CBSE Class 12 Chemistry Electrochemistry NEET JEE Questions Set 02

Single Correct Answer

Question 1. Which of the following statements are correct concerning redox properties?
(i) A metal M for which E^o for the half life reaction Mⁿ⁺ + ne⁻ ⇄ M is very negative will be a good reducing agent.
(ii) The oxidizing power of the halogen decreases from chlorine to iodine.
(iii) The reducing power of hydrogen halides increases from hydrogen chloride to hydrogen iodide
(a) (i), (ii) and (iii)
(b) (i) and (ii)
(c) (i) only
(d) (ii) and (iii) only
(e) (iii) only
Answer:
Ans. (a)
(i) High negative E^o for Mⁿ⁺ + ne⁻ ⇄ M means a high positive value for the reverse reaction, M ⇄ Mⁿ⁺ + ne⁻. This means M is oxidized very easily. Hence, it is a good reducing agent.
(ii) Reduction potentials of halogens are in the order: Cl₂ > Br₂ > I₂. Thus, Cl₂ is reduced most easily and hence is the best oxidizing agent.
(iii) The size of the halide ions is in the order Cl⁻ < Br⁻ < I⁻. Greater the size of the halide ion, more easily it can lose electrons and get oxidized. Thus, I⁻ ions can be oxidized most easily and hence have the greatest reducing power.

Teacher's Note:
1. Oxidising power and reducing power run in opposite directions down the halogen group: Cl₂ best oxidant, I⁻ (HI) best reductant.

Question 2. The standard reduction potentials for Zn²⁺/Zn, Ni²⁺/Ni and Fe²⁺/Fe are −0.76, −0.23 and −0.44 V respectively. The reaction X + Y²⁺ → X²⁺ + Y will be spontaneous when (AIEEE 2012)
(a) X = Zn, Y = Ni
(b) X = Ni, Y = Fe
(c) X = Ni, Y = Zn
(d) X = Fe, Y = Zn
Answer:
Ans. (a)
The reaction X + Y²⁺ → X²⁺ + Y will be spontaneous if E^o_cell is +ve.
(a) Zn + Ni²⁺ → Zn²⁺ + Ni, E^o_cell = E^o_Ni²⁺/Ni − E^o_Zn²⁺/Zn = −0.23 − (−0.76) V = +0.53 V
(b) Ni + Fe²⁺ → Ni²⁺ + Fe, E^o_cell = E^o_Fe²⁺/Fe − E^o_Ni²⁺/Ni = −0.44 − (−0.23) V = −0.21 V
(c) Ni + Zn²⁺ → Ni²⁺ + Zn, E^o_cell = −0.53 V from (a)
(d) Fe + Zn²⁺ → Fe²⁺ + Zn, E^o_cell = E^o_Zn²⁺/Zn − E^o_Fe²⁺/Fe = −0.76 − (−0.44) V = −0.32 V

Teacher's Note:
1. A metal displaces another from solution only if it lies lower (more negative E^o) in the electrochemical series.

Question 3. A button cell used in watches functions as follows:
Zn (s) + Ag₂O (s) + H₂O (l) ⇄ 2Ag (s) + Zn²⁺ (aq) + 2OH⁻ (aq)
The half-cell potentials are
Zn²⁺ (aq) + 2e⁻ → Zn (s); E^o = −0.76 V
Ag₂O (s) + H₂O (l) + 2e⁻ → 2Ag (s) + 2OH⁻ (aq); E^o = 0.34 V
The cell potential will be (AIPMT 2013)
(a) 1.10 V
(b) 0.42 V
(c) 0.84 V
(d) 1.34 V
Answer:
Ans. (a)
E^o_cell = E^o_Cathode − E^o_Anode = 0.34 − (−0.76) V = 1.10 V

Teacher's Note:
1. Use reduction potentials in E_cathode − E_anode; do not change the sign or multiply E^o by coefficients.

Question 4. A solution contains Fe²⁺, Fe³⁺ and I⁻ ions. This solution was treated with iodine at 35°C. E^o for Fe³⁺/Fe²⁺ is +0.77 V and E^o for I₂/2I⁻ is 0.536 V. The favourable redox reaction is (AIPMT Main 2011)
(a) I₂ will be reduced to I⁻
(b) There will be no redox reaction
(c) I⁻ will be oxidised to I₂
(d) Fe²⁺ will be oxidised to Fe³⁺
Answer:
Ans. (c)
The reaction will take place for which E^o_cell is positive.
(a) I₂ + 2e⁻ → 2I⁻, E^o_red = 0.536 V; Fe²⁺ → Fe³⁺ + e⁻, E^o_ox = −0.770 V. \( \therefore \) E^o_cell will be −ve.
(b) Redox reaction will occur.
(c) 2I⁻ → I₂ + 2e⁻, E^o_ox = −0.536 V; Fe³⁺ + e⁻ → Fe²⁺, E^o_red = +0.770 V
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, E^o_cell = +0.164 V. Thus, E^o_cell is +ve. Hence, this reaction will occur.
(d) also will not occur.

Teacher's Note:
1. The couple with the higher E^o (Fe³⁺/Fe²⁺) is reduced and the lower one (I₂/I⁻) runs as oxidation.

Question 5. Standard electrode potentials are
Fe²⁺/Fe, E^o = −0.44 V
Fe³⁺/Fe²⁺, E^o = 0.77 V
Fe²⁺, Fe³⁺ and Fe blocks are kept together, then
(a) Fe³⁺ increases
(b) Fe³⁺ decreases
(c) Fe²⁺/Fe³⁺ remains unchanged
(d) Fe²⁺ decreases
Answer:
Ans. (b)
For the emf to be +ve, the following half-cell reactions will occur.
Fe → Fe²⁺ + 2e⁻, E^o = +0.44 V; 2Fe³⁺ + 2e⁻ → 2Fe²⁺, E^o = +0.77 V
Overall reaction: Fe + 2Fe³⁺ → 3Fe²⁺, E^o_cell = 1.21 V. Thus, Fe³⁺ will decrease.

