CBSE Class 12 Chemistry Electrochemistry NEET JEE Questions Set 03

Single Correct Answer

Question 1. Find the emf of the cell in which the following reaction takes place at 298 K
Ni (s) + 2Ag⁺ (0.001 M) → Ni²⁺ (0.001 M) + 2Ag (s)
[Given that E^o_cell = 1.05 V, \( \frac{2.303RT}{F} \) = 0.059 at 298 K] (NEET 2022)
(a) 1.0385 V
(b) 1.385 V
(c) 0.9615 V
(d) 1.05 V
Answer:
Ans. (c)
For the given cell, applying Nernst equation, \( E_{cell} = E^o_{cell} - \frac{0.059}{n}\log\frac{[\mathrm{Ni^{2+}}]}{[\mathrm{Ag^+}]^2} \)
\( = 1.05 - \frac{0.059}{2}\log\frac{(10^{-3})}{(10^{-3})^2} = 1.05 - \frac{0.059}{2}\log(10)^3 \)
= 1.05 − 0.0295 × 3 = 1.05 − 0.0885 = 0.9615 V

Teacher's Note:
1. Square [Ag⁺] because of the coefficient 2; this makes Q = 10³ even though both ions are 0.001 M.

Question 2. For the following cell
Zn (s) | ZnSO₄ (aq) || CuSO₄ (aq) | Cu (s)
when the concentration of Zn²⁺ is 10 times the concentration of Cu²⁺, the expression for ΔG (in J mol⁻¹) is [F is Faraday constant, R is gas constant, T is temperature, E^o (cell) = 1.1 V] (JEE Advanced 2017)
(a) 2.303 RT + 1.1 F
(b) 1.1 F
(c) 2.303 RT − 2.2 F
(d) −2.2 F
Answer:
Ans. (c)
Cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu
\( \Delta G = \Delta G^o + 2.303RT\log Q = \Delta G^o + 2.303RT\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \)
ΔG^o = −nFE^o_cell = −2F(1.1)
\( \therefore \) ΔG = −2F(1.1) + 2.303 RT log 10 = 2.303 RT − 2.2 F

Teacher's Note:
1. Use ΔG = ΔG^o + 2.303RT log Q directly; it avoids converting E to ΔG twice.

Question 3. Standard free energies of formation (in kJ/mol) at 298 K are −237.2, −394.4 and −8.2 for H₂O (l), CO₂ (g) and pentane (g) respectively. The value of E^o_cell for the pentane-oxygen fuel cell is (AIPMT 2008)
(a) 1.968 V
(b) 2.0968 V
(c) 1.0968 V
(d) 0.0968 V
Answer:
Ans. (c)
The balanced equation for pentane-oxygen cell reaction will be C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O, n = 32 (∵ 16 (O) + 32e⁻ → 16O²⁻)
Δ_rG^o = [5 × Δ_fG^o (CO₂) + 6 Δ_fG^o (H₂O)] − [Δ_fG^o (C₅H₁₂) + 8 Δ_fG^o (O₂)]
= [5 (−394.4) + 6 (−237.2)] − [(−8.2) + 0] = −1972 − 1423.2 + 8.2 = −3387 kJ mol⁻¹
Δ_rG^o = −nFE^o_cell \( \therefore \) −3387000 = −32 × 96500 × E^o_cell or E^o_cell = 1.0968 V

Teacher's Note:
1. The key step is n: 8 O₂ molecules each take 4 electrons, so n = 32.

Question 4. A fuel cell involves combustion of butane at 1 atm and 298 K
C₄H₁₀ (g) + \( \frac{13}{2} \)O₂ (g) → 4CO₂ (g) + 5H₂O (l), ΔG^o = −2746 kJ mol⁻¹
The E^o_cell will be
(a) 0.545 V
(b) 1.09 V
(c) 0.922 V
(d) 0.755 V
Answer:
Ans. (b)
Oxidation number of carbon: in C₄H₁₀ total for 4 C = −10; in 4CO₂ total = +16 (each C = +4).
Change in oxidation number of carbon = 4 (+4) − (−10) = 26
or \( \frac{13}{2} \)O₂ + 26e⁻ → 13O²⁻ (8O²⁻ in 4CO₂ and 5O²⁻ in 5H₂O)
Thus, cell reaction involves 26 electrons, i.e., n = 26
\( E^o_{cell} = -\frac{\Delta G^o}{nF} = \frac{-(-2746)\times1000}{26\times96500} \) = +1.09 V

Teacher's Note:
1. n equals 4 × (number of O₂ molecules) in a combustion: 13/2 × 4 = 26.

Question 5. For a cell involving one electron, E^o_cell = 0.59 V at 298 K. The equilibrium constant for the cell reaction is (Given that \( \frac{2.303RT}{F} \) = 0.059 V at 298 K) (NEET 2019)
(a) 1.0 × 10³⁰
(b) 1.0 × 10²
(c) 1.0 × 10⁵
(d) 1.0 × 10¹⁰
Answer:
Ans. (d)
The relation between standard e.m.f. of a cell (E^o_cell) and equilibrium constant (K) of the cell reaction at 298 K: \( E^o_{cell} = \frac{0.059}{n}\log K_c \)
\( \therefore 0.59 = \frac{0.059}{1}\log K_c \) or log K_c = 10 or K_c = antilog 10 = 1 × 10¹⁰

Teacher's Note:
1. Every 0.059 V of E^o (for n = 1) multiplies K by 10.

