CBSE Class 12 Chemistry Electrochemistry Important Questions Set 03

Very Short Answer Questions (1 Mark)

Question 1. Define specific conductance or conductivity.
Answer:
Specific conductance is defined as conductance of electrolyte when distance between electrodes is 1 cm and area of cross section is 1 cm².

Teacher's Note:
1. Its SI unit is S m⁻¹; in exams S cm⁻¹ is also common (1 S cm⁻¹ = 100 S m⁻¹).

Question 2. What is the effect of temperature on ionic conductance? [CBSE 2019 (56/3/2)]
Answer:
The ionic conductance increases with increase in temperature because the degree of dissociation increases with increase in temperature.

Teacher's Note:
1. Higher temperature also lowers the viscosity of the solvent, so ions move faster; metallic conductance behaves the opposite way.

Question 3. Write the mathematical expression for Kohlrausch’s law.
Answer:
Λ°_m = ν_cΛ°_c + ν_aΛ°_a, where ν_c and ν_a are the number of cations and anions respectively in one formula unit of the compound.

Teacher's Note:
1. For CaCl₂, ν_c = 1 and ν_a = 2 – forgetting the factor 2 is the commonest error.

Question 4. How much charge is required for the reduction of 1 mol of Zn²⁺ to Zn? [CBSE Patna 2015]
Answer:
Zn²⁺ + 2e⁻ → Zn
Quantity of charge required for the reduction of one mole of Zn²⁺ = 2F = 2 × 96500 C = 193000 C

Teacher's Note:
1. Charge needed = (number of electrons in the half reaction) × F per mole of ion.

Question 5. How will the pH of brine (aq. NaCl solution) be affected when it is electrolysed? [NCERT Exemplar] [HOTS]
Answer:
The pH of the solution will increase as NaOH is formed in the electrolytic cell.

Teacher's Note:
1. At the cathode water is reduced to H₂ and OH⁻; the OH⁻ left with Na⁺ makes the solution basic.

Question 6. Value of standard electrode potential for the oxidation of Cl⁻ ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl⁻ oxidised at anode instead of water? [NCERT Exemplar]
Answer:
On electrolysis of aqueous sodium chloride, oxidation of water at anode requires over potential hence Cl⁻ is oxidised instead of water.

Teacher's Note:
1. Overpotential is a kinetic effect: O₂ evolution is slow, so in practice Cl₂ forms even though E° favours O₂.

Question 7. What is primary cell? Give an example. [CBSE Delhi 2008]
Answer:
A primary cell is one in which the redox reaction occurs only once and the cell becomes dead after some time and cannot be used again, e.g., dry cell.

Teacher's Note:
1. Mercury cell is another primary cell; lead storage battery and Ni–Cd cell are secondary (rechargeable).

Question 8. Why does a dry cell become dead after a long time even if it has not been used?
Answer:
Even though not in use, a dry cell becomes dead after some time because the acidic NH₄Cl corrodes the zinc container.

Teacher's Note:
1. NH₄Cl is acidic due to hydrolysis of NH₄⁺, and acid slowly attacks the zinc anode.

Question 9. What is the role of ZnCl₂ in a dry cell?
Answer:
ZnCl₂ combines with the NH₃ produced to form the complex salt [Zn(NH₃)₂Cl₂] otherwise the pressure developed due to NH₃ would crack the seal of the cell.

Teacher's Note:
1. NH₃ is formed at the cathode when NH₄⁺ takes part in the reduction of MnO₂.

Question 10. Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why? [NCERT Exemplar]
Answer:
Ions are not involved in the overall cell reaction of mercury cells.

Teacher's Note:
1. Overall reaction Zn(Hg) + HgO(s) → ZnO(s) + Hg(l) has no ions, so the Nernst quotient stays constant and E stays at about 1.35 V.

Question 11. Write the name of the electrolyte used in (i) fuel cell (ii) mercury cell.
Answer:
(i) Concentrated aqueous KOH solution.
(ii) Moist mercuric oxide (HgO) mixed with KOH.

Teacher's Note:
1. Strictly, in the mercury cell the electrolyte is a paste of KOH and ZnO; HgO mixed with carbon forms the cathode.

Question 12. What is the overall electrochemical reaction taking place in rusting?
Answer:
2Fe(s) + O₂(g) + 4H⁺(aq) → 2Fe²⁺(aq) + 2H₂O(l)

Teacher's Note:
1. Fe²⁺ is then further oxidised by air to form hydrated ferric oxide (rust).

