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ISC Class XII Prelims 2020 : Chemistry (Smt. Sulochanadevi Singhania School, Thane)

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Smt. Sulochanadevi Singhania School, Thane
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2 CHAPTER electrochemistry Syllabus am Redox reactions, conductance in electrolytic solutions, specific and molar conductivity, variations of conductivity with concentration, Kohlrausch s Law, electrolysis and law of electrolysis (elementary idea), dry cell- electrolytic cells and Galvanic cells, lead accumulator, EMF of a cell, standard electrode potential, Nernst equation and its application to chemical cells, Relation between Gibbs energy change and EMF of a cell, fuel cells, corrosion. gr Chapter Analysis D 1 Q. (2 marks) Cell OD 1 Q. (5 marks)# 1 Q. (5 marks)* 1 Q. (3 marks)*# 2018 D/OD 1 Q. (5 marks) ci al Electrode Potential 2017 D OD 1 Q. 1 Q. (3 marks)^ (2 marks) te le 2016 List of Topics An i ke tS ha _o ffi Numerical on emf, Electrode 1 Q. 1 Q. 1 Q. 1 Q. Potential, Molar Conductivity, (3 marks) (2 marks) (3 marks) (5 marks)@ Degree of Dissociation, 1 Q. Standard Gibbs Energy and (3 marks)^ Electrolysis * One question of 5 marks with two choices was asked. *First choice has one numerical of 3 marks and one question of 2 marks on Electrode Potential. #Second choice has one numerical of 3 marks Molar Conductivity and Degree of Dissociation and one question of 2 marks on Electrochemical Cell. ^One question of 3 marks with one numerical of 2 marks on Electrolysis and one question of 1 mark on Secondary Cell was asked. One question of 5 marks with two choices was asked. @First choice has a numerical of 3 marks on Cell Representation and Calculation of E.M.F of a cell and Give Reasons question of 2 marks. Second choice has a numerical of 3 marks on Calculation of E.M.F of a cell and a 2 marks question on Fuel cells. On the basis of above analysis, it can be said that from exam point of view, Cells and numerical on E.M.F., Electrode Potential, Molar Conductivity, Degree of Dissociation, Standard Gibbs Energy and Electrolysis are the most important topics of the chapter. Topic-1 TOPIC - 1 Conductance in Electrolytic Solutions, Specific and Molar Conductivity, Variations of Conductivity with Concentration, Gibbs Energy, Kohlrausch s Law .... P. 30 Revision Notes Topic - 2 Redox Reaction, Electrochemical Cell, Galvanic Cell, EMF of a Cell, Standard Electrode Potential, Nernst Equation .... P. 36 Electrochemistry is the branch of chemistry which deals with the study of the production of electricity from energy released during spontaneous chemical reactions and the use of electrical energy to result in non-spontaneous chemical transformations. Topic - 3 Electrolysis, Laws of Electrolysis, Batteries, Fuel Cells and Corrosion .... P. 44 Conductance in Electrolytic Solutions, Specific and Molar Conductivity, Variations of Conductivity with Concentration, Gibbs Energy, Kohlrausch s Law [ 31 Electrochemistry Electrolytic conduction : The flow of electric current through an electrolytic solution is called electrolytic conduction. Electrolyte : A substance that dissociates in solution to produce ions and hence conducts electricity in dissolved state or molten state. Weak electrolyte H2CO3, CH3COOH, HCN, MgCl2. Strong electrolyte NaCl, HCl, NaOH. Degree of ionisation : It is the ratio of number of ions produced to the total number of molecules of electrolyte. Resistance is defined as the property of given substance to obstruct the flow of charge. It is directly proportional to the length (l) and inversely proportional to its area of cross-section (A). l l or; R = r R A A r : Resistivity or specific resistance. Specific resistivity : If a solution is placed between two parallel electrodes having cross sectional area A and distance l apart, then l , where r is specific resistivity. R=r A te le gr am The unit of specific resistivity is Ohm cm or Ohm m. Conductance : The ease with which current flows through a conductor is called its conductance. It is the reciprocal 1 = A = A K of the resistance. i.e., C = l l R k= C l , A ci tS ha _o ffi al The unit of conductance is Siemens, S or ohm 1. (i.e., 1 s = 1 ohm 1 = 1 W 1) Specific conductivity (electrolytic conductivity) : It is reciprocal of the specific resistivity of an electrolytic solution. It is denoted by k (Kappa). An i ke where C is conductance of solution, l is the distance and A is the area of cross section. Its unit is ohm 1 cm 1 or in S. I. unit S m 1. It depends upon the : (i) Nature of the material (ii) Temperature (iii) Pressure Metallic conductance is the electrical conductance through metal that occurs due to the movement of electrons. It depends upon the : (i) Nature and structure of the metal (ii) Number of valence electrons per atom (iii) Temperature Electrolytic or ionic conductance is the conductance of electricity that occurs due to ions present in the solution. It depends upon the : (i) Nature of electrolyte or interionic attractions (ii) Solvation of ions (iii) Nature of solvent and its viscosity (iv) Temperature Wheatstone bridge helps us to measure R4. R2 R1 G i R4 R3 E R4 = R3R 2 R1 K 32 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII Cell constant (G) : It is the ratio of distance between electrodes to the cross-sectional area between electrodes. l in cm 1 or m 1 Cell constant (G) = A It depends on the : (i) Distance between the electrodes (ii) Area of cross section. Molar conductivity : It is defined as the conducting power of all the ions produced by one gram mole of an electrolyte in a solution. It is denoted by Lm. Lm = 1000 S cm 2 mol - 1 , C gr am where k= Conductivity and C = Concentration of solution. Conductivity decreases with dilution while molar conductivity increases with dilution. Debye Huckel Onsager equation : It is applicable for strong electrolyte : L = L AC1/2, where L = Limiting molar conductivity, L = Molar conductivity, A = Constant and C = Concentration of solution. Kohlrausch s law of independent migration of ions : According to this law, limiting molar conductivity of an electrolyte, at infinite dilution, can be expressed as the sum of contributions from its individual ions. If the molar conductivity of the cations is denoted by + and that of the anions by then the law of independent migration te le of ions is m = v+ + + v or L0 = v+ + + v , where, v+ and v are the number of cations and anions per formula of electrolyte. Application of Kohlrausch s Law : (i) Calculation of molar conductivities of weak electrolyte at infinite dilution. (ii) Calculation of degree of dissociation (a) of weak electrolytes. tS ha _o ffi ci al 0 Degree of dissociation (a) = c m m (iii) Determination of dissociation constant (K) of weak electrolytes : K = c( c ) c 2 = m c 1 m ( m m ) An i ke (iv) Determination of solubility of sparingly soluble salts : K 1000 Solubility = m Know the Formulae Potential difference (V) Resistance ( R ) Current (I) = Resistance (R) = Conductance (C) = K l A A l Specific conductivity (k) = C Cell constant (G) = l Cell constant = R A l A For strong electrolyte, Lm = L m A C L = n+L + + n L Degree of dissociation (a) = c m m 0 [ 33 Electrochemistry K = Solubility = c( m ) c 2 = m ( m m ) 1 K 1000 m Know the Terms Superconductors : Material with a zero resistance. Limiting molar conductivity : Molar conductivity when concentration approaches zero. Electrolyte : Substance which splits into ions in dissolved or molten state by passing electricity. Over voltage : It is the difference between the potential required for the evolution of a gas and its standard reduction potential. Very Short Answer-Objective Type Questions am A [NCERT Exemp. Q. 16, Page 36] gr Ans. Correct option : (c) Explanation : NH 4 Cl NH+4 +Cl - al te le ci A. Multiple Choice Questions: Q. 1. Which of the statements about solutions of electrolytes is not correct? (a) Conductivity of solution depends upon size of ions. (b) Conductivity depends upon viscosity of solution. (c) Conductivity does not depend upon solvation of ions present in solution. (d) Conductivity of solution increases with temperature. R [NCERT Exemp. Q. 7, Page 34] Ans. Correct option : (c) Explanation : Conductivity depends upon solvation of ions present in the solution. So, as the solvation of ions increases, the conductivity will decrease. tS ha _o ffi An i Q. 3. 