Question 6. If the E^o_cell for a given reaction has a negative value, then which of the following gives the correct relationships for the values of ΔG^o and K_eq? (AIPMT Prelim 2011, NEET Phase II 2016)
(a) ΔG^o > 0; K_eq < 1
(b) ΔG^o > 0; K_eq > 1
(c) ΔG^o < 0; K_eq > 1
(d) ΔG^o < 0; K_eq < 1
Answer:
Ans. (a)
ΔG^o = −nFE^o_cell ...(i) and ΔG^o = −RT ln K_eq ...(ii)
From (i), if E^o_cell is negative, ΔG^o will be positive, i.e., ΔG^o > 0
From (ii), ΔG^o can be positive only if K_eq < 1.

Teacher's Note:
1. Remember the trio: E^o > 0 ⇔ ΔG^o < 0 ⇔ K > 1, and all three reverse together.

Question 7. The half-cell reactions for rusting of iron are
2H⁺ + \( \frac{1}{2} \)O₂ + 2e⁻ → H₂O, E^o = +1.23 V,
Fe²⁺ + 2e⁻ → Fe (s), E^o = −0.44 V
ΔG^o (in kJ) for the reaction is
(a) −76
(b) −322
(c) −122
(d) −176
Answer:
Ans. (b)
For emf to be +ve, oxidation should occur at iron electrode.
E_cell = 1.23 + 0.44 V = 1.67 V
ΔG^o = −nFE^o_cell = −2 × 96500 × 1.67 J = −322 kJ.

Teacher's Note:
1. Use n = 2 as written in both half-reactions; E^o is not doubled even if the equation is.

Question 8. For the cell reaction
2Fe³⁺ (aq) + 2I⁻ (aq) → 2Fe²⁺ (aq) + I₂ (aq)
E^o_cell = 0.24 V at 298 K. The standard Gibbs energy (Δ_rG^o) of the cell reaction is (Given that Faraday's constant, F = 96,500 C mol⁻¹) (NEET 2019)
(a) 23.16 kJ mol⁻¹
(b) −46.32 kJ mol⁻¹
(c) −23.16 kJ mol⁻¹
(d) 46.32 kJ mol⁻¹
Answer:
Ans. (b)
Standard Gibbs energy of a reaction (Δ_rG^o) = −nFE^o_cell
For the given reaction, n = 2, E^o_cell = 0.24 V \( \therefore \) Δ_rG^o = −2 × 96500 × 0.24 = −46320 J mol⁻¹ = −46.32 kJ mol⁻¹

Teacher's Note:
1. Two Fe³⁺ each gain one electron, so n = 2; a positive E^o must give a negative ΔG^o.

Question 9. For the given cell:
Cu (s) | Cu²⁺ (C₁ M) || Cu²⁺ (C₂ M) | Cu (s),
change in Gibbs energy (ΔG) is negative, if: (JEE Main 2020)
(a) \( C_2 = \frac{C_1}{\sqrt{2}} \)
(b) \( C_2 = \sqrt{2}\,C_1 \)
(c) C₁ = C₂
(d) C₁ = 2C₂
Answer:
Ans. (b)
The given cell is a concentration cell. The EMF of the cell, as represented, is given by \( E_{cell} = \frac{0.0591}{n}\log\frac{C_2}{C_1} \)
E_cell will be +ve only if C₂ > C₁
As ΔG = −nFE_cell, therefore, ΔG will be −ve only when E_cell is +ve. Hence, ΔG to be negative, C₂ > C₁.

Teacher's Note:
1. A concentration cell works only when the cathode (right) solution is more concentrated than the anode solution.

Question 10. In the electrolysis of which solution, OH⁻ ions are discharged in preference to Cl⁻ ions?
(a) dilute NaCl
(b) very dilute NaCl
(c) fused NaCl
(d) solid NaCl
Answer:
Ans. (b)
In very dilute solution of NaCl, the following reactions take place on electrolysis:
Anode: 2OH⁻ → H₂O + \( \frac{1}{2} \)O₂ + 2e⁻; Cathode: 2H⁺ + 2e⁻ → H₂

Teacher's Note:
1. Because of overpotential, Cl₂ is evolved from concentrated brine; only in very dilute solution does O₂ come off instead.

Question 11. The metal that cannot be obtained by electrolysis of an aqueous solution of its salt is (JEE Main 2014)
(a) Cr
(b) Ag
(c) Ca
(d) Cu
Answer:
Ans. (c)
Out of the given metals, Ca is s-block metal and is highly electropositive. It cannot be obtained by electrolysis of an aqueous solution of its salt.

Teacher's Note:
1. For metals with E^o far below that of water (Ca, Na, Mg, Al), water is reduced first, so they are obtained from molten salts.

Question 12. In the lead-acid battery during charging, the cathode reaction is (AMU Engg. 2015)
(a) formation of PbO₂
(b) formation of PbSO₄
(c) reduction of Pb²⁺ to Pb
(d) decomposition of Pb at the anode
Answer:
Ans. (c)
During charging, the reactions that occur are
At cathode: PbSO₄ (s) + 2e⁻ → Pb (s) + SO₄²⁻ (aq)
At anode: PbSO₄ (s) + 2H₂O → PbO₂ (s) + SO₄²⁻ (aq) + 4H⁺ (aq) + 2e⁻
Thus, Pb²⁺ ions of PbSO₄ are reduced to Pb on the cathode while PbSO₄ is oxidized to PbO₂ at anode.
(Remember. Anode is the electrode on which oxidation takes place, i.e., loss of electrons and cathode is the electrode on which reduction takes place, i.e., gain of electrons. Further, the electrode which acts as anode during discharge acts as cathode during charging and vice-versa)

Teacher's Note:
1. Cathode always means reduction; during charging the Pb plate is the cathode and PbSO₄ on it is reduced back to Pb.