Short Answer Questions (2 Marks)

Question 13. Write the electrode reactions for a H₂–O₂ fuel cell.
Answer:
At anode,
[H₂(g) + 2OH⁻(aq) → 2H₂O(l) + 2e⁻] × 2
At cathode,
O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Net cell reaction,
2H₂(g) + O₂(g) → 2H₂O(l)

Teacher's Note:
1. Multiply the anode reaction by 2 so that the electrons (4e⁻) cancel before adding the half-reactions.

Question 14. At what pH of HCl solution will hydrogen gas electrode show electrode potential of −0.118 V? H₂ gas is passed at 298 K and 1 atm pressure.
Answer:
H⁺ + e⁻ → \( \frac{1}{2} \)H₂
Applying Nernst equation,
\( E_{\text{H}^+/\frac{1}{2}\text{H}_2} = E^\circ_{\text{H}^+/\frac{1}{2}\text{H}_2} - \frac{0.059}{n}\log\frac{1}{[\text{H}^+]} \)
\( -0.118 = 0 - \frac{0.059}{1}\log\frac{1}{[\text{H}^+]} \)
−0.118 = 0.059 log [H⁺]
or   −0.118 = −0.059 pH
pH = 2

Teacher's Note:
1. For a hydrogen electrode at 298 K, E = −0.059 pH; this shortcut lets you check the answer in one line.

Question 15. The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm⁻¹. Calculate its molar conductivity. [CBSE Delhi 2013]
Answer:
\( \Lambda_m = \frac{\kappa\times1000}{C} = \frac{0.025\ \text{S cm}^{-1}\times1000\ \text{cm}^3\ \text{L}^{-1}}{0.2\ \text{mol L}^{-1}} \) = 125 S cm² mol⁻¹

Teacher's Note:
1. The factor 1000 converts litres to cm³; carry units through to see that the answer comes out in S cm² mol⁻¹.

Question 16. 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. [CBSE (AI) 2012]
Answer:
\( \Lambda_m = \frac{\kappa\times1000}{M} \)
\( \kappa = \frac{\Lambda_m\times M}{1000} \)
\( \kappa = \frac{138.9\ \text{S cm}^2\text{mol}^{-1}\times1.5\ \text{mol L}^{-1}}{1000\ \text{cm}^3\ \text{L}^{-1}} \) = 0.20835 S cm⁻¹

Teacher's Note:
1. Rearranging the same formula is all that is needed; a conductivity much larger than 1 S cm⁻¹ would signal a missed factor of 1000.

Question 17. State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution? [CBSE (AI) 2014]
Answer:
Kohlrausch law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of cation and anion of the electrolyte.
In general, if an electrolyte on dissociation gives ν₊ cations and ν₋ anions then its limiting molar conductivity is given by
Λ°_m = ν₊λ°₊ + ν₋λ°₋
where λ°₊ and λ°₋ are the limiting molar conductivities of cations and anions respectively.
Conductivity of a solution decreases with dilution. This is due to the fact that the number of ions per unit volume that carry the current in a solution decreases with dilution.

Teacher's Note:
1. Conductivity decreases but molar conductivity increases on dilution – examiners often test whether you mix these up.

Question 18. Calculate the time to deposit 1.27 g of copper at cathode when a current of 2 A was passed through the solution of CuSO₄.
(Molar mass of Cu = 63.5 g mol⁻¹, 1 F = 96500 C mol⁻¹) [CBSE Allahabad 2015]
Answer:
Cu²⁺ + 2e⁻ → Cu
63.5 g of copper is deposited by 2 × 96500 C.
∴ 1.27 g of copper will be deposited by \( \frac{2\times96500\times1.27}{63.5} \) C = 3860 C
I = 2 A, Q = 3860 C
∴ \( t = \frac{Q}{I} = \frac{3860}{2} \) = 1930 s

Teacher's Note:
1. Cu²⁺ needs 2 F per mole; using 1 F would halve the answer.

Question 19. 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? [CBSE (AI) 2011]
Answer:
Mass of mercury produced at the cathode,
\( m = Z\times I\times t = \frac{M\times I\times t}{n\times F} \)
\( m = \frac{M\ \text{g mol}^{-1}\times2\ \text{A}\times3\times60\times60\ \text{s}}{2\times96500\ \text{C mol}^{-1}} \) = 0.1119 × M g ≈ 0.112 × M g
No. of moles of mercury produced = \( \frac{0.112\times M\ \text{g}}{M\ \text{g mol}^{-1}} \) = 0.112 mol

Teacher's Note:
1. Moles of metal = Q / (nF); the molar mass cancels, so it is not needed for this question.

Question 20. When a current of 0.75 A is passed through a CuSO₄ solution for 25 min, 0.369 g of copper is deposited at the cathode. Calculate the atomic mass of copper.
Answer:
W = Z × I × t
\( 0.369 = \frac{M}{2\times96500}\times0.75\times25\times60 \)   (M = molar mass of copper)
M = 63.3 g/mol

Teacher's Note:
1. Always convert minutes to seconds before using Q = It.

