Please refer to the below Electrochemistry important questions for Class 12 Chemistry. These questions and answers have been prepared as per the latest NCERT Book for Class 12 Chemistry. Students should go through chapter wise Class 12 Chemistry Important Questions designed as per the latest examination pattern issued by CBSE.
Very Short Answer Questions :
Question. A voltaic cell is set up at 25°C with the following half-cells :
Al | Al3+ (0.001 M) and Ni |Ni2+ (0.50 M)
Calculate the cell voltage
[E°Ni2+/Ni = – 0.25V, E°Al3+|Al = – 1.66 V]
Answer :
Question. Calculate the emf for the given cell at 25° C :
Cr|Cr3+ (0.1 M) || Fe2+ (0.01 M)| Fe
Given :
E°Cr3+/Cr = – 0.1 74 V, E°Fe2+/Fe = − 0.44V
Answer :
Question. Calculate the standard cell potential of a galvanic cell in which the following reaction takes place :
2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)
Calculate ΔrG ° and equilibrium constant, K of the above reaction at 25°C.
Given :
E°Cr3+/Cr = – 0.1 74 V, E°Fe2+/Fe = − 0.44V
(F = 96,500 C mol–1)
Answer :
Question. 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.
Answer : Mercury cell is generally used in transistors.
At cathode : HgO + H2O + 2e– → Hg + 2OH–
At anode : Zn(Hg) + 2OH– → ZnO + H2O + 2e–
Question. Calculate the equilibrium constant for the reaction
Fe(s) + Cd2+(aq) ⇌ Fe2+(aq)+ Cd(s)
Given :
[ E°Cd2+|Cd = − 0.40 V, E°Fe2+/Fe = − 0.44]
Answer :
Question. One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO3)2. A voltage of 1.48 V is measured for this cell. Use this information to calculate the concentration of silver nitrate solution used.
E°(Zn2+|Zn) = – 0.76V E°Ag2+/Ag = + 0.80 V
Answer :
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. 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 Thow of current is reversed.
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 :
Zn | Zn2+(aq) || Cu2+(aq) | Cu
Anode Salt Cathode
bridge
Question. Given that the standard electrode potential (E°) of metals are :
K+/K = –2.93 V, Ag+/Ag = 0.80 V,
Cu2+/Cu = 0.34 V,
Mg2+/Mg = –2.37 V, Cr3+/Cr = –0.74 V,
Fe2+ /Fe = –0.44 V.
Arrange these metals in an increasing order of their reducing power
Answer : The reducing power increases with decreasing value of electrode potential. Hence, the order is Ag < Cu < Fe < Cr < Mg < K.
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 :
Question. A voltaic cell is set up at 25°C with the following half cells :
Al/Al3+ (0.001 M) and Ni/Ni2+ (0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
E°Ni2+/Ni = – 0.25 V and E°Al3+/Al = – 1.66 V.
(log 8 × 10–6 = – 5.09)
Answer :
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.
Answer : At anode : Sn2+(aq) → Snn4+(aq) + 2e–] × 5
At cathode : MnO–4(aq) + 8H+(aq) + 5e– → Mn2+(aq) +(aq) + 4H2O(l)] × 2
Net cell reaction :
2MnO–4(aq) + 5Sn2+(aq) + 16H+(aq)→ Mn2+(aq) + 5Sn4+(aq) + 8H2O(l)
E°cell = E°cathode – E°anode = 1.51 V – 0.15 V = 1.36 V.
Since, cell potential is positive therefore the reaction is product favoured.
Question. 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°(Zn2+|Zn) = − 0.25 V and E°Ag+/Ag = 0.80V
Answer :
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)
Answer :
Question. Estimate the minimum potential difference needed to reduce Al2O3 at 500°C. The Gibbs energy change for the decomposition reaction 2/3 Al2O3 → 4/3 Al +O2 is 960 kJ.
(F = 96500 C mol–1)
Answer :
Question. Equilibrium constant (Kc) for the given cell reaction is 10. Calculate E°cell.
