Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Electro Chemistry

Question:

Match List-I with List-II

List-I Type of cell/Batteries

List-II Reaction

(A) Daniell cell

(I) $2H_2(g) + O_2(g) → 2H_2O (l)$

(B) Fuel Cell

(II) $Pb(s) + PbO_2(s) + 2H_2SO_4 (aq) → 2PbSO_4 (s) + 2H_2O (l)$

(C) Primary Batteries

(III) $Zn(s) + Cu^{2+} (aq) → Zn^{2+} (aq) + Cu (s)$

(D) Secondary Batteries

(IV) $Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)$

Choose the correct answer from the options given below:

Options:

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

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

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

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

Correct Answer:

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

Explanation:

The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

List-I Type of cell/Batteries

List-II Reaction

(A) Daniell cell

(III) $Zn(s) + Cu^{2+} (aq) → Zn^{2+} (aq) + Cu (s)$

(B) Fuel Cell

(I) $2H_2(g) + O_2(g) → 2H_2O (l)$

(C) Primary Batteries

(IV) $Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)$

(D) Secondary Batteries

(II) $Pb(s) + PbO_2(s) + 2H_2SO_4 (aq) → 2PbSO_4 (s) + 2H_2O (l)$

(A) Daniell Cell $\rightarrow$ (III)

The Daniell cell is a classic galvanic cell consisting of zinc and copper electrodes. The overall reaction involves the oxidation of zinc and the reduction of copper(II) ions:

$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

(B) Fuel Cell $\rightarrow$ (I)

A fuel cell (specifically the $H_2-O_2$ fuel cell) converts the chemical energy of a fuel directly into electrical energy. The simplest version uses hydrogen and oxygen to produce water:

$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$

(C) Primary Batteries $\rightarrow$ (IV)

Primary batteries are non-rechargeable. A common example is the Mercury cell, used in small devices like hearing aids. Its overall reaction is:

$Zn(Hg) + HgO(s) \rightarrow ZnO(s) + Hg(l)$

(D) Secondary Batteries $\rightarrow$ (II)

Secondary batteries are rechargeable. The most common example is the Lead storage battery used in automobiles. During discharge, the reaction is:

$Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$