0 m ( NH 4 OH ) is equal to ..................... 0 0 0 (a) m ( NH4 OH ) + m ( NH4 Cl) ( HCI ) 0 0 0 (b) m ( NH4 CI ) + m ( NaOH ) ( NaCI ) 0 0 0 (c) m ( NH4 CI ) + m ( NaCI ) ( NaOH ) 0 0 0 (d) m ( NaOH ) + m ( NaCI ) ( NH4 CI ) + NaCl Na +Cl - (i) (ii) NaOH Na++OH - (iii) NH 4 OH NH+4 +OH - (iv) To get equation (iv) + m ( NaOH ) m ( NaCl ) = m ( NH4 OH ) m ( NH4 Cl ) B. Match the following : Q. 1. Match the species given in Column I with those mentioned in Column II. ke Q. 2. Which of the following statement is not correct about an inert electrode in a cell? (a) It does not participate in the cell reaction. (b) It provides surface either for oxidation or for reduction reaction. (c) It provides surface for conduction of electrons. (d) It provides surface for redox reaction. U [NCERT Exemp. Q. 5, Page 34] Ans. Correct option : (d) Explanation : Inert electrode does not participate in redox reaction and acts only as source or sink for electrons. It provides surface either for oxidation or for reduction reaction. (1 mark each) (i) Column I m (a) Scm-1 (ii) ECell (b) m-1 (c) (d) S cm2mol-1 V (iii) (iv) G* Ans. Correct option : (i) (c) (ii) (d) (iii) (a) (iv) (b) Column II [NCERT Exemp. Q. 50, Page 40] Explanation : The SI unit of conductivity is S cm-1. The unit of Ecell is Volt. The SI unit of molar conductivity is S m2 mol-1. The unit G* is m-1. C. Answer the following: Q. 1. In an aqueous solution how does specific conductivity of electrolytes change with addition of water? U [NCERT Exemp. Q. 43, Page 40] Ans. Conductivity decreases because number of ions per unit volume decreases. 1 34 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII Short Answer Type Questions Q. 1. Define the following terms : (i) Fuel cell Answering Tip Q. 2. Define the following terms : (i) Molar conductivity (Lm), (ii) Secondary batteries. R [CBSE OD 2014] = Ans. tS ha _o ffi ke Q. 3. State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution ? R + A&E [CBSE OD 2014] Lm = Lm = 1000k C 1.65 10 4 S cm 1 1000 cm 3 L 1 m/5 cm2 mol 1 0.01 mol L 1 400 200 A Commonly Made Error B Students often reframe the definition or write only the mathematical expression. Write the law as stated. Q. 4. Calculate the degree of dissociation (a) of acetic acid if its molar conductivity (Lm) is 39.05 S cm2 mol-1. Given L (H+) = 349.6 S cm2 mol-1 and L0(CH3COO-) = 40.9 S cm2 mol-1. A [CBSE Delhi Set-1, 2, 3 2017] Ans. L CH3COOH = L CH3COO- + L H+ = 40.9 + 349.6 = 390.5 S cm2/mol Now, a = Lm/L0m = 39.05/390.5 = 0.1 [CBSE Marking Scheme 2017] = 16.5 S cm2 mol 1 1 [CBSE Marking Scheme 2018] Q. 7. The following curve is obtained when molar conductivity (Lm) is plotted against the square root of concentration, c for two electrolytes A and B : An i Ans. Kohlrausch law of independent migration of ions : The law states that limiting molar conductivity of an electrolyte can be represented by the sum of the individual contributions of the anion and cation of the electrolyte. 1 L = n+L + + n L On dilution, the conductivity (k) of the electrolyte decreases as the number of ions per unit volume of solution decreases. [CBSE Marking Scheme 2014] 1 0.025 S cm 1 1000 cm 3 L 1 0.20 mol L 1 = 125 S cm2 mol 1 1 Q. 6. The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 x 10 4 S cm 1. Calculate molar conductivity (Lm) of the solution. A [CBSE Comptt. D/OD 2018] ci Ans. (i) Molar conductivity (Lm) : Molar conductivity is defined as the conductivity due to all the ions produced by dissolving one mole of an electrolyte in solution. 1 (ii) In secondary batteries, the reactions can be reversed the cell reactions imposing a higher voltage than the E.M.F. of cell (external voltage). These batteries can be recharged by passing electric current and used again and again. 1 [CBSE Marking Scheme 2014] am Q. 5. The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm 1. Calculate its molar conductivity. A [CBSE Delhi 2013] Ans. Concentration of solution = 0.20 M Conductivity = 0.025 S cm 1 Molar conductivity Lm Conductivity ( K ) 1000 1 = Concentration of solution gr R [CBSE OD 2014] Ans. (i) Galvanic cells that are designed to convert the energy of combustion of fuels (methane, methanol etc.) directly into electrical energy are called fuel cells. 1 (ii) When the concentration approaches zero, the molar conductivity is known as limiting molar conductivity. It is represented by L m. 1 [CBSE Marking Scheme 2014] Write the working formula in each step followed by value assignment for each entity. Give appropriate unit alongwith the answer. te le o (ii) Limiting molar conductivity ( m ) al (2 marks each) 0.2 0.4 C1/2 (mol L)1/2 (i) How do you account for the increase in the molar conductivity of the electrolyte A on dilution ? (ii) As seen from the graph, the value of limiting molar conductivity (L m) for electrolyte B cannot be obtained graphically. How can this value be obtained ? A&E [CBSE SQP 2016] Ans. (i) As seen from the graph, electrolyte A is a strong electrolyte which is completely ionised in solution. With dilution, the ions are far apart from each other and hence the molar conductivity increases. 1 [ 35 Electrochemistry (ii) To determine the value of limiting molar conductivity for electrolyte B, indirect method based upon Kohlrausch law of independent 1 migration of ions is used. Q. 8. Why on dilution the Lm of CH3COOH increases drastically, while that of CH3COONa increases gradually? A&E [NCERT Exemp. Q. 49, Page 40] Ans. In case of CH 3COOH which is a weak electrolyte, the number of ions increase on dilution due to an increase in degree of dissociation resulting in drastic increase in L m. 1 CH 3COOH+H 2O CH 3COO +H 3O+ In the case of CH3COONa which is a strong electrolyte, the number of ions remains the same but the inter-ionic attraction decreases resulting in gradual increase in Lm. 1 Long Answer Type Questions-I 2o (ii) m (CaSO4) = LoCa2+ + Lo SO4 cm2 mol 1 = 119.0 S + 160.0 S = 279.0 S cm2 mol 1 Commonly Made Error cm2 1 mol 1 1 Students only write the mathematical expression. Answering Tips An i ke Write the law as stated. Stick to the statement as the marks are alloted to that only. Do not forget to mention the units. Q. 2. The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is 4.55 103 ohm. Calculate its molar conductivity. A [CBSE Foreign Set-1, 2, 3 2017] Ans. A = gr Check the compatibility of units. Detailed Answer : A = pr2 = 3.14 0.5 0.5 cm2 = 0.785 cm2 l = 45.5 cm r = R A/l r = 4.55 103W 0.785 cm2/45.5 cm r = 78.5 W cm = 3.14 0.5 0.5 cm2 = 0.785 cm2 l = 45.5 cm G* = I/A = 45.5 cm/0.785 cm2 k = G*/R = 0.0127 S cm-1 1000/0.05 mol/cm3 = 254.77 S cm2 mol-1 Conductivity of 2.5 10 4 M methanoic acid is 5.25 10 5 S cm 1. Given : Lo(H+) = 349.5 S cm2 mol 1 and Lo (HCOO ) = 50.5 S cm2 mol 1. Lm = k 1000/C = [1.27 = 254.77 S cm2 mol-1 1000/0.05 [CBSE Marking Scheme 2017] Lm = S A [CBSE OD 2015] 1000 K S cm 2 mol 1 1 Lm = M Ans. dissociation. = 57.96 cm-1/4.55 x 103 = 1.27 10-2 S cm-1 mol/cm3 = 0.0127 S Q. 3. Calculate the molar conductivity and degree of cm-1] cm-1 = 1/78.5 S cm-1 Molar conductivity, Lm = k 1000/C Conductivity, k = 1/r = 57.96 cm-1 10-2 pr2 am Answering Tip tS ha _o ffi Students often convert centimeter into meter. te le (ii) Calculate limiting molar conductivity of CaSO4 (limiting molar conductivity of calcium and sulphate ions are 119.0 and 160.0 S cm2 mol 1 respectively) R + A [CBSE SQP 2016] Ans. Kohlrausch law of independent migration of ions : (i) The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contribution of the anions and cations of the 1 electrolyte. ci Commonly Made Error al Q. 1. (i) State the law which helps to determine the limiting molar conductivity of weak electrolyte. (3 marks each) 1000 5.25 10 5 S cm 2 mol 1 2.5 10 4 = 210 S cm2mol 1 0m HCOOH = lo HCOO + 1 l o H+ = (50.5 + 349.5) S cm2mol 1 = 400 S cm2mol 1 a = Lm / L m a = 210 / 400 = 0.525 Answering Tip Always mention the working formula followed by substitution of values in it. 1 36 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII Q. 4. When a certain conductance cell was filled with 0.1 M KCl, it has a resistance of 85 ohm at 25 C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 