xSKILLby Infirals
Register

Class 12 Chemistry Part 1

0 of 34 notes read 0%

Reading progress is saved to your INFIRALS account. Sign in to keep this note marked as read.

Previous year questions: Electrochemistry

Electrochemistry32 min readLesson 7 of 7

Q1 (CBSE Board 2026, 3 marks)

State Kohlrausch's law of independent migration of ions. With the help of a curve, explain why the limiting molar conductivity of a weak electrolyte cannot be found by extrapolating the concentration curve as it can for a strong electrolyte.

Answer: Kohlrausch's law states that at infinite dilution every ion migrates on its own, so the limiting molar conductivity of an electrolyte equals the sum of the limiting molar conductivities of its ions, each multiplied by the number of that ion in the formula. For a strong electrolyte the plot of molar conductivity against the square root of concentration is a straight line over the whole low concentration range, and the intercept at zero concentration gives the limiting value. For a weak electrolyte the curve bends sharply upward as concentration approaches zero and never becomes a usable straight line, so the intercept cannot be read. The limiting value is obtained instead by adding and subtracting the limiting values of strong electrolytes so that the wanted ions remain.

Q2 (CBSE Board 2026, 3 marks)

(a) Why is the equilibrium constant of a cell reaction related to E°cell and not to Ecell? (b) Two metals A and B have standard electrode potentials of -0.24 V and +0.80 V. Which one will liberate hydrogen gas from dilute sulphuric acid? (c) Write the cell reaction that occurs in a lead storage battery while it is being charged.

Answer: (a) The standard potential is defined with every species at unit activity, which is exactly the state to which the thermodynamic equilibrium constant refers. A real cell potential changes with the concentrations as the reaction proceeds, so it cannot fix the thermodynamic constant. (b) A metal liberates hydrogen only when its standard reduction potential is lower than that of the hydrogen electrode, which is zero. Metal A has -0.24 V, below the hydrogen couple, so its oxidation can reduce H+ and hydrogen is liberated. Metal B has +0.80 V and cannot reduce H+. (c) Charging reverses the discharge reactions: the lead sulphate at the anode is oxidised back to lead dioxide, and the lead sulphate at the cathode is reduced back to lead. The overall charging reaction is 2PbSO4 plus 2H2O giving Pb plus PbO2 plus 2H2SO4.

Q3 (CBSE Board 2025, 3 marks)

Explain why the electrolysis of an aqueous solution of sodium chloride gives hydrogen gas at the cathode and chlorine gas at the anode. Write the overall reaction. Given E° values: Na+/Na = -2.71 V, H2O/H2 = -0.83 V, Cl2/Cl- = +1.36 V, O2/H2O = +1.23 V.

Answer: At the cathode both Na+ and water can be reduced. The reduction potential of water, -0.83 V, is much higher than that of sodium, -2.71 V, so water is reduced more easily and hydrogen is released. At the anode both chloride and water can be oxidised. The chloride couple has the higher potential, 1.36 V, and the overpotential for oxygen evolution on the electrode raises the practical oxygen potential further, so chloride is oxidised and chlorine is released. The overall change is 2NaCl plus 2H2O giving 2NaOH plus H2 plus Cl2.

Q4 (CBSE Board 2025, 2 marks)

The conductivity of a 0.20 M solution of KCl is 2.48 times 10^-2 S cm^-1. Find its molar conductivity and its degree of dissociation. The limiting ionic conductivities of K+ and Cl- are 73.5 and 76.5 S cm^2 mol^-1. (This is part (b)(i) of Q33.)

Answer: Molar conductivity is 1000 times the conductivity divided by the molarity: 1000 times 2.48 times 10^-2 divided by 0.2 equals 124 S cm^2 mol^-1. The limiting molar conductivity is the sum of the two ionic values, 73.5 plus 76.5, which is 150 S cm^2 mol^-1. The degree of dissociation is the ratio of the two, 124 divided by 150, which is 0.827. The value near one confirms that KCl is a strong electrolyte; the small shortfall comes from interionic attractions at this finite concentration.

Q5 (CBSE Board 2024, 4 marks)

A case study explains that any cell used as a source of electricity is a galvanic cell, that batteries are primary or secondary, and that fuel cells convert the energy of combustion directly into electricity. (a) How do primary batteries differ from secondary batteries? (b) The mercury cell has a steady potential of 1.35 V during its life. Give the reason. (c) Write the electrode reactions and the overall reaction of the lead storage battery while it is being recharged.

Answer: (a) A primary battery runs its reaction only once, so after use it cannot be recharged and is discarded. A secondary battery is rechargeable because its discharge products can be converted back into the original reactants by passing current in the reverse direction. (b) The overall reaction of the mercury cell consumes no ions whose concentration changes and produces no species that changes the electrolyte composition, so the potential stays constant until the active material is used up. (c) On charging the positive plate is oxidised back to lead dioxide and the negative plate is reduced back to lead: the anode reaction is PbSO4 plus 2H2O giving PbO2 plus SO4(2-) plus 4H+ plus 2 electrons, and the cathode reaction is PbSO4 plus 2 electrons giving Pb plus SO4(2-). The overall reaction is 2PbSO4 plus 2H2O giving Pb plus PbO2 plus 2H2SO4.

