CBSE Class 12 Chemistry Electrochemistry Important Questions Set 09

Very Short Answer Questions (1 Mark)

Question 1. Write the reaction taking place at the cathode when the lead storage battery is in use.
Answer:
PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)

Teacher's Note:
1. During discharge both electrodes get coated with PbSO₄ and H₂SO₄ is used up, so the density of the acid falls.

Question 2. How much electricity in terms of Faraday is required to carry out the reduction of one mole of PbO₂ ?
Answer:
From the above equation, 1 mole of PbO₂ requires 2 Faradays.

Teacher's Note:
1. Pb changes from +4 in PbO₂ to +2 in PbSO₄, a gain of 2 electrons per Pb.

Question 3. What is the molarity of sulphuric acid before discharge ?
Answer:
Before discharge, the acid in a lead storage battery is 38% H₂SO₄ by mass, with density 1.294 g cm⁻³.
Molarity = \( \frac{38/98}{100/1.294}\times 1000 \) = 5.02 M

Teacher's Note:
1. Volume of 100 g of solution = mass/density = 100/1.294 cm³; multiply by 1000 to convert to per litre.

Question 4. Lead storage battery is considered a secondary cell. Why ?
Answer:
It is a secondary cell because it can be recharged after use.

Teacher's Note:
1. On charging, the cell acts as an electrolytic cell and the discharge reactions are reversed.

Question 5. Write the products of electrolysis when dilute sulphuric acid is electrolysed using Platinum electrodes ?
Answer:
H₂(g) at cathode and O₂(g) at anode.

Teacher's Note:
1. With dilute acid, oxidation of water (giving O₂) is preferred over oxidation of SO₄²⁻ at the anode.

Question 6. Which cells were used in the Apollo space programme ? What was the product used for ?
Answer:
H₂ – O₂ fuel cell. The product H₂O was used for drinking by the astronauts.

Teacher's Note:
1. The fuel cell thus supplied both electricity and drinking water to the spacecraft.

Question 7. Define fuel cell ? (CBSE 2017)
Answer:
A device in which the heat produced as a result of the combustion of a fuel (like H₂ in presence of O₂) is converted into electricity is called a fuel cell.

Teacher's Note:
1. Strictly, the chemical energy of combustion is converted directly into electrical energy, with the reactants supplied continuously.

Question 8. How does H₂ – O₂ fuel cell operate ?
Answer:
At anode : 2H₂(g) + 4OH⁻(aq) → 4H₂O + 4e⁻
At cathode : O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall reaction : 2H₂(g) + O₂(g) → 2H₂O(l).

Teacher's Note:
1. Add the two half reactions and cancel the OH⁻ and H₂O terms to check that the overall reaction is simply combustion of hydrogen.

Question 9. State two advantages of H₂ – O₂ fuel cell over ordinary cell.
Answer:
(i) They do not cause any pollution.
(ii) They have high efficiency of 60–70%.

Teacher's Note:
1. A third advantage: the fuel cell runs continuously as long as the reactants are supplied, unlike an ordinary cell which runs down.

Question 10. Write the name of the electrolyte used in fuel cell.
Answer:
Concentrated aqueous KOH solution.

Teacher's Note:
1. The H₂ and O₂ gases are bubbled through porous carbon electrodes into this KOH solution.

Question 11. Why is chromium used for coating iron ?
Answer:
Chromium is a non-corroding metal which forms a protective layer on iron.

Teacher's Note:
1. Chromium itself gets a thin, tough oxide film that stops further attack, so a chrome plating keeps air and moisture away from iron.

Question 12. What would happen if the protective tin coating over an iron bucket is broken in some places ?
Answer:
In such a case, iron corrodes faster than it does in the absence of tin as oxidation potential of Fe is greater than that of Sn.

Teacher's Note:
1. At the crack, iron becomes the anode and tin the cathode of a small cell, so iron is oxidised preferentially.

Question 13. Rusting of iron is quicker in saline water than in ordinary water. Give reason.
Answer:
In saline water, the presence of Na⁺ and Cl⁻ ions increases the conductance of the solution in contact with the metal surface. This accelerates the formation of Fe²⁺ ion and hence that of rust, Fe₂O₃.xH₂O.

