CBSE Class 12 Chemistry Electrochemistry NEET JEE Questions Set 01

Single Correct Answer

Question 1. What current is to be passed for 0.25 s for deposition of a certain weight of metal which is equal to its electrochemical equivalent?
(a) 4 A
(b) 100 A
(c) 200 A
(d) 2 A
Answer:
Ans. (a)
Electrochemical equivalent is the weight deposited by 1 coulomb.
Q = I × t \( \therefore \) 1 = I × 0.25 or I = 4 A.

Teacher's Note:
1. The electrochemical equivalent is the mass deposited by 1 coulomb, so the charge passed here must be exactly 1 C.

Question 2. In an experiment 0.04 F was passed through 400 ml of 1 M solution of NaCl. What would be the pH of the solution after the electrolysis?
(a) 8
(b) 10
(c) 13
(d) 6
(e) 9
Answer:
Ans. (c)
NaCl + H₂O →(Electrolysis) NaOH + \( \frac{1}{2} \)H₂ (at cathode) + \( \frac{1}{2} \)Cl₂ (at anode)
400 ml of 1 M NaCl solution contains NaCl \( = \frac{1}{1000}\times400 \) mol = 0.4 mol
As Na⁺ + e⁻ → Na →(H₂O) NaOH + \( \frac{1}{2} \)H₂, 1 F produces 1 mol of NaOH
\( \therefore \) 0.04 F produces NaOH = 0.04 mol. Thus, 400 ml of the solution now contain 0.04 mol of NaOH
\( \therefore \) Molar concentration of NaOH sol. \( = \frac{0.04}{400}\times1000 \) = 0.1 M
[OH⁻] = 0.1 M, [H⁺] = 10⁻¹³ M \( \therefore \) pH = 13.

Teacher's Note:
1. In brine electrolysis each faraday makes one mole of OH⁻; divide by the volume in litres before taking pOH.

Question 3. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H₂ gas at the cathode is (1 Faraday = 96500 C mol⁻¹) (IIT 2008)
(a) 9.65 × 10⁴ sec
(b) 19.3 × 10⁴ sec
(c) 28.95 × 10⁴ sec
(d) 38.6 × 10⁴ sec
Answer:
Ans. (b)
NaCl + aq → Na⁺ + Cl⁻, H₂O ⇄ H⁺ + OH⁻; H⁺ + e⁻ → \( \frac{1}{2} \)H₂
Thus, 0.5 mole of H₂ is liberated by 1 F = 96500 C
\( \therefore \) 0.01 mole of H₂ will be liberated by charge \( = \frac{96500}{0.5}\times0.01 \) = 1930 C
Q = I × t or \( t = \frac{Q}{I} = \frac{1930\,\mathrm{C}}{10\times10^{-3}\,\mathrm{A}} \) = 193000 sec = 19.3 × 10⁴ sec

Teacher's Note:
1. Convert milliampere to ampere (10 mA = 0.01 A) before dividing; forgetting this is the usual slip.

Question 4. One faraday of electricity is passed through molten Al₂O₃, aqueous solution of CuSO₄ and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al, Cu and Na deposited at the respective cathodes is (Kerala PET 2010)
(a) 2 : 3 : 6
(b) 6 : 2 : 3
(c) 6 : 3 : 2
(d) 1 : 2 : 3
(e) 3 : 6 : 2
Answer:
Ans. (a)
Al³⁺ + 3e⁻ → Al; Cu²⁺ + 2e⁻ → Cu; Na⁺ + e⁻ → Na
Thus, 1 F will deposit \( \frac{1}{3} \) mole of Al, \( \frac{1}{2} \) mole of Cu and 1 mole of Na. Hence, molar ratio \( = \frac{1}{3} : \frac{1}{2} : 1 \) = 2 : 3 : 6.

Teacher's Note:
1. In series the same charge flows through every cell, so moles deposited are inversely proportional to the charge on the ion.

Question 5. 9.65 C electric current is passed through fused anhydrous magnesium chloride. The magnesium metal thus obtained is completely converted into a Grignard reagent. The number of moles of the Grignard reagent obtained is (Karnataka CET 2010)
(a) 5 × 10⁻⁴
(b) 1 × 10⁻⁴
(c) 5 × 10⁻⁵
(d) 1 × 10⁻⁵
Answer:
Ans. (c)
Mg²⁺ + 2e⁻ → Mg
2 F, i.e., 2 × 96500 C deposit Mg = 1 mole
\( \therefore \) 9.65 C will deposit Mg \( = \frac{1}{2\times96500}\times9.65 \) = 5 × 10⁻⁵ mole
RX + Mg → RMgX
1 mole of Mg forms G.R. = 1 mole
\( \therefore \) 5 × 10⁻⁵ mole of Mg will form G.R. = 5 × 10⁻⁵ mole.

Teacher's Note:
1. Each Mg atom gives one RMgX, so moles of Grignard reagent equal moles of Mg, which is charge/(2F).

