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Electrochemistry question

2022 · 28 Jul · Shift 1 · Q3
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  5. /2022 · 28 Jul · Shift 1 · Q3

Electrochemistry question

2022 · 28 Jul · Shift 1 · Q3

JEE MainChemistryElectrochemistryMCQ+4 / −1

Match List - I with List - II.

List - I List - II
(A) Cd(s)+2Ni(OH)3(s)→CdO(s)+2Ni(OH)2(s)+H2O(l)Cd(s) + 2Ni{(OH)_3}(s) \to CdO(s) + 2Ni{(OH)_2}(s) + {H_2}O(l)Cd(s)+2Ni(OH)3​(s)→CdO(s)+2Ni(OH)2​(s)+H2​O(l) (I) Primary battery
(B) Zn(Hg)+HgO(s)→ZnO(s)+Hg(l)Zn(Hg) + HgO(s) \to ZnO(s) + Hg(l)Zn(Hg)+HgO(s)→ZnO(s)+Hg(l) (II) Discharging of secondary battery
(C) 2PbSO4(s)+2H2O(l)→Pb(s)+PbO2(s)+2H2SO4(aq)2PbS{O_4}(s) + 2{H_2}O(l) \to Pb(s) + Pb{O_2}(s) + 2{H_2}S{O_4}(aq)2PbSO4​(s)+2H2​O(l)→Pb(s)+PbO2​(s)+2H2​SO4​(aq) (III) Fuel cell
(D) 2H2(g)+O2(g)→2H2O(l)2{H_2}(g) + {O_2}(g) \to 2{H_2}O(l)2H2​(g)+O2​(g)→2H2​O(l) (IV) Charging of secondary battery

Choose the correct answer from the options given below:

  1. A
    (A)−(I),(B)−(II),(C)−(III),(D)−(IV)(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})(A)−(I),(B)−(II),(C)−(III),(D)−(IV)
  2. B
    (A)−(IV),(B)−(I),(C)−(II),(D)−(III)(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{III})(A)−(IV),(B)−(I),(C)−(II),(D)−(III)
  3. C
    (A)−(II),(B)−(I),(C)−(IV),(D)−(III)(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{III})(A)−(II),(B)−(I),(C)−(IV),(D)−(III)
  4. D
    (A)−(II),(B)−(I),(C)−(III),(D)−(IV)(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})(A)−(II),(B)−(I),(C)−(III),(D)−(IV)
View written solutionFree

Correct answer: C

  1. Identify each cell/reaction type from List-I

We match each reaction with the correct description in List-II.

  1. Reaction (A)

Cd(s)+2Ni(OH)3(s)→CdO(s)+2Ni(OH)2(s)+H2O(l)Cd(s) + 2Ni(OH)_3(s) \to CdO(s) + 2Ni(OH)_2(s) + H_2O(l)Cd(s)+2Ni(OH)3​(s)→CdO(s)+2Ni(OH)2​(s)+H2​O(l)

This is the overall reaction of the nickel-cadmium cell during discharge.

  • Reactants are being consumed to produce products spontaneously.
  • Ni-Cd is a secondary battery.

So,

(A)→(II)Discharging of secondary battery(A) \to (II) \quad \text{Discharging of secondary battery}(A)→(II)Discharging of secondary battery

  1. Reaction (B)

Zn(Hg)+HgO(s)→ZnO(s)+Hg(l)Zn(Hg) + HgO(s) \to ZnO(s) + Hg(l)Zn(Hg)+HgO(s)→ZnO(s)+Hg(l)

This is the reaction of the mercury cell.

  • Mercury cell is a primary battery.
  • It is not rechargeable in standard classification.

So,

(B)→(I)Primary battery(B) \to (I) \quad \text{Primary battery}(B)→(I)Primary battery

  1. Reaction (C)

2PbSO4(s)+2H2O(l)→Pb(s)+PbO2(s)+2H2SO4(aq)2PbSO_4(s) + 2H_2O(l) \to Pb(s) + PbO_2(s) + 2H_2SO_4(aq)2PbSO4​(s)+2H2​O(l)→Pb(s)+PbO2​(s)+2H2​SO4​(aq)

In lead-acid battery:

  • During discharge, Pb+PbO2+2H2SO4→2PbSO4+2H2OPb + PbO_2 + 2H_2SO_4 \to 2PbSO_4 + 2H_2OPb+PbO2​+2H2​SO4​→2PbSO4​+2H2​O
  • The given reaction is the reverse of discharge.

Hence, it represents charging of secondary battery.

So,

(C)→(IV)Charging of secondary battery(C) \to (IV) \quad \text{Charging of secondary battery}(C)→(IV)Charging of secondary battery

  1. Reaction (D)

2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \to 2H_2O(l)2H2​(g)+O2​(g)→2H2​O(l)

This is the overall reaction of the hydrogen-oxygen fuel cell.

So,

(D)→(III)Fuel cell(D) \to (III) \quad \text{Fuel cell}(D)→(III)Fuel cell

  1. Final matching

Thus the correct matching is:

(A)−(II),(B)−(I),(C)−(IV),(D)−(III)(A)-(II),\quad (B)-(I),\quad (C)-(IV),\quad (D)-(III)(A)−(II),(B)−(I),(C)−(IV),(D)−(III)

This corresponds to Option C.

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