 kJ/mol ∆Gº = − 2.303 RT log Kc ∆Gº 45.55 × 103 = log K c = = 7.983 −2.303RT 2.303 × 8.314 × 298 Kc = antilog (7.983) = 9.616 × 107 For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 33 Q.11. The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissociation and dissociation constant. Given Λº (H+) = 349.6 S cm2 mol-1, Λº (HCOO−) = 54.6 S cm2 mol-1. Ans. Λºm (HCOOH) = Λºm (H+) + Λºm (HCOO−) = 349.6 + 54.6 S cm2 mol-1 = 404.2 S cm2 mol-1 Λºm = 46.1 S cm2 mol-1 HCOOH  HCOO− + H+ Λcm 46.1 α = = = 0.114 Λ m 404.2 º Initial conc. C mol L-1 0 0 At equil. C(1 – α) Cα Cα 0.025 × ( 0.114 ) 2 Cα 2 Ka = = 1−α 1 − 0.114 = 3.67 × 10 −4 Q.12. Calculate the standard cell potentials of galvanic cells in which the following reaction take place : 2Cr (s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd (s) Also calculate ∆Gº and equilibrium constant of the reaction. Ans. E0cell = E0cathode − E0anode = − 0.40 − (− 0.74) = 0.34 V ∆Gº = − nFE0cell = − 6 × 96500 × 0.34 = − 196860 = − 196860 J mol-1 = − 196.86 kJ/mol − ∆Gº = 2.303 RT log Kc 196860 = 2.303 × 8.314 × 298 log Kc Or log Kc = 34.5014 Kc = antilog 34.5014 = 3.192 × 1034 Q.13. Calculate the potential of the following cell Sn4+ (1.5 M) + Zn → Sn2+ (0.5 M) + Zn2+ (2M). Given : E0 Sn4+ /Sn 2+ = 0.13V, E0 Zn2+ / Zn = −0.76V Will the cell potential ↑ or ↓ if the concentration of Sn4+ is increased ? θ 0.0591 Sn 2+   Zn 2+  Ans. Ecell = E − log Sn 4+  [ Zn ] cell n For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 34 | Chemistry-XII Electrochemistry Concepts 0.0591 0.5 × 2 = 0.89 − log 2 1.5 × 1 0.0591 1 = 0.89 − log 2 1.5 = 0.895 V On increasing the concentration of Sn4+, EMF of the cell will increase. Q.14. Eº (Cu2+/Cu) and Eº (Ag+/Ag) is + 0.337 V and + 0.799 V respectively. Make a cell whose EMF is +ve. If the concentration of Cu2+ is 0.01M and Ecell at 25ºC is zero, calculate the concentration of Ag+. Ans. Cu is more reactive than silver, so that the cell is as Cu/Cu2+ (0.01M) || Ag+ (C)/Ag or cell reaction Cu + 2Ag+ → Cu2+ + 2Ag  Cu 2+  [ Ag ] 2 0.0591 Ecell = E º − log [Cu ]  Ag +  cell 2 n 0.0591 ( 0.01) × 12 = E º cell − log 2 n 1 ×  Ag +  Or [Ag+] = 1.47 × 10−9 M Q.15. Calculate the potential of the cell at 298 K : Cd/Cd2+ (0.1M) || H+ (0.2M)/Pt, H2 (0.5 atm) Given Eº for Cd2+/Cd = − 0.403 V, R = 8.314 J-1 mol-1, F = 96500 C mol-1. Ans. The cell reaction is Cd + 2H+ (0.2M) → Cd2+ (0.1M) + H2 (0.5 atm) Eºcell = 0 – (– 0.403) = + 0.403 V 2.303RT Cd 2+  × PH2 Ecell = 0.403 − log [ Cd ]  H +  2 nF 2.303 × 8.314 × 298 0.1 × 0.5 = 0.403 − log 2 × 96500 ( 0.2 ) 2 Ecell = 0.403 – 0.003 = 0.40 V Q.16. The electrical resistance of a column of 0.05M NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity. For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 35 Ans. Diameter = 1 cm, radius = 0.5 cm Area = πr2 = 3.14 × (0.5)2 = 0.785 cm2 R×A 5.55 × 103 × 0.785 ρ = = l 50 = 87.135 ohm cm 1 1 Conductivity (k)= = = 0.01148 ohm −1 cm −1 ρ 87.135 = 0.01148 ohm cm K × 1000 0.01148 × 1000 = Molar conductivity Λm c = = 29.6 S cm2 mol–1 M 0.05 Q.17. Conductivity of saturated solution of BaSO4 at 315 K is 3.648 × 10–6 ohm–1 cm–1 and that of water is 1.25 × 10–6 ohm–1 