Question 21. How many molecules of chlorine should be deposited from molten sodium chloride in one minute by a current of 300 milliampere?
Answer:
Q = I × t
Here, I = \( \frac{300}{1000} \) = 0.3 A; t = 60 s; Q = 0.3 A × 60 s = 18 C
2Cl⁻ → Cl₂ + 2e⁻
2 × 96500 C deposit Cl₂ = 1 mol
∴ 18 C will deposit Cl₂ = \( \frac{1\ \text{mol}\times18\ \text{C}\times6.022\times10^{23}}{2\times96500\ \text{C}} \) molecules
= 5.616 × 10¹⁹ molecules

Teacher's Note:
1. One Cl₂ molecule needs 2 electrons, so 2F (not 1F) liberates one mole of Cl₂.

Question 22. Chromium metal is electroplated using an acidic solution containing CrO₃ according to the following equation:
CrO₃(aq) + 6H⁺ + 6e⁻ → Cr(s) + 3H₂O
Calculate how many grams of chromium will be electroplated by 24,000 coulombs. How long will it take to electroplate 1.5 g chromium using 12.5 A current?
[Atomic mass of Cr = 52 g mol⁻¹, 1 F = 96500 C mol⁻¹]
Answer:
6 × 96500 C deposit Cr = 52 g
∴ 24,000 C will deposit Cr = \( \frac{52\ \text{g}}{6\times96500\ \text{C}} \) × 24,000 C = 2.155 g
52 g of Cr is deposited by 6 × 96500 C.
∴ 1.5 g of Cr require = \( \frac{6\times96500\ \text{C}}{52\ \text{g}} \) × 1.5 g = 16701.9 C
Q = I × t ⇒ \( t = \frac{Q}{I} \)
∴ Time for which the current is passed = \( \frac{16701.9\ \text{C}}{12.5\ \text{A}} \) = 1336.15 s

Teacher's Note:
1. Cr in CrO₃ is in the +6 state, so 6 F are needed per mole of Cr – read n from the half equation.

Question 23. Following reactions may occur at cathode and anode during electrolysis of aqueous sodium chloride. What products will be held at anode and cathode? Use given E° values to justify your answer.
Cathode: Na⁺(aq) + e⁻ → Na(s)   E° = −2.71 V
H⁺(aq) + e⁻ → \( \frac{1}{2} \)H₂(g)   E° = 0.00 V
Anode: Cl⁻(aq) → \( \frac{1}{2} \)Cl₂(g) + e⁻   E° = +1.36 V
2H₂O(aq) → O₂(g) + 4H⁺ + 4e⁻   E° = +1.23 V [CBSE 2019 (56/5/2)]
Answer:
H₂(g) is produced due to greater E° value of H⁺ ion and Cl₂(g) is produced due to over-potential of oxygen.

Teacher's Note:
1. Cathode: higher E° wins. Anode: lower E° should win, but overpotential of O₂ reverses the choice.

Question 24. Define the following terms:
(i) Molar conductivity (Λ_m)   (ii) Secondary batteries [CBSE (AI) 2014]
Answer:
(i) Molar conductivity, Λ_m of a solution at a dilution V is defined as the conductance of all the ions produced from one gram mole of the electrolyte dissolved in V cm³ of the solution when the electrodes are one centimetre apart and the area of the electrodes is so large that the whole of the solution is contained between them.
Λ_m = κ × V   or   \( \Lambda_m = \frac{\kappa\times1000}{\text{Molarity}} \)
where, κ is the conductivity and V is the volume of the solution containing 1 mole of the electrolyte.
(ii) Secondary batteries are those batteries which can be recharged by passing electric current through them and hence can be used over again e.g., lead storage battery.

Teacher's Note:
1. Give the formula along with the definition of Λ_m; it often earns the full mark.

Question 25. Write electrode reactions taking place in (i) Ni–Cd cell, (ii) Lead Acid Accumulator.
Answer:
(i) Cd(s) + 2OH⁻(aq) → Cd(OH)₂(s) + 2e⁻ (at anode)
NiO₂(s) + 2H₂O + 2e⁻ → Ni(OH)₂(s) + 2OH⁻(aq) (at cathode)
(ii) Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻ (at anode)
PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l) (at cathode)

Teacher's Note:
1. In the lead accumulator PbSO₄ forms at both electrodes during discharge, so the H₂SO₄ concentration falls.