Teacher's Note:
1. Rusting is electrochemical, so anything that makes the water a better electrolyte (salt, dissolved CO₂, acids) speeds it up.

Short Answer Questions (2 Marks)

Question 14. How Nernst equation can be applied in the calculation of equilibrium constant of any cell reaction ?
Answer:
For the cell reaction, Nernst equation is E_cell = E°_cell − \( \frac{2.303\,RT}{nF} \) log Q.
At equilibrium, E_cell = 0 and Q = K_c.
∴ 0 = E°_cell − \( \frac{2.303\,RT}{nF} \) log K_c or E°_cell = \( \frac{2.303\,RT}{nF} \) log K_c
At 298 K, E°_cell = \( \frac{0.059}{n} \) log K_c or log K_c = \( \frac{nE^\circ_{cell}}{0.059} \)
Thus, knowing E°_cell, K_c can be calculated.

Teacher's Note:
1. Even a small positive E°_cell gives a large K_c; e.g. 1.10 V with n = 2 gives K_c ≈ 10³⁷.

Question 15. Derive the relationship between Gibb’s free energy change and the cell potential.
Answer:
In a galvanic cell, the electrical work done = charge passed × potential.
If n moles of electrons are transferred, charge passed = nF coulombs. If E_cell is the emf of the cell, maximum work done by the cell, W_max = nFE_cell.
The maximum useful work obtained from a reaction is equal to the decrease in Gibbs free energy, i.e., W_max = −Δ_rG.
∴ Δ_rG = −nFE_cell
Under standard conditions, Δ_rG° = −nFE°_cell.

Teacher's Note:
1. Use F = 96500 C mol⁻¹ and E in volts to get ΔG in J mol⁻¹; divide by 1000 for kJ.

Question 16. Derive relationship between Gibbs energy and equilibrium constant of a reaction.
Answer:
From thermodynamics, ΔG = ΔG° + RT ln Q.
At equilibrium, ΔG = 0 and Q = K.
∴ 0 = ΔG° + RT ln K or ΔG° = −RT ln K = −2.303 RT log K
Also, ΔG° = −nFE°_cell. Hence, nFE°_cell = 2.303 RT log K or E°_cell = \( \frac{2.303\,RT}{nF} \) log K.

Teacher's Note:
1. Negative ΔG° means K > 1, i.e. products are favoured at equilibrium.

Question 17. What are the difficulties in the construction of a fuel cell ? What are its advantages over the other cells ?
Answer:
Difficulties : (i) The catalysts used (Pt, Pd) are very costly. (ii) The electrodes must be porous and resistant to the corrosive electrolyte. (iii) Storage and handling of the gaseous fuel (H₂) is difficult and hazardous. (iv) The reactions are slow, so special electrodes and conditions are needed.
Advantages : (i) High efficiency (about 60–70%) as compared to thermal plants (about 40%). (ii) They do not cause pollution. (iii) They work continuously as long as the reactants are supplied and do not need recharging.

Teacher's Note:
1. High cost of the catalyst is the main reason fuel cells are not yet in everyday use.

Question 18. What is sacrificial protection of iron from corrosion ? Give one example.
Answer:
In this method, the iron object is covered with (or connected to) a more electropositive metal like zinc or magnesium. The more active metal is oxidised (corroded) in preference to iron, i.e., it sacrifices itself to protect iron. This is called sacrificial protection.
Example : Galvanised iron (iron coated with zinc); also, blocks of magnesium connected to buried iron pipes or ship hulls.

Teacher's Note:
1. The sacrificial metal acts as the anode, so iron becomes the cathode and is not oxidised.

Question 19. Explain ‘Iron is galvanised for protecting it from rusting’.
Answer:
Galvanisation is the process of coating iron with a thin layer of zinc. Zinc is more electropositive than iron (E°_Zn²⁺/Zn = −0.76 V, E°_Fe²⁺/Fe = −0.44 V). Hence, zinc gets oxidised in preference to iron. Even if the zinc coating is broken at some places, zinc continues to act as anode and iron does not rust. Moreover, the zinc layer itself gets covered with a protective film of basic zinc carbonate.

