CBSE Class 12 Chemistry Electrochemistry Case Study Questions Set 02

Case Study 1

Products of electrolysis depend on the nature of material being electrolysed and the type of electrodes being used. If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert electrodes.
Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells.

Case Study 2

Question 6. Write the relationship between specific conductivity and equivalent conductivity.
Answer:
Λ = κ × \( \frac{1000}{N} \)

Teacher's Note:
1. Here N is the normality; with molarity M in place of N the same formula gives molar conductivity.

Question 7. What is the effect of decreasing concentration on the molar conductivity of a weak electrolyte?
Answer:
The molar conductivity of a weak electrolyte increases with decrease in concentration.

Teacher's Note:
1. The rise is steep for a weak electrolyte because its degree of dissociation increases sharply on dilution.

Question 8. Write the mathematical expression for Kohlrausch’s law in terms of equivalent conductivities.
Answer:
Λ°_eq = λ°₊ + λ°₋

Teacher's Note:
1. In terms of molar conductivities the stoichiometric numbers appear: Λ°_m = ν₊λ°₊ + ν₋λ°₋.

Question 9. What is meant by limiting molar conductivity?
Answer:
The molar conductivity of a solution at infinite dilution is called limiting molar conductivity.

Teacher's Note:
1. It is written Λ°_m and corresponds to concentration approaching zero.

Question 10. Which equation gives the relationship between equivalent or molar conductivity and concentration of a strong electrolyte?
Answer:
Debye-Huckel Onsager equation
Λ = Λ° − A√c

Teacher's Note:
1. The intercept of the straight line Λ_m vs √c gives Λ°_m and the slope is −A.

Case Study 3

Redox reactions play an important role in chemistry. Whenever a redox reaction takes place directly in a single beaker, chemical energy in the form of heat is produced. By suitable means, it is possible to bring about the redox reactions indirectly so as to convert the chemical energy into electrical energy. A device used to convert the chemical energy produced in a redox reaction into electrical energy is called an electrochemical cell. If a redox reaction is allowed to take place in such a way that oxidation half reaction takes place in one beaker and the reduction half reaction in another beaker, the electrons given out by the former will be taken by the latter and the current will flow. The two portions of the cell are called half cells. The values of standard redox potential (E°) of two half cell reactions decides in which way the reaction will proceed. A redox reaction is feasible when the substance having higher reduction potential gets reduced and the one having lower reduction potential gets oxidised. For example, In Daniell cell, zinc goes into solution and copper gets deposited.

Question 11. Formulate the galvanic cell in which the following reaction takes place:
Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)
Answer:
The cell is represented as
Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)

Teacher's Note:
1. The species oxidised (anode) is always written on the left and the cathode on the right, separated by || for the salt bridge.

Question 12. Is it safe to stir AgNO₃ solution with a copper spoon? Why or why not?
Given: E°_Ag⁺/Ag = 0.80 volt and E°_Cu²⁺/Cu = 0.34 volt
Answer:
No, it is not safe to stir AgNO₃ solution with copper, because copper is more reactive than Ag and can displace Ag from AgNO₃ solution.

Teacher's Note:
1. Check with E°_cell = 0.80 − 0.34 = +0.46 V for Cu + 2Ag⁺ → Cu²⁺ + 2Ag; positive means spontaneous.

Question 13. Two half cell reactions of an electrochemical cell are given below:
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l),   E° = +1.51 V
Sn²⁺(aq) → Sn⁴⁺(aq) + 2e⁻,   E° = +0.15 V
Construct the redox reaction from the two half cell reactions and predict if this reaction favours formation of reactants or products shown in the equation.
Answer:
The redox reaction will be
2MnO₄⁻(aq) + 16H⁺(aq) + 5Sn²⁺(aq) → 2Mn²⁺(aq) + 5Sn⁴⁺(aq) + 8H₂O(l)
E°_cell = E°_cathode − E°_anode = 1.51 V − 0.15 V = 1.36 V, i.e., it will be positive.
Hence, reaction will favour formation of products.

Teacher's Note:
1. Multiply the half reactions so that electrons cancel (×2 and ×5), but never multiply the E° values.

Question 14. State the factors that influence the value of cell potential of the following cell:
Mg(s) | Mg²⁺(aq) || Ag⁺(aq) | Ag(s)
Answer:
The factors that influence the value of cell potential are concentration of Mg²⁺ and Ag⁺ ions in the solution and temperature.