Question 6. A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At wt. = 177). The oxidation state of the metal in the metal salt is
(a) +1
(b) +2
(c) +3
(d) +4
Answer:
Ans. (c)
Q = 2 × 5 × 60 × 60 = 36000 C
Thus, 36000 C deposit metal = 22.2 g
\( \therefore \) 96500 C will deposit metal \( = \frac{22.2}{36000}\times96500 \) g = 59.5 g
Thus, Eq. wt. of the metal = 59.5
Valency or ox. state = At. wt./Eq. wt. = 177/59.5 = 3.

Teacher's Note:
1. The mass deposited by one faraday is the equivalent weight; atomic mass divided by it gives the charge on the ion.

Question 7. Same quantity of electricity was passed through solutions of salts of elements A, B and C with atomic weights 7, 27 and 48 respectively. The masses of A, B and C deposited were 2.1 g, 2.7 g and 7.2 g respectively. The valencies of A, B and C respectively are
(a) 3, 2 and 1
(b) 1, 2 and 3
(c) 1, 3 and 2
(d) 2, 3 and 2
Answer:
Ans. (c)
Suppose valencies of A, B and C are a, b and c respectively. Then their equivalent weights will be (At. wt/valency) 7/a, 27/b and 48/c. When same quantity of electricity is passed, masses deposited are in the ratio of their equivalent weights, i.e.,
\( \frac{\text{Mass of A}}{\text{Eq. wt of A}} = \frac{\text{Mass of B}}{\text{Eq. wt of B}} = \frac{\text{Mass of C}}{\text{Eq. wt of C}} \)
\( \therefore \frac{2.1}{7/a} = \frac{2.7}{27/b} = \frac{7.2}{48/c} \) or 0.3 a = 0.1 b = 0.15 c
or 3a = b = 1.5c. Thus, if b = 1, then a = \( \frac{1}{3} \), c = \( \frac{1}{1.5} = \frac{2}{3} \)
\( \therefore \) Ratio a : b : c = \( \frac{1}{3} : 1 : \frac{2}{3} \) = 1 : 3 : 2.

Teacher's Note:
1. With equal charge, equivalents deposited are equal, so mass × valency/atomic mass is the same for every element.

Question 8. Al₂O₃ is reduced by electrolysis at low potentials and high currents. If 4.0 × 10⁴ amperes of current is passed through molten Al₂O₃ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. At. mass of Al = 27 g mol⁻¹)
(a) 8.1 × 10⁴ g
(b) 2.4 × 10⁵ g
(c) 1.3 × 10⁴ g
(d) 9.0 × 10³ g
Answer:
Ans. (a)
Al³⁺ + 3e⁻ → Al
Quantity of electricity passed = (4.0 × 10⁴ A) × (6 × 60 × 60 s) = 864 × 10⁶ coulombs
3 F, i.e., 3 × 96500 C produce Al = 1 mole, i.e., 27 g
\( \therefore \) 864 × 10⁶ C will produce Al \( = \frac{27}{3\times96500}\times864\times10^{6} \) g = 8.058 × 10⁴ g = 8.1 × 10⁴ g.

Teacher's Note:
1. Always convert hours to seconds before multiplying by current; Q in coulombs is then compared with 3F per mole of Al.

Question 9. The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O₂ at STP will be (AIPMT 2014)
(a) 5.4 g
(b) 10.8 g
(c) 54.0 g
(d) 108.0 g
Answer:
Ans. (d)
By Faraday's second law, \( \frac{W_{Ag}}{W_{O_2}} = \frac{E_{Ag}}{E_{O_2}} \)
Weight of 5600 mL of O₂ at STP \( = \frac{32}{22400}\times5600 \) = 8 g
\( \therefore \frac{W_{Ag}}{8} = \frac{108}{8} \) or W_Ag = 108 g

Teacher's Note:
1. 5600 mL of O₂ is 1/4 mol, which needs 1 F; one faraday deposits one mole (108 g) of silver.

Question 10. The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on the electron = 1.60 × 10⁻¹⁹ C) (NEET Phase II 2016)
(a) 6 × 10²³
(b) 6 × 10²⁰
(c) 3.75 × 10²⁰
(d) 7.48 × 10²³
Answer:
Ans. (c)
Charge flowing during electrolysis = I × t = 1 × 60 = 60 coulombs
But charge flowing = number of electrons flowing × charge on each electron
\( \therefore \) 60 = n × 1.60 × 10⁻¹⁹ or n = \( \frac{60}{1.60\times10^{-19}} \) = 37.5 × 10¹⁹ = 3.75 × 10²⁰

Teacher's Note:
1. Number of electrons = total charge / charge of one electron; no Faraday constant is needed here.

Question 11. The number of Faradays (F) required to produce 20 g of calcium from molten CaCl₂ (Atomic mass of Ca = 40 g mol⁻¹) is (NEET 2020, Phase 1)
(a) 2
(b) 3
(c) 4
(d) 1
Answer:
Ans. (d)
CaCl₂ → Ca²⁺ + 2Cl⁻, Ca²⁺ + 2e⁻ → Ca
To produce 1 mole of Ca (40 g), Faradays required = 2
\( \therefore \) To produce 20 g of Ca, Faradays required = 1

Teacher's Note:
1. Faradays needed = moles of metal × charge on its ion: here 0.5 × 2 = 1.