Question 13. 0.05 F charge is passed through a lead storage battery. In the anodic half cell, the amount of lead sulphate precipitated is (Molar mass of PbSO₄ = 303 g mol⁻¹) (JEE Main 2019)
(a) 22.8 g
(b) 15.2 g
(c) 7.6 g
(d) 11.4 g
Answer:
Ans. (c)
The anodic half cell reaction is Pb (s) + SO₄²⁻ (aq) → PbSO₄ (s) + 2e⁻
For 2 F current passed, PbSO₄ deposited = 303 g (Molar mass of PbSO₄ = 303 g mol⁻¹)
For 0.05 F current passed, PbSO₄ deposited \( = \frac{303}{2}\times0.05 \) g = 7.6 g

Teacher's Note:
1. Only the anode is asked: 1 mol PbSO₄ forms per 2 F there (another mole forms at the cathode).

Question 14. In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
CH₃OH (l) + \( \frac{3}{2} \)O₂ (g) → CO₂ (g) + 2H₂O (l)
At 298 K, standard Gibbs energies of formation for CH₃OH (l), H₂O (l) and CO₂ (g) are −166.2, −237.2 and −394.4 kJ mol⁻¹ respectively. If standard enthalpy of combustion of methanol is −726 kJ mol⁻¹, efficiency of the fuel cell will be (AIEEE 2009)
(a) 80%
(b) 87%
(c) 90%
(d) 97%
Answer:
Ans. (d)
Δ_rG = −394.4 + 2 (−237.2) − (−166.2) = −702.6 kJ mol⁻¹
Efficiency \( = \frac{\Delta_r G}{\Delta H} = \frac{-702.6}{-726}\times100 \) = 97%.

Teacher's Note:
1. Fuel-cell efficiency is ΔG/ΔH: the useful electrical work divided by the total heat the fuel could release.

Question 15. Rust is a mixture of
(a) FeO and Fe(OH)₂
(b) FeO and Fe(OH)₃
(c) Fe₂O₃ and Fe(OH)₃
(d) Fe₃O₄ and Fe(OH)₃
Answer:
Ans. (c)
Rust is a mixture of Fe₂O₃ and Fe(OH)₃ (hydrated ferric oxide, Fe₂O₃·xH₂O).

Teacher's Note:
1. Rust contains iron in the +3 state; Fe²⁺ formed at the anode is further oxidised by air.

Question 16. Zinc can be coated on iron to produce galvanised iron but the reverse is not possible. It is because (NEET Phase II 2016)
(a) zinc is lighter than iron
(b) zinc has lower melting point than iron
(c) zinc has lower negative electrode potential than iron
(d) zinc has higher negative electrode potential than iron
Answer:
Ans. (d)
Zinc has higher negative electrode potential than iron, so iron cannot be coated on zinc.

Teacher's Note:
1. Zinc protects iron sacrificially because it is more easily oxidised (more negative E^o); a coating of iron could not protect zinc.

Question 17. A Daniell cell is set up by dipping a zinc rod weighing 100 g in 1 litre of 1.0 M CuSO₄ solution. How long would the cell run if it delivers a steady current of 1.0 ampere? (Atomic masses: Cu = 63.5, Zn = 65)
(a) 82.47 hrs
(b) 53.61 hrs
(c) 41.23 hrs
(d) 26.80 hrs
Answer:
Ans. (b)
1 L of 1.0 M CuSO₄ solution contains Cu²⁺ = 1 mole; 100 g Zn = \( \frac{100}{65} \) mole
Thus, in the cell reaction Zn + Cu²⁺ → Zn²⁺ + Cu, Cu²⁺ ions are the limiting reagent. The cell will run till all the Cu²⁺ ions are deposited, i.e., 1 mole of Cu is deposited. Quantity of electricity required for deposition of 1 mole of Cu = 2 F
Time, \( t = \frac{2\times96500\,\mathrm{C}}{1\,\mathrm{A}} \) = 2 × 96500 s \( = \frac{2\times96500}{3600} \) hr = 53.61 hr

Teacher's Note:
1. Find the limiting reagent first: 1 mol Cu²⁺ is less than 1.54 mol Zn, so Cu²⁺ decides how long the cell runs.

Question 18. For the half-cell reaction,
2H₂O + 2e⁻ → H₂ + 2OH⁻, E^o = −0.8277 V at 298 K. Autoprotolysis constant of water calculated from this value will be
(a) 1 × 10⁻¹⁰
(b) 1 × 10⁻¹²
(c) 1 × 10⁻¹³
(d) 1 × 10⁻¹⁴
Answer:
Ans. (d)
Autoprotolysis constant of H₂O = [H⁺][OH⁻] = K_w
Redn. half reaction: 2H₂O + 2e⁻ → H₂ + 2OH⁻, E^o = −0.8277 V
Ox. half reaction: H₂ → 2H⁺ + 2e⁻, E^o = 0.0
Cell reaction: 2H₂O ⇄ 2H⁺ + 2OH⁻, E^o = −0.8277 V
\( E = E^o - \frac{0.0591}{2}\log[\mathrm{H^+}]^2[\mathrm{OH^-}]^2 \). At equilibrium, E = 0. Hence, \( E^o = \frac{0.0591}{2}\log[\mathrm{H^+}]^2[\mathrm{OH^-}]^2 \)
\( \therefore \) −0.8277 = 0.0591 log [H⁺][OH⁻] = 0.0591 log K_w
or log K_w = −14 or K_w = 10⁻¹⁴

Teacher's Note:
1. Combining this half-reaction with the SHE gives the ionisation of water, so E^o = 0.0591 log K_w.