Teacher's Note:
1. Galvanisation is a case of sacrificial (cathodic) protection, unlike tin plating which only gives barrier protection.

Question 20. What is corrosion ? Explain any four factors affecting corrosion. (Chhatisgarh Board 2011)
Answer:
Corrosion is the slow eating away of the surface of a metal due to the attack of atmospheric gases and moisture.
Factors : (i) Reactivity of the metal : more active metals (lower reduction potential) corrode more readily.
(ii) Presence of impurities : impurities in the metal set up small electrochemical cells and speed up corrosion.
(iii) Presence of moisture and air : corrosion needs both water and oxygen; it is faster in humid air.
(iv) Presence of electrolytes : salts (as in sea water) and acidic gases like CO₂, SO₂ increase the conductance of water and speed up corrosion.
(Strains in the metal and higher temperature also increase corrosion.)

Teacher's Note:
1. Iron rusts fastest in coastal and industrial areas, where salt and acidic gases are present.

Question 21. Describe any two of the techniques used for preventing corrosion of metals.
Answer:
(i) Barrier protection : The surface of the metal is covered with paint, oil, grease, varnish or a coating of a less reactive metal (like tin, chromium) so that it does not come in contact with air and moisture.
(ii) Sacrificial (cathodic) protection : The metal (iron) is coated with or connected to a more electropositive metal like Zn or Mg. The more active metal is oxidised in preference to iron and protects it even if the coating is broken (e.g., galvanisation).
(Other methods : alloying, e.g., stainless steel; and use of anti-rust solutions.)

Teacher's Note:
1. Barrier protection fails once the coating is scratched; sacrificial protection keeps working.

Question 22. We can use aluminium in place of zinc for cathodic protection of rusting. Comment.
Answer:
The statement is correct. Aluminium (E°_Al³⁺/Al = −1.66 V) is more electropositive than iron (E°_Fe²⁺/Fe = −0.44 V), even more than zinc (−0.76 V). Hence, when connected to iron, aluminium is oxidised in preference to iron and acts as the sacrificial anode. Thus it can be used for cathodic protection of iron.

Teacher's Note:
1. Any metal with E° lower than −0.44 V (Mg, Al, Zn) can protect iron cathodically.

Question 23. Iron does not rust even if the zinc coating is broken in a galvanised iron pipe but rusting occurs much faster if the tin coating over iron is broken. Explain.
Answer:
Zinc is more electropositive whereas tin is less electropositive than iron.
When the zinc coating is broken, zinc (E° = −0.76 V) still acts as the anode and is oxidised in preference to iron, so iron does not rust.
When the tin coating is broken, iron (E° = −0.44 V) has a lower reduction potential than tin (E° = −0.14 V). So iron acts as the anode and tin as the cathode of the cell formed; iron is oxidised faster than it would be without the tin coating.

Teacher's Note:
1. Tin gives only barrier protection; zinc gives sacrificial protection.

Question 24. State reasons for the following :
(i) Rusting of iron is said to be an electrochemical phenomenon
(ii) For a weak electrolyte, its molar conductance in dilute solutions increases sharply as its concentration in solution is decreased.
Answer:
(i) During rusting, small electrochemical cells are set up on the surface of iron: at the anodic area iron is oxidised (Fe → Fe²⁺ + 2e⁻) and at the cathodic area oxygen is reduced (O₂ + 4H⁺ + 4e⁻ → 2H₂O), with water containing dissolved CO₂ acting as electrolyte. Since oxidation and reduction take place at different spots with flow of electrons, as in a galvanic cell, rusting is an electrochemical phenomenon.
(ii) A weak electrolyte is only partially dissociated. On dilution, its degree of dissociation increases sharply, so the number of ions produced from one mole of the electrolyte increases. Hence its molar conductance increases sharply in dilute solutions.

Teacher's Note:
1. For (ii), the cause is increase in the number of ions (degree of dissociation), not increase in ionic mobility as for strong electrolytes.