Question 12. On electrolysis of dil. sulphuric acid using Platinum (Pt) electrodes, the product obtained at anode will be (NEET 2020, Phase 1)
(a) Oxygen gas
(b) H₂S gas
(c) SO₂ gas
(d) Hydrogen gas
Answer:
Ans. (a)
In H₂SO₄ solution, H₂SO₄ ⇄ 2H⁺ + SO₄²⁻; H₂O ⇄ H⁺ + OH⁻
At cathode: 2H⁺ + 2e⁻ → H₂
At anode: OH⁻ ions have lower discharge potential than SO₄²⁻ ions, hence, 4OH⁻ → 2H₂O + O₂

Teacher's Note:
1. Sulphate ions are very hard to oxidise, so water (OH⁻) is oxidised at the anode and O₂ is evolved.

Question 13. The resistance of a 0.10 M weak acid HA in a conductivity cell is 2.0 × 10³ ohm. The cell constant of the cell is 0.78 cm⁻¹ and Λ₀ of the acid is 390 S cm² mol⁻¹.
Consider the following statements:
1. pH of the acid solution = 3
2. pK_a of the acid = 5
3. Degree of dissociation of the acid = 0.01
Which of the statements given above are correct? (IAS Prelim 2010)
(a) 1 and 2 only
(b) 1 and 3 only
(c) 2 and 3 only
(d) 1, 2 and 3
Answer:
Ans. (d)
κ = Conductance (G) × Cell constant (G*) \( = \frac{1}{2.0\times10^{3}}\times0.78 \) = 3.9 × 10⁻⁴ Ω⁻¹ cm⁻¹
\( \Lambda = \frac{1000\,\kappa}{\text{Molarity}} = \frac{1000\times3.9\times10^{-4}}{0.10} \) = 3.9 S cm² mol⁻¹
\( \therefore \alpha = \frac{\Lambda}{\Lambda_0} = \frac{3.9}{390} \) = 10⁻² = 0.01
HA ⇄ H⁺ + A⁻; [H⁺] = Cα = (0.10)(10⁻²) = 10⁻³ M ∴ pH = 3
\( K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]} = \frac{10^{-3}\times10^{-3}}{0.1} \) = 10⁻⁵, pK_a = 5. Hence all three statements are correct.

Teacher's Note:
1. Chain the steps: R → κ → Λ_m → α → [H⁺] → K_a; one slip early spoils every later answer.

Question 14. The sequence of ionic mobility or conductivity in the aqueous solution is (AIPMT 2008, JEE Main 2021)
(a) K⁺ > Na⁺ > Rb⁺ > Cs⁺
(b) Cs⁺ > Rb⁺ > K⁺ > Na⁺
(c) Rb⁺ > K⁺ > Cs⁺ > Na⁺
(d) Na⁺ > K⁺ > Rb⁺ > Cs⁺
Answer:
Ans. (b)
Smaller the size of an ion, more strongly it is hydrated and less is the ionic mobility. The sizes of the unhydrated ions increase in the order: Na⁺ < K⁺ < Rb⁺ < Cs⁺
The sizes of the hydrated ions will be in the order: Cs⁺ < Rb⁺ < K⁺ < Na⁺.
Hence, their ionic mobilities will be in the order: Cs⁺ > Rb⁺ > K⁺ > Na⁺

Teacher's Note:
1. In water it is the size of the hydrated ion that matters: the smallest bare ion carries the biggest water shell and moves slowest.

Question 15. Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 S m⁻¹. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m² mol⁻¹ is (JEE Main 2014)
(a) 5 × 10²
(b) 5 × 10⁻⁴
(c) 5 × 10⁻³
(d) 5 × 10³
Answer:
Ans. (b)
Case I: C = 0.2 M, R = 50 Ω, κ = 1.4 S m⁻¹
\( \kappa = \text{Conductance}\times\text{Cell const.} = \frac{1}{R}\times\text{Cell const.} \)
\( \therefore \) Cell constant = κ × R = (1.4 S m⁻¹)(50 Ω) = 70 m⁻¹
Case II: C = 0.5 M, R = 280 Ω, Cell const. = 70 m⁻¹
\( \kappa = \frac{1}{R}\times\text{Cell constant} = \frac{1}{280\,\Omega}\times70\,\mathrm{m^{-1}} \) = 0.25 S m⁻¹
Molar conductivity (in S m² mol⁻¹) \( = \frac{\kappa\,(\mathrm{S\,m^{-1}})}{\text{Molarity in mol m}^{-3}} = \frac{0.25\,\mathrm{S\,m^{-1}}}{0.5\,\mathrm{mol\,L^{-1}}\times1000\,\mathrm{L\,m^{-3}}} \)
= 5 × 10⁻⁴ S m² mol⁻¹

Teacher's Note:
1. In SI units concentration must be in mol m⁻³ (1 M = 1000 mol m⁻³) when κ is in S m⁻¹.