Question 19. A lead storage battery has been used for one month (30 days) at the rate of one hour per day by drawing a constant current of 2 amperes. H₂SO₄ consumed by the battery is
(a) 1.12 mole
(b) 2.24 mole
(c) 3.36 mole
(d) 4.48 mole
Answer:
Ans. (b)
The reactions occurring during discharge are:
At anode: Pb (s) + SO₄²⁻ (aq) → PbSO₄ (s) + 2e⁻
At cathode: PbO₂ (s) + 4H⁺ (aq) + SO₄²⁻ (aq) + 2e⁻ → PbSO₄ (s) + 2H₂O (l)
Overall reaction: Pb (s) + PbO₂ (s) + 2H₂SO₄ → 2PbSO₄ (s) + 2H₂O (l)
Quantity of electricity consumed = (2 A) (30 × 3600 s) = 216000 C
2 × 96500 coulombs consume H₂SO₄ = 2 moles \( \therefore \) 216000 coulombs will consume H₂SO₄ \( = \frac{1}{96500}\times216000 \) = 2.24 moles.

Teacher's Note:
1. In a lead storage battery one mole of H₂SO₄ is used up for every faraday drawn.

Question 20. Given the data at 25°C,
Ag + I⁻ → AgI + e⁻; E^o = 0.152 V
Ag → Ag⁺ + e⁻; E^o = −0.800 V
What is the value of log K_sp for AgI? (2.303 RT/F = 0.059 V)
(a) −37.83
(b) −16.13
(c) −8.12
(d) +8.612
Answer:
Ans. (b)
Ag + I⁻ → AgI + e⁻, E^o = 0.152 V; Ag⁺ + e⁻ → Ag, E^o = 0.800 V
Ag⁺ + I⁻ → AgI, E^o_cell = 0.952 V
At equilibrium, \( E^o = \frac{2.303RT}{F}\log K_c \). But \( K_c = \frac{[\mathrm{AgI}]}{[\mathrm{Ag^+}][\mathrm{I^-}]} = \frac{1}{K_{sp}} \)
\( \therefore 0.952 = -\frac{2.303RT}{F}\log K_{sp} \) = −0.059 log K_sp or log K_sp = −16.13

Teacher's Note:
1. The precipitation reaction has K = 1/K_sp; forgetting this sign flip gives +16.13.

Question 21. The Gibbs energy for the decomposition of Al₂O₃ at 500°C is as follows:
\( \frac{2}{3} \)Al₂O₃ → \( \frac{4}{3} \)Al + O₂, Δ_rG = +966 kJ mol⁻¹
The potential difference needed for electrolytic reduction of Al₂O₃ at 500°C is at least (AIEEE 2010)
(a) 2.5 V
(b) 5.0 V
(c) 4.5 V
(d) 3.0 V
Answer:
Ans. (a)
\( \frac{2}{3} \)Al₂O₃ → \( \frac{4}{3} \)Al + O₂, Δ_rG = +966 kJ mol⁻¹
Thus, \( \frac{2}{3} \) × (3O²⁻), i.e., 2O²⁻ → O₂, n = 4
ΔG = −nFE \( \therefore \) 966 × 10³ = −4 × 96500 × E or E = −2.50 V
Thus, minimum potential difference required = 2.50 V.

Teacher's Note:
1. The minimum electrolysis voltage equals ΔG/(nF) for the non-spontaneous reaction; n comes from 4/3 Al³⁺ taking 4 electrons.

Question 22. For the cell reaction
2H₂ (g) + O₂ (g) → 2H₂O (l), Δ_rS^o_{298 K} = −0.32 kJ K⁻¹
What is the value of Δ_fH^o (H₂O)?
Given: O₂ (g) + 4H⁺ (aq) + 4e⁻ → 2H₂O (l), E^o = 1.23 V
(a) −189.71 kJ mol⁻¹
(b) −285.08 kJ mol⁻¹
(c) −379.42 kJ mol⁻¹
(d) −570.16 kJ mol⁻¹
Answer:
Ans. (b)
For the given cell reaction involving formation of 2 moles of H₂O, ΔG^o = −nFE^o_cell = −4 × 96500 × 1.23 J = −474780 J = −474.78 kJ (n = 4 for the given cell reaction)
ΔG^o = ΔH^o − TΔS^o or ΔH^o = ΔG^o + TΔS^o = −474.78 kJ + 298 K (−0.32 kJ K⁻¹) = −474.8 − 95.36 kJ = −570.16 kJ
This is enthalpy change for the formation of 2 moles of H₂O.
\( \therefore \) Enthalpy change for the formation of 1 mole of H₂O (Δ_fH^o) \( = -\frac{570.16}{2} \) = −285.08 kJ

Teacher's Note:
1. Δ_fH^o is per mole of product, so halve the value obtained for 2 mol of water; option (d) is the trap.

Question 23. AgNO₃ (aq) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. The plot of conductance (Λ) versus the volume of AgNO₃ is (IIT 2011)(a) (P)
(b) (Q)
(c) (R)
(d) (S)
Answer:
Ans. (d)
[K⁺ (aq) + Cl⁻ (aq)] + Ag⁺ (aq) + NO₃⁻ (aq) → K⁺ (aq) + NO₃⁻ (aq) + AgCl (s)
Initially the conductance is due to K⁺ and Cl⁻ ions. On adding AgNO₃, precipitate of AgCl is formed. Thus, some Cl⁻ ions are replaced by NO₃⁻ ions. As both have nearly the same conductance, the conductance remains almost constant. After the end point, addition of AgNO₃ will give Ag⁺ and NO₃⁻ ions in the solution. Hence, conductance will increase rapidly.