ohms. Calculate the molar conductance of the electrolyte at this concentration. G = R K = 85 1.29 10 2 cm 1 = 109.6 10 2 cm 1 = 1 1.09 W 96 = 1.13 10 2 W 1 cm 1 1000 K \ Molar conductance Lm = M = 1 1 1.13 10 2 1 cm 1 1000 cm 3 L 1 0.052 mol L 1 = 217.307 W 1 cm2 mol 1 1 Answering Tip Separate marks are alloted for formula, so remember to write it. Also check compatibility of units. am [Specific conductance of 0.1 M KCl = 1.29 10 2 ohm 1 cm 1] A [CBSE Comptt. OD 2012] Ans. Calculation of cell constant Conductivity of 0.1 M KCl = 1.29 10 2 W 1 cm 1 Resistance = 85 ohm Conductivity, 1 Cell constant (G) K= R = 1.09 cm 1 Conductivity of unknown electrolyte K, 1 G = R te le gr Topic-2 tS ha _o ffi Revision Notes ci al Redox Reaction, Electrochemical Cell, Galvanic Cell, EMF of a Cell, Standard Electrode Potential, Nernst Equation An i ke Redox reaction : A chemical reaction in which oxidation and reduction both processes takes place is known as redox reaction. Oxidation is a process in which any substance loses one or more electrons while reduction is the process in which one or more electrons are gained by another substance. Electrochemical cell : A device in which the redox reaction is carried indirectly and chemical energy is converted to electrical energy. It is also called galvanic cell or voltaic cell. Redox couple : It is defined as having together the oxidised and reduced form of a substance taking part in an oxidation or reduction half reaction. Galvanic cell or Voltaic cell : It consists of two metallic electrodes dipped in electrolytic solutions. Electrical energy is produced as a result of chemical reaction which takes place in this cell. Daniell cell : It is the improved form of galvanic cell. It consists of zinc rod dipped in ZnSO4 solution which acts as the oxidation half-cell. In the reduction half-cell, the copper vessel itself acts as cathode while the saturated solution of CuSO4 acts as the electrolyte. Both solutions are kept apart by taking zinc sulphate in porous pot and putting it in a copper sulphate solution. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Cell is represented as, Zn(s) |Zn2+(aq) (C1)| | Cu2+(aq) (C2)| Cu(s) Salt Bridge and its function : It is an inverted U-shaped glass tube which contains a suitable salt in the form of a thick paste made in agar-agar. It performs following functions (i) It completes inner cell circuit. (ii) It prevents transference of electrolyte from one half-cell to the other. (iii) It maintains the electrical neutrality of the electrolytes in the two half-cells. Electrode Potential : It is the potential difference set up between the metal and solution of its own ions. It shows the tendency of an electrode to either lose or gain electrons. When the concentration of all the species involved in a half-cell is unity, then the electrode potential is called standard electrode potential. Standard Electrode Potential : Electrode potential at 25 C, 1 bar pressure and 1 M solution is known as standard electrode potential (E ). The standard electrode potential of any electrode can be measured by connecting it to standard hydrogen electrode (SHE). SHE has a standard potential at all temperatures. It consists of a platinum foil coated with platinum black dipped into an aqueous solution in which the H+ = 1 M at 25 C and 1 bar pressure. The potential difference between the two electrodes of a galvanic cell is called the cell potential (measured in volts). It is also called the emf of the cell when no current is flowing through the circuit. [ 37 Electrochemistry An i ke tS ha _o ffi ci al te le gr am EMF of the cell : Electromotive force is also called emf (denoted and measured in volts). It is the voltage developed by any source of electrical energy such as battery or dynamo. It is generally defined as the potential for a source in a circuit. Ecell = Ecathode Eanode In terms of standard oxidation electrode potential : E cell = E anode E cathode, where E cathode = standard electrode potential of cathode and E anode = standard electrode potential of anode Standard oxidation potential : It is the potential difference when given electrode is in contact with its ions having 1 molar concentration, undergoes oxidation when coupled with standard hydrogen electrode. The higher value of E for a half-cell indicates that reductant in the half-cell is stronger reducing agent than hydrogen gas. The series in which all the standard electrode potentials have been arranged in order of increasing value is known as electrochemical series. The characteristics of the electrochemical series are : (i) The reducing agent with lowest E value, considered as the strongest reducing agent, is placed well above (top) of the electrochemical series. The oxidising agent with highest E value, considered as the strongest oxidising agent, is placed below (bottom) of the electrochemical series. (ii) Oxidising power increases with increase in E value. (iii) Any metal of lower E value can reduce other metal of higher E value. Similarly, any metal of higher E value can oxidise other metal of lower E value. Applications of electrochemical series : Important applications are as follows (i) In comparing relative reducing or oxidising power of the elements. (ii) In predicting the feasibility of spontaneity of a redox reaction. (iii) In predicting whether a metal can evolve hydrogen from an acid or not. (iv) Comparison of reactivities of metals and non-metals. (v) Determination of standard Gibbs energy change. (vi) Determination of equilibrium constant. Nernst equation : If the concentration of species in the electrode reaction is not equal to 1 M, then we use Nernst equation. For a general electrode, Mn+(aq) + ne M(s) the Nernst equation can be written as [M] RT 0 E n+ ln (M / M) = E (Mn+ / M) nF Mn+( aq ) where E = Standard electrode potential, R = Gas constant and is 8.31 JK 1 mol 1, T = Temperature (K), n = Number of moles of electrons and F = Faraday (96500 C), At equilibrium, E cell = 0.059 log K c n Kc = Equilibrium constant M Kc = [Mn + ] For the cell with the net reaction, ne aA + bB mM + nN the Nernst equation at 298 K can be written as Ecell = E cell M m N n 0.059 log n A a B b where E cell = E cathode E anode Gibbs energy : DG = nFE cell for cell reaction to be spontaneous, DG must be negative, Calculations of DrG and DrG : DrG = nF E cell and DrG = nF Ecell We also know that Gibbs energy change is equal to the useful work done. 38 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII For cell reaction to be spontaneous, DG must be negative. DG = 2.303 RT log K. Know the Formulae Nernst equation : Ecell = Ecathode Eanode E cell = E cathode E anode Ecell = E cell 2.303 RT [C]c [ D]d log nF [ A]a [ B]b Ecell = E cell 0.0591 [C]c [ D]d log at 298 K n [ A]a [ B]b Ecell = DrG = nFE cell DrG = 2.303 RT log KC. te le gr Know the Terms am C 2.303 RT log 2 where C2 > C1 nF C1 al Concentration cell : A cell in which both electrodes and electrolytic solutions are of same substance but the solution of electrolyte in which they dip have different concentrations. Gibbs energy : Energy associated with a chemical reaction that can be used to do work. ci Very Short Answer-Objective Type Questions tS ha _o ffi (a) Pt(s)H 2 |(g, 0.1bar)|H+ (aq., 1 M)||Cu 2+(aq., 1M)|Cu (b) Pt(s)|H2 (g, 1bar)|H+(aq., 1 M)||Cu 2+(aq., 2 M)|Cu (c) Pt(s)|H2 (g, 1bar)|H+ (aq., 1 M)||Cu 2+(aq., 2 M)|Cu + 2+ (d) Pt(s)|H2 (g, 1bar)|H (aq., 0.1 M)||Cu (aq., 1 M)|Cu U [NCERT Exemp. Q. 1, Page 33] Ans. Correct option : (c) Explanation : When copper electrode is connected to standard hydrogen electrode, it acts as cathode and its standard electrode potential can be measured by the following way : E = E R E L = E R 0 = E R + 2+ Pt(s)| H 2 (g,1bar)| H (aq.,1M)||Cu (aq.,1M)|Cu will measure standard electrode potential of copper electrode. So, to calculate the standard electrode potential of the given cell it is coupled with the standard The graph of E Mg 2+ /Mg vs. log [Mg 2+ ] is (a) (b) log [Mg2+] (c) (d) Emg2+|Mg log [Mg2+] Emg2+|Mg ke An i Q. 2. Which cell will measure standard electrode potential of copper electrode? hydrogen electrode in which pressure of hydrogen gas is one bar and H+ ion in the solution is one molar and also the concentrations of the oxidized and the reduced forms of the species in the right hand half-cell is unity. Q. 3. Electrode potential for Mg electrode varies according to the equation : 0.059 1 E Mg 2 + /Mg =E Mg 2+ /Mg log . 