Short Answer Questions (3 Marks)

Question 26. (a) 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 Gibb’s energy of the cell reaction. (Given: 1 F = 96,500 C mol⁻¹)
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1 F = 96,500 C mol⁻¹)
Answer:
(a) ΔG° = −nFE°_cell
n = 2
ΔG° = −2 × 96500 C/mol × 0.236 V
= −45548 J/mol = −45.548 kJ/mol
(b) Q = It = 0.5 × 2 × 60 × 60 = 3600 C
96500 C = 6.023 × 10²³ electrons
3600 C = 2.25 × 10²² electrons
Detailed Answer:
(a) 2Fe³⁺ + 2e⁻ → 2Fe²⁺
2I⁻ → I₂ + 2e⁻
For the given cell reaction, n = 2.
ΔG° = −nFE°_cell = −2 × 96500 × 0.236 = −45548 J mol⁻¹ = −45.55 kJ mol⁻¹
(b) I = 0.5 A
t = 2 hours = 2 × 60 × 60 s = 7200 s
Q = It = 0.5 × 7200 = 3600 coulombs
A flow of 96500 C is equal to flow of 1 mole of electrons i.e., 6.023 × 10²³ electrons.
∴ 3600 C is equivalent to \( \frac{6.023\times10^{23}}{96500} \) × 3600 electrons
= 2.246 × 10²² electrons

Teacher's Note:
1. Convert hours to seconds before using Q = It; leaving t in hours is the most common error in such questions.

Question 27. Calculate Δ_rG° and log K_c for the following reaction at 298 K.
2Cr(s) + 3Fe²⁺(aq) → 2Cr³⁺(aq) + 3Fe(s)
[(E°_cell = 0.30 V), 1F = 96500 C mol⁻¹] [CBSE Comptt. OD Set-2 2017]
Answer:
Δ_rG° = −nFE°_cell, n = 6
= −6 × 96500 C/mol × 0.30 V
= −173700 J/mol = −173.7 kJ/mol
E°_cell = 0.059 V/n × log K_c
log K_c = 0.30 V × 6/0.059 V = 30.5

Teacher's Note:
1. A positive E°_cell gives a negative ΔG° and a large K_c – all three tell you the reaction is spontaneous.

Question 28. Calculate e.m.f. of the following cell at 298 K:
2Cr(s) + 3Fe²⁺ (0.1M) → 2Cr³⁺ (0.01M) + 3Fe(s)
E°(Cr³⁺/Cr) = −0.74 V
E°(Fe²⁺/Fe) = −0.44 V. [CBSE Delhi Set-1, 2 & 3, 2016]
Answer:
E°_cell = E°_cathode − E°_anode
= (−0.44) − (−0.74) V = 0.30 V
E_cell = E°_cell − \( \frac{0.0591}{n} \)\( \log\frac{[\text{Products}]}{[\text{Reactants}]} \)
= 0.30 − \( \frac{0.0591}{6} \)\( \log\frac{[\text{Cr}^{3+}]^{2}}{[\text{Fe}^{2+}]^{3}} \)
= 0.30 − \( \frac{0.0591}{6} \)\( \log\frac{[0.01]^{2}}{[0.1]^{3}} \)
= 0.30 − \( \left(-\frac{0.0591}{6}\right) \)
= 0.3098 V

Teacher's Note:
1. Here the log term is negative (10⁻⁴/10⁻³ = 10⁻¹), so E_cell comes out slightly larger than E°_cell.

Question 29. Calculate the emf of the following cell at 25° C:
Fe | Fe²⁺ (0.001 M) || H⁺ (0.01 M) | H₂(g) (1bar) | Pt(s)
E° (Fe²⁺/Fe) = −0.44 V   E° (H⁺/H₂) = 0.00 V [CBSE Delhi 2015]
Answer:
Cell reaction is
Fe(s) + 2H⁺(aq) → Fe²⁺ (aq) + H₂(g)
E°_cell = 0.00 − (−0.44) = 0.44 V
E_cell = E°_cell − \( \frac{0.0591}{2} \)\( \log\frac{[\text{Fe}^{2+}]}{[\text{H}^{+}]^{2}} \)
= 0.44 V − \( \frac{0.0591}{2} \) V \( \log\frac{0.001}{(0.01)^{2}} \)
= 0.44 V − 0.02955 V
E_cell = 0.41045 V

Teacher's Note:
1. Gases at 1 bar and pure solids do not appear in the Nernst log term; only the ion concentrations do.