A(s) + B2+(aq) ⇌ A2+(aq) + B(s)
Answer :
Question. In the button cell, widely used in watches, the following reaction takes place.
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH–(aq)
Determine E° and ΔG° for the reaction.
(Given : E°Ag+/Ag = + 0.80 V, E°Zn2+/Zn = – 0.76 V)
Answer :
Question. Calculate the emf of the following cell at 298 K :
Fe(s)|Fe2+(0.001 M)||H+(1 M)|H2(g) (1 bar), Pt(s)
(Given E°cell = + 0.44 V)
Answer :
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 :
Short Answer Questions :
Question. Predict the products of electrolysis obtained at the electrodes in each if the electrodes used are of platinum?
(i) An aqueous solution of AgNO3.
(ii) An aqueous solution of H2SO4.
Answer : (i) At cathode : The following reduction reactions compete to take place at the cathode.
Ag+(aq) + e– → Ag(s) ; E° = 0.80 V
H+(aq) + e– → 1/2 H2(g) ; E° = 0.00 V
The reaction with a higher value of E° takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.
Since, Pt electrodes are inert, the anode is not attacked by NO–3 ions. Therefore, OH– or NO–3 ions can be oxidized at the anode. But OH– ions having a lower discharge potential get preference and decompose to liberate O2.
Question. Define the following term :
Kohlrausch’s law of independent migration of ions.
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 λ°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
α = Λm /Λ°m
where Λm is the molar conductivity and Λ°m is the limiting molar conductivity.
Question. Following reactions occur at cathode during the electrolysis of aqueous copper (II) chloride solution :
Cu2+(aq) + 2e– → Cu(s) E° = +0.34 V
H+(aq) + e– →1/2H2(g) E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
Answer : The species that get reduced at cathode is the one which have higher value of standard reduction potential. Hence, the reaction that will occur at cathode is Cu2+(aq) + 2e– → Cu(s)
Question. State the Faraday’s first law of electrolysis.
Answer : Faraday’s fiirst law of electrolysis : During electrolysis the amount of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte i.e.,
w a Q or w ∝ I × t [∴ Q = I × t]
w = Z × I × t
where, Z is a constant of proportionality known as electrochemical equivalent of the substance deposited.
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. What are fuel cells? Explain the electrode reactions involved in the working of H2 — O2 fuel cell.
Answer : Fuel cells : Those galvanic cells which give us direct electrical energy by the combustion of fuels like hydrogen, methane, methanol etc. are called fuel cells.
The reactions taking place in hydrogen – oxygen fuel 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(s)
∴ Quantity of charge required for reduction of 1 mole of Cu2+ = 2F = 2 × 96500 = 193000 C
Question. The resistance of 0.01 M NaCl solution at 25°C is 200 Ω. The cell constant of the conductivity cell used is unity. Calculate the molar conductivity of the solution.
Answer :
Question. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with change in temperature.
Answer : The reciprocal of resistivity is known as specific conductance or simply conductivity. It is denoted by k (kappa).
k = 1/p or k = G X l/a
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.
Variation of conductivity and molar conductivity with temperature : Both increase with increase in temperature as degree of ionisation increases.
Question. What is corrosion? Explain the electrochemical theory of rusting of iron and write the reactions involved in the rusting of iron.
Answer : Corrosion is the process of slowly eating away of the metal due to attack of the atmospheric gases on the surface of the metal resulting into the formation of compounds such as oxides, sulphides, carbonates, etc. The corrosion of iron is called rusting.
Electrochemical theory of rusting of iron : According to electrochemical theory of rusting the impure iron surface behaves like small electrochemical cell. In this any point of iron acts as anode and other iron surface acts as cathode. Moisture having dissolved CO2 or O2 acts as an electrolyte. The reactions are given below.
At anode :
Question. How much charge is required for the reduction of 1 mol of Zn2+ to Zn?
Answer :
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
Molar conductivity Λcm = k×1000/Molarity
= 0.025×1000/0.20 = 125 S cm2 mol–1
Question. The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10–3 S cm–1?
