List - I List - II (P) H 3 PO 4 + NaOH → Na 2 HPO 4 + H 2 O (1) M (Q) Pb + PbO 2 + H 2 SO 4 → 2 PbSO 4 + H 2 O (2) 2 M (R) KMnO 4 + HCl → MnCl 2 + Cl 2 + KCl + H

Question

List - I List - II (P) H ​ 3 ​ PO ​ 4 ​ + NaOH → Na 2 ​ HPO 4 ​ + H 2 ​ O (1) M (Q) Pb + PbO 2 ​ + H ​ 2 ​ SO ​ 4 ​ → 2 PbSO 4 ​ + H 2 ​ O (2) 2 M ​ (R) KMnO 4 ​ + HCl ​ → MnCl 2 ​ + Cl 2 ​ + KCl + H 2 ​ O (3) 3 M ​ (S) H ​ 2 ​ SO ​ 4 ​ + NaI → Na 2 ​ SO 4 ​ + I 2 ​ + H 2 ​ S + H 2 ​ O (4) 8 5 M ​ (5) 5 8 M ​ The correct match is :

TestHub · Accepted Answer

Explain Question: Concept: Based on eq. wt. Solution:

(P) $\underline{H}_{3} \underline{PO}_{4} + NaOH \to Na_{2} HPO_{4} + H_{2} O$ $nf = 2$ Eq.wt $= \frac{M}{nf} = \frac{M}{2}$

(Q) $Pb + PbO_{2} + \underline{H}_{2} \underline{SO}_{4} \to 2 PbSO_{4} + H_{2} O$ $nf = 1 Eq.wt = \frac{M}{1}$

(R) $2 KMnO_{4} + 16 \underline{HCl} ⟶ 2 MnCl_{2} + 5 Cl_{2} + 2 KCl + 8 H_{2} O$ $nf = 5/8$ $Eq.wt = \frac{8 M}{5}$

(S) $5 \underline{H}_{2} \underline{SO}_{4} + 8 NaI \to 4 Na_{2} SO_{4} + 4 I_{2} + H_{2} S + 4 H_{2} O$ $nf = 8/5$ $Eq. wt = \frac{5 M}{8}$

Final answer: The correct sequence is P-2, Q-1, R-5, S-4

List - I	List - II
(P) $\underline{H}_{3} \underline{PO}_{4} + NaOH \to Na_{2} HPO_{4} + H_{2} O$	(1) M
(Q) $Pb + PbO_{2} + \underline{H}_{2} \underline{SO}_{4} \to 2 PbSO_{4} + H_{2} O$	(2) $\frac{M}{2}$
(R) $KMnO_{4} + \underline{HCl} \to MnCl_{2} + Cl_{2} + KCl + H_{2} O$	(3) $\frac{M}{3}$
(S) $\underline{H}_{2} \underline{SO}_{4} + NaI \to Na_{2} SO_{4} + I_{2} + H_{2} S + H_{2} O$	(4) $\frac{5 M}{8}$
	(5) $\frac{8 M}{5}$

Chemistry - Redox Reactions Question with Solution | TestHub

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JEE Advanced 2027 (Class 12)