Question 30. A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO₃)₂ solution and metallic plate of lead in 0.02 M Pb(NO₃)₂ solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell.
(Given: E°_Zn²⁺/Zn = −0.76 V; E°_Pb²⁺/Pb = −0.13 V)
Answer:
Anode reaction: Zn(s) → Zn²⁺(aq) + 2e⁻
Cathode reaction: Pb²⁺(aq) + 2e⁻ → Pb(s)
Cell representation:
Zn(s)|Zn²⁺(aq)||Pb²⁺(aq)|Pb(s)
According to Nernst equation:
E_cell = E°_cell − \( \frac{0.0591}{2} \)\( \log\frac{[\text{Zn}^{2+}]}{[\text{Pb}^{2+}]} \)
E_cell = [−0.13 − (−0.76)] − \( \frac{0.0591}{2} \)\( \log\frac{0.1}{0.02} \)
= 0.63 − 0.02955 × log 5
= 0.63 − 0.02955 × 0.6990
= 0.63 − 0.0206 = 0.6094 V

Teacher's Note:
1. The metal with the more negative E° (Zn) is the anode and is written on the left of the cell representation.

Question 31. When a steady current of 2A was passed through two electrolytic cells A and B containing electrolytes ZnSO₄ and CuSO₄ connected in series, 2 g of Cu were deposited at the cathode of cell B. How long did the current flow? What mass of Zn was deposited at cathode of cell A?
[Atomic mass: Cu = 63.5 g mol⁻¹, Zn = 65 g mol⁻¹; 1F = 96500 C mol⁻¹]
Answer:
Zn²⁺(aq) + 2e⁻ → Zn(s)   (2 mol electrons : 1 mol Zn)
Cu²⁺(aq) + 2e⁻ → Cu(s)   (2 mol electrons : 1 mol Cu)   (2 gm given)
The charge Q on a mole of electrons, Q = nF
Calculation of time for the flow of current:
n = 2 mol
Q = 2 × 96500 C mol⁻¹ = 193000 C
Molar mass of Cu = 63.5 gm mol⁻¹
∵ 63.5 gm of Cu is deposited by electric charge = 193000 C
∴ 2 gm of Cu is deposited by electric charge = \( \frac{193000}{63.5} \) × 2 = 6078.74 C
Let 2 A of current be passed for time t, quantity of electricity used = 2A × t = 6078.74 C
or, t = \( \frac{6078.74\ \text{C}}{2} \) = 3039.37 s
= 50 min 39 s
Calculation of mass of Zn deposited:
\( \frac{W_1}{W_2} \) = \( \frac{E_1}{E_2} \) = \( \frac{\text{Equivalent weight of Zn}}{\text{Equivalent weight of Cu}} \)
= \( \frac{\text{Molar mass of Zn / Charge on Zn}}{\text{Molar mass of Cu / Charge on Cu}} \)
Amount of Zn deposited:
= 2 × \( \frac{65/2}{63.5/2} \) = 2.0472 g

Teacher's Note:
1. Cu²⁺ needs 2 F per mole, so 63.5 g of Cu requires 2 × 96500 C, not 96500 C – always take n from the electrode reaction.

Question 32. (a) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO₃ for 15 minutes.
(Given: Molar mass of Ag = 108 g mol⁻¹, 1 F = 96500 C mol⁻¹)
(b) Define fuel cell.
Answer:
(a) m = ZIt
= \( \frac{108\times2\times15\times60}{1\times96500} \)
= 2.01 g (or any other correct method)
(b) Cells that convert the energy of combustion of fuels directly into electrical energy.
Detailed Answer:
(a) t = 900 s
Charge = Current × Time = 2 × 900 = 1800 C
According to the reaction
Ag⁺ (aq) + e⁻ → Ag(s)
We require 1 F to deposit 1 mol or 108 g of Ag
For 1800 C, the mass of Ag deposited will be = \( \frac{108\times1800}{1\times96500} \) = 2.0145 g
(b) Fuel cell is the name given to the galvanic cells which are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy.

Teacher's Note:
1. Electrochemical equivalent Z = M/(nF); for Ag⁺ n = 1, but for Cu²⁺ or Zn²⁺ remember to divide by 2.

Question 33. The electrolysis of a metal salt solution was carried out by passing a current of 4 A for 45 minutes. It resulted in deposition of 2.977 g of a metal. If atomic mass of the metal is 106.4 g mol⁻¹, calculate the charge on the metal cation.
Answer:
Let the charge on the metal ion = n⁺
Reduction half-reaction,
Mⁿ⁺ + ne⁻ → M
(1 mol)   (n mol)   (106.4 g)
Quantity of electricity required for depositing 106.4 g of metal = n × 96500 C
Quantity of electricity required for depositing 2.977 g of metal = \( \frac{n\times96500\times2.977}{106.4} \) = n × 2700
Quantity of electricity passed = 4 × 45 × 60 C = 10800 C
Applying law of conservation of charge,
10800 = n × 2700
n = \( \frac{10800}{2700} \) = 4
Charge on metal ion = +4

Teacher's Note:
1. The charge on the ion must come out as a whole number – if you get a fraction, recheck the time conversion (minutes to seconds).