Answer : Here, conductivity (k) = 0.146 × 10–3 S cm–1,
resistance (R) = 1500 Ω
Cell constant = Conductivity/Conductance
= Conductivity × Resistance
= k × R ∴ conductance = 1/resistance
= 0.146 × 10–3 × 1500 = 0.219 cm–1
Question. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5.0 ampere for 20 minutes. What mass of nickel will be deposited at the cathode?
(Given : At. mass of Ni = 58.7 g mol–1, 1F = 96500 C mol–1)
Answer : Given : Current I = 5 A; t = 20 × 60 s, w = ?
Question. The conductivity of 0.001 M acetic acid is 4 × 10–5 S/cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm2/mol.
Answer :
Question. Define molar conductivity of a substance and describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such change explained?
Answer : Strong electrolyte : The molar conductivity of strong electrolyte decreases slightly with the increase in concentration. This increase is due to increase in attraction as a result of greater number of ions per unit volume. With dilution the ions are far apart, interionic attractions become weaker and conductance increases.
Weak electrolyte : When the concentration of weak electrolyte becomes very low, its degree of ionisation rises sharply. There is sharp increase in the number of ions in the solution. Hence the molar conductivity of a weak electrolyte rises steeply at low concentration.
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
Question. State Kohlrausch’s law of independent migration of ions. How can the degree of dissociation of acetic acid in a solution be calculated from its molar conductivity data?
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 λ°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
α = Λm /Λ°m
where Λm is the molar conductivity and Λ°m is the limiting molar conductivity.
Question. Predict the products of electrolysis in each of the following :
(i) An aqueous solution of AgNO3 with platinum electrodes.
(ii) An aqueous solution of H2SO4 with platinum electrodes.
Answer : (i) At cathode : The following reduction reactions compete to take place at the cathode.
Ag+(aq) + e– → Ag(s) ; E° = 0.80 V
H+(aq) + e– → 1/2 H2(g) ; E° = 0.00 V
The reaction with a higher value of E° takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.
Since, Pt electrodes are inert, the anode is not attacked by NO–3 ions. Therefore, OH– or NO–3 ions can be oxidized at the anode. But OH– ions having a lower discharge potential get preference and decompose to liberate O2.
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 λ°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
α = Λm /Λ°m
where Λm is the molar conductivity and Λ°m is the limiting molar conductivity.
Question. Explain with examples the terms weak and strong electrolytes.
Answer : Weak electrolytes : The electrolytes which are not completely dissociated into ions in solution are called weak electrolytes e.g., CH3COOH, NH4OH, HCN, etc.
Strong electrolytes : The electrolytes which are completely dissociated into ions in solution are called strong electrolytes. e.g., HCl, KCl, NaOH, NaCl, etc.
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 .e., ions present per cm3 of the solution becomes less and less on dilution.
Question. The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol–1.
Calculate the conductivity of this solution.
Answer :
Question. State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu.
Answer : Faraday’s fiirst law of electrolysis : During electrolysis the amount of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte i.e.,
w a Q or w ∝ I × t [∴ Q = I × t]
w = Z × I × t
where, Z is a constant of proportionality known as electrochemical equivalent of the substance deposited.
The electrode reaction is Cu2+ + 2e– → Cu(s)
∴ Quantity of charge required for reduction of 1 mole of Cu2+ = 2F = 2 × 96500 = 193000 C
Question. Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
Answer : Λm M = k X 1000/M in CGS units
Λm = k X 10−3/M in SI units
where k is the conductivity, M is the molar concentration and Λm is molar conductivity.
k = 1/R X l/A
where k is the conductivity R is resistance and l/A is the cell constant
Question. How much electricity in terms of Faradays is required to produce 20 g of calcium from molten CaCl2?
Answer : Reaction for production of Ca from molten CaCl2 :
CaCl2 → Ca2+ + 2Cl–
Ca2+ + 2e– → Ca
2F 40 g
Electricity required to produce 40 g = 2 F
∴ Electricity required to produce 20 g = 0.5 × 2 F = 1 F