Question 16. The molar conductivity of a conductivity cell filled with 10 moles of NaCl in 20 mL solution is Λ_m1 and that of 20 moles of NaCl in 80 mL solution in another identical cell is Λ_m2. The conductivities exhibited by these two cells are same. The relationship between Λ_m2 and Λ_m1 is (JEE Main 2022)
(a) Λ_m2 = 2Λ_m1
(b) Λ_m2 = Λ_m1/2
(c) Λ_m2 = Λ_m1
(d) Λ_m2 = 4Λ_m1
Answer:
Ans. (a)
\( \Lambda_{m_1} = \frac{\kappa_1\times1000}{\text{Molarity}} = \frac{\kappa_1\times1000}{10\,\mathrm{mol}/0.02\,\mathrm{L}} \) = 2κ₁
\( \Lambda_{m_2} = \frac{\kappa_2\times1000}{20\,\mathrm{mol}/0.08\,\mathrm{L}} \) = 4κ₂
Given κ₁ = κ₂ \( \therefore \frac{\Lambda_{m_1}}{2} = \frac{\Lambda_{m_2}}{4} \) or Λ_m2 = 2Λ_m1

Teacher's Note:
1. With equal conductivity, molar conductivity is inversely proportional to concentration: 500 M vs 250 M gives a factor of 2.

Question 17. Ionic mobility of Ag⁺ is (λ_Ag⁺ = 5 × 10⁻⁴ ohm⁻¹ cm² eq⁻¹)
(a) 5.2 × 10⁻⁹
(b) 2.4 × 10⁻⁹
(c) 1.52 × 10⁻⁹
(d) 8.25 × 10⁻⁹
Answer:
Ans. (a)
\( \text{Ionic mobility} = \frac{\text{Ionic conductance}}{96500} = \frac{5\times10^{-4}}{96500} \) = 5.18 × 10⁻⁹ cm/sec, i.e., ≈ 5.2 × 10⁻⁹.

Teacher's Note:
1. Ionic mobility (speed per unit potential gradient) is the ionic conductance divided by the faraday.

Question 18. Degree of dissociation of pure water is 1.9 × 10⁻⁹. Molar ionic conductances of H⁺ and OH⁻ ions at infinite dilution are 200 S cm² mol⁻¹ and 350 S cm² mol⁻¹ respectively. Molar conductance of water is
(a) 3.8 × 10⁻⁷ S cm² mol⁻¹
(b) 5.7 × 10⁻⁷ S cm² mol⁻¹
(c) 9.5 × 10⁻⁷ S cm² mol⁻¹
(d) 1.045 × 10⁻⁶ S cm² mol⁻¹
Answer:
Ans. (d)
Degree of dissociation \( (\alpha) = \frac{\Lambda_m^c}{\Lambda_m^0} \)
\( \therefore \Lambda_m^c = \alpha\times\Lambda_m^0 \) = (1.9 × 10⁻⁹) × (200 + 350) S cm² mol⁻¹
= 1.9 × 550 × 10⁻⁹ = 1.045 × 10⁻⁶ S cm² mol⁻¹.

Teacher's Note:
1. First get Λ⁰_m of water by adding the two ionic values (Kohlrausch), then multiply by α.

Question 19. Λ^o_eq for BaCl₂, H₂SO₄ and HCl are x₁, x₂ and x₃ S cm² eq⁻¹ respectively. If conductivity of saturated BaSO₄ solution is y S cm⁻¹, then K_sp for BaSO₄ is
(a) \( \frac{10^6\,y^2}{2(x_1+x_2-2x_3)} \)
(b) \( \frac{10^9\,y^3}{8(x_1+x_2-2x_3)^3} \)
(c) \( \frac{10^3\,y}{2(x_1+x_2-2x_3)} \)
(d) \( \frac{10^6\,y^2}{4(x_1+x_2-2x_3)^2} \)
Answer:
Ans. (d)
Λ^o_eq (BaSO₄) = Λ^o_eq (BaCl₂) + Λ^o_eq (H₂SO₄) − 2Λ^o_eq (HCl) = x₁ + x₂ − 2x₃
\( \Lambda^o_{eq}(\mathrm{BaSO_4}) = \frac{1000\times\kappa}{\text{Normality (Solubility in g eq L}^{-1})} \)
\( \therefore \) Solubility in g eq L⁻¹ \( = \frac{1000\times y}{x_1+x_2-2x_3} \); Solubility in mol L⁻¹ \( = \frac{1000\,y}{2(x_1+x_2-2x_3)} \)
BaSO₄ ⇄ Ba²⁺ + SO₄²⁻; K_sp for BaSO₄ = [Ba²⁺][SO₄²⁻] = s × s = s²
\( = \left[\frac{1000\,y}{2(x_1+x_2-2x_3)}\right]^2 = \frac{10^6\,y^2}{4(x_1+x_2-2x_3)^2} \)

Teacher's Note:
1. Equivalent-based values give solubility in g eq L⁻¹; divide by 2 for a 2+/2− salt before squaring for K_sp.