Teacher's Note:
1. Iron metal reduces Fe³⁺ to Fe²⁺ (comproportionation); this is why iron nails are added to keep FeSO₄ solutions from oxidising.

Question 6. Cr₂O₇²⁻ + I⁻ → I₂ + Cr³⁺
E^o_cell = 0.79 V, E^o_Cr2O7²⁻ = 1.33 V, E^o_I2 = ?
(a) 0.54 V
(b) −0.054 V
(c) +0.18 V
(d) −0.18 V
Answer:
Ans. (a)
In the given reaction, I⁻ has been oxidized to I₂ and Cr₂O₇²⁻ ions have been reduced to Cr³⁺
\( \therefore \) E^o_cell = E^o_Cr2O7²⁻ − E^o_I2, i.e., 0.79 = 1.33 − E^o_I2 or E^o_I2 = 0.54 V

Teacher's Note:
1. Identify which species is reduced (cathode) before writing E_cathode − E_anode.

Question 7. An aqueous solution containing one mole per litre of each Cu(NO₃)₂, AgNO₃, Hg₂(NO₃)₂ and Mg(NO₃)₂ is being electrolysed using inert electrodes. The values of the standard electrode potentials in volts (reduction potentials) are
Ag | Ag⁺ = 0.80, 2Hg | Hg₂²⁺ = +0.79, Cu | Cu²⁺ = +0.34, Mg | Mg²⁺ = −2.37
With increasing voltage, the sequence of deposition of metals on cathode will be
(a) Ag, Hg, Cu, Mg
(b) Mg, Cu, Hg, Ag
(c) Ag, Hg, Cu
(d) Cu, Hg, Ag
(e) Cu, Hg, Ag, Mg
Answer:
Ans. (c)
Higher the reduction potential, more easily is the metal deposited (Mⁿ⁺ + ne⁻ → M). Mg²⁺ ions will not be deposited because H₂O is reduced more easily than Mg²⁺ ions.

Teacher's Note:
1. In aqueous electrolysis, ions of very reactive metals (Mg, Na, Al) are never deposited; water is reduced to H₂ instead.

Question 8. The EMF of a cell formed by combining a particular electrode with standard calomel electrode is found to be 0.344 V and calomel electrode is found to act as cathode. If the same electrode is combined with standard hydrogen electrode, the EMF of the cell will be (Given standard reduction potential, E^o_calomel = +0.244 V)
(a) 0.344 V
(b) 0.244 V
(c) 0.588 V
(d) 0.100 V
Answer:
Ans. (d)
E_cell = E^o_cathode − E^o_anode
0.344 = E^o_calomel − E^o_X, 0.344 = 0.244 − E^o_X or E^o_X = −0.100 V
As E^o_H⁺/H2 = 0, E^o_cell with SHE = 0.100 V.

Teacher's Note:
1. First find the electrode's own E^o (here −0.100 V); the emf against SHE is just its magnitude.

Question 9. Given below are the half cell reactions:
Mn²⁺ + 2e⁻ → Mn, E^o = −1.18 V
2(Mn³⁺ + e⁻ → Mn²⁺), E^o = +1.51 V
The E^o for 3Mn²⁺ → Mn + 2Mn³⁺ will be (JEE Main 2014)
(a) −0.33 V, the reaction will occur
(b) −2.69 V, the reaction will not occur
(c) −2.69 V, the reaction will occur
(d) −0.33 V, the reaction will not occur
Answer:
Ans. (b)
The given cell reaction can be obtained from half-cell reactions as under:
Mn²⁺ + 2e⁻ → Mn, E^o = −1.18 V; 2Mn²⁺ → 2Mn³⁺ + 2e⁻, E^o = −1.51 V
Overall reaction: 3Mn²⁺ → Mn + 2Mn³⁺, E^o = −1.18 V + (−1.51) = −2.69 V
As E^o_cell is −ve, reaction will not occur.

Teacher's Note:
1. Electrons cancel here (2 in each half), so the E^o values add directly; a negative E^o_cell means non-spontaneous.

Question 10. Given below are half cell reactions:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, E^o_{Mn²⁺/MnO4⁻} = −1.510 V
\( \frac{1}{2} \)O₂ + 2H⁺ + 2e⁻ → H₂O, E^o_O2/H2O = +1.223 V
Will the permanganate ion, MnO₄⁻ liberate O₂ from water in the presence of an acid? (NEET 2022)
(a) Yes, because E^o_cell = +0.287 V
(b) No, because E^o_cell = −0.287 V
(c) Yes, because E^o_cell = +2.733 V
(d) No, because E^o_cell = −2.733 V
Answer:
Ans. (a)
We have to check whether the cell formed from the following two half-cell reactions will have +ve or −ve cell voltage.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, E^o_Red = +1.510 V
H₂O → \( \frac{1}{2} \)O₂ + 2H⁺ + 2e⁻, E^o_Ox = −1.223 V
Overall E^o_cell = E^o_Red + E^o_Ox = +1.510 + (−1.223) V = +0.287 V
As it is +ve, cell reaction will take place, i.e., O₂ will be liberated and E^o_cell = +0.287 V

Teacher's Note:
1. Note that the value −1.510 V is given for Mn²⁺/MnO₄⁻ (oxidation); the reduction potential of MnO₄⁻/Mn²⁺ is +1.510 V.