2 [Mg 2+ ] Emg2+|Mg A. Multiple Choice Questions: Q. 1. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ___________. (a) Cell potential (b) Cell emf (c) Potential difference (d) Cell voltage R [NCERT Exemp. Q. 4, Page 34] Ans. Correct option : (b) Emg2+|Mg (1 mark each) log [Mg2+] log [Mg2+] A [NCERT Exemp. Q. 2, Page 33] Ans. Correct option : (b) 0.059 0 log Mg 2 + Explanation : E Mg 2+ /Mg = E Mg 2+ / Mg 2 Compare this equation with the equation of straight line y = mx + c. The graph of E Mg 2+ /Mg vs. log [Mg2+] is a straight line with a positive slope and intercept E Mg 2+ /Mg. Q. 4. Using the data given below find strongest reduction agent. [ 39 Electrochemistry E Cr O2 /Cr3+ =1.33 V, E Cl 2 7 2 /C1 At cathode, =1.36 V 1 H 2 + OH 2 At anode, two reactionsarepossible. H 2O + e = 0.74 V E 2+ =1.51V, E Cr 3+ /Cr MnO4 /Mn (a) Cl (b) Cr (c) Cr3+ (d) Mn2+ U [NCERT Exemp. Q. 8, Page 34] Ans. Correct option : (b) Explanation : The negative value of standard reduction potential for Cr3+ to Cr means that the redox couple is a stronger reducing agent. Q. 5. In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode? (a) Na+ (aq)+e Na ( s) ; E Cell = 2.71 V 1 H 2 ( g ) ; E Cell = 0.00 V 2 ( 1 H (E 2 Na + e Na E 2 Cell Ans. At equilibrium, that is, when the cell is completely discharged. 1 Q. 2. Depict the galvanic cell in which the cell reaction is Cu + 2Ag+ 2Ag + Cu2+. A [NCERT Exemp. 33, Page 39] Ans. Cu|Cu2+||Ag+|Ag 1 Q. 3. Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes A and B in the electrolytic cell? U [NCERT Exemp. 36, Page 39] ci ) = 0.00 V ) An i Zinc plate (A) Electrode On the basis of their standard reduction electrode potential (E ) values, which reaction is feasible at the cathode and why ? (ii) Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration ? U + R [CBSE Delhi 2015] Ans. (i) Ag+(aq) + e Ag(s); E = + 0.80 V. 1 H+(aq) + e H 2 (g) ; E = 0.00 V. 2 On the basis of their standard reduction potential (E ) values, cathode reaction is given by the one with higher E values. Solution ZnSO4 CuSo4 Solution (B) Electrode Electrolytic cell Ans. A will have negative polarity; B will have positive polarity. 1 Short Answer Type Questions Q. 1. (i) Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution : Ag+(aq) + e Ag(s) E = +0.80 V 1 + E = 0.00 V H (aq) + e H2 (g) 2 Salt bridge Copper plate = 2.71V ke H+ + e Cell U [NCERT Exemp. 30, Page 38] tS ha _o ffi + te le A [NCERT Exemp. Q. 17, Page 36] Ans. Correct option : (b) Explanation : During electrolysis H 2O H + + OH Q. 1. Under what condition is ECell = 0 or rG=0? gr 1 (d) Cl (aq) Cl2 (g)+e ; E Cell = 1.36 V 2 NaCl Na + + Cl C. Answer the following: al (c) H+ (aq)+e am ( b ) 2H 2O(l) O 2 ( g )+4H+ (aq) + 4e ; E Cell =1.23V 1 Cl Cl 2 + e ; E Cell = 1.36 V 2 2 H 2O O2 + 4 H + + 4 e ; E Cell = 1.23 V (2 marks each) Thus Ag+(aq) + e Ag(s) reaction will be more feasible at cathode. 1 (ii) Limiting molar conductivity : When the concentration approaches zero, the molar conductivity is known as limiting molar conductivity. It is represented by m. The conductivity decreases with decrease in concentration due to decrease in the no. of ions that carry the current in a solution. 1 [CBSE Marking Scheme 2015] Q. 2. In a galvanic cell, the following cell reactions occurs: Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s) E0cell = +1.56 V (i) Is the direction of flow of electrons from zinc to silver or silver to zinc? (ii) How will concentration of Zn2+ ions and Ag+ ions be affected when the cell functions? A&E [CBSE Foreign Set-1, 2, 3 2017] 40 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII Ans. (i) Zinc to silver 1 (ii) Concentration of Zn2+ ions will increase and Ag+ ions will decrease. 1 [CBSE Marking Scheme 2017] 1 H+ (aq) + e H2 ( g) : E0 = 0.00 V 2 On the basis of their standard electrode potential values, which reaction is feasible at cathode and why? A&E [CBSE Comptt. OD Set-1, 2 2017] Ans. Ag + ( aq ) + e Ag( s) 1 Because it has higher reduction potential. 1 = 2 96500 C mol 1 1.1V = 212300 J mol 1 or = 212.3 kJ mol 1 Q. 6. Calculate the emf of the following cell at 298 K Cr(s)/Cr3+ (0.1M)//Fe2+ (0.01M)/Fe(s) [Given: E cell = + 0.30 V] OR The conductivity of 10-3 mol/L acetic acid at 25 C is 4.1 10 5 S cm 1. Calculate its degree of dissociation if L0m for acetic acid at 25 C is 390.5 S cm2 mol-1. A [CBSE SQP 2017] Ans. 2Cr(s) + 3 Fe2+(aq.) 3Fe(s) + 2Cr3+ (aq.) n = 6 [CBSE Marking Scheme 2017] te le ECell = (1 F = 96500 C mol 1) A [CBSE OD 2013; NCERT] al 2 ECell = 0.26 V OR m = m = 1000 C 1000 cm 3 L 1 4.1 10 5 S cm 1 10 3 mol L 1 = 41 S cm2 mol 1 = c m 0m = 41 = 0.105 390.5 [CBSE Marking Scheme 2017] Q.7. Suggest a list of metals that are extracted electrolytically. C [NCERT] Ans. Metals that are on the top of the reactivity series such as sodium, potassium, calcium, lithium, magnesium, aluminium are extracted electrolytically. 2 Long Answer Type Questions-I Q. 1. Consider the following reaction : Cu(s) + 2Ag+(aq) 2Ag(s) + Cu2+(aq) (i) Depict the galvanic cell in which the given reaction takes place. (ii) Give the direction of flow of current. (iii) Write the half-cell reactions taking place at cathode and anode. U [CBSE Comptt. Delhi/OD 2018] 2 Cr 3 + 2.303RT log 3 nF Fe2+ 10 1 0.059 = 0.30 V V log 3 6 10 2 ECell tS ha _o ffi ke An i Ans. (i) Since Ti4+/Ti3+ has lower reduction potential than Fe3+/Fe2+, it cannot be reduced in comparison with Fe3+/Fe2+ ions. Hence Ti4+ cannot oxidise Fe2+ to Fe3+. As the value of reduction potential increases the stability of +2 oxidation increases. Therefore correct order of stability is Cr3+/Cr2+ < Fe3+/Fe2+ < Mn3+/Mn2+ [CBSE Marking Scheme 2018] Q. 5. The standard electrode potential (E ) for Daniell is +1.1 V. Calculate the DG for the reaction. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) o ECell gr ci Detailed Answer: As reaction with higher value of standard electrode potential occurs at cathode, Ag gets reduced. So, the reaction occurring at cathode is 2 Ag + ( aq ) + e Ag( s) Q. 4. (i) On the basis of the standard electrode potential values stated for acid solutions, predict whether Ti4+ species may be used to oxidise Fe(II) to Fe(III) Ti4+ + e Ti3+ E = +0.01V Fe3+ + e Fe2+ E = +0.77V (ii) Based on the data arrange Fe2+, Mn2+ and Cr2+ in the increasing order of stability of +2 oxidation state. (Give a brief reason) E Cr3+/Cr2+ = 0.4V E Mn3+/Mn2+ = +1.5V E Fe3+/Fe2+ = +0.8V A&E [CBSE SQP 2018-2019] 2 [CBSE Marking Scheme 2013] am Q. 3. Following reactions can occur at cathode during the electrolysis of aqueous silver nitrate solution using Pt electrodes: Ag+ (aq) + e Ag( s) : E0 = 0.80 V DG = n FE cell Ans. (3 marks each) Ans. (i) Cu(s) / Cu2+(aq) ll Ag+ (aq) / Ag(s) 1 (ii) Current will flow from silver to copper electrode in the external circuit. 1 (iii) Cathode: 2Ag+(aq) + 2e 2Ag(s) Anode: Cu(s) Cu2+ (aq) + 2e 1 [CBSE Marking Scheme 2018] [ 41 Electrochemistry 0.0591 [0.01]2 log 6 [0.1]3 1 0.059 = 0.30 6 = 0.3098 V 1 [CBSE Marking Scheme 2016] 0.0591 [ Fe 2 + ] log + 2 2 [H ] Ecell = E cell 0.0591 0.001 V log = 0.44 V 2 ( 0.01)2 am = 0.44 V 0.02955 V = 0.41045 V 3 al Q. 6. A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E Zn2+/Zn = 0.76 V; E Pb2+/Pb = 0.13V) A [CBSE SQP 2016] ci (i) Students often get confused between the weak electrolyte and strong electrolyte. They sometimes make mistake in identifying the correct answer. = 0.30 Q. 5. Calculate the emf of the following cell at 25 C : Fe | Fe2+ (0.001 M) || H+ (0.01 M) | H2(g) (1bar) | Pt(s) E (Fe2+ | Fe) = 0.44 V E (H+ | H2) = 0.00 V A [CBSE Delhi 2015] Ans. Cell reaction is Fe(s) + 2H+(aq) Fe2+ (aq) + H2(g) E cell = 0.00 ( 0.44) = 0.44 V Commonly Made Error gr Ans. (i) B is a strong electrolyte. 1 A strong electrolyte is already dissociated into ions, but on dilution interionic forces are overcome, ions are free to move. So there is slight increase in molar conductivity on dilution. 1 (ii) On anode water should get oxidised in preference to Cl-, but due to overvoltage/overpotential Cl- is oxidised in preference to water. 