Question 34. One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. The other half-cell consists of a zinc electrode in a 0.10 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 ions in the solution.
(Given: E°_Zn²⁺/Zn = −0.763 V, E°_Ag⁺/Ag = +0.80 V) [CBSE (F) 2010]
Answer:
Electrochemical cell
Zn(s) | Zn²⁺ (0.10 M) || Ag⁺ (conc.) | Ag(s)
E°_cell = E°_R − E°_L = E°_Ag⁺/Ag − E°_Zn²⁺/Zn
= 0.80 V − (−0.763) V = 1.563 V
We know that,
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log\frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} \)
\( 1.48 = 1.563 - \frac{0.0591}{2}\log\frac{[0.10]}{[\text{Ag}^+]^2} \)
\( \log\frac{[0.10]}{[\text{Ag}^+]^2} = \frac{0.083}{0.02955} = 2.8087 \)
Or \( \frac{[0.10]}{[\text{Ag}^+]^2} \) = antilog 2.8087 = 643.7
\( [\text{Ag}^+]^2 = \frac{0.10}{643.7} = 1.553\times10^{-4} \)
[Ag⁺] = 1.247 × 10⁻² M

Teacher's Note:
1. Remember to square [Ag⁺] (from 2Ag⁺ in the balanced equation) and to take the square root at the end.

Question 35. 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
\( E_{\text{Zn}^{2+}/\text{Zn}} = E^\circ_{\text{Zn}^{2+}/\text{Zn}} - \frac{0.0591}{2}\log\frac{1}{[\text{Zn}^{2+}]} \)
As 0.1 M ZnSO₄ solution is 95% dissociated, this means that in the solution,
\( [\text{Zn}^{2+}] = \frac{95}{100}\times0.1\ \text{M} = 0.095\ \text{M} \)
\( E_{\text{Zn}^{2+}/\text{Zn}} = -0.76 - \frac{0.0591}{2}\log\frac{1}{0.095} \)
= −0.76 − 0.02955 (log 1000 − log 95)
= −0.76 − 0.02955 (3 − 1.9777) = −0.76 − 0.03021 = −0.79021 V

Teacher's Note:
1. Use the actual ion concentration (α × C), not the salt concentration, in the Nernst equation.

Long Answer Questions (5 Marks)

Question 36. (a) Calculate E_cell for the following reaction at 298K:
2Al(s) + 3Cu²⁺ (0.01M) → 2Al³⁺ (0.01M) + 3Cu(s)
Given: E°_cell = 1.98 V
(b) Using the E° values of A and B, predict which is better for coating the surface of iron [E°_(Fe²⁺/Fe) = −0.44V] to prevent corrosion and why?
Given: E°_(A²⁺/A) = −2.37V: E°_(B²⁺/B) = −0.14V [CBSE Outside Delhi Set-1, 2 & 3, 2018]
Answer:
(a) E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[\text{Al}^{3+}]^2}{[\text{Cu}^{2+}]^3} \)
E_cell = 1.98 V − \( \frac{0.0591}{6} \) log \( \frac{(0.01)^2}{(0.01)^3} \)
E_cell = 1.98 V − \( \frac{0.0591}{6} \) log 10²
E_cell = 1.98 V − \( \frac{0.0591}{6} \) × 2 log 10   [∵ log 10 = 1]
E_cell = 1.98 V − \( \frac{0.0591\text{ V}}{6} \) × 2
E_cell = 1.98 V − 0.0197 V
E_cell = 1.9603 V
(b) A, because its E° value is more negative.

Teacher's Note:
1. When the reaction quotient Q is greater than 1, E_cell comes out smaller than E°_cell; use this as a quick sign check.
2. A metal with a more negative E° than iron corrodes in preference to iron, so it protects iron as a sacrificial coating.

Question 37. (a) The conductivity of 0.001 mol L⁻¹ solution of CH₃COOH is 3.905 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity and degree of dissociation (α).
Given λ°(H⁺) = 349.6 S cm² mol⁻¹ and λ°(CH₃COO⁻) = 40.9 S cm² mol⁻¹
(b) Define electrochemical cell. What happens if external potential applied becomes greater than E°_cell of electrochemical cell? [CBSE Outside Delhi Set-1, 2 & 3, 2016]
Answer:
(a) Λ_m = κ × 1000/C
= \( \frac{3.905\times10^{-5}\times1000}{0.001} \)
= 39.05 S cm²/mol
CH₃COOH → CH₃COO⁻ + H⁺
Λ°_CH3COOH = λ°_CH3COO⁻ + λ°_H⁺
= 40.9 + 349.6
Λ°_CH3COOH = 390.5 S cm²/mol
α = \( \frac{\Lambda_m}{\Lambda_m^0} \) = \( \frac{39.05}{390.5} \) = 0.1
(b) Device used for the production of electricity from energy released during spontaneous chemical reaction and the use of electrical energy to bring about a chemical change.
The reaction gets reversed / It starts acting as an electrolytic cell & vice – versa.