Teacher's Note:
1. In precipitation titrations, ask which ion replaces which: Cl⁻ replaced by NO₃⁻ of similar mobility keeps the conductance flat until the end point.

Question 24. The electrode potentials for Cu²⁺ (aq) + e⁻ → Cu⁺ (aq) and Cu⁺ (aq) + e⁻ → Cu (s) are +0.15 V and +0.50 V respectively. The value of E^o_Cu²⁺/Cu will be (AIPMT 2011)
(a) 0.150 V
(b) 0.500 V
(c) 0.325 V
(d) 0.650 V
Answer:
Ans. (c)
Cu²⁺ + e⁻ → Cu⁺, ΔG₁ = −nFE^o = −1 × F × 0.15
Cu⁺ + e⁻ → Cu, ΔG₂ = −1 × F × 0.50
Cu²⁺ + 2e⁻ → Cu, ΔG₃ = −2 × F × E^o_Cu²⁺/Cu; ΔG₃ = ΔG₁ + ΔG₂
−2F E^o_Cu²⁺/Cu = −0.15F + (−0.50F) or −2E^o_Cu²⁺/Cu = −0.15 − 0.50 = −0.65
or E^o_Cu²⁺/Cu = \( \frac{0.65}{2} \) = 0.325 V

Teacher's Note:
1. When adding two half-reactions to get a third half-reaction, add ΔG (nE), not E; the answer is the electron-weighted average.

Question 25. How long (approximate) should water be electrolysed by passing through 100 amperes current so that oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u) (JEE Main 2018)
(a) 6.4 hours
(b) 0.8 hour
(c) 3.2 hours
(d) 1.6 hours
Answer:
Ans. (c)
B₂H₆ + 3O₂ → B₂O₃ + 3H₂O
Molar mass of B₂H₆ = 2 × 10.8 + 6 × 1 = 27.6 g mol⁻¹
27.66 g of B₂H₆ ≈ 1 mole of B₂H₆. It requires 3 moles of O₂ for complete burning.
H₂O → H₂ + \( \frac{1}{2} \)O₂; \( \frac{1}{2} \) mole of O₂ requires 2 faradays \( \therefore \) 3 moles of O₂ will require 12 faradays
12 × 96500 = i × t = 100 × t or \( t = \frac{12\times96500}{100} \) s \( = \frac{12\times96500}{100\times3600} \) hrs = 3.2 hrs

Teacher's Note:
1. Each mole of O₂ from water needs 4 F; 3 mol O₂ therefore needs 12 F.

Question 26. Consider the change in oxidation state of bromine corresponding to different emf values as shown in the diagram below:
BrO₄⁻ —(1.82 V)→ BrO₃⁻ —(1.5 V)→ HBrO —(1.595 V)→ Br₂ —(1.0652 V)→ Br⁻
Then the species under disproportionation is (NEET 2018)
(a) Br
(b) BrO₄⁻
(c) BrO₃⁻
(d) HBrO
Answer:
Ans. (d)
HBrO (+1) → Br₂ (0), E^o_HBrO/Br2 = 1.595 V; HBrO (+1) → BrO₃⁻ (+5), E^o_BrO3⁻/HBrO = 1.5 V
E^o_cell for the disproportionation of HBrO, E^o_cell = E^o_HBrO/Br2 − E^o_BrO3⁻/HBrO = 1.595 − 1.5 = 0.095 V = +ve
Hence, option (d) is correct.

Teacher's Note:
1. A species disproportionates when the E^o on its right (reduction) is greater than the E^o on its left in a Latimer diagram.

Question 27. The standard electrode potential E^o and its temperature coefficient \( \left(\frac{dE^o}{dT}\right) \) for a cell are 2 V and −5 × 10⁻⁴ V K⁻¹ at 300 K respectively. The cell reaction is
Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)
The standard reaction enthalpy (Δ_rH^o) at 300 K in kJ mol⁻¹ is (Use R = 8 J K⁻¹ mol⁻¹ and F = 96,000 C mol⁻¹) (JEE Main 2019)
(a) −412.8
(b) −384.0
(c) 206.4
(d) 192.0
Answer:
Ans. (a)
\( \Delta_r H^o = -nFE^o + nFT\left(\frac{\Delta E^o}{\Delta T}\right) \)
= −2 × 96000 × 2 + 2 × 96000 × 300 × (−5 × 10⁻⁴)
= −384000 − 28800 = −412800 J mol⁻¹ = −412.8 kJ mol⁻¹

Teacher's Note:
1. ΔS^o = nF(dE^o/dT); with a negative temperature coefficient, ΔH^o is more negative than ΔG^o.