1 [CBSE Marking Scheme 2017] te le Q. 2. (i) Solutions of two electrolytes A and B are diluted. The limiting molar conductivity of B increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer. (ii) The products of electrolysis of aqueous NaCl at the respective electrodes are: Cathode: H2 Anode: Cl2 and not O2. Explain. A&E [CBSE SQP 2017] tS ha _o ffi Q. 3. Calculate DrG0 and log Kc for the following reaction at 298 K. 2Cr( s) + 3Fe2+( aq) 2Cr 3+( aq) + 3Fe( s) [(E cell = 0.30 V), IF = 96500C mol-1] A [CBSE Comptt. OD Set-2 2017] -nFE cell, n = 6 = -6 96500 C/mol 0.30 V = -173700 J/mol = -173.7 kJ/mol 1 E cell = 0.059V/n log Kc log Kc = 0.30 V 6/0.059V = 30.5 1 [CBSE Marking Scheme 2017] ke Ans. DrG0 = An i Answering Tip Always write the working formula followed by the value substitution for each entity. Do not forget to mention units wherever required. Q. 4. Calculate e.m.f. of the following cell at 298 K : 2Cr(s) + 3Fe2+ (0.1M) 2Cr3+ (0.01M) + 3Fe(s) E (Cr3+ | Cr) = 0.74 E (Fe2+ | Fe) = 0.44 V. A [CBSE Delhi 2016] Ans. E cell = E cathode E anode = ( 0.44) ( 0.74) V = 0.30 V 0.0591 [ Products] log E = E n [ Reactants] = 0.30 0.0591 [Cr 3+ ]2 log 6 [ Fe 2+ ]3 Ans. Anode reaction : Zn(s) Zn2+(aq) + 2e Cathode reaction : Pb2+(aq) + 2e Pb(s) Cell representation : Zn(s)/Zn2+(aq)||Pb2+(aq)/Pb(s) According to Nernst equation : Zn 2+ 0.0591 log 2+ Pb 2 0.0591 0.1 log Ecell = [ 0.13 ( 0.76)] 2 0.02 = 0.63 0.02955 log 5 = 0.63 0.02955 0.6990 = 0.63 0.0206 = 0.6094 V Ecell = E cell Commonly Made Errors The cell representation is given incorrectly by many candidates. The calculation of emf of the cell by using Nernst equation is incorrect, in some cases. Answering Tip Do more practice of cell representation and numerical based on Nernst equation. 42 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII 0 0 [ E Ni2+ / Ni = 0.25 V, E Ag+ / Ag = 0.80 V] log 10 1 = 1 A [CBSE Foreign 2012] Ans. (i) Ni(s) Ni2+(aq) + 2e 2Ag+(aq) + 2e 2 Ag(s) E cell = E anode 1 = 1.05 0.0591 0.1 log 2 2 (1) 0.0591 (log 10 1) 2 = 1.05 0.295 ( 1) = 1.05 + 0.0295 = 1.0795 V Commonly Made Errors Answering Tip Do more practice of cell reaction and numerical based on Nernst equation. Long Answer Type Questions-II ci al (5 marks each) Given : E (A2+/A) = 2.37V : E (B2+/B) = 0.14V A OR (i) The conductivity of 0.001 mol L 1 solution of CH3COOH is 3.905 10 5 S cm 1. Calculate its molar conductivity and degree of dissociation (a). Given L (H+) = 349.6 S cm2 mol 1 and L (CH3COO ) = 40.9 S cm2 mol 1 (ii) Define electrochemical cell. What happens if external potential applied becomes greater than E cell of electrochemical cell ? A + R [CBSE OD 2016] An i ke tS ha _o ffi Q. 1. (i) What is limiting molar conductivity? Why there is steep rise in the molar conductivity of weak electrolyte on dilution? (ii) Calculate the emf of the following cell at 298 K: Mg(s)|Mg2+ (0.1 M)||Cu2+ (1.0 10-3 M)|Cu(s) [Given = E Cell = 2.71 V] R + A [CBSE OD 2016] Ans. (i) When concentration approaches zero, the molar conductivity is known as limiting molar conductivity. 1 The change in Lm with dilution is due to the increase in the degree of dissociation and consequently the number of ions in the total volume of the solution that contains 1 mol of electrolyte, hence Lm increases steeply. 1 0 (ii) ECell = ECell 2+ 0.059 [ Mg ] log n [Cu 2 + ] = 2.71V 1 0.059 0.1 log 2 0.001 0.059 = 2.71V log 10 2 2 = 2.651 V 1 1 Commonly Made Error Students often write incorrect formula or concentrations in the working formula. Practice numericals with different cell reactions. Q. 2. (i) Calculate E cell for the following reaction at 298K : 4Al(s) + 3Cu2+ (0.01M) 2Al3+ (0.01M) + 3Cu(s) Given : Ecell = 1.98 V (ii) Using the E values of A and B, predict which is better for coating the surface of iron [E (Fe2+/Fe) = 0.44V] to prevent corrosion and why ? 1 The cell reaction is given incorrectly by many students. In the calculation of emf of the cell by using Nernst equation, students ignore the power of concentration terms. = 0.80 ( 0.25) = 1.05 V = E cell gr E cathode te le Ni(s) + 2 Ag+(aq) Ni2+(aq) + 2Ag(s) Ni 2+ Ecell = E cell 0.0591 log 2 + 2 Ag (ii) am Q. 7. A strip of nickel metal is placed in a 1 molar solution of Ni(NO3)2 and a strip of silver metal is placed in a one molar solution of AgNO3. An electrochemical cell is created when the two solution are connected by a salt bridge and the two strips are connected by wires to a voltmeter. (i) Write the balanced equation for the overall reaction occurring in the cell and calculate the cell potential. (ii) Calculate the cell potential, E at 25 C for the cell, if the initial concentration of Ni(NO3)2 is 0.100 molar and the initial concentration of AgNO3 is 1.00 molar. Ans. (i) Al(s) | Al3+(aq) (0.01M) || Cu2+(aq) (0.01M)|Cu(s) LHE [Al(s) Al3+(aq) + 3e )] 2 (Oxidation at anode) RHE [Cu2+(aq) + 2e Cu(s)] 3 (Reduction at cathode) \ n = 6 0.0591 [ Al 3 + ]2 1 log n [Cu2 + ]3 Ecell = E cell E cell = Ecell + = 1.98 + 0.0591 ( 0.01)2 log 1 6 ( 0.01)3 = 1.98 + 0.0591 log 102 6 = 1.98 0.0591 2 log 10 6 + 0.0591 [ Al 3 + ]2 log n [Cu2 + ]3 [ 43 Electrochemistry 0.0591 2 [Q log 10 = 1] 6 = 1.98 + 0.0197 = 1.9997 V 1 (ii) A is better than B because its E value is more negative. 1+1 Commonly Made Error Students often make error in identifying oxidation reaction and reduction reaction from a cell representation. OR (i) C = 0.001 mol L 1, k = 3.905 10 5 S cm 1 Molar conductivity k 1000 lm = C n = 2 cell = 0.46 V Cu 2 + 0.059 1 log 2 n + Ag 0.1 = 0.46 0.059 log ( ) 1 2 2 (0.001) 0.059 = 0.46 V V log 10 5 2 = 0.46 V 0.059 V 5 2 = 0.46 V 0.1475 V = 0.3125 V 1 am Commonly Made Error 1 Students often make mistake in writing half-cell reactions. Also, in some cases calculation errors are seen while solving Nernst equation as the students miss out the power of concentration terms. Q. 4. Consider the given below figure (inside) and answer the following questions : (a) Cell A has Ecell=2V and Cell B has Ecell=1.1V which of the two cells A or B will act as electrolytic cell. Which electrode reactions will occur in this cell? ci 1 tS ha _o ffi 39.05 = 390.5 Eo Ecell = Eocell 3.95 10 5 S cm 1 1000 cm 3 L 1 = 0.001 mol L 1 = 39.05 S cm2 mol 1 or 1 cm3 mol 1 l (H+) = 349.6 S cm2 mol 1, l (CH3COO ) = 40.9 S cm3 mol 1 CH3COOH CH3COO + H+ L (CH3COOH) = L CH3COO + L H+ = (40.9 + 349.6) S cm2 mol 1 = 390.5 S cm2 mol 1 m a = o m \ gr = 1.98 + te le al An i ke = 0.1 1 (ii) Electrochemical cell is a device used for the production of electricity from energy released during spontaneous chemical reaction and use electrical energy to bring about the chemical change. 1 If the external potential applied becomes greater than E cell of electrochemical cell, the reaction gets reversed. It starts acting as an electrolytic cell and vice-versa. 1 [CBSE Marking Scheme 2016] Q. 3. A voltaic cell is set up at 25 C with the half cells Ag+ (0.001 M) Ag and Cu2+ (0.10 M) Cu. What should be its cell potential ? [E = 0.46 V, log 105 = 5] A [CBSE Foreign 2012] 1 Ans. Cu + 2Ag+ Cu2+ + 2Ag Half cell reactions : Cathode (reduction) : 2Ag+ (0.001 M) + 2e 2Ag(s) Anode (oxidation) : Cu(s) Cu2+ (0.10 M) + 2e 1 (b) If cell A has Ecell = 0.5 V and Cell B has Ecell = 1.1 V then what will be the reactions at anode and cathode? Salt bridge Copper plate 2+ Zinc plate Zn (aq) 2+ Cu (aq) Cell A (A) Electrode (B) Electrode Cell B [NCERT Exemp. Q. 66, Page 43] A Ans. (a) Cell B will act as electrolytic cell as it has lower EMF Therefore, the electrode reactions will be : Zn2+ + 2e Zn at cathode Cu Cu2+ + 2e at anode. 