Question 11. A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of the electrode would be (AIPMT 2013)
(a) 0.059 V
(b) 0.59 V
(c) 0.118 V
(d) 1.18 V
Answer:
Ans. (b)
For hydrogen electrode, reduction reaction is H⁺ + e⁻ → \( \frac{1}{2} \)H₂
\( E_{\mathrm{H^+/\frac{1}{2}H_2}} = E^o_{\mathrm{H^+/\frac{1}{2}H_2}} - \frac{0.0591}{n}\log\frac{p_{\mathrm{H_2}}^{1/2}}{[\mathrm{H^+}]} = 0 - \frac{0.0591}{1}\log\frac{1}{10^{-10}} \) = −0.591 V
(pH = 10 means [H⁺] = 10⁻¹⁰ M)
\( \therefore \) Oxidation potential = 0.591 V

Teacher's Note:
1. For a hydrogen electrode at 1 atm, E_red = −0.059 pH, so the oxidation potential is +0.059 pH.

Question 12. How much will the reduction potential of a hydrogen electrode change when its solution initially at pH = 0 is neutralized to pH = 7?
(a) Increases by 0.059 V
(b) Decreases by 0.059 V
(c) Increases by 0.41 V
(d) Decreases by 0.41 V
Answer:
Ans. (d)
H⁺ + e⁻ → \( \frac{1}{2} \)H₂
\( E_{red} = E^o_{red} - \frac{0.0591}{1}\log\frac{1}{[\mathrm{H^+}]} \) = E^o_red + 0.0591 log [H⁺]
E^o_red = 0. When pH = 0, [H⁺] = 10⁰ = 1 M \( \therefore \) E_red = 0
When pH = 7, [H⁺] = 10⁻⁷ M, E_red = 0.0591 log 10⁻⁷ = (0.0591)(−7) = −0.4137
Thus, E_red decreases by 0.41 V.

Teacher's Note:
1. Each unit rise in pH lowers the hydrogen electrode's reduction potential by about 0.059 V at 298 K.

Question 13. Calculate the reduction potential of a half-cell consisting of platinum electrode immersed in 2.0 M Fe²⁺ and 0.02 M Fe³⁺.
Given E^o_{Fe³⁺/Fe²⁺} = 0.771 V. (J & K CET 2010)
(a) 0.653 V
(b) 0.889 V
(c) 0.683 V
(d) 2.771 V
Answer:
Ans. (a)
Fe³⁺ + e⁻ → Fe²⁺
\( E_{red} = E^o_{\mathrm{Fe^{3+}/Fe^{2+}}} - \frac{0.0591}{1}\log\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]} = 0.771 - 0.0591\log\frac{2.0}{0.02} \)
= 0.771 − 0.0591 × log 100 = 0.771 − 0.0591 × 2 = 0.653 V.

Teacher's Note:
1. In the Nernst quotient for a reduction, the reduced form (Fe²⁺) goes in the numerator.

Question 14. E₁, E₂ and E₃ are the emf values of the three galvanic cells respectively
(i) Zn | Zn²⁺ (1 M) || Cu²⁺ (0.1 M) | Cu
(ii) Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu
(iii) Zn | Zn²⁺ (0.1 M) || Cu²⁺ (1 M) | Cu
Which one of the following is true? (Karnataka CET 2010, 2011)
(a) E₂ > E₃ > E₁
(b) E₃ > E₂ > E₁
(c) E₁ > E₂ > E₃
(d) E₁ > E₃ > E₂
Answer:
Ans. (b)
Zn + Cu²⁺ → Zn²⁺ + Cu, \( E_{cell} = E^o_{cell} - \frac{0.0591}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \)
Thus, higher the value of [Zn²⁺]/[Cu²⁺], lower is the emf of the cell. \( \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \) for (i) = \( \frac{1}{0.1} \) = 10, for (ii) = \( \frac{1}{1} \) = 1, for (iii) = \( \frac{0.1}{1} \) = 0.1.
Hence, E₃ > E₂ > E₁.

Teacher's Note:
1. Raising the reactant ion (Cu²⁺) or lowering the product ion (Zn²⁺) increases the emf of a Daniell cell.

Question 15. The cell, Zn | Zn²⁺ (1 M) || Cu²⁺ (1 M) | Cu, (E_cell = 1.10 V), was allowed to be completely discharged at 298 K. The relative concentration of Zn²⁺ to Cu²⁺, \( \left(\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}\right) \) is (AIEEE 2007)
(a) 9.65 × 10⁴
(b) antilog 24.08
(c) 37.3
(d) 10^37.3
Answer:
Ans. (d)
The cell reaction is: Zn + Cu²⁺ → Zn²⁺ + Cu. When the cell is completely discharged, E_cell = 0.
Hence, \( E^o_{cell} = \frac{0.0591}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \), i.e., \( 1.10 = \frac{0.0591}{2}\log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \)
or \( \log\frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \) = 37.3 or \( \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} \) = 10^37.3.

Teacher's Note:
1. A completely discharged cell is at equilibrium: E_cell = 0 and the concentration ratio equals K.

Question 16. An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO₃ and the volume made to 100 ml. A silver electrode was dipped in the solution and the EMF of the cell set up
Pt (s), H₂ (g) | H⁺ (1 M) || Ag⁺ (aq) | Ag (s)
was 0.62 V. If E^o_cell = 0.80 V, what is the percentage of Ag in the alloy? [At 25°C, 2.303 RT/F = 0.06] (Kerala PET 2007)
(a) 25
(b) 2.50
(c) 10
(d) 1
(e) 5
Answer:
Ans. (d)
The cell reaction is H₂ + 2Ag⁺ → 2H⁺ + 2Ag
\( E_{cell} = E^o_{cell} - \frac{2.303RT}{2F}\log\frac{1}{[\mathrm{Ag^+}]^2} \); 0.62 = 0.80 + 0.06 log [Ag⁺]
or log [Ag⁺] = \( \frac{-0.18}{0.06} \) = −3 \( \therefore \) [Ag⁺] = antilog (−3) = 1.0 × 10⁻³ M = (1.0 × 10⁻³) × 108 g L⁻¹ = 0.108 g L⁻¹
\( \therefore \) Amount of Ag present in 100 ml solution = 0.0108 g
\( \therefore \) % of Ag = \( \frac{0.0108}{1.08}\times100 \) = 1%

Teacher's Note:
1. Get [Ag⁺] from the Nernst equation, then convert to grams in the actual volume (100 mL, not 1 L) before finding the percentage.