Question 28. Calculate the standard cell potential (in V) of the cell in which the following reaction takes place
Fe²⁺ (aq) + Ag⁺ (aq) → Fe³⁺ (aq) + Ag (s)
Given that E^o_Ag⁺/Ag = x V; E^o_Fe²⁺/Fe = y V; E^o_Fe³⁺/Fe = z V (JEE Main 2019)
(a) x − z
(b) x − y
(c) x + 2y − 3z
(d) x + y − z
Answer:
Ans. (c)
Ag⁺ + e⁻ → Ag, E^o = x V \( \therefore \) ΔG^o₁ = −Fx ...(i)
Fe²⁺ + 2e⁻ → Fe, E^o = y V \( \therefore \) ΔG^o₂ = −2Fy ...(ii)
Fe³⁺ + 3e⁻ → Fe, E^o = z V \( \therefore \) ΔG^o₃ = −3Fz ...(iii)
Aim: Fe²⁺ + Ag⁺ → Fe³⁺ + Ag. Adding eqns. (i) and (ii) and subtracting eqn. (iii), we get Fe²⁺ + Ag⁺ → Fe³⁺ + Ag, ΔG = −FE^o_cell (n = 1)
\( \therefore \) ΔG = ΔG^o₁ + ΔG^o₂ − ΔG^o₃; −FE^o_cell = −Fx − 2Fy − (−3Fz) or E^o_cell = x + 2y − 3z

Teacher's Note:
1. Combine the reactions through ΔG = −nFE so that the electron numbers are carried along correctly.

One or More Correct Answers

Question 29. In a galvanic cell, the salt bridge (JEE Advanced 2014)
(a) does not participate chemically in the cell reaction
(b) stops the diffusion of ions from one electrode to another
(c) is necessary for the occurrence of the cell reaction
(d) ensures mixing of the two electrolytic solutions
Answer:
Ans. (a, b)
Salt bridge does not participate chemically in the cell reaction because the amount of electrolyte in the bridge remains unchanged. It keeps the solutions in the two half-cells electrically neutral. It simply allows the passage of ions through it. It prevents transference or diffusion of ions from one half-cell to the other.

Teacher's Note:
1. The salt bridge completes the circuit and keeps each half-cell neutral; the reaction itself can occur without it (e.g., Zn in CuSO₄).

Question 30. Some standard electrode potentials at 298 K are given below:
Pb²⁺/Pb = −0.13 V, Ni²⁺/Ni = −0.24 V
Cd²⁺/Cd = −0.40 V, Fe²⁺/Fe = −0.44 V
To a solution containing 0.001 M of X²⁺ and 0.1 M of Y²⁺, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X. The correct combination(s) of X and Y, respectively, is (are) (Given: Gas constant, R = 8.314 J K⁻¹; Faraday constant, F = 96500 C mol⁻¹) (JEE Advanced 2021)
(a) Cd and Ni
(b) Cd and Fe
(c) Ni and Pb
(d) Ni and Fe
Answer:
Ans. (a, b, c)
(a) Cd + Ni²⁺ → Cd²⁺ + Ni
\( E_{cell} = 0.40 + (-0.24) - \frac{0.0591}{2}\log\frac{0.001}{0.1} = 0.16 + \frac{0.0591}{2}\times2 \) = 0.16 + 0.059 = 0.22 V (+ve)
Similarly, test for (b), (c) and (d). We get (b) = 0.02 V (+ve), (c) = 0.17 V (+ve), (d) = −0.14 V (−ve).

Teacher's Note:
1. X dissolves, so X must be the anode: check that E^o_Y − E^o_X + 0.059 is positive for each pair.

Passage-Based Questions

Question 31. Comprehension 3 (IIT 2010): The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is
M(s) | M⁺ (aq; 0.05 molar) || M⁺ (aq; 1 molar) | M(s)
For the above electrolytic cell, the magnitude of the cell potential | E_cell | = 70 mV.
If the 0.05 molar solution of M⁺ is replaced by a 0.0025 molar M⁺ solution, then the magnitude of the cell potential would be
(a) 35 mV
(b) 70 mV
(c) 140 mV
(d) 700 mV
Answer:
Ans. (c)
From the above equation, \( \frac{2.303RT}{F}\log\frac{1}{0.05} \) = 70 mV = 70 × 10⁻³ V = 0.07 V
\( \therefore \frac{2.303RT}{F} \)(1.301) = 0.07 or \( \frac{2.303RT}{F} = \frac{0.07}{1.301} \) = 0.0538
For the new concentration, \( E_{cell} = \frac{2.303RT}{F}\log\frac{1}{0.0025} \) = 0.0538 log 400 = 0.0538 × 2.6021 = 0.140 V = 140 mV

Teacher's Note:
1. Notice 0.0025 = (0.05)², so log of the ratio doubles and the emf doubles to 140 mV.

Question 32. Comprehension 4 (IIT 2012): The electrochemical cell shown below is a concentration cell: M | M²⁺ (saturated solution of sparingly soluble salt MX₂) || M²⁺ (0.001 mol dm⁻³) | M. The emf of the cell depends on the difference in the concentration of M²⁺ ions in the two electrodes. The emf of the cell at 298 K is 0.059 V.
The solubility product (K_sp; mol³ dm⁻⁹) of MX₂ at 298 K based on the information available for the given concentration cell is (take 2.303 × R × 298/F = 0.059 V)
(a) 1 × 10⁻¹⁵
(b) 4 × 10⁻¹⁵
(c) 1 × 10⁻¹²
(d) 4 × 10⁻¹²
Answer:
Ans. (b)
The given concentration cell is M | M²⁺ (Saturated = C₁) || M²⁺ (0.001 M = C₂) | M
EMF of concentration cell is \( E_{cell} = \frac{2.303RT}{nF}\log\frac{C_2}{C_1} = \frac{0.059}{n}\log\frac{C_2}{C_1} \)
\( \therefore 0.059 = \frac{0.059}{2}\log\frac{0.001}{C_1} \) or \( 2 = \log\frac{10^{-3}}{C_1} \) or log 10⁻³ − log C₁ = 2
or −3 − log C₁ = 2 or log C₁ = −5 or C₁ = 10⁻⁵ M
For the salt MX₂, MX₂ ⇄ M²⁺ (s) + 2X⁻ (2s)
K_sp = S (2S)² = 4S³ = 4 × (10⁻⁵)³ = 4 × 10⁻¹⁵

Teacher's Note:
1. [M²⁺] in the saturated solution equals the solubility s, but [X⁻] = 2s, so K_sp = 4s³.