Question 25. (i) For a weak electrolyte, molar conductance in dilute solution increases sharply as its concentration in solution is decreased. Give reason.
(ii) Write overall cell reaction for lead storage battery when the battery is being charged.
Answer:
(i) A weak electrolyte is only partially dissociated. On dilution, its degree of dissociation increases sharply, so the number of ions in solution from one mole of the electrolyte increases. Hence, molar conductance increases sharply.
(ii) 2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq)

Teacher's Note:
1. The charging reaction is just the discharge reaction written in reverse.

Question 26. Explain (Definition and Formula)
(a) Kohlrausch law (b) Faraday’s first law of electrolysis. (MP Board 2012)
Answer:
(a) Kohlrausch’s law : The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte. Λ°_m = ν₊λ°₊ + ν₋λ°₋, where ν₊ and ν₋ are the number of cations and anions in the formula unit.
(b) Faraday’s first law : The amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. w ∝ Q or w = ZQ = Z I t, where Z is the electrochemical equivalent, I is current in amperes and t is time in seconds.

Teacher's Note:
1. In w = ZIt time must be in seconds; convert minutes or hours first.

Short Answer Questions (3 Marks)

Question 27. What is a Battery ? Describe the mercury cell.
Answer:
A battery is a combination of two or more galvanic cells connected in series; in practice, any cell or set of cells used as a source of electrical energy is called a battery.
Mercury cell : anode is zinc-mercury amalgam, cathode is a paste of HgO and carbon, electrolyte is a paste of KOH and ZnO.
At anode : Zn(Hg) + 2OH⁻ → ZnO(s) + H₂O + 2e⁻
At cathode : HgO(s) + H₂O + 2e⁻ → Hg(l) + 2OH⁻
Overall : Zn(Hg) + HgO(s) → ZnO(s) + Hg(l). Its potential is about 1.35 V and remains constant during its life.

Teacher's Note:
1. A useful battery should be light, compact and give a voltage that does not change much during use.

Question 28. What are primary cells ? How does a dry cell function ?
Answer:
Primary cells are those in which the redox reaction occurs only once and the cell becomes dead after some time; they cannot be recharged. e.g., dry cell, mercury cell.
Dry cell : It consists of a zinc container (anode) and a graphite rod (cathode) surrounded by powdered MnO₂ and carbon. The space between is filled with a moist paste of NH₄Cl and ZnCl₂ (electrolyte).
At anode : Zn(s) → Zn²⁺(aq) + 2e⁻
At cathode : MnO₂(s) + NH₄⁺(aq) + e⁻ → MnO(OH)(s) + NH₃(g)
NH₃ forms the complex [Zn(NH₃)₄]²⁺ with Zn²⁺. The cell gives about 1.5 V.

Teacher's Note:
1. The dry cell has a short shelf life because the acidic NH₄Cl slowly corrodes the zinc container even when not in use.

Question 29. What are secondary cells ? Describe the Nickel-Cadmium cell.
Answer:
Secondary cells are those which can be recharged by passing current through them in the opposite direction, so that they can be used again and again. e.g., lead storage battery, nickel-cadmium cell.
Nickel-cadmium cell : Anode is cadmium, cathode is a metal grid containing Ni(OH)₃ and the electrolyte is KOH solution.
At anode : Cd(s) + 2OH⁻ → CdO(s) + H₂O + 2e⁻
At cathode : 2Ni(OH)₃(s) + 2e⁻ → 2Ni(OH)₂(s) + 2OH⁻
Overall : Cd(s) + 2Ni(OH)₃(s) → CdO(s) + 2Ni(OH)₂(s) + H₂O(l)
It has a longer life than the lead storage cell but is more expensive to manufacture.

Teacher's Note:
1. On charging, the reverse reactions take place and the cell is restored to its original state.

Question 30. Explain discharging and recharging of lead storage battery. (HP Board 2011)
Answer:
A lead storage battery consists of a lead anode and a grid of lead packed with PbO₂ as cathode, dipping in 38% H₂SO₄ solution.
Discharging (when the battery is in use) :
At anode : Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
At cathode : PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
Overall : Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)
H₂SO₄ is used up, so the density of the acid falls.
Recharging : When a current from an external source is passed through the battery, it acts as an electrolytic cell and the above reactions are reversed :
2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq)
PbSO₄ on the anode is converted back to Pb and that on the cathode to PbO₂, and the density of the acid is restored.