Teacher's Note:
1. These are the variables in the Nernst equation: E_cell = E°_cell − \( \frac{RT}{2F} \) ln \( \frac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2} \).

Question 15. Can E°_cell or Δ_rG° for cell reaction ever be equal to zero?
Answer:
No, E°_cell or Δ_rG° for cell reaction can never be zero.
ΔG° = −nFE°
E° is never zero hence ΔG° will also be not equal to zero.

Teacher's Note:
1. Do not confuse E°_cell with E_cell: E_cell does become zero when the cell reaction reaches equilibrium.

Case Study 4

Molar conductivity of strong electrolytes is found to vary with concentration according to Debye-Huckel Onsager equation. By plotting molar conductance (Λ_m) versus square root of concentration, the molar conductance of the strong electrolytes at infinite dilution (Λ°_m) called ‘limiting molar conductance’ can be found out by extrapolation of the linear plot (near high dilutions) to zero concentration. However, in case of weak electrolytes, such an extrapolation to zero concentration cannot be done as at high dilutions, the plot for weak electrolyte is almost parallel to the molar conductance axis. Hence, to find the value of Λ°_m of weak electrolytes, studies were made by Kohlrausch who put forward a law known after his name as Kohlrausch’s law. Kohlrausch’s law helps to calculate the Λ°_m values of weak electrolytes from the Λ°_m values of suitable strong electrolytes. Further, knowing the Λ_m^c value of the weak electrolyte at a particular concentration c and also using the Λ°_m value of that weak electrolyte, the degree of dissociation (α) of the weak electrolyte at concentration c can be calculated. Knowing the value of c and α, equilibrium constant (K_c) of the weak electrolyte can be calculated.

Question 16. Write Debye-Huckel Onsager equation in the most common form. What do different symbols signify ?
Answer:
Debye-Huckel Onsager equation is Λ_m^c = Λ°_m − A\( \sqrt{c} \)
where Λ_m^c = molar conductivity at concentration c mol L⁻¹
Λ°_m = molar conductivity at infinite dilution called limiting molar conductivity.
A = constant depending upon the type of electrolyte.

Teacher's Note:
1. A depends on the charges of the ions (NaCl, CaCl₂ and MgSO₄ types have different slopes) as well as on solvent and temperature.

Question 17. The molar conductances of NaOH, NaCl and BaCl₂ at infinite dilution are 2.481 × 10⁻², 1.265 × 10⁻² and 2.800 × 10⁻² S m² mol⁻¹ respectively. Calculate Λ°_m for Ba(OH)₂.
Answer:
Λ°_m Ba(OH)₂ = Λ°_m (BaCl₂) + 2Λ°_m (NaOH) − 2Λ°_m (NaCl)
= (2.800 × 10⁻²) + 2(2.481 × 10⁻²) − 2(1.265 × 10⁻²) S m² mol⁻¹
= (2.800 + 4.962 − 2.530) × 10⁻² S m² mol⁻¹
= 5.232 × 10⁻² S m² mol⁻¹

Teacher's Note:
1. Check by ions: Ba²⁺ + 2Cl⁻ + 2Na⁺ + 2OH⁻ − 2Na⁺ − 2Cl⁻ leaves exactly Ba²⁺ + 2OH⁻.

Teacher's Note:
1. The ionic values λ°(H⁺) = 349.6 and λ°(HCOO⁻) = 54.6 S cm² mol⁻¹ are not in the question; in the exam they are always supplied, so use the values given.

Question 19. A strong electrolyte is already completely ionized in the solution. Then why its molar conductivity increases on dilution ?
Answer:
At high concentration, ions are close to each other and, therefore, there are strong interionic attractions. On dilution, ions go far apart, therefore, interionic attractions decrease, i.e., ions become free to move. Hence, conductance increases.

Teacher's Note:
1. For a strong electrolyte the number of ions stays the same on dilution; only their mobility increases.

Case Study 5

Any cell or battery that we use as a source of electrical energy is basically a galvanic cell. But every galvanic cell is not suitable for commercial use. For a battery to be of practical use, it should be reasonably light, compact and its voltage should not vary much during its use. Batteries are broadly classified as ‘Primary batteries’ and ‘Secondary batteries.’ Primary batteries are those in which the redox reaction occurs only once and it becomes dead after a period of time and cannot be reused. Two common examples of this type are dry cell and mercury cell. Secondary batteries are those which can be recharged by passing an electric current through them and hence can be used over and again. Two well known examples of this type are lead storage battery and nickel-cadmium storage cell. Another source of electrical energy is from fuel cells. Their advantage over the thermal plants is that they do not cause pollution and are comparatively simple to operate. The reactants are fed continuously and products are removed continuously. They are the devices which convert the energy produced during the combustion of fuel, like hydrogen, methane, methanol etc. directly into electrical energy. One such cell which has been very successful and was used in Apollo moon flights is H₂–O₂ fuel cell.