cm–1. Ionic conductance of Ba2+ and SO42– are 110 and 136.6 ohm–1 cm2 mol–1 respectively. Calculate the solubility of BaSO4 in g/L. Ans. Λºm (BaSO4) = Λºm Ba2+ + Λºm SO42− = 110 + 136.6 = 246.6 ohm–1 cm–1 KBaSO4 = KBaSO4 (solution) – Kwater = 3.648 × 10–6 – 1.25 × 10–6 = 2.398 × 10–6 S cm–1 K × 1000 2.398 × 10−6 × 1000 = Λ cm = = 9.72 × 10–6 mol/L Solubility 246.6 Solubility = 9.72 × 10–6 × 233 = 2.26 × 10–3 g/L LONG ANSWER TYPE QUESTIONS (5 Marks) Q. 1. Conductivity of 0.00241M acetic acid is 7.896 × 10−5 S cm-1. Calculate its molar conductivity and if Λºm for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant ? k ×1000 Ans. Λºm = M 7.896 × 10−5 S cm −1 × 1000 cm3 L−1 = 0.00241 mol L−1 = 32.76 S cm2 mol-1 Λm 32.76 α = = = 8.39 × 10-2 Λ º m 390.5 For more FREE DOWNLOADS, visit www.aspirationsinstitute.com 36 | Chemistry-XII Electrochemistry Concepts ( ) 3 Cα 2 0.00241 × 8.39 × 10−2 Ka = = 1−α 1 − 8.39 × 10−2 = 1.86 × 10−5 Q. 2. Three electrolytic cells A, B, C containing solution of ZnSO4, AgNO3 and CuSO4 respectively all connected in series. A steady current of 1.5 amperes was passed through then until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow ? What mass of copper and of zinc were deposited ? Ans. Ag+ + e− → Ag (s) 108 g of silver is deposited by 96500 C. 96500 × 1.45 1.45 g silver is deposited by = 108 = 1295.6 C Q =I×t 1295.6 = 1.5 × t 12956 t = = 863 s 1.5 In cell A, the electrode reaction is 5.6 Zn2+ + 2e− → Zn 2F of electricity deposit Zn = 65.3 g 65.3 × 1295.6 1295.6 of electricity deposit Zn = 2 × 96500 = 0.438 g In cell C, the electrode reaction is Cu2+ + 2e− → Cu (s) 2F of electricity deposit Cu = 63.5 g 63.5 × 1295.6 1295.6 of electricity deposit Cu = 2 × 96500 = 0.426 g For more FREE DOWNLOADS, visit www.aspirationsinstitute.com Electrochemistry Concepts | 37 Q. 3. (a) Define Kohlraush’s law. (b) Suggest a way to determine the Λºm for CH3COOH. (c) The Λºm for sodium acetate, HCl, NaCl are 91.0, 425.9 and 126.4 S cm2 mol-1 respectively at 298 K. Calculate Λºm for CH3COOH. Ans. (a) The molar conductivity at a infinite dilution for a given salt can be expressed as the sum of the individual contribution from the ions of electrolyte. (b) Λº CH3COOH = ? Λº CH3COO− + Λº H+ = Λº CH3COO− + Λº Na+ + Λº H+ + Λº Cl− – Λº Na+ – Λº Cl-...(i) Λºm CH3COOH = Λº CH3COONa + Λº HCl – Λº NaCl (c) Λºm CH3COOH = Λº CH3COONa + Λº HCl – Λº NaCl = 91.0 + 425.9 – 126.4 = 390.5 S cm2 mol-1 Q. 4. (a) Define weak and strong electrolytes. (b) The Eθ values corresponding to the following two reduction electrode processes are : (i) Cu+/Cu = 0.52 V (ii) Cu2+/Cu+ = 0.16 V Formulate the galvanic cell for their combination. Calculate the cell potential and ∆Gº for the cell reaction. Ans. (a) Weak electrolyte : The substance which partially ionized in solution is known as weak electrolyte. Example : NH4OH. Strong electrolyte : The substance which completely ionized in solution is known as strong electrolyte. Example : NaCl. (b) Cu+ + e− → Cu Cu+ → Cu2+ + e− Overall cell reaction : 2Cu+ → Cu + Cu2+ Cu+/Cu2+||Cu+/Cu Eθcell = 0.52 – 0.16 = 0.36 V ∆Gº = – nFEθcell = – 1 × 96500 × 0.36 = – 34740 J mol-1 Source: EDUDEL