Question 20. Equivalent conductivity at infinite dilution for sodium potassium oxalate, (COO⁻)₂Na⁺ K⁺, will be (given, molar conductivities of oxalate, K⁺ and Na⁺ ions at infinite dilution are 148.2, 50.1, 73.5 S cm² mol⁻¹ respectively) (West Bengal JEE Engg. 2013)
(a) 271.8 S cm² eq⁻¹
(b) 67.95 S cm² eq⁻¹
(c) 543.6 S cm² eq⁻¹
(d) 135.9 S cm² eq⁻¹
Answer:
Ans. (d)
Λ^o_m (Sod. pot. oxalate) = λ^o_m (Na⁺) + λ^o_m (K⁺) + λ^o (oxalate) = (50.1 + 73.5 + 148.2) S cm² mol⁻¹ = 271.8 S cm² mol⁻¹
\( \Lambda^o_{eq} = \frac{\Lambda^o_m}{\text{Total charge on cations or anions (n factor)}} = \frac{271.8}{2} \) = 135.9 S cm² eq⁻¹

Teacher's Note:
1. Equivalent conductivity = molar conductivity divided by the total positive (or negative) charge per formula unit, here 2.

Question 21. The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. What is the dissociation constant of acetic acid? Choose the correct option
[Λ^o_H⁺ = 350 S cm² mol⁻¹, Λ^o_CH3COO⁻ = 50 S cm² mol⁻¹] (NEET 2021)
(a) 1.75 × 10⁻⁴ mol L⁻¹
(b) 2.50 × 10⁻⁴ mol L⁻¹
(c) 1.75 × 10⁻⁵ mol L⁻¹
(d) 2.50 × 10⁻⁵ mol L⁻¹
Answer:
Ans. (c)
Λ_m = 20 S cm² mol⁻¹; Λ^o_m = Λ^o_CH3COO⁻ + Λ^o_H⁺ = 50 + 350 = 400 S cm² mol⁻¹
\( \alpha = \frac{\Lambda_m}{\Lambda_m^o} = \frac{20}{400} = \frac{1}{20} \)
\( K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2 = 7\times10^{-3}\times\left(\frac{1}{20}\right)^2 = 7\times10^{-3}\times\frac{1}{4}\times10^{-2} \) = 1.75 × 10⁻⁵ mol L⁻¹

Teacher's Note:
1. For a weak acid use α = Λ_m/Λ^o_m, then Ostwald's law K_a ≈ Cα² when α is small.

Question 22. The reaction taking place in the cell
Pt | H₂(g) (1 atm) | HCl (1.0 M) | AgCl | Ag is (IAS Prelim 2010)
(a) AgCl + (1/2) H₂ → Ag + H⁺ + Cl⁻
(b) Ag + H⁺ + Cl⁻ → AgCl + (1/2) H₂
(c) 2Ag⁺ + H₂ → 2Ag + 2H⁺
(d) 2Ag + 2H⁺ → 2Ag⁺ + H₂
Answer:
Ans. (a)
LHS electrode (oxidation half reaction): \( \frac{1}{2} \)H₂ → H⁺ + e⁻
RHS electrode (reduction half reaction): AgCl + e⁻ → Ag + Cl⁻
Adding the two half reactions, we get AgCl + \( \frac{1}{2} \)H₂ → Ag + H⁺ + Cl⁻.

Teacher's Note:
1. In cell notation the left electrode is the anode (oxidation) and the right one the cathode; the Ag electrode here is AgCl/Ag, not Ag⁺/Ag.

Question 23. Consider the cell shown in the figure: a Zn rod dipping in 1 M ZnSO₄ solution and a Cu rod dipping in 1 M CuSO₄ solution, connected by a salt bridge, with an external source of variable emf (E_ext) applied across the two rods.E^o_Cu²⁺/Cu = +0.34 V, E^o_Zn²⁺/Zn = −0.76 V
Identify the incorrect statement from the options given below for the above cell: (JEE Main 2020)
(a) If E_ext = 1.1 V, no flow of e⁻ or current occurs
(b) If E_ext < 1.1 V, Zn dissolves at anode and Cu deposits at cathode
(c) If E_ext > 1.1 V, e⁻ flow from Cu to Zn
(d) If E_ext > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode
Answer:
Ans. (d)
E^o_Cell = E^o_Cu²⁺/Cu − E^o_Zn²⁺/Zn = 0.34 − (−0.76) = 1.10 V
\( \therefore \) If E = 1.1 V, no current will flow.
If E > 1.1 V, cell will act as electrolytic cell and the electrons will flow from Cu to Zn (Zn is deposited and Cu dissolves).
If E < 1.1 V, cell acts as galvanic cell so that Zn dissolves and Cu deposits. Hence statement (d) is incorrect.