Question 17. The potential of the cell for the reaction
M (s) + 2H⁺ (1 M) → H₂ (g) (1 atm) + M²⁺ (0.1 M)
is 1.500 V. The standard reduction potential for M²⁺/M (s) couple is
(a) 0.1470 V
(b) 1.470 V
(c) −1.47 V
(d) none of these
Answer:
Ans. (c)
The cell is M | M²⁺ (0.1 M) || H⁺ (1 M) | H₂ (1 atm)
\( E_{cell} = E^o_{cell} - \frac{0.0591}{2}\log[\mathrm{M^{2+}}] \) \( \therefore \) \( 1.50 = E^o_{cell} - \frac{0.0591}{2}\times\log(0.1) \) = E^o_cell + 0.02955
or E^o_cell = 1.50 − 0.02955 = 1.47045 V
But E^o_cell = E^o_{H⁺/\( \frac{1}{2} \)H2} − E^o_M²⁺/M, i.e., 1.47 = 0 − E^o_M²⁺/M or E^o_M²⁺/M = −1.47 V.

Teacher's Note:
1. A metal that releases H₂ from 1 M acid with positive E^o_cell must have a negative reduction potential.

Question 18. Consider the following cell reaction:
2Fe (s) + O₂ (g) + 4H⁺ (aq) → 2Fe²⁺ (aq) + 2H₂O (l), E^o = 1.67 V
At [Fe²⁺] = 10⁻³ M, P(O₂) = 0.1 atm and pH = 3, the cell potential at 25°C is (IIT 2011)
(a) 1.47 V
(b) 1.77 V
(c) 1.87 V
(d) 1.57 V
Answer:
Ans. (d)
Applying Nernst equation to the given reaction, \( E_{cell} = E^o_{cell} - \frac{0.0591}{n}\log\frac{[\mathrm{Fe^{2+}}]^2}{P_{\mathrm{O_2}}\times[\mathrm{H^+}]^4} \)
For the given reaction, n = 4 and pH = 3 means [H⁺] = 10⁻³ M
\( \therefore E_{cell} = 1.67 - \frac{0.0591}{4}\log\frac{(10^{-3})^2}{0.1\times(10^{-3})^4} = 1.67 - \frac{0.0591}{4}\log10^{7} \) = 1.67 − 0.10 = 1.57 V.

Teacher's Note:
1. Count n from the balanced equation (2 Fe atoms each losing 2e⁻ = 4); solids and liquid water stay out of Q.

Question 19. Value of E_{H2O/H2 (1 atm) Pt} at 298 K would be
(a) −0.207 V
(b) +0.207 V
(c) −0.414 V
(d) +0.414 V
Answer:
Ans. (c)
For water at 298 K, [H⁺] = 10⁻⁷ M. Reduction reaction is: H⁺ + e⁻ → \( \frac{1}{2} \)H₂
\( \therefore E = -\frac{RT}{F}\ln\frac{p_{\mathrm{H_2}}^{1/2}}{[\mathrm{H^+}]} = -0.0591\log\frac{p_{\mathrm{H_2}}^{1/2}}{[\mathrm{H^+}]} = -0.0591\log\frac{1}{10^{-7}} \) = −0.4137 = −0.414 V

Teacher's Note:
1. Pure water has pH 7, so the hydrogen electrode potential is −0.059 × 7 ≈ −0.41 V, a value often used for reduction of water.

Question 20. The pressure of H₂ required to make the potential of H₂-electrode zero in pure water at 298 K is (AMU Engg. 2015, NEET Phase I 2016)
(a) 10⁻¹⁰ atm
(b) 10⁻⁴ atm
(c) 10⁻¹⁴ atm
(d) 10⁻¹² atm
Answer:
Ans. (c)
The reduction reaction for H₂-electrode is 2H⁺ (aq) + 2e⁻ → H₂ (g)
\( E_{\mathrm{H^+/H_2}} = E^o_{\mathrm{H^+/H_2}} - \frac{0.0591}{2}\log\frac{p_{\mathrm{H_2}}}{[\mathrm{H^+}]^2} \). In pure water at 298 K, [H⁺] = 10⁻⁷ M
\( \therefore 0 = 0 - \frac{0.0591}{2}\log\frac{p_{\mathrm{H_2}}}{(10^{-7})^2} \) or \( \log\frac{p_{\mathrm{H_2}}}{10^{-14}} = 0 \) or \( \frac{p_{\mathrm{H_2}}}{10^{-14}} = 1 \) (∵ log 1 = 0)
or p_H2 = 10⁻¹⁴ atm

Teacher's Note:
1. E = 0 means the quotient p_H2/[H⁺]² must equal 1; with [H⁺] = 10⁻⁷, p_H2 = 10⁻¹⁴.

Question 21. For the electrode Cu/Cu²⁺, log [Cu²⁺] (along X-axis) is plotted against E_red (along Y-axis). The plot obtained is shown in the fig. The electrode potential of the half cell Cu | Cu²⁺ (0.1 M) will be(a) \( -0.34 + \frac{0.0591}{2} \) V
(b) 0.34 V
(c) \( 0.34 + \frac{0.0591}{2} \) V
(d) \( -0.34 - \frac{0.0591}{2} \) V
Answer:
Ans. (a)
The electrode reaction (reduction reaction) is Cu²⁺ + 2e⁻ → Cu
\( E_{red} = E^o_{red} + \frac{0.0591}{2}\log[\mathrm{Cu^{2+}}] \)
Thus, a plot of log [Cu²⁺] vs E_red will have intercept = E^o_red = 0.34 V (Given in the plot)
We have to calculate oxidation potential when [Cu²⁺] = 0.1 M
\( E_{ox} = E^o_{ox} - \frac{0.0591}{2}\log(0.1) = -0.34 - \frac{0.0591}{2}(-1) = -0.34 + \frac{0.0591}{2} \)

Teacher's Note:
1. The half cell written as Cu | Cu²⁺ refers to oxidation, so its potential is the negative of the reduction potential at that concentration.