Teacher's Note:
1. Check units before dividing: κ in S cm⁻¹ × 1000 / C in mol L⁻¹ gives Λ_m in S cm² mol⁻¹.

Question 38. (a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO₄ and ZnSO₄ until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y.
(Molar mass: Fe = 56 g mol⁻¹, Zn = 65.3 g mol⁻¹, 1 F = 96500 C mol⁻¹)
(b) In the plot of molar conductivity Λ_m vs. square root of concentration (C^½), following curve obtained for two electrolytes A and B:
Answer the following:
(i) Predict the nature of electrolytes A and B:
(ii) What happens on extrapolation of Λ_m to concentration approaching zero for electrolytes A and B? [CBSE Delhi Set 1, 2019]

Answer:
(a) m = Z I t
2.8 = \( \frac{56\times2\times t}{2\times96500} \)
t = 4825 s / 80.417 min
\( \frac{m_1}{m_2} \) = \( \frac{E_1}{E_2} \)
\( \frac{2.8}{m_{Zn}} \) = \( \frac{56}{2} \) × \( \frac{2}{65.3} \)
m_Zn = 3.265 g
(b) (i) A- Strong electrolyte, B-weak electrolyte
(ii) Λ°_m for weak electrolytes cannot be obtained by extrapolation while Λ°_m for strong electrolytes can be obtained as intercept.
Detailed Answer:
(a) Charge required to deposit 2.8 g of Fe
moles of Fe = \( \frac{\text{Given mass}}{\text{molar mass}} \) = \( \frac{2.8\text{ g}}{56\text{ g mol}^{-1}} \) = 0.05 mol
2 F charge is required to discharge 1 mol of Fe²⁺ ions as Fe, therefore deposition of 0.05 mol Fe will need
0.05 × 2 = 0.1 F = 0.1 F × \( \frac{96500\text{ C}}{\text{F}} \) = 9650 C
The quantity of charge is related to current as Q = It
Therefore, the time needed to deposit 2.8 g Fe is:
t = \( \frac{Q}{I} \) = \( \frac{9650\text{ C}}{2\text{ A}} \) = 4825 s
So, the current flowed through the cells for 4825 seconds.
The amount of Zn deposited in cell Y can be calculated using Faraday’s second law:
\( \frac{\text{mass of Zn}}{\text{mass of Fe}} \) = \( \frac{\text{Eq. wt of Zn}}{\text{Eq. wt of Fe}} \) = \( \frac{\text{molar mass of Zn / charge on zinc ion}}{\text{molar mass of Fe / charge on iron ion}} \)
mass of Zn = 2.8 g × \( \frac{65.3\text{ g}/2}{56\text{ g}/2} \) = 3.265 g ≈ 3.3 g
Therefore, the mass of Zn deposited in cell Y in the same time is 3.3 g.
(b) (i) Molar conductivity of strong electrolytes increases linearly as the square root of the concentration decreases; therefore, electrolyte A is a strong electrolyte. Molar conductivity of weak electrolytes increases non-linearly as square root of concentration decreases; therefore, electrolyte B is a weak electrolyte.
(ii) As concentration of strong electrolyte approaches zero, the molar conductivity of the plot intercepts the molar conductivity axis, giving the limiting value of molar conductivity Λ°_m. The plot of molar conductivity of weak electrolyte tends to infinity as its concentration approaches zero; it does not intersect the molar conductivity axis.

