Please refer to Electrochemistry Class 12 Chemistry Important Questions with solutions provided below. These questions and answers have been provided for Class 12 Chemistry based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 12 Chemistry for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 12.
Class 12 Chemistry Important Questions Electrochemistry
Very Short Answer Questions
Question. Represent the galvanic cell in which the reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) takes place.
Answer. Representation of the galvanic cell for the given reaction is :
Question. What is the necessity to use a salt bridge in a Galvanic cell?
Answer. The salt bridge allows the movement of ions from one solution to the other without mixing of the two solutions. Moreover, it helps to maintain the electrical neutrality of the solutions in the two half cells.
Question. Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration?
Answer. The limiting molar conductivity of an electrolyte is defined as its molar conductivity when the concentration of the electrolyte in the solution approaches zero. Conductivity of an electrolyte decreases with dilution because the number of current carrying particles i.e., ions present per cm3 of the solution becomes less and less on dilution.
Question. Define : Fuel cell
Answer. Those galvanic cells which give us direct electrical energy by the combustion of fuels like hydrogen, methane, methanol etc. are called fuel cells.
Question. Name the type of cell which was used in Apollo space programme for providing electrical power.
Answer. H2 → O2 fuel cell was used in Appollo space programme.
Question. Using the E° values of A and B predict which is better for coating the surface of iron
Answer. Metals of lower electrode potential value when connected with iron protect it from oxidation and prevent corrosion. Hence, coating of metal A having lower electrode potential will be better than B which has higher E°(B2+/B) = – 0.14 V.
Question. Define electrochemical cell. What happens if external potential applied becomes greater than E°cell of electrochemical cell?
Answer. the device which converts the chemical energy liberated during the chemical reaction to electrical energy is called electrochemical cell. If external potential applied becomes greater than E°cell of electrochemical cell then the cell behaves as an electrolytic cell and the direction of flow of current is reversed.
Question. Calculate ΔrG° for the reaction :
Mg(s) + Cu(2+aq) → Mg(2+aq ) + Cu(s)
Given E°cell = +2.71 V, 1 F = 96500 C mol–1
Answer. (a) Given : E°cell = 2.71 V
For the reaction,
Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
n = 2, ΔrG° = ?
Using formula, ΔrG° = – nFE°cell
ΔrG° = –2 × 96500 C mol–1 × 2.71 V
or ΔrG° = 523.03 kJ mol–1
Question. Equilibrium constant (Kc) for the given cell reaction is 10. Calculate E°cell.
Question. The standard electrode potential (E°) for Daniell cell is +1.1 V. Calculate the ΔG° for the reaction.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
(1 F = 96500 C mol–1)
Answer. Here n = 2, E°cell = 1.1 V
F = 96500 C mol–1
ΔrG° = –nFE°cell
ΔrG° = – 2 × 1.1 × 96500 = – 212300 J mol–1
= – 212.3 kJ mol–1
Question. A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?
Answer. Current (I) = 1.5 A
Time (t) = 10 min = 10 × 60 = 600 s
Quantity of electricity passed = I × t
= (1.5 A) × (600 s) = 900 C
Copper is deposited as :
Cu2+ + 2e– → Cu(s)
2 × 96500 C of current deposit copper = 63.56 g
∴ 900 C of current will deposit copper
Question. Explain why electrolysis of aqueous solution aof NaCl gives H2 at cathode and Cl2 at anode.
Write overall reaction.
Answer. Aqueous NaCl ionises as
NaCl → Na+ + Cl–
(i) Following reactions are possible at cathode,
2Na+(aq) + 2e– → 2Na(s); E° = – 2.71 V
2H2O(l) + 2e– → H2(g) + 2OH–(aq); E° = – 0.83 V
The reaction with higher E° value will take place, hence H2 is produced at cathode.
(ii) Following reactions may take place at anode
The reaction with lower E° value will take place, but due to over voltage. Cl2 is liberated at anode.