Question 33. Comprehension 4 (IIT 2012): The electrochemical cell shown below is a concentration cell: M | M²⁺ (saturated solution of sparingly soluble salt MX₂) || M²⁺ (0.001 mol dm⁻³) | M. The emf of the cell depends on the difference in the concentration of M²⁺ ions in the two electrodes. The emf of the cell at 298 K is 0.059 V.
The value of ΔG (kJ mol⁻¹) for the given cell is (take 1 F = 96500 C mol⁻¹)
(a) −5.7
(b) 5.7
(c) 11.4
(d) −11.4
Answer:
Ans. (d)
ΔG = −nFE_cell = −2 × 96500 × 0.059 J mol⁻¹ = −11387 J mol⁻¹ = −11.38 kJ mol⁻¹ ≈ −11.4 kJ mol⁻¹

Teacher's Note:
1. Use n = 2 for M²⁺; a working cell with positive emf always has negative ΔG.

Matrix Match Questions

Teacher's Note:
1. Only metals with negative E^o (above hydrogen in the series) displace H₂ from dilute acids.

Teacher's Note:
1. Remember the dry cell as the only acidic-type electrolyte here (NH₄Cl/ZnCl₂); the others use alkaline KOH.

Integer and Numerical Answer Questions

Question 36. The cell potential for the following cell
Pt | H₂ (g) | H⁺ (aq) || Cu²⁺ (0.01 M) | Cu (s)
is 0.576 V at 298 K. The pH of the solution is .......... (Nearest integer) (Given: E^o_Cu²⁺/Cu = 0.34 V and \( \frac{2.303RT}{F} \) = 0.06 V) (JEE Main 2021)
Answer:
Ans. 5
H₂ (g) + Cu²⁺ (aq) → 2H⁺ (aq) + Cu (n = 2)
\( E_{cell} = E^o_{cell} - \frac{0.06}{2}\log\frac{[\mathrm{H^+}]^2}{[\mathrm{Cu^{2+}}]} \); \( 0.576 = 0.34 - 0.03\log\frac{[\mathrm{H^+}]^2}{0.01} \)
0.576 − 0.34 = −0.03 log [H⁺]² + 0.03 log 10⁻²
or 0.236 = 0.06 pH − 0.06 or 0.06 pH = 0.296
or pH \( = \frac{0.296}{0.06} \) = 4.93 ≈ 5

Teacher's Note:
1. −0.03 log [H⁺]² = +0.06 pH; keep track of the 0.01 M Cu²⁺ term, which subtracts 0.06 V.

Question 37. Consider an electrochemical cell
A (s) | Aⁿ⁺ (aq, 2 M) || B²ⁿ⁺ (aq, 1 M) | B (s)
The value of ΔH^o for the cell reaction is twice that of ΔG^o at 300 K. If the e.m.f of the cell is zero, the ΔS^o (in J K⁻¹ mol⁻¹) of the cell reaction per mole of B formed at 300 K is ........... (Given ln (2) = 0.7, R (universal gas constant) = 8.3 J K⁻¹ mol⁻¹. H, S and G are enthalpy, entropy and Gibbs energy respectively) (numerical value, two decimal places) (JEE Advanced 2018)
Answer:
Ans. −11.62
For the given cell, A (s) → Aⁿ⁺ (aq) + ne⁻] × 2; B²ⁿ⁺ (aq) + 2ne⁻ → B
Overall reaction: 2A (s) + B²ⁿ⁺ (aq) → 2Aⁿ⁺ (aq) + B
\( E = E^o - \frac{RT}{2nF}\ln\frac{[\mathrm{A^{n+}}]^2}{[\mathrm{B^{2n+}}]} \); \( 0 = E^o - \frac{RT}{2nF}\ln\frac{(2)^2}{1} \) or \( E^o = \frac{RT}{2nF}\ln4 \)
\( \Delta G^o = -2nFE^o = -2nF\frac{RT}{2nF}\ln4 = -RT\ln4 \)
ΔG^o = ΔH^o − TΔS^o. But ΔH^o = 2ΔG^o (Given) \( \therefore \) ΔG^o = 2ΔG^o − TΔS^o or TΔS^o = ΔG^o
or \( \Delta S^o = \frac{\Delta G^o}{T} = \frac{-RT\ln4}{T} = -R\ln4 \) = −8.3 × 2 × 0.7 = −11.62 J K⁻¹ mol⁻¹ (ln 4 = ln 2² = 2 ln 2 = 2 × 0.7)

Teacher's Note:
1. Zero emf means the cell is at equilibrium under these concentrations, so ΔG^o = −RT ln Q with Q = 4.