Question 20. Why mercury cell gives a constant voltage of 1.35 V throughout its life ?
Answer:
It involves the reaction between Zn and HgO in presence of KOH. The OH⁻ ions used in the reaction at anode (Zn + 2OH⁻ → ZnO + H₂O + 2e⁻) are reproduced in the reaction at cathode (HgO + H₂O + 2e⁻ → Hg + 2OH⁻). Thus, overall reaction does not involve any ion whose concentration may change. Hence, it gives a constant voltage throughout its life.

Teacher's Note:
1. Overall reaction Zn(Hg) + HgO → ZnO + Hg has only solids and a liquid, so the Nernst log term is constant.

Question 21. Write the reactions occurring at the anode and cathode of a lead storage battery during discharge.
Answer:
At anode : Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
At cathode : PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O

Teacher's Note:
1. Both electrodes end up as PbSO₄ during discharge, which is what makes recharging possible.

Question 22. Write the reactions occurring at the anode and cathode of the H₂–O₂ fuel cell and the overall reaction. What is its voltage ?
Answer:
At anode : 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
At cathode : O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall reaction : 2H₂(g) + O₂(g) → 2H₂O(l)
Voltage of H₂–O₂ fuel cell = 0.9 V

Teacher's Note:
1. The theoretical emf is 1.23 V (see Q8); the working voltage of about 0.9 V is lower because of losses at the electrodes.

Question 23. The theoretical EMF of H₂–O₂ fuel cell is 1.23 V and the heat of combustion of H₂ in O₂ is −285.8 kJ mol⁻¹. What is the theoretical efficiency of the H₂–O₂ fuel cell ?
Answer:
ΔG = −nFE°_cell = −(2)(96500 C mol⁻¹)(1.23 V)
= −237390 CV mol⁻¹ = −237390 J mol⁻¹ = −237.390 kJ mol⁻¹ (1 CV = 1 J)
η = \( \frac{\Delta G}{\Delta H}\times 100 = \frac{-237.39\ \text{kJ mol}^{-1}}{-285.8\ \text{kJ mol}^{-1}}\times 100 \) = 83%

Teacher's Note:
1. Use n = 2 because the heat of combustion is per mole of H₂ (H₂ + ½O₂ → H₂O involves 2 electrons).

Case Study 6

When a D.C. voltage from a battery is applied between the two electrodes dipped in the solution or melt of an electrolyte, the ions of the electrolyte move towards the opposite electrodes. The process is called as ‘electrolysis’. The process is of great importance in laboratory as well as in chemical industry. For example, pure copper is obtained from impure copper by this process. Many metals like Na, Mg, Al etc. are produced on a large scale by this process. If a number of ions are present in the solution, the ion with lowest discharge potential is liberated at the electrode. Similarly, the products of electrolysis depend upon whether the electrodes are attacked by the ions or the electrodes are inert. Quantitatively, the amounts of the substances liberated at the electrodes during electrolysis depend upon the amount of quantity of electricity passed. These studies were made by Faraday. Alternatively, the products of electrolysis can also be predicted by comparing the standard electrode potentials of the probable reactions occurring at the electrode. However, in case of liberation of gases at the electrodes, some overvoltage is required.

Question 24. What will be the products of electrolysis of CuSO₄ solution using (i) copper electrodes (ii) platinum electrodes ?
Answer:
(i) With Cu electrodes : Cu at cathode, no gas at anode.
At cathode : Cu²⁺(aq) + 2e⁻ → Cu(s) ; At anode : Cu(s) → Cu²⁺(aq) + 2e⁻ (the copper anode dissolves)
(ii) With platinum electrodes : Cu at cathode, O₂ at anode.
At cathode : Cu²⁺(aq) + 2e⁻ → Cu(s) ; At anode : 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻

Teacher's Note:
1. An active (copper) anode is itself oxidised, which is the basis of electrolytic refining of copper.

Question 25. What is overvoltage ?
Answer:
Over voltage is the difference between the potential required for the evolution of the gas and its standard reduction potential.

Teacher's Note:
1. Overvoltage is the reason Cl₂, not O₂, is obtained at the anode in electrolysis of aqueous NaCl.