Teacher's Note:
1. When the opposing external emf exceeds the cell emf, every process is reversed: the cell becomes electrolytic.

Question 24. When measured against a standard calomel electrode, an electrode is found to have a standard reduction potential of 0.100 V. If standard reduction potential of calomel electrode is +0.244 V and it acts as anode, the standard electrode potential of the same electrode against standard hydrogen electrode will be
(a) −0.144 V
(b) +0.100 V
(c) −0.344 V
(d) −0.100 V
Answer:
Ans. (b)
Standard electrode potential is fixed quantity for a given electrode. Hence, its standard reduction potential against SHE will remain same viz. 0.100 V (EMF of the cell formed will change).

Teacher's Note:
1. A standard reduction potential is by definition referred to SHE; only the emf of the cell changes when the reference electrode changes.

Question 25. Quinhydrone electrode is sometimes used to find the pH of a solution. It is based on the following electrode reaction:
Quinone, C₆H₄O₂ (s) + 2H⁺ (aq) + 2e⁻ ⇄ Hydroquinone, C₆H₄(OH)₂ (s)
Its standard electrode potential is 0.70 V. If in a particular solution, the electrode potential is found to be 0.58 V, the pH of the solution is
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
Ans. (a)
For the given electrode reaction (reduction reaction), \( E = E^o - \frac{0.0591}{2}\log\frac{1}{[\mathrm{H^+}]^2} \)
= E^o + 0.0591 log [H⁺] = E^o − 0.0591 pH
\( \therefore \) 0.58 = 0.70 − 0.0591 pH or pH = \( \frac{0.12}{0.059} \) = 2.0.

Teacher's Note:
1. Solids (quinone, hydroquinone) do not appear in Q, so the potential depends only on [H⁺] and falls by 0.059 V per pH unit.

Question 26. Given E^o_Cr³⁺/Cr = −0.74 V, E^o_{MnO4⁻/Mn²⁺} = 1.51 V, E^o_{Cr2O7²⁻/Cr³⁺} = 1.33 V, E^o_Cl/Cl⁻ = 1.36 V
Based on the data given above, strongest oxidizing agent will be (JEE Main 2013)
(a) MnO₄⁻
(b) Cl⁻
(c) Cr³⁺
(d) Mn²⁺
Answer:
Ans. (a)
Higher the reduction potential, more easily it is reduced and hence stronger is the oxidizing agent. As reduction potential of MnO₄⁻/Mn²⁺ is maximum, hence it is strongest oxidizing agent.

Teacher's Note:
1. Pick the oxidised form (left side) of the couple with the highest E^o; Cl⁻ and Mn²⁺ are reduced forms and cannot be oxidants here.

Question 27. Given E^o_Cl2/Cl⁻ = 1.36 V, E^o_Cr³⁺/Cr = −0.74 V, E^o_{Cr2O7²⁻/Cr³⁺} = 1.33 V, E^o_{MnO4⁻/Mn²⁺} = 1.51 V
Among the following, the strongest reducing agent is (JEE Main 2017)
(a) Cr³⁺
(b) Cl⁻
(c) Cr
(d) Mn²⁺
Answer:
Ans. (c)
Greater the standard oxidation potential (or lesser is the standard reduction potential), more easily the substance is oxidized and hence stronger is the reducing agent. As E^o_Cr³⁺/Cr (i.e., standard reduction potential) is lowest, hence, Cr will be strongest reducing agent.

Teacher's Note:
1. The strongest reducing agent is the reduced form (right side) of the couple with the most negative E^o.

Question 28. Small quantities of solutions of compounds TX, TY and TZ are put into separate test tubes containing X, Y and Z solutions. TX does not react with any of these. TY reacts with both X and Z. TZ reacts with X. The decreasing order of oxidation of the anions X⁻, Y⁻, Z⁻ is (DCE 2009)
(a) Y⁻, Z⁻, X⁻
(b) Z⁻, X⁻, Y⁻
(c) Y⁻, X⁻, Z⁻
(d) X⁻, Z⁻, Y⁻
Answer:
Ans. (a)
TX does not react with any solution. This means that X⁻ is least easily oxidized.
TY reacts with both X and Z. This means Y⁻ is oxidized by both. (Y⁻ + X → Y + X⁻, Y⁻ + Z → Y + Z⁻)
TZ reacts with X only (Z⁻ + X → Z + X⁻). This means Z⁻ is less easily oxidized than Y⁻. Thus, decreasing order of oxidation of anions: Y⁻, Z⁻, X⁻

Teacher's Note:
1. An anion that reacts with more of the free elements is the one most easily oxidised; rank by the number of reactions.