Question. How many coulombs are required to reduce 1 mole Cr2O72– to Cr3+?
Answer. The given reaction is
Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O
one mole Cr2O72– requires 6 mol of electrons for reduction. Hence, quantity of electricity required
= 6 mol × 96500 C mol–1= 5.79 × 105 coulomb
Question. How many moles of mercury will be produced by electrolysing 1.0 M Hg(NO3)2 solution with a current of 2.00 A for 3 hours?
Answer. Quantity of electricity passed
Q = I × t = 2.0 A × 3 × 60 × 60 s = 21600 C
Hg2+ + 2e– → Hg
2 F 1 mol
2 × 96500 C electricity produces 1 mole Hg
Question. What type of battery is mercury cell? Why is it more advantageous than dry cell?
Answer. Mercury cell is a primary battery. Hence, it can be used only once and cannot be recharged.
Advantage : The cell potential remains constant during its life time. Hence, it is useful for devices requiring constant current e.g., hearing aids and watches.
Question. What type of a battery is the lead storage battery?
Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it.
Answer. Lead storage battery is a secondary cell.
Cell reactions during operation
At anode : Pb(s) + SO2–4(aq) → PbSO4(s) + 2e–
At cathode :
PbO2(s) + SO2–4(aq) + 4H+(aq) + 2e– → PbSO4(s) + 2H2O(l)
Overall reaction :
Pb(s) + PbO2(s) + 2H2SO2–4(aq) → 2PbSO4(s) + 2H2O(l)
Question. 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?
Answer. The lead storage battery is most important secondary cell. The cell reactions when the battery is in use :
Pb(s) + SO2–4(aq) → PbSO4(s) + 2e–
At cathode :
PbO2(s) + SO2–4(aq) + 4H+(aq) + 2e– → PbSO4(s) + 2H2O(l)
The overall cell reaction is
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
Following reaction takes place during recharging
At cathode : PbSO4(s) + 2e– → Pb(s) + SO2–4(aq)
At anode : PbSO4(s)+ 2H2O(l) → PbO2(s) + SO2–4(aq)+ 4H+(aq) + 2e–
Net reaction : 2PbSO4(s) + 2H2O(l) → Pb(s) + PbO2(s) + 2H2SO4(aq)
Question. From the given cells :
Lead 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 invertors?
(iv) Which cell does not have long life?
Answer. (i) Mercury cell is used for low current devices like watches and hearing aids.
(ii) The hydrogen oxygen fuel cell was used in Apollo space programme.
(iii) Lead storage battery is used in automobiles and invertors.
(iv) Dry cell
Question. How much charge is required for the reduction of 1 mole of Cu2+ to Cu?
Answer. The electrode reaction is Cu2+ + 2e– → Cu
∴ Quantity of charge required for reduction of
1 mole of Cu2+ = 2F = 2 × 96500 = 193000 C
Question. Determine the value of equilibrium constant (Kc) and DG° for the following reactions :
Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s), E° = 1.05 V
(1 F = 96500 C mol–1)
Answer. Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s), E° = 1.05 V Here, n = 2
Question. Calculate the emf for the given cell at 25° C :
Cr|Cr3+ (0.1 M) || Fe2+ (0.01 M)| Fe
Answer. E°cell = E°right – E°left = – 0.44 – (– 0.74) = 0.30 V
Question. Define : Secondary batteries
Answer. Secondary batteries : The batteries which can be recharged again and again are called as secondary batteries. e.g., lead storage battery.
Question. On the basis of standard electrode potential values stated for acid solutions, predict whether Ti4+ species may be used to oxidise FeII to FeIII.
Ti4+ + e– → Ti3+ ; + 0.01
Fe3+ + e– → Fe2+ ; + 0.77
Answer. Because standard electrode potential of Ti4+/Ti3+ is less than that of Fe3+/Fe2+ so, it cannot
oxidise FeII to FeIII.