Question 22. For the calomel electrode, Hg, Hg₂Cl₂ | Cl⁻ (aq), electrode potentials measured at different Cl⁻ ion concentration are plotted against log [Cl⁻]. The variation is correctly represented by the plot
(a) Plot (a)
(b) Plot (b)
(c) Plot (c)
(d) Plot (d)
Answer:
Ans. (a)
The half cell reaction of the calomel electrode (written as reduction reaction) is Hg₂Cl₂ (s) + 2e⁻ → 2Hg (l) + 2Cl⁻ (aq)
\( E = E^o - \frac{0.0591}{2}\log[\mathrm{Cl^-}]^2 \) or E = E^o − 0.0591 log [Cl⁻]
Thus, plot of E vs log [Cl⁻] is linear with a negative slope (i.e., E decreases linearly with log [Cl⁻])

Teacher's Note:
1. Plotting against log [Cl⁻] (not [Cl⁻]) gives a straight line; Cl⁻ is a product of reduction, so the slope is negative.

Question 23. The following cell is found to have EMF equal to zero
Pt, H₂ (x atm) | 0.01 M H⁺ || 0.1 M H⁺ | H₂ (y atm), Pt
The ratio x/y is
(a) 0.01
(b) 0.1
(c) 10
(d) 100
Answer:
Ans. (a)
For LHS electrode, 2H⁺ (0.01 M) + 2e⁻ → H₂ (x atm), \( E_{LHS} = \frac{0.0591}{2}\log\frac{x}{(0.01)^2} \)
For RHS electrode, 2H⁺ (0.1 M) + 2e⁻ → H₂ (y atm), \( E_{RHS} = \frac{0.0591}{2}\log\frac{y}{(0.1)^2} \)
E_cell = E_RHS − E_LHS = 0 (Given) \( \therefore \) E_LHS = E_RHS
log (x × 10⁴) = log (y × 10²), log x + 4 = log y + 2 or log x − log y = −2 or \( \log\frac{x}{y} = \bar{2} \)
or \( \frac{x}{y} \) = 10⁻² = 0.01

Teacher's Note:
1. Zero emf means both electrodes have the same p_H2/[H⁺]²; the 100-fold larger [H⁺]² on the right needs a 100-fold larger pressure.

Question 24. The e.m.f. of the cell
Zn | Zn²⁺ (0.01 M) || Fe²⁺ (0.001 M) | Fe
at 298 K is 0.2905 volt. Then the value of equilibrium constant for the cell reaction is
(a) \( e^{\frac{0.32}{0.0295}} \)
(b) \( 10^{\frac{0.32}{0.0295}} \)
(c) \( 10^{\frac{0.26}{0.0295}} \)
(d) \( 10^{\frac{0.32}{0.0591}} \)
Answer:
Ans. (b)
Zn + Fe²⁺ → Zn²⁺ + Fe (n = 2)
\( E = E^o - \frac{0.0591}{n}\log Q \); \( \therefore 0.2905 = E^o - \frac{0.0591}{2}\log\frac{0.01}{0.001} \) or E^o = 0.2905 + 0.0295 = 0.32 volt
\( E^o = \frac{0.0591}{n}\log K_{eq} \) \( \therefore 0.32 = \frac{0.0591}{2}\log K_{eq} \) \( \therefore K_{eq} = 10^{\frac{0.32}{0.0295}} \)

Teacher's Note:
1. First recover E^o from the Nernst equation using the actual concentrations, then use log K = nE^o/0.0591.

Question 25. Given at 25°C,
[Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃, E^o = 0.02 V
Ag⁺ + e⁻ → Ag, E^o = 0.80 V
The order of magnitude of the equilibrium constant of the reaction [Ag(NH₃)₂]⁺ ⇄ Ag⁺ + 2NH₃ will be
(a) 10⁻⁸
(b) 10⁻¹⁰
(c) 10⁻¹²
(d) 10⁻¹⁴
Answer:
Ans. (d)
[Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃, E^o = 0.02 V; Ag → Ag⁺ + e⁻, E^o = −0.80 V
[Ag(NH₃)₂]⁺ ⇄ Ag⁺ + 2NH₃, E^o_cell = −0.78 V
\( E^o_{cell} = \frac{0.0591}{n}\log K \) \( \therefore -0.78 = \frac{0.0591}{1}\log K \) or \( \log K = \frac{-0.78}{0.0591} \) = −13.2
K = antilog (−13.2) = 6.3 × 10⁻¹⁴, i.e., of the order of 10⁻¹⁴

Teacher's Note:
1. The dissociation constant of a complex comes from combining two reduction half-reactions; a very negative E^o means a very stable complex.

Question 26. 2Hg → Hg₂²⁺, E^o = +0.855 V
Hg → Hg²⁺, E^o = +0.799 V
Equilibrium constant for the reaction Hg + Hg²⁺ → Hg₂²⁺ at 27°C is
(a) 89
(b) 82.3
(c) 79
(d) none of these
Answer:
Ans. (c)
E^o for the given reaction = 0.855 − 0.799 = 0.056 V
\( E^o_{cell} = \frac{0.0591}{2}\log K_c \) or \( \log K_c = \frac{0.056\times2}{0.059} \) = 1.8983
K_c = Antilog 1.8983 = 79.12 ≈ 79

Teacher's Note:
1. For the comproportionation, take the difference of the two E^o values and use n = 2 in log K = nE^o/0.059.

