Redox Reactions | JEE Main Chemistry Chapter Wise Question Papers

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Q 1 :

1 mole of PbS is oxidised by ''X" moles of $O_{3}$ to get "Y'' moles of $O_{2} .$ X + Y = _________ . [2024]

(8)

$PbS + 4 O_{3} \to {PbSO}_{4} + 4 O_{2}$

$4$ $(O_{3} \to O_{2} + \underset{N a s c e n t o x y g e n}{[O])}$

$\frac{PbS + 4 [O] \to {PbSO}_{4}}{PbS + 4 O_{3} \to 4 O_{2} + {PbSO}_{4}}$

$X = Y = 4$

$X + Y = 8$

Q 2 :

Only 2 mL of ${KMnO}_{4}$ solution of unknown molarity is required to reach the end point of a titration of 20 mL of oxalic acid (2 M) in acidic medium. The molarity of ${KMnO}_{4}$ solution should be ____________ M. [2024]

(8)

$K \overset{+ 7}{Mn} O_{4} + H_{2} \overset{+ 3}{C_{2}} O_{4} \to {Mn}^{2 +} + \overset{+ 4}{C} O_{2}$ (unbalanced)

$(n f a c t o r = 5) (n f a c t o r = 1 \times 2 = 2)$

$n factor of {KMnO}_{4}$

$n_{{KMnO}_{4}} = change in O . N . of Mn \times no . of Mn atoms in 1 molecule of {KMnO}_{4}$

$= 5 \times 1 = 5$

$n factor of H_{2} C_{2} O_{4}$

$n H_{2} C_{2} O_{4}$ = change in O. N. of C $\times$ no. of C atoms in 1 molecule of $H_{2} C_{2} O_{4}$

$= 1 \times 2 = 2$

By law of equivalence:

Equivalents of ${KMnO}_{4}$ = Equivalents of $H_{2} C_{2} O_{4}$

$n factor of {KMnO}_{4} \times$ moles of ${KMnO}_{4}$ = $n factor of H_{2} C_{2} O_{4}$ $\times$ moles of $H_{2} C_{2} O_{4}$

$5 \times$ millimoles of ${KMnO}_{4} = 2 \times$ millimoles of $H_{2} C_{2} O_{4}$

$5 \times M_{{KMnO}_{4}} \times V_{{KMnO}_{4}} (in mL) = 2 \times M_{H_{2} C_{2} O_{4}} \times V_{H_{2} C_{2} O_{4}} (in mL)$

$5 \times M_{{KMnO}_{4}} \times 2 mL = 2 \times 2 M \times 2 mL$

$M_{{KMnO}_{4}} = 8 M$

Q 3 :

$M n O_{4}^{-} + x e^{-} \to M n O_{4}^{2 -}$

[IMAGE 219] $per mole of M n O_{4}^{-}; x, y and z are respectively:$

1, 2, 3
1, 5, 3
1, 3, 5
5, 3, 1

(2)

[IMAGE 220]

Topic Question Set

Q 1 :

1 mole of PbS is oxidised by ''X" moles of $O_{3}$ to get "Y'' moles of $O_{2} .$ X + Y = _________ . [2024]

Q 2 :

Only 2 mL of ${KMnO}_{4}$ solution of unknown molarity is required to reach the end point of a titration of 20 mL of oxalic acid (2 M) in acidic medium. The molarity of ${KMnO}_{4}$ solution should be ____________ M. [2024]

Q 3 :

$M n O_{4}^{-} + x e^{-} \to M n O_{4}^{2 -}$

[IMAGE 219] $per mole of M n O_{4}^{-}; x, y and z are respectively:$

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Topic Question Set

Q 1 : 1 mole of PbS is oxidised by ''X" moles of O3 to get "Y'' moles of O2. X + Y = _________ . [2024]

Q 2 : Only 2 mL of KMnO4 solution of unknown molarity is required to reach the end point of a titration of 20 mL of oxalic acid (2 M) in acidic medium. The molarity of KMnO4 solution should be ____________ M. [2024]

Q 3 : MnO4-+xe-→MnO42- [IMAGE 219] per mole of MnO4-; x, y and z are respectively:

Want Us to Email you About Special Offers & Updates?

Q 1 :

1 mole of PbS is oxidised by ''X" moles of $O_{3}$ to get "Y'' moles of $O_{2} .$ X + Y = _________ . [2024]

Q 2 :

Only 2 mL of ${KMnO}_{4}$ solution of unknown molarity is required to reach the end point of a titration of 20 mL of oxalic acid (2 M) in acidic medium. The molarity of ${KMnO}_{4}$ solution should be ____________ M. [2024]

Q 3 :

$M n O_{4}^{-} + x e^{-} \to M n O_{4}^{2 -}$

[IMAGE 219] $per mole of M n O_{4}^{-}; x, y and z are respectively:$