Question. Define the term degree of dissociation. Write an expression that relates the molar conductivity of a weak electrolyte to its degree of dissociation.
Answer. The fraction of the total number of molecules present in solution as ions is known as degree of dissociation.
Molar conductivity (λm) = aλ°m where λ°m is the molar conductivity at infinite dilution.
Question. The resistance of 0.01 M NaCl solution at 25°C is 200 W. The cell constant of the conductivity cell used is unity. Calculate the molar conductivity of the solution.
Question. The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm–1. Calculate its molar conductivity.
Answer. Here, k = 0.025 S cm–1, Molarity = 0.20 M
Question. Two half-reactions of an electrochemical cell are given below :
MnO–4(aq) + 8H+(aq) + 5e– → Mn2+ (aq) + 4H2O(l), E° = + 1.51V
Sn2+(aq) → Sn4+(aq) + 2e–, E° = + 0.15 V
Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured
Question. A zinc rod is dipped in 0.1 M solution of ZnSO4.
The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential.
[E°Zn2+/Zn = – 0.76 V]
Answer. The electrode reaction written as reduction reaction is
Zn2+ + 2e– → Zn (n = 2)
Applying Nernst equation, we get,
= – 0.76 – 0.02955 (log 1000 – log 95)
= – 0.76 – 0.0295 (3 – 1.9777) = – 0.79021 V
Question. State Kohlrausch’s law of independent migration of ions. Write its one application.
Answer. Kohlrausch’s law of independent migration of ions : It states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. If λ°Na+ and λl°Cl– are limiting molar conductivities of the sodium and chloride ions respectively then the limiting molar conductivity for sodium chloride is given by λ°m (NaCl) = λ°Na+ λ°Cl–
Kohlrausch’s law helps in the calculation of degree of dissociation of weak electrolyte like acetic acid.
The degree of dissociation a is given by
where λm is the molar conductivity and λ°m is the limiting molar conductivity.
Question. Define the following term :
Answer. Molar Conductivity : Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one cm apart and the area of the electrodes is so large that the whole solution is contained between them.
Λm = kV
It units is S cm2 mol–1
Short Answer Questions
Question. The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity.
Answer. Given : Diameter = 1 cm, length = 50 cm
R = 5.5 × 103 ohm, M = 0.05 M
r = ? k = ? Λm = ?
Area of the column,
Question. When a certain conductance cell was filled with 0.1 M KCl, it has a resistance of 85 ohms at 25°C. When the same cell was ffilled 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.
[Specific conductance of 0.1 M KCl = 1.29 × 10–2 ohm–1 cm–1]
Answer. k = 1.29 × 10–2 ohm–1 cm–1
Question. Calculate the strength of the current required to deposit 1.2 g of magnesium from molten
MgCl2 in 1 hour.
[1 F = 96,500 C mol–1 ;
Atomic mass : Mg = 24.0]
Answer. Reaction for deposition of Mg from molten MgCl2 :
Question. A solution of CuSO4 is electrolysed for 16 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?
Answer. According to the reaction :
Cu2+(aq) + 2e– → Cu(s)
Charge = Current × time = 1.5 amp × 16 × 60 s = 1440 C
2 × 96500 C electricity deposits = 63.5 g Cu
Question. Calculate the standard cell otential of the galvanic cell in which the following reaction takes place :
Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Calculate the DrG° and equilibrium constant of the reaction also,
Answer. E°cell = +0.80 V – 0.77 V = +0.03 V
ΔrG° = –nFE°cell = –1 × 96500 × 0.03
= –2895 J mol–1 = –2.895 kJ mol–1
ΔG° = – 2.303 RT log Kc
– 2895 = –2.303 × 8.314 × 298 × log K or log Kc = 0.5074
Kc = Antilog (0.5074) = 3.22
Question. The cell in which the following reaction occurs :
2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(s) has E°cell = 0.236 V at 298 K, Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
(Antilog of 6.5 = 3.162 × 106 ; of 8.0 = 10 × 108 ; of 8.5 = 3.162 × 108)
Question. Calculate the potential for half-cell containing 0.10 M K2Cr2O7(aq) , 0.20 M Cr3+(aq) and 1.0 × 10–4 M H+(aq). The half cell reaction is :
Cr2O2–7(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l) and the standard electrode potential is given as
E° = 1.33 V.