Teacher's Note:
1. During recharging the electrode that was the anode becomes the cathode (Pb²⁺ reduced to Pb) and vice versa.

Question 31. What is a fuel cell ? Give the construction and working of a fuel cell.
Answer:
A galvanic cell in which the chemical energy of combustion of fuels like hydrogen, methane, methanol, etc. is directly converted into electrical energy is called a fuel cell.
Construction (H₂–O₂ fuel cell) : It consists of two porous carbon electrodes impregnated with a suitable catalyst (finely divided Pt or Pd). The electrolyte is concentrated aqueous KOH (or NaOH) solution. Hydrogen and oxygen are bubbled through the anode and cathode respectively.
Working :
At anode : 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
At cathode : O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall : 2H₂(g) + O₂(g) → 2H₂O(l)
The cell runs continuously as long as the reactants are supplied.

Teacher's Note:
1. The catalyst on the porous electrodes is needed to make the electrode reactions fast enough.

Question 32. Describe the construction of a H₂–O₂ fuel cell and the reactions taking place in it.
Answer:
It consists of two porous carbon electrodes containing a suitable catalyst (finely divided Pt or Pd), dipping in concentrated aqueous KOH solution. H₂ gas is bubbled through the anode and O₂ gas through the cathode.
At anode : 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
At cathode : O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall : 2H₂(g) + O₂(g) → 2H₂O(l)

Teacher's Note:
1. The only product is water, which makes this cell pollution-free.

Question 33. Leclanche cell, Lead storage cell and Fuel cell are galvanic cells having different uses
(i) Among these Leclanche cell is a primary cell and Lead storage cell is a secondary cell. Write any two differences between primary cells and secondary cells.
(ii) What is a fuel cell?
(iii) Write the overall cell reaction in H₂–O₂ fuel cell. (Kerala Board 2012)
Answer:
(i) 1. In a primary cell, the reaction occurs only once and the cell cannot be recharged, whereas a secondary cell can be recharged by passing current through it and used again.
2. Primary cells become dead after some time (short life), whereas secondary cells can be used for a long time through many charge-discharge cycles.
(ii) A fuel cell is a galvanic cell in which the chemical energy of combustion of a fuel (like H₂, CH₄, CH₃OH) is directly converted into electrical energy.
(iii) 2H₂(g) + O₂(g) → 2H₂O(l)

Teacher's Note:
1. A fuel cell is neither primary nor secondary in the usual sense: it is never recharged, it is continuously fed.

Question 34. Give a brief account of corrosion and its mechanism. (Pb. Board 2011, J & K Board 2012)
Answer:
Corrosion is the slow eating away (deterioration) of the surface of a metal due to the attack of atmospheric gases and moisture, converting it into oxide, carbonate, sulphide, etc. Rusting of iron is the most common example.
Mechanism (electrochemical theory) : A small electrochemical cell is set up on the surface of iron. A spot where iron is oxidised acts as anode, and another spot acts as cathode; water containing dissolved CO₂ (H₂CO₃) acts as electrolyte.
At anode : 2Fe(s) → 2Fe²⁺ + 4e⁻ ; E°_Fe²⁺/Fe = −0.44 V
At cathode : O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) ; E° = 1.23 V
Overall : 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l) ; E°_cell = 1.67 V
The Fe²⁺ ions are further oxidised by atmospheric oxygen to Fe³⁺ ions, which form hydrated ferric oxide (rust) : 4Fe²⁺ + O₂ + 4H₂O → 2Fe₂O₃ + 8H⁺ ; Fe₂O₃ + xH₂O → Fe₂O₃.xH₂O

Teacher's Note:
1. The H⁺ ions consumed at the cathode come from H₂CO₃; this is why both water and air are needed for rusting.

Question 35. What is electrochemical theory of rusting ? Explain. (HP Board 2011)
Answer:
According to the electrochemical theory, rusting takes place through the formation of small electrochemical cells on the surface of iron. A point where iron is oxidised acts as anode and another point acts as cathode; water with dissolved CO₂ and O₂ acts as electrolyte.
At anode : 2Fe(s) → 2Fe²⁺ + 4e⁻
At cathode : O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)
Overall : 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l) ; E°_cell = 1.67 V
Fe²⁺ is further oxidised by air to Fe³⁺, which appears as rust, Fe₂O₃.xH₂O.