Question 26. A current of 4 ampere was passed for 1.5 hours through a solution of copper sulphate when 3.2 g of copper was deposited. Calculate the current efficiency.
Answer:
Quantity of electricity passed, Q = I × t = 4 A × (1.5 × 60 × 60) s = 21600 C
Cu²⁺ + 2e⁻ → Cu ; 2 × 96500 C deposit 63.5 g of Cu
∴ 21600 C should deposit \( \frac{63.5}{2\times 96500}\times 21600 \) = 7.107 g of Cu
Actual mass deposited = 3.2 g
Current efficiency = \( \frac{3.2}{7.107}\times 100 \) = 45.0%

Teacher's Note:
1. Current efficiency = actual mass / theoretical mass × 100; convert hours to seconds before using Q = It.

Question 27. A current is passed through two cells connected in series. The first cell contains X(NO₃)₃(aq) and the second cell contains Y(NO₃)₂(aq). The relative atomic masses of X and Y are in the ratio 1 : 2. Calculate the ratio of mass of X to that of Y liberated.
Answer:
By Faraday’s second law, \( \frac{\text{Mass of X}}{\text{Mass of Y}} = \frac{\text{Eq. mass of X}}{\text{Eq. mass of Y}} \)
X is present as X³⁺ and Y as Y²⁺. If atomic mass of X = a, atomic mass of Y = 2a.
Eq. mass of X = \( \frac{a}{3} \), Eq. mass of Y = \( \frac{2a}{2} \) = a
∴ \( \frac{\text{Mass of X}}{\text{Mass of Y}} = \frac{a/3}{a} = \frac{1}{3} \), i.e., 1 : 3

Teacher's Note:
1. Read the charges from the formulas: three NO₃⁻ means X³⁺, two means Y²⁺.

Case Study 7

A number of devices or gadgets require electrical energy. One of the sources of this energy is ‘galvanic cell’. A galvanic cell is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy. The most common galvanic cell is Daniell cell based on the redox reaction, Zn + CuSO₄ → ZnSO₄ + Cu. The redox reaction consists of two half reactions, one is oxidation-half and the other is reduction-half. The arrangement is set-up consisting of electrodes connected by a wire for flow of electrons from one electrode to the other and flow of ions through a salt bridge from one solution to the other. The tendency of any electrode to lose or gain electrons is measured with respect to a standard hydrogen electrode taken as a reference electrode and whose electrode potential is taken as zero under standard conditions (1 M concentration, 1 bar pressure). All these standard electrode potentials (expressed as standard reduction potentials) are arranged in the increasing order. The series thus obtained is called ‘electrochemical series’. This series helps to calculate the standard EMF of a cell choosing any two electrodes as well comparing relative oxidizing and reducing powers or relative activities of metals etc. The EMF of the cell or electrode potential of the electrodes depends upon the concentrations of the electrolytes used in the electrodes and temperature. These studies were made by Nernst. We may have a cell in which the electrodes are same but differ only in the concentrations of the electrolytes taken in them. Such cells are called ‘concentration cells’.

Case Study 8

When a cell reaction takes place in two half-cells, in one of the electrodes, metal ions from the solution are deposited in one of electrode (reduction half reaction occurs) whereas in the other half-cell, metal from the electrode passed into the solution as metal ions (oxidation half reaction occurs). Thus, concentration of metal ions in one of the electrodes keeps on decreasing whereas in the other it keeps on increasing. Consequently, the electrode potential of one of the electrodes keeps on decreasing whereas that of the other electrode keeps on increasing. Ultimately, a stage is reached when the electrode potentials of the two electrodes become equal and the current stops flowing, i.e., EMF of the cell becomes zero. The cell reaction is said to have attained equilibrium : Applying Nernst equation to the cell reaction in equilibrium, the relationship between standard EMF of the cell (E°_cell) and equilibrium constant (K_c) can be found. Another important aspect is that when electrical energy is produced in a cell reaction or when electrical work is done, this is at the cost of Gibbs free energy (ΔG) of the reaction which decreases. Thus, there exits a relationship between decrease in Gibbs energy of a cell reaction (−ΔG_r) and the electrical work done (nFE_cell). As E°_cell is related to equilibrium constant (K_c) and E°_cell is also related to decrease in free energy of the cell reaction (ΔG°), therefore, ΔG° is related directly to equilibrium constant of the cell reaction. Thus, the three quantities, E°_cell, ΔG° and K_c are inter-related with each other. These help in a number of calculations.