Answer. For half cell reaction,
Cr2O2–7(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l)
Question. Calculate the emf of following cell at 298 K :
Mg(s) | Mg2+(0.1 M) || Cu2+(0.01 M)|Cu(s)
[Given : E°cell = + 2.71 V, 1 F = 96500 C mol–1]
Answer. The cell reaction can be represented as :
Mg(s) + Cu(2+(aq) → Mg2+(aq) + Cu(s)
Given: Ecell , T = +2.71° V, T= 298 K
According to the Nernst equation :
Question. Estimate the minimum potential difference needed to reduce Al2O3 at 500°C. the Gibbs energy change for the decomposition reaction
Question. Calculate the degree of dissociation of acetic acid at 298 K, given that :
Lm (CH3COOH) = 11.7 S cm2 mol–1
L°m(CH3COO–) = 49.9 S cm2 moc
L°m(H+) = 349.1 S cm2 mol–1
Answer. According to Kohlrauch’s law,
Λ°CH3COOH = λ°CH3COO– + λ°H+
Long Answer Questions
Question. the resistance of a conductivity cell when filled with 0.05 M solution of an electrolyte X is 100 ohms at 40°C. the same conductivity cell filled with 0.01 M solution of electrolyte Y has a resistance of 50 ohms. The conductivity of 0.05 M solution of electrolyte X is 1.0 × 10–4 S cm–1.
(i) Cell constant
(ii) Conductivity of 0.01 M Y solution
(iii) Molar conductivity of 0.01 M Y solution
Answer. For electrolyte X :
Molarity = 0.05 M
resistance = 100 ohms
conductivity = 1.0 × 10–4 S cm–1
For electrolyte Y :
Molarity = 0.01 M
resistance = 50 ohms
conductivity = ?
(i) Cell constant, G*
= conductivity (k) × resistance (R)
= 100 × 1 × 10–4 = 10–2 cm–1
∴ Conductivity of solution Y is
Question. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer. The reciprocal of resistivity is known as specific conductance or simply conductivity. It is denoted by k (kappa).
Hence, conductivity of a solution is defined as the conductance of a conductor of 1 cm length and having 1 sq. cm as the area of cross section. Alternatively, it may be defined as conductance of one centimetre cube of the solution of the electrolyte. Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from 1 mole of the electrolyte dissolved in V cm3 of the solution. It is represented by λm. λm = kV Variation of conductivity and molar conductivity with concentration : Conductivity always decreases with decrease in concentration, for both weak and strong electrolytes. Because the number of ions per unit volume that carry the current in a solution decreases on dilution.
Molar conductivity increases with decrease in concentration. Because that total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in k on dilution of a solution is more than compensated by increase in its volume.
40. Calculate e.m.f and ΔG for the following cell
Mg(s) | Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu(s)
41. Calculate the standard electrode potential of Ni2+/Ni electrode if emf of the cell
Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1M) | Cu(s) is 0.059 V.
[Given : ECu2+ /Cu = + 0.34 V] °
42. Calculate the cell emf and ΔrG° for the cell reaction at 25°C
Zn(s) | Zn2+ (0.1 M) || Cd2+ (0.01 M)| Cd(s)
75. (a) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte.
(b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 W. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10–3 S cm–1?
76. (a) State Kohlrausch’s law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch’s law.
(b) Calculate L°m for acetic acid.
Λ °m(HCl) = 426 S cm2 mol–1
Λ °m(NaCl) = 126 S cm2 mol–1
Λ °m(CH3COONa) = 91 S cm2 mol–1