One or More Correct Answers

Question 29. Which of the following statements are not correct?
(a) Same quantity of electricity deposits more of iron from ferric sulphate solution than from ferrous sulphate solution
(b) Electrochemical equivalent of an element can be obtained by dividing its equivalent weight by 96,500
(c) 1 Faraday always liberates 1 mole of the substance at the electrode
(d) A 60 watt bulb emits 60 Joules of energy per second.
Answer:
Ans. (a, c)
(a) is wrong because Fe²⁺ + 2e⁻ → Fe, Fe³⁺ + 3e⁻ → Fe. One faraday from Fe²⁺ will deposit \( \frac{1}{2} \) mol of Fe and from Fe³⁺, it will deposit \( \frac{1}{3} \) mol of Fe.
(c) is wrong because 1 F liberates one gram equivalent of the substance.

Teacher's Note:
1. One faraday always gives one equivalent, not one mole; the number of moles depends on the charge of the ion.

Passage-Based Questions

Question 30. Comprehension 1: There are two principal types of electrochemical cells. A galvanic cell is an electrochemical cell that produces electricity as a result of spontaneous reaction occurring inside it. An electrolytic cell is an electrochemical cell in which a non-spontaneous reaction is driven by an external source of current. Any redox reaction may be expressed in terms of two half-reactions which are conceptual reactions showing the loss and gain of electrons. Each half reaction has a definite value of standard electrode potential. The overall reaction is represented by a universally accepted method. Knowing the standard electrode potentials of the half reactions, the standard EMF of the cell can be calculated. The standard EMF further helps in the calculation of free energy change, equilibrium constant of the cell reaction as well as parameters like solubility product of a sparingly soluble salt. A cell can also be set up in which the two electrodes may be of the same type, e.g., both may be hydrogen electrodes but the concentration of H⁺ ions in the two solutions may be different. Such cells are called concentration cells.
The reaction \( \frac{1}{2} \)H₂ (g) + AgCl (s) → H⁺ (aq) + Cl⁻ (aq) + Ag (s) occurs in the galvanic cell
(a) Ag | AgCl (s) | KCl (aq) | AgNO₃ (aq) | Ag
(b) Pt | H₂ (g) | HCl (aq) | AgNO₃ (aq) | Ag
(c) Pt | H₂ (g) | HCl (aq) | AgCl (s) | Ag
(d) Pt | H₂ (g) | KCl (aq) | AgCl (s) | Ag
Answer:
Ans. (c)
H₂ has been oxidized and AgCl (Ag⁺) has been reduced. Hydrogen electrode on left has to be Pt | H₂ | H⁺.

Teacher's Note:
1. Write the anode (oxidation, here H₂ → H⁺) on the left; the solution must contain both H⁺ and Cl⁻, i.e., HCl.

Question 31. Comprehension 2: A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% H₂SO₄ by mass having a density of 1.294 g mL⁻¹. The battery holds 3.5 L of the acid. During the discharge of the battery, the density H₂SO₄ falls from 1.294 g mL⁻¹ to 1.139 g mL⁻¹ which is 20% H₂SO₄ by mass.
The reaction occurring at the anode during charging is
(a) Pb²⁺ + 2e⁻ → Pb
(b) Pb²⁺ + SO₄²⁻ → PbSO₄
(c) Pb → Pb²⁺ + 2e⁻
(d) PbSO₄ + 2H₂O → PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻
Answer:
Ans. (d)
At anode, oxidation occurs. PbSO₄ is oxidized to PbO₂ according to reaction (d).

Teacher's Note:
1. During charging the reactions of discharge are reversed; the anode is where oxidation of PbSO₄ to PbO₂ occurs.

Question 32. Comprehension 2: A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% H₂SO₄ by mass having a density of 1.294 g mL⁻¹. The battery holds 3.5 L of the acid. During the discharge of the battery, the density H₂SO₄ falls from 1.294 g mL⁻¹ to 1.139 g mL⁻¹ which is 20% H₂SO₄ by mass.
Moles of sulphuric acid lost during discharge is
(a) 9.88
(b) 8.88
(c) 7.88
(d) 6.88
Answer:
Ans. (a)
Mass of the solution before discharge = 3500 mL × 1.294 g mL⁻¹ = 4529 g
Mass of H₂SO₄ present in this solution \( = \frac{39}{100}\times4529 \) g = 1766.31 g
Mass of the solution after discharge = 3500 mL × 1.139 g mL⁻¹ = 3986.5 g
Mass of H₂SO₄ present in this solution \( = \frac{20}{100}\times3986.5 \) g = 797.3 g
Loss in mass of H₂SO₄ during discharge = 1766.31 − 797.3 = 969.01 g
Moles of H₂SO₄ lost during discharge \( = \frac{969.01}{98} \) = 9.88786

Teacher's Note:
1. Mass % applies to the mass of solution, so find solution mass from density × volume first.

Matrix Match Questions

Teacher's Note:
1. Only the ratio [Zn²⁺]/[Cu²⁺] matters, not the absolute concentrations.

Teacher's Note:
1. For each conductometric titration ask which ion is being replaced and by what; H⁺ and OH⁻ are far more conducting than other ions.