Question 27. For the following electrochemical cell at 298 K
Pt (s) | H₂ (g, 1 bar) | H⁺ (aq, 1 M) || M⁴⁺ (aq) | M²⁺ (aq) | Pt (s)
E_cell = 0.092 V when \( \frac{[\mathrm{M^{2+}}(aq)]}{[\mathrm{M^{4+}}(aq)]} = 10^x \)
Given E^o_{M⁴⁺/M²⁺} = 0.151 V, 2.303 \( \frac{RT}{F} \) = 0.059 V
The value of x is (JEE Advanced 2016)
(a) −2
(b) −1
(c) 1
(d) 2
Answer:
Ans. (d)
The reactions occurring in the cell are, at anode: H₂ (g) → 2H⁺ (aq) + 2e⁻; at cathode: M⁴⁺ (aq) + 2e⁻ → M²⁺ (aq)
Overall reaction: H₂ (g) + M⁴⁺ (aq) → M²⁺ (aq) + 2H⁺ (aq)
\( E_{cell} = E^o_{cell} - \frac{0.059}{2}\log\frac{[\mathrm{M^{2+}}][\mathrm{H^+}]^2}{[\mathrm{M^{4+}}]} \); E^o_cell = E^o_{M⁴⁺/M²⁺} − E^o_H⁺/H2 = 0.151 − 0 = 0.151 V
\( \therefore 0.092 = 0.151 - \frac{0.059}{2}\log(10^x\times1^2) \) or 0.0295 log 10^x = 0.151 − 0.092 = 0.059
or \( \log10^x = \frac{0.059}{0.0295} \) = 2 \( \therefore \) 10^x = 10² \( \therefore \) x = 2

Teacher's Note:
1. Since E_cell is less than E^o_cell, the product ratio [M²⁺]/[M⁴⁺] must be greater than 1, so x is positive.

Question 28. In the electrochemical cell:
Zn | ZnSO₄ (0.01 M) || CuSO₄ (1.0 M) | Cu,
the e.m.f of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ is changed to 0.01 M, the e.m.f changes to E₂. From the following, which one is the relationship between E₁ and E₂? (Given, \( \frac{RT}{F} \) = 0.059) (NEET 2017)
(a) E₁ = E₂
(b) E₁ < E₂
(c) E₁ > E₂
(d) E₂ = 0 ≠ E₁
Answer:
Ans. (c)
The cell reaction is Zn + CuSO₄ → ZnSO₄ + Cu, n = 2
\( E = E^o_{cell} - \frac{2.303RT}{2F}\log\frac{[\mathrm{ZnSO_4}]}{[\mathrm{CuSO_4}]} \)
In 1st case, \( E_1 = E^o_{cell} - \frac{2.303RT}{2F}\log\frac{0.01}{1} \); In 2nd case, \( E_2 = E^o_{cell} - \frac{2.303RT}{2F}\log\frac{1}{0.01} \)
Evidently, E₁ > E₂.

Teacher's Note:
1. A small [Zn²⁺]/[Cu²⁺] ratio makes log Q negative and raises the emf above E^o.

One or More Correct Answers

Question 29. For the cell, Tl | Tl⁺ (0.001 M) || Cu²⁺ (0.1 M) | Cu, E_cell at 25°C is 0.83 V. This can be increased
(a) by increasing [Cu²⁺]
(b) by increasing [Tl⁺]
(c) by decreasing [Cu²⁺]
(d) by decreasing [Tl⁺]
Answer:
Ans. (a, d)
2Tl + Cu²⁺ → 2Tl⁺ + Cu, \( E_{cell} = E^o_{cell} - \frac{0.0591}{2}\log\frac{[\mathrm{Tl^+}]^2}{[\mathrm{Cu^{2+}}]} \)
Hence E_cell increases when [Cu²⁺] is increased or [Tl⁺] is decreased.

Teacher's Note:
1. Like Le Chatelier's principle: more reactant ion or less product ion pushes the cell reaction forward and raises the emf.

Question 30. Which of the following are correct?
(a) Electrolysis of dilute NaOH solution gives H₂ at cathode and O₂ at anode.
(b) Electrolysis of sulphuric acid (dilute or concentrated) gives H₂ at cathode and O₂ at anode.
(c) Electrolysis of aqueous KF solution gives fluorine at the anode.
(d) Oxidation of copper anode occurs in the electrolysis of aqueous copper sulphate solution using copper electrodes.
Answer:
Ans. (a, d)
(b) is wrong because electrolysis of concentrated H₂SO₄ does not give O₂ at the anode (peroxodisulphuric acid is formed).
(c) is wrong because electrolysis of aqueous KF solution gives O₂ at the anode and not F₂.
(a) and (d) are correct: water is electrolysed in dilute NaOH, and an active Cu anode itself dissolves (Cu → Cu²⁺ + 2e⁻).

Teacher's Note:
1. Products depend on concentration and electrode material: an attackable anode such as Cu is oxidised in preference to water.

Passage-Based Questions

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 amount of charge in coulombs used up by the battery is nearly
(a) 954180
(b) 477090
(c) 95418
(d) 47709
Answer:
Ans. (a)
During the discharge reaction, 4 moles of H⁺ ions, i.e., 2 moles of H₂SO₄ require 2 F, i.e., 2 × 96500 coulombs
\( \therefore \) 9.88786 moles will require \( = \frac{2\times96500\times9.88786}{2} \) = 954178 coulombs

Teacher's Note:
1. In the lead storage battery 1 mol H₂SO₄ is consumed per faraday, so charge = moles lost × 96500 C.

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.
The number of ampere-hour for which the battery must have been used is
(a) 2650.5
(b) 265.05
(c) 26.505
(d) 2.6505
Answer:
Ans. (b)
\( \text{Ampere-hour} = \frac{\text{Coulombs}}{3600\ \text{s}} = \frac{954178}{3600} \) = 265.05

Teacher's Note:
1. 1 ampere-hour = 3600 coulombs; this is the unit in which battery capacity is usually stated.

