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Question: Match the entries of Column-I with Column-ll and choose the correct option. Column-I (Electrolytes)...

Match the entries of Column-I with Column-ll and choose the correct option.

Column-I (Electrolytes)

Column-Il (Product of electrolysis)

(A) Dilute solution of HCl(aq)

(B) Dllute solution of NaCl(aq)

(C) Concentrated solution of NaCl(aq)'

(D) Concentrated solution of AgNO3_3(aq)

A

A. (A - i), (B - i, ii), (C - iv), (D - ii)

B

B. (A - ⅰ), (B – iv), (C - iii), (D - iii)

C

C. (A-i), (B - i, ii), (C - ii, iii), (D - i, iv)

D

D. (A - ⅰ), (B - i, iii), (C - iv), (D - i)

Answer

C. (A-i), (B - i, ii), (C - ii, iii), (D - i, iv)

Explanation

Solution

The process of electrolysis involves the reduction of cations or water at the cathode and the oxidation of anions or water at the anode. The species that gets reduced or oxidized depends on their standard electrode potentials and, in some cases, their concentration and overpotential effects.

Let's analyze each electrolyte:

(A) Dilute solution of HCl(aq)

  • Species present: H⁺, Cl⁻, H₂O
  • At Cathode (Reduction): H⁺ ions are preferentially reduced over water in acidic solution. 2H+(aq)+2e−→H2(g)2H^+(aq) + 2e^- \rightarrow H_2(g) Product: H₂ gas (Matches Column-II: II)
  • At Anode (Oxidation): Water is preferentially oxidized over dilute Cl⁻ ions due to lower standard oxidation potential. 2H2O(l)→O2(g)+4H+(aq)+4e−2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- Product: O₂ gas (Matches Column-II: I)
  • Therefore, (A) matches (I) and (II). Option C lists (A - i).

(B) Dilute solution of NaCl(aq)

  • Species present: Na⁺, Cl⁻, H₂O
  • At Cathode (Reduction): Water is preferentially reduced over Na⁺ ions (E°(Na⁺/Na) = -2.71 V, E°(H₂O/H₂) = -0.83 V). 2H2O(l)+2e−→H2(g)+2OH−(aq)2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) Product: H₂ gas (Matches Column-II: II)
  • At Anode (Oxidation): Water is preferentially oxidized over dilute Cl⁻ ions (E°(Cl₂/Cl⁻) = +1.36 V, E°(O₂/H₂O) = +1.23 V). 2H2O(l)→O2(g)+4H+(aq)+4e−2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- Product: O₂ gas (Matches Column-II: I)
  • Therefore, (B) matches (I) and (II). Option C lists (B - i, ii).

(C) Concentrated solution of NaCl(aq)

  • Species present: Na⁺, Cl⁻, H₂O
  • At Cathode (Reduction): Water is preferentially reduced over Na⁺ ions. 2H2O(l)+2e−→H2(g)+2OH−(aq)2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) Product: H₂ gas (Matches Column-II: II)
  • At Anode (Oxidation): In concentrated Cl⁻ solutions, due to high concentration of Cl⁻ ions and the overpotential for O₂ evolution on common electrode materials, Cl⁻ ions are preferentially oxidized over water. 2Cl−(aq)→Cl2(g)+2e−2Cl^-(aq) \rightarrow Cl_2(g) + 2e^- Product: Cl₂ gas (Matches Column-II: III)
  • Therefore, (C) matches (II) and (III). Option C lists (C - ii, iii).

(D) Concentrated solution of AgNO₃(aq)

  • Species present: Ag⁺, NO₃⁻, H₂O
  • At Cathode (Reduction): Ag⁺ ions are preferentially reduced over water (E°(Ag⁺/Ag) = +0.80 V, E°(H₂O/H₂) = -0.83 V). Ag+(aq)+e−→Ag(s)Ag^+(aq) + e^- \rightarrow Ag(s) Product: Ag metal deposited (Matches Column-II: IV)
  • At Anode (Oxidation): Nitrate (NO₃⁻) ions are very difficult to oxidize. Water is preferentially oxidized. 2H2O(l)→O2(g)+4H+(aq)+4e−2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^- Product: O₂ gas (Matches Column-II: I)
  • Therefore, (D) matches (I) and (IV). Option C lists (D - i, iv).

Summary of Matches:

  • (A) Dilute solution of HCl(aq) →\rightarrow (I) O₂ evolved at anode, (II) H₂ evolved at cathode
  • (B) Dilute solution of NaCl(aq) →\rightarrow (I) O₂ evolved at anode, (II) H₂ evolved at cathode
  • (C) Concentrated solution of NaCl(aq) →\rightarrow (II) H₂ evolved at cathode, (III) Cl₂ evolved at anode
  • (D) Concentrated solution of AgNO₃(aq) →\rightarrow (I) O₂ evolved at anode, (IV) Ag deposited at cathode

Comparing these with the given options, option C provides the correct matches.

The final answer is C\boxed{C}

Explanation of the solution: The products of electrolysis depend on the relative standard electrode potentials of the species present and, importantly, on their concentrations and overpotential effects. For dilute HCl, H⁺ is reduced at the cathode (H₂) and H₂O is oxidized at the anode (O₂). For dilute NaCl, H₂O is reduced at the cathode (H₂) and H₂O is oxidized at the anode (O₂). For concentrated NaCl, H₂O is reduced at the cathode (H₂) but Cl⁻ is oxidized at the anode (Cl₂) due to high concentration and overpotential. For concentrated AgNO₃, Ag⁺ is reduced at the cathode (Ag) and H₂O is oxidized at the anode (O₂).