Integer and Numerical Answer Questions

Question 35. Three litres of 0.5 M K₂Cr₂O₇ solution have to be completely reduced in the acidic medium. The number of faradays of electricity required will be (single digit integer)
Answer:
Ans. 9
3 L of 0.5 M K₂Cr₂O₇ solution will contain 0.5 × 3 = 1.5 moles of K₂Cr₂O₇. The reduction reaction is
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Thus, 1 mole of Cr₂O₇²⁻ requires 6 F of electricity. \( \therefore \) 1.5 mole will require electricity = 9 F.

Teacher's Note:
1. Each dichromate ion takes 6 electrons (two Cr going from +6 to +3).

Question 36. In the Mg–Al cell, the number of electrons involved in the cell reaction is (single digit integer)
Answer:
Ans. 6
The cell reaction is 3Mg + 2Al³⁺ → 3Mg²⁺ + 2Al, n = 6.

Teacher's Note:
1. n is the LCM of the two ionic charges (2 and 3) once the equation is balanced.

Question 37. For the Mg-Ag cell, how many times the difference between the EMF of the cell and its standard EMF will change if concentration of Mg²⁺ ions is changed from 0.1 M to 0.01 M and that of Ag⁺ ions is changed from 0.5 M to 0.25 M? (single digit integer)
Answer:
Ans. 2
Mg + 2Ag⁺ → Mg²⁺ + 2Ag, \( E_{cell} = E^o_{cell} - \frac{0.0591}{n}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^+}]^2} \)
or \( E^o_{cell} - E_{cell} = \frac{0.0591}{n}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^+}]^2} \)
\( \therefore \) In 1st case, \( E^o_{cell} - E_{cell} = \frac{0.0591}{2}\log\frac{0.1}{(0.5)^2} \)
In 2nd case, \( E^o_{cell} - E_{cell} = \frac{0.0591}{2}\log\frac{0.01}{(0.25)^2} = \frac{0.0591}{2}\log\frac{(0.1)^2}{[(0.5)^2]^2} = \frac{0.0591}{2}\times2\log\frac{0.1}{(0.5)^2} \) = 2 times.

Teacher's Note:
1. Look for a pattern: the new Q is the square of the old Q, so the log term exactly doubles.

Question 38. All the energy released from the reaction X → Y, Δ_rG^o = −193 kJ mol⁻¹ is used for oxidizing M⁺ as M⁺ → M³⁺ + 2e⁻, E^o = −0.25 V. Under standard conditions, the number of moles of M⁺ oxidized when one mole of X is converted to Y is (F = 96500 C mol⁻¹) (JEE Advanced 2015)
Answer:
Ans. 4
X → Y, Δ_rG^o = −193 kJ mol⁻¹ ...(i)
M⁺ → M³⁺ + 2e⁻, E^o = −0.25 V
For this reaction, ΔG^o = −nFE^o = −2 × 96500 × (−0.25) = 48250 J mol⁻¹ = 48.25 kJ mol⁻¹
i.e., 48.25 kJ of energy is used for oxidizing 1 mole of M⁺ to M³⁺
\( \therefore \) No. of moles of M⁺ oxidized by 193 kJ of energy \( = \frac{193}{48.25} \) = 4

Teacher's Note:
1. The non-spontaneous oxidation needs +48.25 kJ per mole; divide the energy available by this amount.

Assertion–Reason Questions

Question 39. Statement-1: Specific conductivity of an electrolytic solution decreases with dilution whereas molar conductivity increases with dilution.
Statement-2: Specific conductivity is the conductance of a specific amount of the electrolyte whereas molar conductivity is for 1 mole of the electrolyte.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Correct Statement-2. Specific conductivity is the conductance of 1 cm³ of the solution whereas molar conductivity is the conductance of a solution containing 1 mole of the electrolyte.

Teacher's Note:
1. κ refers to a fixed volume (1 cm³) of solution, not a fixed amount of electrolyte; that is why it falls on dilution.

Question 40. Statement-1: Iron is protected from corrosion by connecting magnesium metal with it.
Statement-2: Iron acts as a cathode and magnesium as anode which gradually disappears.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Statement-2 is the correct explanation of statement-1.

Teacher's Note:
1. In sacrificial (cathodic) protection the more active metal is oxidised instead of iron and must be replaced from time to time.

Question 41. Statement-1: The cell constant of a conductivity cell depends upon the nature of the material of the electrodes.
Statement-2: The electrodes of the cell are coated with platinum black to avoid polarisation effects.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (d)
Correct Statement-1. The cell constant of a conductivity cell depends upon the distance between the electrodes and the area of their cross-section and not on the material of the electrodes.

Teacher's Note:
1. Cell constant = l/a is purely geometric.

Question 42. Statement-1: When acidified zinc sulphate solution is electrolysed between zinc electrodes, it is zinc that is deposited at the cathode and no H₂ gas is evolved.
Statement-2: The electrode potential of zinc is more negative than that of hydrogen as the overvoltage for the H₂ evolution on zinc is quite large.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation of Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (a)
Statement-2 is the correct explanation of statement-1.

Teacher's Note:
1. Overvoltage can reverse the order predicted by E^o values: high hydrogen overvoltage on zinc lets Zn be deposited instead.