Question 33. 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.
For the above cell
(a) E_cell < 0; ΔG > 0
(b) E_cell > 0; ΔG < 0
(c) E_cell < 0; ΔG^o > 0
(d) E_cell > 0; ΔG^o < 0
Answer:
Ans. (b)
For the given concentration cell, \( E_{cell} = \frac{2.303RT}{nF}\log\frac{C_2}{C_1} = \frac{2.303RT}{F}\log\frac{1}{0.05} \) = +ve, i.e., E_cell > 0.
For the cell reaction to be spontaneous, ΔG < 0.

Teacher's Note:
1. For a concentration cell E^o_cell = 0 and ΔG^o = 0, so options with ΔG^o are ruled out at once.

Matrix Match Questions

Teacher's Note:
1. Add E^o values directly only for a complete cell reaction (electrons cancel); for a new half-reaction combine nE^o values.

Teacher's Note:
1. Anode and cathode are defined by oxidation and reduction; their signs are opposite in galvanic and electrolytic cells.

Integer and Numerical Answer Questions

Question 36. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If \( \lambda^o_{X^-} \approx \lambda^o_{Y^-} \), the difference in their pK_a values, pK_a (HX) − pK_a (HY) is (consider degree of ionization of both acids to be << 1) (JEE Advanced 2015)
Answer:
Ans. 3
Suppose Λ_m (HX) is represented by Λ_m1 and Λ_m (HY) by Λ_m2. Then given that \( \Lambda_{m_1} = \frac{1}{10}\Lambda_{m_2} \)
HX ⇄ H⁺ + X⁻, \( K_a = \frac{[\mathrm{H^+}][\mathrm{X^-}]}{[\mathrm{HX}]} \). Representing K_a of HX by K_a1, then as K_a = Cα² and \( \alpha = \frac{\Lambda_m}{\Lambda_m^o} \), we have \( K_{a_1} = C_1\left(\frac{\Lambda_{m_1}}{\Lambda^o_{m_1}}\right)^2 \) ...(i)
Similarly, HY ⇄ H⁺ + Y⁻; \( K_{a_2} = C_2\left(\frac{\Lambda_{m_2}}{\Lambda^o_{m_2}}\right)^2 \) ...(ii)
As \( \lambda^o_{X^-} = \lambda^o_{Y^-} \), \( \Lambda^o_{HX} = \Lambda^o_{HY} \) or \( \Lambda^o_{m_1} = \Lambda^o_{m_2} \)
From eqns. (i) and (ii), \( \frac{K_{a_1}}{K_{a_2}} = \frac{C_1}{C_2}\left(\frac{\Lambda_{m_1}}{\Lambda_{m_2}}\right)^2 = \frac{0.01}{0.1}\left(\frac{1}{10}\right)^2 \) = 0.001 = 10⁻³
or log K_a1 − log K_a2 = log 10⁻³ = −3 or −log K_a1 − (−log K_a2) = 3 or pK_a1 − pK_a2 = 3

Teacher's Note:
1. Since Λ^o_m is the same for both acids, the ratio of α equals the ratio of Λ_m.

Question 37. The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of Pt electrodes. The distance between the electrodes is 120 cm with an area of cross-section of 1 cm². The conductance of this solution was found to be 5 × 10⁻⁷ S. The pH of the solution is 4. The value of the limiting molar conductivity (Λ^o_m) of the weak monobasic acid in aqueous solution is Z × 10² S cm² mol⁻¹. The value of Z is (JEE Advanced 2017)
Answer:
Ans. 6
Conc. C = 0.0015 M, Conductance, G = 5 × 10⁻⁷ S, l = 120 cm, a = 1 cm²
\( \kappa = G\times\frac{l}{a} = 5\times10^{-7}\times\frac{120}{1} \) = 6 × 10⁻⁵ S cm⁻¹
\( \Lambda_m^c = \frac{\kappa\times1000}{\text{Molarity}} = \frac{6\times10^{-5}\times1000}{0.0015} \) = 40 S cm² mol⁻¹
HA ⇄ H⁺ + A⁻; [H⁺] = Cα = 10⁻⁴ M (∵ pH = 4) \( \therefore \alpha = \frac{10^{-4}}{C} = \frac{10^{-4}}{0.0015} \)
\( \alpha = \frac{\Lambda_m^c}{\Lambda_m^o} \) \( \therefore \frac{10^{-4}}{0.0015} = \frac{40}{\Lambda_m^o} \) or \( \Lambda_m^o = \frac{40\times0.0015}{10^{-4}} \) = 6 × 10² S cm² mol⁻¹, so Z = 6

Teacher's Note:
1. Two routes to α meet here: from pH (α = [H⁺]/C) and from conductance (α = Λ_m/Λ^o_m).

Question 38. For the electrochemical cell
Mg (s) | Mg²⁺ (aq, 1 M) || Cu²⁺ (aq, 1 M) | Cu (s)
the standard e.m.f of the cell is 2.70 V at 300 K. When the concentration of Mg²⁺ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is ........... (Given, \( \frac{F}{R} \) = 11500 K V⁻¹ where F is the Faraday constant and R is the gas constant, ln (10) = 2.30) (JEE Advanced 2018)
Answer:
Ans. 10.00
Mg → Mg²⁺ + 2e⁻; Cu²⁺ + 2e⁻ → Cu; Mg + Cu²⁺ → Mg²⁺ + Cu, n = 2
For the given cell, \( E = E_0 - \frac{RT}{2F}\ln\frac{1}{1} = E_0 \) = 2.70 V
When concentration of Mg²⁺ is changed to x M, \( 2.67 = 2.70 - \frac{300}{2\times11500}\ln\frac{x}{1} \)
\( 0.03 = \frac{300}{2\times11500}\ln x \) or ln x = 2.3 or x = 10

Teacher's Note:
1. Use RT/F = T/(F/R) = 300/11500 V directly; ln x = 2.30 = ln 10.