2 (b) Now cell B acts as galvanic cell as it has higher EMF and will push electrons into cell A . The electrode reaction will be : At anode : Zn Zn2+ + 2e At cathode : Cu2+ + 2e Cu 2 44 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII Topic-3 Electrolysis, Laws of Electrolysis, Batteries, Fuel Cells and Corrosion Revision Notes tS ha _o ffi ci al te le gr am Electrolysis is the process of decomposition of an electrolyte when electric current is passed through either its aqueous solution or molten (fused) state. This process takes place in electrolytic cell. Faraday s first law of electrolysis : The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte. m = Z I t, where Z = Electrochemical equivalent Faraday s second law of electrolysis : Amount of various substances liberated by the same quantity of electricity passed through the electrolytic solution is proportional to their chemical equivalent weights. w w = E1 E2 Products of electrolysis depend on (i) Physical state of material. (ii) Types of electrode being used. Battery is a combination of galvanic cells in series and used as a source of electrical energy. Types of batteries : (i) Primary battery are non-chargeable batteries such as Lechlanche cell and Dry cell. (ii) Secondary battery are chargeable cells involving reversible reaction. Example, Lead storage battery and Nickel-cadmium cells. Dry cell (Leclanche cell) : The anode consists of a zinc container and the cathode is a graphite electrode surrounded by powdered MnO2 and C. The space is filled with paste of NH4Cl and ZnCl2. ke + (Cathode) Pitch seal Graphite (carbon) Cathode (with metal cap) An i MnO2 + C Paste of NH4Cl + ZnCl2 Zinc Anode (Anode) Fig. 1 : A dry cell At anode : Zn(s) Zn2+(aq) + 2e . At cathode : MnO2(s) + NH+ 4(aq)+ 2e MnO(OH) + NH3 + The net reaction : Zn +NH 4(aq)+ MnO2 Zn2+ + MnO(OH) +NH3 Lead storage battery : Anode Spongy lead Cathode Lead packed with lead dioxide Electrolyte Aqueous solution of H2SO4 Anode Lead H2SO4 Lead Dioxide (PbO2) Fig. 2 : Storage battery Cathode Lead (Pb) Dil. H2SO4 [ 45 Electrochemistry Discharge reaction of cell : At anode : Following reaction takes place at anode : Pb(s) +SO42 (aq) PbSO4(s) +2e Reaction at cathode : PbO2 filled in lead grid gets reduced to Pb2+ ions which combines with SO42 ions to form PbSO4(s). Complete cathode reaction is as follows : PbO2(s) + 4H+(aq) + SO42 (aq) + 2e PbSO4(s) + 2H2O(l) Complete cell reaction : Pb(s) + PbO2(s) + 2H2SO4(aq) 2PbSO4(s) + 2H2O(l) Recharge reaction of cell : It changes the direction of electrode reaction PbSO4 accumulated at cathode gets reduced to Pb. At cathode, PbSO4 + 2e Pb(s) + SO42 (aq) At anode, PbSO4 gets oxidised to PbO2 PbSO4(s) + 2H2O PbO2(s) + 4H+(aq) + SO42 (aq) + 2e Complete cell reaction would be as follows : charge te le gr am 2PbSO4(s) + 2H2O(l) Pb(s) + PbO2(s) + 2H2SO4(aq) Conventions for representing the galvanic cell : (i) Double vertical line is used for salt bridge. Left hand side of the double line is anode and the cathode is on the right hand side. (ii) A single vertical line is used to separate metal and the electrolytic solution. (iii) If there is no metallic surface involved, we write Pt. ci al Example : Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) tS ha _o ffi Fuel cells : Electrical cells that are designated to convert the energy from the combustion of fuels such as hydrogen, carbon monoxide or methane directly into electrical energy are called fuel cells. In the cell : Anode : [H2(g) + 2OH (aq) 2H2O (l) + 2e ] 2 Cathode : O2(g) + 2H2O(l) + 4e 4OH (aq) An i ke Net reaction : 2H2(g) + O2(g) 2H2O(l). Corrosion : The process of slow conversion of metals into their undesirable compounds (usually oxide) by reaction with moisture and other gases present in the atmosphere. Rusting of iron : 1 O2(g) Fe2+(aq) + H2O(l) Fe(s) + 2H+(aq) + 2 2Fe2+(s) + 1 O2(g) + 2H2O(l) Fe2O3(s) + 4H+ 2 Fe2O3 + xH2O Fe2O3.xH2O Rust Prevention of corrosion : (i) Barrier protection : By covering the surface with paint or a thin film of grease or by electroplating. (ii) Sacrificial protection : By galvanization. (iii) Alloying. Know the Formulae Q = It m = ZIt Know the Terms Electrolytic cell : A cell in which the electrical energy is used to carry out a non-spontaneous reaction. 46 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII Faraday constant : It is the quantity of electricity. 1 F = 96500 C Electrosynthesis : It is a method of producing substance through non-spontaneous reaction carried by electrolysis. Tinning : Coating of iron with tin. Very Short Answer-Objective Type Questions (c) PbSO4 cathode is oxidised to Pb. (d) PbSO4 anode is oxidised to PbO2 U [NCERT Exemp. Q. 15, Page 35] Ans. Correct option : (a) Explanation : While charging the lead storage battery the reaction occurring on cell is reversed and lead sulphate (PbSO4) on anode and cathode is converted into (lead) Pb and lead dioxide (PbO2), respectively. The electrode reactions are as follows : At anode : PbSO4 (s)+2H 2O PbO2 (s)+SO42 Overall reaaction : 2 PbSO4 (s)+2H 2O Pb(s) +PbO2 (s)+ 4H+ (aq.)+2SO24 (aq.) B. Match of the following : Q. 1. Match the items of Column I and Column II. Column II (i) Lead storage battery ke Column I (ii) Mercury cell (b) An i (a) maximum efficiency prevented by galvanisation (iii) Fuel cell (c) gives steady potential (iv) (d) Pb is anode, PbO2 is cathode Rusting (ii) Mercury cell : The cell potential is approximately 1.35 V and remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life time. (iii) Fuel cell runs continuously as long as the reactants are supplied. It produces electricity with an efficiency of about 70% compared to thermal plants whose efficiency is about 40%. (iv) Rusting of iron is an example of corrosion. Corrosion results in huge material loss resulting in damage to buildings, ships, bridges, machinery, etc. Galvanization is a process of coating of zinc over iron, so as to protect it from rusting. tS ha _o ffi +4H++2e (Oxidation) Cathode : PbO2 (s) + S O24 (aq q) + 4 H+ (aq)+2e PbSO4 (s)+2H 2O(l) Overall cell reaction : Pb(ss)+PbO 2 (s)+2H 2SO4 (aq) 2PbSO4 (s)+2H 2O(l ) ci At cathode : PbSO4 (s)+2e Pb(s)+SO42 (aq) (Reduction) am (b) PbSO4 cathode is reduced to Pb. gr (a) PbSO4 anode is reduced to Pb. te le Ans. (i) (d) (ii) (c) (iii) (a) (iv) (b) Explanation : (i) Lead storage battery : The cell reactions when the battery is in use are given below : Anode : Pb(s)+SO24 (aq) PbSO4 (s) + 2 e al A. Multiple choice Questions: Q. 1. While charging the lead storage battery________. (1 mark each) [NCERT Exemp. Q. 52, Page 40] C. Answer the following: Q. 1. What are secondary cell? R [NCERT Exemplar] Ans. Those cell which are rechargeable, i.e., in which 1 products can convert back into reactants. Q. 2. Suggest two materials other than hydrogen that can be used as fuels in fuel cells. U [NCERT Exemplar] Ans. Methane and methanol can be used as fuels in fuel cells. 1 Short Answer Type Questions Q. 1. Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell. U [CBSE OD Set-1 2017] Ans. Mercury cell. 1 Anode: Zn(Hg) + 2OH- ZnO(s) + H2O + 2e Cathode: HgO + H2O + 2e- Hg(l) + 2OH- [CBSE Marking Scheme 2017] (2 marks each) Commonly Made Error The cell reaction is given incorrectly by many students. They often fail to balance the reaction. Answering Tip Do more practice of cell reaction. [ 47 Electrochemistry Q. 2. Write the name of the cell which is generally used in inverters. Write the reactions taking place at the anode and the cathode of this cell. U [CBSE OD Set-2 2017] Commonly Made Error Students generally mix up the reactions at cathode and anode. Ans. Lead storage battery 1 Anode: Pb(s) + SO42-(aq) PbSO4(s) + 2e- Cathode: PbO2 + SO42-(aq) + 4H++ 2e- PbSO4(s) + 2H2O(l) [CBSE Marking Scheme 2017] Answering Tip Learn the reactions taking place at cathode and anode carefully. 2 [Topper's Answer 2017] al Q. 5. A current of 1.50 A was passed through an electrolytic cell containing AgNO3 solution with inert electrodes. The weight of silver deposited was 1.50 g. How long did the current flow ? (Molar mass of Ag = 108 g mol 1, 1F = 96500 C mol 1). A [CBSE Comptt. D/OD 2018] Commonly Made Error tS ha _o ffi ci Q. 3. Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell. U [CBSE OD Set-3 2017] Ans. Dry cell/Leclanche cell 1 Anode: Zn(s) Zn2+ + 2e- Cathode: MnO2 + NH4+ + e- MnO(OH) + NH3 [CBSE Marking Scheme 2017] te le gr am OR Answering Tip ke The cell reaction is given incorrectly by many students. They often fail to balance the reaction. An i Do more practice of cell reaction. Q. 4. From the given cells : L ead storage cell, Mercury cell, Fuel cell and Dry cell. Answer the following : (i) Which cell is used in hearing aids ? (ii) Which cell was used in Apollo Space Programme ? (iii) Which cell is used in automobiles and inverters ? (iv) Which cell does not have long life ? R [CBSE Delhi 2016] Ans. (i) Mercury cell, (ii) Fuel cell, (iii) Lead storage cell, (iv) Dry cell. 4 [CBSE Marking Scheme 2016] Ans. Quantity of charge required to deposit 108 g of silver = 96500 C Quantity of charge required to deposit 1.50 g of 96500 silver = 1.50 = 1340.28 C 108 Time taken = 1 (or by any other suitable method) [CBSE Marking Scheme 2018] Q. 6. (i) Calculate D G for the reaction Mg(s) + Cu2+(aq) r Mg2+(aq) + Cu(s) Given : E0cell = + 2.7 V, 1 F = 96500 C mol 1 (ii) Name the type of cell which was used in Apollo space programme for providing electrical power. A + R [CBSE OD 2014] Ans. (i) Given, E Cell = +2.71V and F = 96500 C mol 1, n = 2 (from the given reaction) DrG = n F E0cell DrG = 2 96500C mol 1 2.71V = 523030 J/mol or 523.03 KJ / mol 1 (ii) Hydrogen oxygen fuel Cell / fuel cell. [CBSE Marking Scheme 2014] 1 Long Answer Type Questions-I Q. 1. (i) What are fuel cells ? Explain the electrode reactions involved in the working of H2 O2 fuel cell. Q 1340.28 = = 893.5 s I 1.50 (3 marks each) (ii) Represent the galvanic cell in which the reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) takes place. R + U [CBSE Comptt. Delhi/OD 2013] 48 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII Ans. (i) The cells which convert chemical energy of a fuel directly into electrical energy is known as fuel cells. 1 The electrode reactions are : Anode : [H2(g) + 2OH (aq) 2H2O(l) + 2e ] 2 Cathode : O2(g) + 2H2O(l) + 4e 4OH (aq) Net reaction : 2H2(g) + O2(g) 2H2O(l) 1 (ii) Cu ZnSO4 Solution am Zn + CuSO4 Solution 1 Galvanic Cell 2+ 2+ May be represented as Zn(s)|Zn (aq)||Cu (aq)|Cu(s) Commonly Made Error = 4.03 g 1 Calculation of thickness: Let the thickness of silver deposited be x cm. Mass = Volume Density = Area Thickness Density (Volume = Area thickness) 4.03 g = 900 (cm2) x (cm) 10.5 (g cm 3) 4.03 cm = 4.26 10 4 cm. 1 x= 900 10.5 ci Students often make errors in writing the half-cell reactions and cell representations. gr e Salt bridge te le V e Q. 4. Silver is uniformly electro-deposited on a metallic vessel of surface area of 900 cm2 by passing a current of 0.5 ampere for 2 hours. Calculate the thickness of silver deposited. [Given : the density of silver is 10.5 g cm 3 and atomic mass of Ag = 108 amu.] A [CBSE Comptt. OD 2013] Ans. Calculation of mass of Ag deposited : The electrode reaction is Ag+ + e Ag The quantity of electricity passed = Current Time = 0.5 (amp.) 2 60 60 (sec) = 3600 C. 1 From the electrode reaction, it is clear that 96500 C of electricity deposit Ag = 108 g 108 3600 3600 C of electricity will deposit Ag = 96500 al Ans. (i) m = Zit 108 2 15 60 1 = 1 96500 = 2.01 g (or any other correct method) (ii) Cells that converts the energy of combustion of fuels directly into electrical energy. 1 [CBSE Marking Scheme 2017] tS ha _o ffi Answering Tip Understand the concept clearly and practice to write cell reactions and representations. An i ke Q. 2. Write the reactions taking place at cathode and anode in lead storage battery when the battery is in use. What happens on charging the battery ? U [CBSE Comptt. OD 2012] Ans. The cell reactions when the battery is in use are given below : Anode : Pb(s) + SO42 (aq) PbSO4(s) + 2e Cathode : PbO2(s) + SO42 (aq) + 4H+(aq) + 2e PbSO4(s) + 2H2O(l) i.e., overall cell reaction consisting of cathode and anode reactions is : Pb(s) + PbO2(s) + 2H2SO4(aq) 2PbSO4(s) + 2H2O(l) 1 On charging the battery, the electrode reactions are reverse of those that occur during discharge. 1 Commonly Made Error Answering Tip Always write working formula followed by the step substituting the value of each entity. Do not forget to mention the units. Q. 5. (i) The cell in which the following reaction occurs : 2Fe3+(aq) + 2I-(aq) 2Fe2+(aq) + I2(s) has Eocell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given: 1F = 96,500 C mol-1) A [CBSE Comptt. OD 2012; NCERT] (ii) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F = 96,500 C mol-1) A [CBSE OD Set-1, 2, 3 2017] Students often get confused between reaction taking place on cathode and anode. Ans. (i) DG = -nFE cell Q. 3. (i) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes. (Given: Molar mass of Ag = 108 g mol 1 1F = 96500 C mol 1) (ii) Define fuel cell. A + R [CBSE Delhi Set-1, 2 2017] n = 2 DG = -2 96500 C/mol 0.236 V =-45548 J/mol =-45.548 kJ/mol (ii) Q = lt = 0.5 2 60 60 = 3600 C [ 49 Electrochemistry Answering Tip 96500 C = 6.023 1023 electrons 3600 C = 2.25 1022 electrons 1 [CBSE Marking Scheme 2017] Always write working formula followed by the step substituting the value of each entity. Do not forget to mention the units. tS ha _o ffi ci al te le gr am OR ke An i Q. 6. (i) How many coulombs are required to reduce 1 mole Cr2O72 to Cr3+ ? (ii) The conductivity of 0.001 M acetic acid is 4 10 5 S/cm. Calculate the dissociation constant of acetic acid if L m for acetic acid is 390 S cm2 mol 1. A [CBSE Comptt. OD 2012] + 3+ + 7H O Ans. (i) Cr2O2 7 + 14 H + 6e 2Cr 2 2 One mole Cr2O7 requires 6 moles of electrons for reduction. Thus, quantity of electricity required = 6 96,500 = 579000 Coulomb 1 = 5.79 105 Coulomb (ii) C = 0.001 M k = 4 10 5 S cm 1 Lm = Lm = 1000 k C m = 40 = 0.1025 390 a = a = 0.1025 0.103 CH COO H+ 3 Ka = [CH3COOH ] C H3COOH CH3COO + H+ C (1 a) Ca Ca C 2 C C = (1 ) C (1 ) Ka = = = 1.18 10 5 0.001 ( 0.103)2 1 0.103 Answering Tips 1000 cm 3 L 1 4 10 cm 5S cm L 1 0.001 moll [Topper's Answer 2017] 3 1 Lm = 40 S cm2 mol 1 L = 390 S cm2 mol 1 a = degree of dissociation of CH3COOH Always write working formula followed by the step substituting the value of each entity. Do not forget to mention the units. Remember to put the power of concentration terms as mentioned in stoichiometry of the balance chemical equation 1 50 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII Q. 7. What is corrosion ? Explain the electro-chemical theory of rusting of iron and write the reactions involved in the rusting of iron. R + U [CBSE Comptt. Delhi 2012] Ans. The process of slowly eating away of the metal due to attack of the moisture and atmospheric gases on the surface of the metal resulting into the formation of compound such as oxides, sulphides, carbonates, sulphates etc., is called corrosion. 1 The electrochemical phenomenon of rusting of iron can be described as : At anode : Fe(s) undergoes oxidation to releases electrons. Fe(s) Fe2+(aq) + 2e At cathode : O2(g) + 4H+ + 4e 2H2O(l) Electrons released at anode move to another metal and reduce oxygen in presence of H+. It is available from H2CO3 formed from the dissolution of CO2 from air into water. H+ in water may be available also through dissolution of other acidic oxides from the atmosphere. 1 This site behaves as cathode 1 Net reaction : Fe(s) + 2 H+(aq) + O2(g) Fe2+(aq) 2 + H2O(l) Fe2+ again gets oxidised to form rust. 1 O2(g) + 2H2O(l) Fe2O3(s) + 4H+ 2Fe2+(s) + 2 Rust 1 ci log Kc = 15.59 1 DrG = n FE cell 1 = 2 96500 C mol 1 0.46V = 88,780 J mol 1 = 88.78 kJ mol 1 1 [CBSE Marking Scheme 2016] te le (ii) The resistance of 0.01 M NaCl solution at 25 C is 200 W. The cell constant of the conductivity cell is unity. Calculate the molar conductivity of the solution. OR (i) What are fuel cells ? Give an example of a fuel cell. (ii) Calculate the equilibrium constant (log Kc) and DrG for the following reaction at 298 K. Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) Given E cell = 0.46 V and IF = 96500 C mol 1 R + U + A [CBSE Comptt. OD 2016] al (5 marks each) gr Q. 1. (i) What are the two classifications of batteries ? What is the difference between them ? am Long Answer Type Questions-II tS ha _o ffi Answering Tip An i ke Ans. (i) Classification of batteries : (a) Primary batteries (b) Secondary batteries 1 Primary batteries are non-chargeable batteries whereas secondary batteries are rechargeable. 1 k 1000 S cm2 mol 1 1 Lm = C 1 l R A k = = 1 1 S cm 1 200 = 1 S cm 1 200 Lm = 1 1000 S cm2 mol 1 200 0.01 = 500 S cm2 mol 1 1 OR (i) Fuel cells are the cells which converts energy of combustion of fuel directly into electricity. Example H2 O2 fuel cell. 1 nE cell (ii) log Kc = 0.059 1 = 2 0.46 V 0.92 = 0.059 0.059 While solving the numerical, write the working formula followed by the data in it. Write proper unit. Q. 2. (i) Define the following terms : (a) Molar conductivity (Lm) (b) Secondary batteries (c) Fuel cell (ii) State the following laws : (a) Faraday first law of electrolysis (b) Kohlrausch s law of independent migration of ions R OR (i) Define the term degree of dissociation. Write an expression that relates the molar conductivity of a weak electrolyte to its degree of dissociation. (ii) For the cell reaction Ni(s) | Ni2+(aq)|| Ag+(aq) | Ag(s) Calculate the equilibrium constant at 25 C. How much maximum work would be obtained by operation of this cell ? E Ni2+/Ni= 0.25 V and E Ag+/Ag= 0.80 V. R + U + A [CBSE Comptt. Delhi 2015] Ans. (i) (a) Molar conductivity of a solution at a given concentration is the conducting power of all the ions produced by 1 mol of an electrolyte. 