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

Question 39. (a) The conductivity of 0.001 mol L⁻¹ acetic acid is 4.95 × 10⁻⁵ S cm⁻¹. Calculate the dissociation constant if Λ°_m for acetic acid is 390.5 S cm² mol⁻¹.
(b) Write Nernst equation for the reaction at 25°C:
2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s)
(c) What are secondary batteries? Give an example.
Answer:
(a) Λ_m = \( \frac{\kappa}{C} \) = \( \frac{4.95\times10^{-5}\text{ S cm}^{-1}}{0.001\text{ mol L}^{-1}} \) × \( \frac{1000\text{ cm}^3}{\text{L}} \)
= 49.5 S cm²mol⁻¹
α = \( \frac{\Lambda_m}{\Lambda_m^0} \) = \( \frac{49.5\text{ S cm}^2\text{mol}^{-1}}{390.5\text{ S cm}^2\text{mol}^{-1}} \) = 0.127
K = \( \frac{C\alpha^2}{(1-\alpha)} \) = \( \frac{0.001\text{ mol L}^{-1}\times(0.127)^2}{1-0.127} \)
= 1.8 × 10⁻⁵ mol L⁻¹
(If K = Cα², then K = 1.6 × 10⁻⁵ mol L⁻¹)
(b) E_cell = E°_cell − \( \frac{0.059}{6} \) log \( \frac{[\text{Al}^{3+}]^2}{[\text{Cu}^{2+}]^3} \)
(c) Batteries which are rechargeable
Example- Lead storage, Ni-Cd batteries (Or any other one example).

Teacher's Note:
1. For a weak acid with small α, K ≈ Cα² is an acceptable shortcut, but the full form Cα²/(1 − α) is more accurate when α is above about 5%.

Question 40. (a) For the reaction
2AgCl (s) + H₂ (g) (1 atm) → 2Ag(s) + 2H⁺ (0.1 M) + 2Cl⁻(0.1 M),
ΔG° = − 43600 J at 25°C.
Calculate the e.m.f. of the cell.
[log 10⁻ⁿ = −n]
(b) Define fuel cell and write its two advantages.
Answer:
(a) ΔG° = − nFE°
−43600 = − 2 × 96500 × E°
E° = 0.226 V
E = E° − 0.059/2 log ([H⁺]² [Cl⁻]² / [H₂])
= 0.226 − 0.059/2 log[ (0.1)² × (0.1)² ] / 1
= 0.226 − 0.059 /2 log 10⁻⁴
= 0.226 + 0.118 = 0.344 V
(b) Cells that convert the energy of combustion of fuels (like hydrogen, methane, methanol etc.) directly into electrical energy are called fuel cells.
Advantages: High efficiency, non polluting (or any other suitable advantage)
Detailed Answer
(a) ΔG° = − 43600 J
No. of electrons (n) = 2
F = 96500 C
∵ ΔG° = − nFE°
− 43600 = − 2 × 96500 × E°
∴ E° = \( \frac{-43600}{-2\times96500} \) = 0.226 V
For the reaction,
2AgCl(s) + H₂(g) (1 atm) → 2Ag(s) + 2H⁺(0.1M) + 2Cl⁻(0.1M)
From Nernst equation
E_cell = E°_cell − \( \frac{0.059}{2} \) log \( \frac{[\text{H}^+]^2[\text{Cl}^-]^2}{[\text{H}_2]} \)
(∵ Concentration of solids are taken as unity)
= 0.226 − \( \frac{0.059}{2} \) log \( \frac{(0.1)^2(0.1)^2}{1} \)
= 0.226 − \( \frac{0.059}{2} \) log 10⁻⁴
= 0.226 + 0.0295 log 10⁴
= 0.226 + 0.118
= 0.344 V
(b) Fuel cell: The cell which converts chemical energy of a fuel directly into electrical energy is called fuel cell.
Advantages of fuel cell:
(i) Pollution free working
(ii) High efficiency

Teacher's Note:
1. A negative ΔG° always gives a positive E°; if your sign comes out negative, recheck ΔG° = −nFE°.
2. Gases at 1 atm and solids enter the Nernst quotient as 1, so only [H⁺] and [Cl⁻] matter here.

Question 41. The electrochemical cell given alongside converts the chemical energy released during the redox reaction
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
to electrical energy. It gives an electrical potential of 1.1 V when concentration Zn²⁺ and Cu²⁺ ions is unity. State the direction of flow of current and also specify whether zinc and copper are deposited or dissolved at their respective electrodes when:
(i) an external opposite potential of less than 1.1 V is applied.
(ii) an external potential of 1.1 V is applied.
(iii) an external potential of greater than 1.1 V is applied. [HOTS]

Answer:
(i) Reaction continues to take place.
Electrons flow from Zn electrode to copper electrode, hence current flows from Cu to Zn. Zn dissolves and copper deposits at their respective electrodes.
(ii) The reaction stops and no current flows.
A state of equilibrium is achieved and no change is observed at zinc and copper electrodes.
(iii) Reaction takes place in opposite directions.
Electrons flow from copper electrode to zinc electrode and hence current flows from Zn to Cu. Zinc deposits and copper dissolves at their respective electrodes. The cell functions as an electrolytic cell.

Teacher's Note:
1. Current always flows opposite to the electron flow; compare E_ext with 1.1 V to decide which way electrons are pushed.