Teacher's Note:
1. E°_cell = 1.23 − (−0.44) = 1.67 V; its positive value shows rusting is spontaneous.

Question 36. What is corrosion ? Give mechanism (electrochemical phenomenon) of rusting of iron. What do you understand by sacrificial, cathodic and barrier protection of corrosion ?
Answer:
Corrosion is the slow eating away of the surface of a metal due to the attack of atmospheric gases and moisture.
Mechanism : Small electrochemical cells are set up on the iron surface, with water containing dissolved CO₂ as electrolyte.
At anode : 2Fe(s) → 2Fe²⁺ + 4e⁻
At cathode : O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l)
Overall : 2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l) ; E°_cell = 1.67 V. Fe²⁺ is further oxidised by air to Fe₂O₃.xH₂O (rust).
Sacrificial protection : The iron object is coated with a more electropositive metal (like Zn) which is corroded in preference to iron.
Cathodic protection : The iron object (pipe, tank, ship hull) is connected by a wire to a block of a more active metal like Mg or Zn. The active metal acts as anode and is oxidised, while iron acts as cathode and is protected.
Barrier protection : The iron surface is covered with paint, oil, grease or a less reactive metal so that air and moisture cannot reach it.

Teacher's Note:
1. In both sacrificial and cathodic protection iron is made the cathode; the difference is only coating versus a separate connected block.

Long Answer Questions (5 Marks)

Question 37. (a) State Kohlrausch law of independent migration of ions. (b) What is a primary battery ? Give one example.
(c) Three electrolytic cells A, B and C containing electrolytes ZnSO₄, AgNO₃ and CuSO₄ respectively were connected in series. A steady current of 1.5 A was passed through them. 1.45 g of Ag were deposited at the cathode of cell B. (i) How long did the current flow ? (ii) What mass of copper and zinc were deposited ?
(At mass of Cu = 63 u, Zn = 65.3 u, Ag = 108 u) (Assam Board 2013)
Answer:
(a) The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte : Λ°_m = ν₊λ°₊ + ν₋λ°₋.
(b) A primary battery is one in which the redox reaction occurs only once and the battery cannot be recharged after it is used up. Example : dry cell (or mercury cell).
(c) (i) Ag⁺ + e⁻ → Ag ; 108 g of Ag is deposited by 96500 C.
∴ 1.45 g Ag is deposited by \( \frac{96500}{108}\times 1.45 \) = 1295.6 C
t = \( \frac{Q}{I} \) = \( \frac{1295.6}{1.5} \) = 863.7 s
(ii) By Faraday’s second law, \( \frac{w_{Cu}}{w_{Ag}} = \frac{E_{Cu}}{E_{Ag}} \) ; E_Cu = 63/2 = 31.5, E_Ag = 108, E_Zn = 65.3/2 = 32.65
w_Cu = \( \frac{31.5}{108}\times 1.45 \) = 0.423 g
w_Zn = \( \frac{32.65}{108}\times 1.45 \) = 0.438 g

Teacher's Note:
1. In cells connected in series the same charge passes through each, so masses deposited are in the ratio of equivalent masses (atomic mass/charge on ion).

Question 38. (a) How do you explain with the help of a graph, the increase in the value of molar conductivity with dilution in case of strong and weak electrolyte?
(b) Calculate the e.m.f. of the cell at 298 K in which the following reaction takes place :
Ni(s) + 2Ag⁺(aq) [0.002 M] → Ni²⁺(aq) [0.160 M] + 2Ag(s)
(Given E°_cell = 1.05 V) (Assam Board 2013)