Q6 (CBSE Board 2025, 5 marks)

(a) Calculate Ecell at 25 degrees Celsius for Zn(s) + Pb2+(0.02 M) giving Zn2+(0.1 M) + Pb(s). Given E° for Zn2+/Zn is -0.76 V and for Pb2+/Pb is -0.13 V. (b) State Faraday's first law of electrolysis and find the electricity in Faraday needed to reduce one mole of MnO4- to Mn2+.

Answer: (a) The standard cell potential is the cathode potential minus the anode potential, which is -0.13 minus (-0.76), equal to 0.63 V. Two electrons are transferred. The reaction quotient is the zinc ion concentration divided by the lead ion concentration, 0.1 divided by 0.02, which is 5. The Nernst correction is 0.0591 divided by 2 times log 5, which is 0.0207 V. Therefore Ecell is 0.63 minus 0.0207, or about 0.61 V. (b) Faraday's first law states that the mass of a substance liberated or deposited at an electrode is directly proportional to the quantity of charge passed through the electrolyte. The half reaction MnO4- plus 8H+ plus 5 electrons giving Mn2+ plus 4H2O needs five electrons per ion, so one mole of permanganate requires five Faraday.

Q7 (CBSE SQP 2021-22, 3 marks)

For the reaction 2Al(s) + 3Cd2+(0.1 M) giving 3Cd(s) + 2Al3+(0.01 M), the standard cell potential is 1.260 V. Represent the cell and calculate Ecell.

Answer: Aluminium is oxidised at the anode and cadmium ions are reduced at the cathode, so the cell is written Al | Al3+(0.01 M) || Cd2+(0.1 M) | Cd. Six electrons are transferred in the balanced reaction. The reaction quotient is the squared aluminium ion concentration divided by the cubed cadmium ion concentration, which is (0.01)^2 divided by (0.1)^3, equal to 0.1. With log Q equal to -1, the Nernst term is minus 0.0591 divided by 6 times (-1), which is +0.00985 V. Therefore Ecell is 1.260 plus 0.00985, or about 1.27 V, slightly above the standard value because the product ions are very dilute.

Q8 (CBSE SQP 2023-24, 3 marks)

(a) Can an electrochemical cell be constructed from two half cells both made of zinc in zinc sulphate solution? Explain. (b) Calculate the limiting molar conductivity of the chloride ion given that the limiting molar conductivity of MgCl2 is 258.6 S cm^2 mol^-1 and that of the magnesium ion is 106 S cm^2 mol^-1. (c) A conductivity cell has a cell constant of 0.146 cm^-1 and the resistance of a 0.01 M solution is 1000 ohm. Find the conductivity.

Answer: (a) No. A cell needs two electrodes with different potentials so that electrons can flow. Two zinc electrodes in the same zinc sulphate solution have identical potentials, the cell potential is zero, and no current flows. (b) Kohlrausch's law gives the limiting molar conductivity of MgCl2 as the sum of the magnesium value and twice the chloride value. Therefore twice the chloride value is 258.6 minus 106, which is 152.6, and the chloride value is 76.3 S cm^2 mol^-1. (c) Conductivity is the cell constant divided by the resistance, 0.146 divided by 1000, which is 1.46 times 10^-4 S cm^-1.

Q9 (CBSE SQP 2024-25, 3 marks)

The equilibrium constant of the reaction Sn4+(aq) + Al(s) giving Al3+ plus Sn2+(aq) is 4.617 times 10^184 at 25 degrees Celsius. (a) Calculate the standard emf of the cell. (b) Find the standard potential of the Al3+/Al half cell if the Sn4+/Sn2+ potential is 0.15 V. Given log(4.617 times 10^184) = 184.6644.

Answer: (a) The relation between the standard potential and the equilibrium constant at 298 K is E°cell = 0.0591 divided by n times log Kc. The balanced reaction transfers six electrons: two aluminium atoms lose three electrons each and three tin(IV) ions gain two electrons each. Therefore E°cell is 0.0591 divided by 6 times 184.6644, which is 1.82 V. (b) The tin couple is the cathode and aluminium is the anode, so E°cell equals E°(Sn4+/Sn2+) minus E°(Al3+/Al). Rearranging gives E°(Al3+/Al) = 0.15 minus 1.82, which is -1.67 V. The strongly negative value is consistent with aluminium being a strong reducing agent.

Q10 (CBSE SQP 2025-26, 3 marks)

Write the Nernst equation for (i) Ni(s) + Cu2+(aq) giving Ni2+(aq) + Cu(s), (ii) Al(s) + FeSO4(aq) giving Al2(SO4)3(aq) + Fe(s), and (iii) Mg(s) | Mg2+(aq) || Ag+(aq) | Ag(s).

Answer: (i) Two electrons are transferred, so E = E° minus 0.0591 divided by 2 times log of the nickel ion concentration divided by the copper ion concentration. (ii) The balanced reaction is 2Al plus 3Fe2+ giving 2Al3+ plus 3Fe, with six electrons transferred, so E = E° minus 0.0591 divided by 6 times log of the squared aluminium ion concentration divided by the cubed iron(II) concentration. (iii) The overall reaction is Mg plus 2Ag+ giving Mg2+ plus 2Ag, so E = E° minus 0.0591 divided by 2 times log of the magnesium ion concentration divided by the square of the silver ion concentration. In each case pure solids do not appear in the quotient.