Question 38. Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is
H₂ (g) + \( \frac{1}{2} \)O₂ (g) → H₂O (l)
The work derived from the cell on the consumption of 1.0 × 10⁻³ mol of H₂ (g) is used to compress 1.00 mol of a monoatomic ideal gas in a thermally insulated container. What is the change on temperature (in K) of the ideal gas?
The standard reduction potentials of the two half cell are given below:
O₂ (g) + 4H⁺ (aq) + 4e⁻ → 2H₂O (l), E^o = 1.23 V
2H⁺ (aq) + 2e⁻ → H₂, E^o = 0.00 V
Use F = 96500 C mol⁻¹, R = 8.314 J mol⁻¹ K⁻¹ (numerical value, two decimal places) (JEE Advanced 2020)
Answer:
Ans. 13.32
For the given cell reaction, n = 2
ΔG^o = −nFE^o_cell; work available = 2 × 96500 × 1.23 × 10⁻³ × 0.70 = 166.173 J
w = −ΔG^o = 166.173 J
From thermodynamics (studied in class 11), \( w = \frac{nR(\Delta T)}{\gamma - 1} \). For monoatomic gas, \( \gamma = \frac{5}{3} \), so γ − 1 = \( \frac{2}{3} \)
As 1 mol of gas is to be compressed, n = 1 \( \therefore 166.173 = \frac{1\times8.314\times\Delta T}{2/3} \) or ΔT = 13.32 K

Teacher's Note:
1. In an insulated container q = 0, so all the work goes into internal energy: w = nC_vΔT with C_v = 3R/2.

Question 39. The reduction potential (E^o, in V) of MnO₄⁻ (aq) | Mn (s) is ..........
(Given: E^o_{MnO4⁻(aq)/MnO2(s)} = 1.68 V; E^o_{MnO2(s)/Mn²⁺(aq)} = 1.21 V; E^o_{Mn²⁺(aq)/Mn(s)} = −1.03 V) (numerical value, two decimal places) (JEE Advanced 2022)
Answer:
Ans. 0.77
MnO₄⁻ (+7) + 3e⁻ → MnO₂ (+4), E^o₁ = 1.68 V
MnO₂ (+4) + 2e⁻ → Mn²⁺ (+2), E^o₂ = 1.21 V
Mn²⁺ (+2) + 2e⁻ → Mn (0), E^o₃ = −1.03 V
MnO₄⁻ (+7) + 7e⁻ → Mn (0), E^o₄ = ?
ΔG^o₄ = ΔG^o₁ + ΔG^o₂ + ΔG^o₃; −7FE^o₄ = −3FE^o₁ − 2FE^o₂ − 2FE^o₃
7E^o₄ = 3E^o₁ + 2E^o₂ + 2E^o₃; \( E^o_4 = \frac{3\times1.68 + 2\times1.21 - 2\times1.03}{7} \) = 0.77 V

Teacher's Note:
1. Weight each step's E^o by its number of electrons and divide by the total (7).

Question 40. Question stem (JEE Advanced 2021): At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 × 10² S cm² mol⁻¹. At 298 K, for an aqueous solution of the acid, the degree of dissociation is α and the molar conductivity is y × 10² S cm² mol⁻¹. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y × 10² S cm² mol⁻¹.
The value of α is .......... (numerical value, two decimal places)
Answer:
Ans. 0.22
\( \alpha_1 = \frac{\Lambda_m^c}{\Lambda_m^o} = \frac{y\times10^2}{4\times10^2} = \frac{y}{4} = \alpha \) ...(i)
\( \alpha_2 = \frac{3y\times10^2}{4\times10^2} = 3\alpha_1 = 3\alpha \)
HA ⇄ H⁺ + A⁻; C(1 − α), Cα, Cα; \( K_a = \frac{C\alpha^2}{1-\alpha} \). As temperature is constant, K_a will be constant.
\( \frac{C_1\alpha_1^2}{1-\alpha_1} = \frac{C_2\alpha_2^2}{1-\alpha_2} \); \( \frac{C\alpha^2}{1-\alpha} = \frac{(C/20)(3\alpha)^2}{1-3\alpha} \) or \( \frac{1}{1-\alpha} = \frac{9/20}{1-3\alpha} \)
or \( \frac{1}{1-\alpha} = \frac{9}{20(1-3\alpha)} \) or 20 − 60α = 9 − 9α or 51α = 11
or \( \alpha = \frac{11}{51} \) = 0.2156 = 0.22

Teacher's Note:
1. Dilution changes α but not K_a; equate the two K_a expressions and do not use the approximation 1 − α ≈ 1 here.

Assertion–Reason Questions

Question 41. Assertion: In the electrolysis of aqueous NaCl, Na is preferentially discharged at the mercury cathode forming sodium amalgam.
Reason: It is due to the fact that hydrogen has high overvoltage at mercury cathode. (AIIMS 2007)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A.

Teacher's Note:
1. This is the basis of the Castner-Kellner process: high hydrogen overvoltage on Hg allows Na to be discharged as amalgam.

Question 42. Assertion: The cell potential of mercury cell is 1.35 V which remains constant.
Reason: In mercury cell, the electrolyte is a paste of KOH and HgO. (AIIMS 2008)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. The cell reaction, Zn + HgO → ZnO + Hg does not involve any ion whose concentration may change (OH⁻ ions consumed in one half reaction are produced in the second half reaction).

Teacher's Note:
1. A cell voltage stays constant when the overall reaction contains no dissolved species, so Q does not change during use.

Question 43. Assertion: As a lead storage battery gets discharged, density of the electrolyte present in it decreases.
Reason: Lead and lead dioxide both react with sulphuric acid to form lead sulphate. (AIIMS 2011)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A because sulphuric acid is consumed due to formation of lead sulphate and hence density decreases.

Teacher's Note:
1. Measuring the acid density with a hydrometer is a simple way to check the state of charge of a lead storage battery.

Question 44. Assertion: Galvanised iron does not rust.
Reason: Zinc has a more negative electrode potential than iron.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
R is the correct explanation of A (More negative electrode potential of zinc means higher oxidation potential. Hence, zinc is more easily oxidized than iron).

Teacher's Note:
1. Even if the zinc layer is scratched, zinc is oxidised in preference to iron, so the iron remains protected.