Answer:
(a) (Graph above) For a strong electrolyte (KCl), Λ_m increases slowly with dilution and the plot of Λ_m vs √C is nearly a straight line (Λ_m = Λ°_m − A√C). The strong electrolyte is completely ionised at all concentrations; on dilution the inter-ionic attractions decrease, so the ions move faster.
For a weak electrolyte (CH₃COOH), Λ_m increases very sharply at low concentrations. This is because the degree of dissociation increases on dilution, so the number of ions in solution increases.
(b) n = 2. E_cell = E°_cell − \( \frac{0.0591}{2}\log\frac{[Ni^{2+}]}{[Ag^+]^2} \) = 1.05 − \( \frac{0.0591}{2}\log\frac{0.160}{(0.002)^2} \)
= 1.05 − 0.02955 log (4 × 10⁴) = 1.05 − 0.02955 × 4.602 = 1.05 − 0.136 = 0.914 V

Teacher's Note:
1. Write the cell reaction first to read off n = 2; a common slip is to use n = 1 because one Ag⁺ gains one electron.

Question 39. (a) 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.
(b) Calculate the emf of the 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] (CBSE 2014)
Answer:
(a) Faraday’s first law : The amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte, i.e., w = Z I t.
Cu²⁺ + 2e⁻ → Cu. Hence, reduction of 1 mol of Cu²⁺ requires 2 F.
(b) Cell reaction : Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s), n = 2
E_cell = E°_cell − \( \frac{0.059}{2}\log\frac{[Mg^{2+}]}{[Cu^{2+}]} \) = 2.71 − \( \frac{0.059}{2}\log\frac{0.1}{0.01} \) = 2.71 − 0.0295 = 2.6805 V

Teacher's Note:
1. A tenfold ratio inside the log changes E_cell by just 0.059/n V, a quick way to check such answers.

Teacher's Note:
1. Conductivity and molar conductivity move in opposite directions on dilution; examiners often test exactly this contrast.

Question 41. (a) The e.m.f. of the following cell at 298 K is 0.1745 V
Fe(s)/Fe²⁺(0.1 M)//H⁺(x M)/H₂(g)(1 bar)/Pt(s)
Given E°_Fe²⁺/Fe = −0.44 V. Calculate the H⁺ ion concentration of the solution at the electrode when hydrogen is being produced.
(b) Aqueous solution of copper sulphate and silver nitrate are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different ? Explain your answer. (CBSE Sample Paper 2019)
Answer:
(a) Fe + 2H⁺ → Fe²⁺ + H₂
E_cell = E°_cell − \( \frac{0.0591}{n}\log\frac{[Fe^{2+}]}{[H^+]^2} \)
E°_cell = E°_H⁺/H2 − E°_Fe²⁺/Fe = 0 − (−0.44) = 0.44 V
∴ 0.1745 = 0.44 − \( \frac{0.0591}{2}\log\frac{0.1}{[H^+]^2} \)
On solving it gives log [H⁺] = −5 or [H⁺] = 10⁻⁵ M
(b) The mass of copper and silver deposited at the cathode will be different. This is because when the same quantity of electricity is passed through the electrolytic solutions, the masses deposited are in the ratio of their equivalent masses (Cu = 63.5/2 = 31.75, Ag = 108).

Teacher's Note:
1. When the same charge passes, moles of electrons are equal, so moles of Ag deposited are twice the moles of Cu deposited.

Question 42. (a) Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 × 10⁻⁵ S cm⁻¹.
Given λ°_H⁺ = 349.6 S cm² mol⁻¹, λ°_CH3COO⁻ = 40.9 S cm² mol⁻¹
(b) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases while that of ‘A’ increases to a much larger extent comparatively. Which of the two is a strong electrolyte ? Justify your answer. (CBSE Sample Paper 2019)
Answer:
(a) Λ°(CH₃COOH) = 349.6 + 40.9 = 390.5 S cm² mol⁻¹
Λ_m^c = \( \frac{\kappa\times 1000}{c} = \frac{(8.0\times 10^{-5})(1000)}{0.0024} \) = 33.33 S cm² mol⁻¹
α = \( \frac{\Lambda_m^c}{\Lambda_m^\circ} = \frac{33.33}{390.5} \) = 0.085
(b) B is a strong electrolyte because limiting molar conductivity of a strong electrolyte increases gradually while that of a weak electrolyte increases rapidly with dilution.

Teacher's Note:
1. The question says “limiting” molar conductivity, but what increases on dilution is the molar conductivity Λ_m; Λ°_m itself is a fixed value.