Question 39. The magnitude of the change in oxidizing power of the MnO₄⁻/Mn²⁺ couple is x × 10⁻⁴ V if the H⁺ concentration is decreased from 1 M to 10⁻⁴ M at 25°C (Assume concentration of MnO₄⁻ and Mn²⁺ to be same on change in H⁺ concentration). The value of x is .......... (Rounded off to the nearest neighbour) Given: 2.303 RT/F = 0.059 (JEE Main 2021)
Answer:
Ans. 3776
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Change in oxidizing power means that we have to find change in the value of reduction potential (E₂ − E₁) when [H⁺] changes from 1 M to 10⁻⁴ M.
In 1st case: \( E_1 - E^o = \frac{0.059}{5}\log\frac{[\mathrm{Mn^{2+}}]}{[\mathrm{H^+}]_1^8[\mathrm{MnO_4^-}]} \); In 2nd case: \( E_2 - E^o = \frac{0.059}{5}\log\frac{[\mathrm{Mn^{2+}}]}{[\mathrm{H^+}]_2^8[\mathrm{MnO_4^-}]} \)
As [Mn²⁺] and [MnO₄⁻] remain the same, \( E_2 - E_1 = \frac{0.059}{5}\log\frac{[\mathrm{H^+}]_2^8}{[\mathrm{H^+}]_1^8} = \frac{0.059}{5}\log\left(\frac{10^{-4}}{1}\right)^8 = \frac{0.059}{5}\times(-32) \)
or \( |E_2 - E_1| = \frac{0.059}{5}(32) \) = x × 10⁻⁴ or 0.3776 = x × 10⁻⁴ or 3776 × 10⁻⁴ = x × 10⁻⁴ or x = 3776

Teacher's Note:
1. [H⁺] enters to the 8th power, so lowering it by 10⁴ changes the potential by (0.059/5) × 32 V.

Question 40. 5.0 mmol dm⁻³ aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 cm⁻¹. The molar conductivity of this solution is .......... mS m² mol⁻¹ (Round off to the nearest integer) (JEE Main 2021)
Answer:
Ans. 14
Specific conductivity (κ) = Obsd conductance (G) × Cell const. (G*) = (0.55 × 10⁻³ S) × (1.3 × 10² m⁻¹) = (55 × 10⁻⁵ S) × (1.3 × 10² m⁻¹) = 55 × 1.3 × 10⁻³ S m⁻¹
\( \Lambda_m = \frac{\kappa\,(\mathrm{S\,m^{-1}})}{\text{Molarity}\,(\mathrm{mol\,m^{-3}})} = \frac{55\times1.3\times10^{-3}\,\mathrm{S\,m^{-1}}}{5\,\mathrm{mol\,m^{-3}}} \) (5 mmol dm⁻³ = 5 mol m⁻³)
= 11 × 1.3 × 10⁻³ S m² mol⁻¹ = 14.3 × 10⁻³ S m² mol⁻¹ = 14.3 mS m² mol⁻¹
Nearest integer = 14

Teacher's Note:
1. Convert cell constant to m⁻¹ (1.3 cm⁻¹ = 130 m⁻¹) and mmol dm⁻³ to mol m⁻³ (they are equal).

Assertion–Reason Questions

Question 41. Statement-1: Electrolysis of an aqueous solution of KI gives I₂ at the anode but that of KF gives O₂ at the anode and not F₂.
Statement-2: F₂ is more reactive than I₂.
(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. (b)
Correct explanation. I⁻ ions have much higher oxidation potential than that of water while F⁻ ions have much lower oxidation potential than that of water.

Teacher's Note:
1. The anode product is decided by comparing oxidation potentials with that of water, not by the reactivity of the halogen.

Question 42. Assertion: Electrolysis of molten CaH₂ produces hydrogen gas at anode.
Reason: In CaH₂, hydrogen is present in the form of H⁻. (AIIMS 2014)
(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)
CaH₂ →(Melt) Ca²⁺ + 2H⁻
At anode: H⁻ → H + e⁻; H + H → H₂
Hence, R is the correct explanation of A.

Teacher's Note:
1. Hydride ion is an anion, so it moves to the anode and is oxidised to H₂ there; this is evidence for ionic hydrides.

Question 43. Assertion: On dilution, the equivalent as well as molar conductivity of the solution increases.
Reason: With dilution, the number of current carrying particles per cm³ increases. (AIIMS 2009)
(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. (c)
Correct R. On dilution, the number of current carrying particles per cm³ decreases and hence specific conductivity decreases. Λ_eq and Λ_m increase with dilution because Λ_eq = sp. conductivity × volume of solution containing 1 g eq. of the electrolyte and Λ_m = sp. conductivity × volume of solution containing 1 mole of the electrolyte and volume increases much more than decrease in specific conductivity.

Teacher's Note:
1. Ions per cm³ fall on dilution (so κ falls), but the volume holding one mole rises faster, so Λ_m rises.

Question 44. Assertion: According to Kohlrausch law, the molar conductivity of a strong electrolyte at infinite dilution is sum of molar conductivities of its ions.
Reason: The current carried by cation and anion is always equal. (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. (d)
Correct A. The molar conductivity of any electrolyte (weak or strong) at infinite dilution is sum of molar conductivities of the ions multiplied by the respective number of ions in one formula unit.
Correct R. The current carried by cations and anions is not equal.

Teacher's Note:
1. Kohlrausch's law applies to weak electrolytes too and uses the stoichiometric numbers, e.g., Λ^o(CaCl₂) = λ^o(Ca²⁺) + 2λ^o(Cl⁻).