1 (b) Secondary battery can be recharged by passing current through it in opposite direction so that 1 it can be used again. (c) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells. 1 [ 51 Electrochemistry (ii) (a) Faraday s first law of electrolysis states that the amount of chemical reaction which occurs at any electrode during electrolysis by current is proportional to the quantity of electricity passed through the electrolyte (solution or melt). 1 (b) According to Kohlrausch law of independent migration of ions limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. 1 [CBSE Marking Scheme 2015] Students re-frame the statement or only give the statement. am 0m An i ke gr tS ha _o ffi E cell = E Ag+ / Ag E Ni 2+ / Ni = 0.80V 0.25V = 0.55V 1 ci c Degree of dissociation (a) = m Q. 3. (i) Calculate E cell for the following reaction at 298 K: 2Cr(s) + 3Fe2+ (0.01M) 2Cr3+(0.01M) + 3Fe(s) Given : Ecell = 0.261 V (ii) Using the E values of A and B, predict which one is better for coating the surface of iron [E (Fe2+/Fe) = -0.44 V] to prevent corrosion and why? Given: E (A2+/A) = -2.37 V: E (B2+/B) = -0.14 V te le (i) Degree of dissociation is the ratio of molar conductivity at a specific concentration to the molar conductivity at infinite solution. 1 al OR Ans. Answering Tip While solving the numerical, write the working formula followed by the data in it. Write proper unit. Write the laws as stated. (ii) 2 0.55 0.059 log Kc = 18.644 K c = Antilog 18.644 K c = 4 1018 G = nFE cell = 2 96500 C mol 1 0.55V = 106,150 J mol 1 Max. work = + 106150 J mol 1 or 106.150 kJ mol 1 1 [CBSE Marking Scheme 2015] = Commonly Made Error Answering Tip nEocell log Kc = 0.059 A [CBSE OD Set-2 2016] 52 ] al te le gr am Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII Detailed Answer : (i) Nernst Equation : 0.059 [Prod.] o Ecell Ecell log n [React.] 0.059 [Prod.] o Ecell Ecell log n [React.] 0.059 [Cr 3+ ]2 log 6 [Fe2+ ]3 0.261 0.059 (10 -2 )2 log 6 (10 -2 )3 An i 0.261 0.059 log 10 2 6 0.059 2 0.0261 6 0.0261 + 0.01966 0.28068 V 0.281 V 0.0261 3 (ii) As corrosion is a phenomenon of oxidation of iron considering the oxidation potential of all the elements is essential. Element with higher oxidation potential than Fe will oxidise faster than iron preventing corrosion in iron. Oxidation potential of Fe = 0.44 V Oxidation potential of A = 2.37 V Oxidation potential of B = 0.14 V As A has higher oxidation potential than iron, it can be used for coating the surface of iron. 2 3+2 [Topper s Answer 2016] Commonly Made Error Students often write incorrect formula of concentrations in the working formula. Practice numericals with different cell reactions. ke tS ha _o ffi ci Q. 4. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K : Sn (s) | Sn2+ (0.004 M) || H+ (0.020 M) | H2 (g) (1 bar) | Pt (s) (Given : E Sn2+/ Sn = - 0.14V) (b) Give reasons : (i) On the basis of E values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl. (ii) Conductivity of CH3COOH decreases on dilution. OR (a) For the reaction 2AgCl (s) + H2 (g) (1 atm) 2Ag(s)+2H+ (0.1 M)+2Cl (0.1 M), G = 43600 J at 25 C. Calculate the e.m.f. of the cell. [log 10 n = n] (b) Define fuel cell and write its two advantages. R + U + A [CBSE Delhi/OD 2018] [ 53 Electrochemistry (ii) (a) Electrons flow from Zn to Ag plate. (b) Zn as anode and Ag acts as cathode (c) Cell will stop functioning (d) Concentration of Zn2+ ions will increase and that of Ag+ ions will decrease. , (e) No change [CBSE Marking Scheme 2017] Ans. (a) Sn + 2H+ Sn2+ + H2 (Equation must be balanced) 1 Sn 2 + 0.059 E = E log + 2 2 [H ] = 0.14 0.0295 log 10 = 0.11 V / 0.1105 V 1 (b) (i) Due to overpotential/ Overvoltage of O2 1 (ii) The number of ions per unit volume decreases. 1 OR (a) G = nFE 43600 = 2 96500 E E = 0.226 V E = E 0.059/2 log ( [H+]2[Cl ]2/ [H2] ) = 0.226 0.059/2 log[ (0.1)2 (0.1)2] / 1 = 0.226 -0.059 /2 log 10 4 1 = 0.226 + 0.118 = 0.344 V (Deduct half mark if unit is wrong or not written) 1 (b) Cells that convert the energy of combustion of fuels (like hydrogen, methane, methanol, etc.) directly into electrical energy are called fuel cells. Advantages : High efficiency, non polluting (or any other suitable advantage) + [CBSE Marking Scheme 2018] Q. 5. (i) When a bright silver object is placed in the solution of gold chloride, it acquires a golden tinge but nothing happens when it is placed in a solution of copper chloride. Explain the behaviour of silver. Zinc plate Zn2+(aq.) An i ke tS ha _o ffi ci Q. 6. (a) A cell is prepared by dipping a zinc rod in 1M zinc sulphate solution and a silver electrode in 1M silver nitrate solution. The standard electrode potential given: E Zn2+ / Zn = 0.76 V, E Ag+ / Ag = + 0.80 V What is the effect of increase in concentration of Zn2+ on the Ecell ? (b) Write the products of electrolysis of aqueous solution of NaCl with platinum electrodes. (c) Calculate e.m.f. of the following cell at 298 K: Ni(s) / Ni2+ (0.01 M) // Cu2+ (0.1M) / Cu (s) [Given E Ni2+/ Ni = 0.25 V , E Cu2+/Cu = + 0.34 V ) A+U Write the overall cell reaction. OR (a) Apply Kohlrausch law of independent migration of ions, write the expression to determine the limiting molar conductivity of calcium chloride. (b) Given are the conductivity and molar conductivity of NaCl solutions at 298K at different concentrations: am ( 0.02 )2 gr ( 0.004 ) te le = [0 ( 0.14 )] 0.295 log al Salt bridge Silver plate Ag2+(aq.) (ii) Consider the figure given above and answer the following questions : (a) What is the direction of flow of electrons? (b) Which is anode and which is cathode? (c) What will happen if the salt bridge is removed? (d) How will concentration of Zn2+ and Ag+ ions be affected when the cell functions? (e) How will concentration of these ions be affected when the cell becomes dead? A&E [CBSE Comptt. Delhi Set-1, 2, 3 2017] Ans. (i) E value of silver is lower than that of gold, hence silver displaces gold which gets deposited on the silver object. 1 E value of copper is lower than that of silver, hence silver cannot displace copper from its solution. 1 Concentration Conductivity M Scm 1 0.100 106.74 10 4 0.05 55.53 Molar conductivity S cm2 mol 1 106.7 10 4 111.1 115.8 0.02 23.15 10 4 Compare the variation of conductivity and molar conductivity of NaCl solutions on dilution. Give reason. (c) 0.1 M KCl solution offered a resistance of 100 ohms in a conductivity cell at 298 K. If the cell constant of the cell is 1.29 cm 1, calculate the molar conductivity of KCl solution. C + A & E + A [CBSE SQP 2018-2019] Ans. (a) Ecell decreases. 1 (b) Anode: Cl2 Cathode : H2 (c) Cu2+(aq.) + Ni(s) Ni2+(aq.) + Cu(s) E cell = E cathode E anode E cell = 0.34 ( 0.25) E cell = 0.59 V Ecell = E cell Ecell = 0.59 2+ 2.303RT [ Ni ] log nF [Cu 2 + ] 0.059 [0.01] log 2 [0.1] Ecell = 0.6195 V + 54 ] Oswaal CBSE Chapterwise & Topicwise Question Bank, Chemistry, Class XII OR (a) m(CaCl2) = l Ca2+ + 2l Cl 1 (b) Conductivity of NaCl decreases on dilution as the number of ions per unit volume decreases. 1 Whereas molar conductivity of NaCl increases on dilution as on dilution the interionic interactions are overcome and ions are free to move. 1 (c) G* = kR 1.29 = 0.0129 S cm 1 k = 100 1000 C m = m = m = 129 S cm2 mol 1 [CBSE Marking Scheme 2018] 1000 cm 3 L 1 0.0129 S m 1 0.1 mol L 1 oswaal learning tools For Suggested Online Videos Or Scan the Code te le gr Or Scan the Code Visit : https://qrgo.page.link/XXZbN Visit : https://qrgo.page.link/EoVdZ al Or Scan the Code Or Scan the Code tS ha _o ffi ci am Visit : https://qrgo.page.link/A1PiL Visit : https://qrgo.page.link/Qwums To learn from NCERT Prescribed Videos Visit : https://qrgo.page.link/3mFtJ An i ke Or Scan the Code Visit : https://qrgo.page.link/bpVk5 Or Scan the Code

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