Redox Reactions | JEE Main Chemistry Chapter Wise Question Papers

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 :

500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M ${KMnO}_{4}$ solution in basic medium. The liberated iodine was titrated with standard 0.1 M ${Na}_{2} S_{2} O_{3}$ solution in the presence of starch indicator till the blue color disappeared. The volume (in L) of ${Na}_{2} S_{2} O_{3}$ consumed is _____. (Nearest integer) [2026]

(3)

${MnO}_{4}^{-} + I^{-} \to {MnO}_{2} + I_{2}$

$I_{2} + S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + I^{-}$

${gram eq of KMnO}_{4} = {gram eq of Na}_{2} S_{2} O_{3}$

$0.2 \times \frac{500}{1000} \times 3 = 0.1 \times V \times 1$

$V = 3 L$

Q 4 :

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

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

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

1

2

3

4

(2)

Q 5 :

In order to oxidise a mixture of 1 mole each of ${FeC}_{2} O_{4}$ , ${Fe}_{2} {(C_{2} O_{4})}_{3}$ , ${FeSO}_{4}$ and ${Fe}_{2} {({SO}_{4})}_{3}$ in acidic medium, the number of moles of ${KMnO}_{4}$ required is [2026]

3
2
5
7

1

2

3

4

(2)

Equivalents of ${KMnO}_{4}$ (oxidizing agent) = Equivalents of $({FeC}_{2} O_{4} + {Fe}_{2} {(C_{2} O_{4})}_{3} + {FeSO}_{4})$ (reducing agents)

Number of equivalents = $n$ factor $\times$ number of moles

$5 \times {moles of KMnO}_{4} = 3 \times {moles of FeC}_{2} O_{4}$ $+ 6 \times {moles of Fe}_{2} {(C_{2} O_{4})}_{3} + 1 \times {moles of FeSO}_{4}$

$5 \times {moles of KMnO}_{4} = 3 \times 1 + 6 \times 1 + 1 \times 1$

${Moles of KMnO}_{4} = 2$

Q 6 :

500 mL of 0.2 M ${MnO}_{4}^{-}$ solution in basic medium when mixed with 500 mL of 1.5 M $KI$ solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard $x$ M thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, then the value of $x$ is _______. [2026]

(1)

Equivalents of ${MnO}_{4}^{-} = n$ factor of ${MnO}_{4}^{-}$ in slightly basic medium $\times M_{{MnO}_{4}^{-}} \times V_{{MnO}_{4}^{-}} = 3 \times 0.2 \times 500 = 300$ meq

Equivalents of $KI = n$ factor of $KI \times M_{KI} \times V_{KI} = 1 \times 1.5 \times 500 = 750$ meq

As ${MnO}_{4}^{-}$ is the limiting reagent

Equivalents of ${MnO}_{4}^{-}$ = Equivalents of hypo

300 meq = $n$ factor of $S_{2} O_{3}^{2 -}$ in basic medium $\times M_{S_{2} O_{3}^{2 -}} \times V_{S_{2} O_{3}^{2 -}}$

$3 \times 0.2 \times 500 = 1 \times x \times 300$

$x = 1$

Q 7 :

200 cc of $x \times 10^{- 3} M$ potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here $x =$ _______. [2026]

(375)

$\begin{matrix} {Cr}_{2} O_{7}^{2 -} & + & {Fe}^{2 +} & \to & {Cr}^{3 +} & + & {Fe}^{3 +} \\ (n = 6) & (n = 1) \end{matrix}$

$m_{e q} (K_{2} {Cr}_{2} O_{7}) = m_{e q} (mohr salt)$

$n factor (K_{2} {Cr}_{2} O_{7}) \times M_{K_{2} {Cr}_{2} O_{7}} \times V_{K_{2} {Cr}_{2} O_{7}} = n factor$ $(mohr salt) \times M_{Mohr salt} \times V_{Mohr salt}$

$6 \times x \times 10^{- 3} \times 200 = 1 \times 0.6 \times 750$

$x = 375$

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 4 :

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

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

Q 5 :

In order to oxidise a mixture of 1 mole each of ${FeC}_{2} O_{4}$ , ${Fe}_{2} {(C_{2} O_{4})}_{3}$ , ${FeSO}_{4}$ and ${Fe}_{2} {({SO}_{4})}_{3}$ in acidic medium, the number of moles of ${KMnO}_{4}$ required is [2026]

Q 7 :

200 cc of $x \times 10^{- 3} M$ potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here $x =$ _______. [2026]

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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 4 : MnO4-+xe-→MnO42- per mole of MnO4-; x, y and z are respectively:

Q 5 : In order to oxidise a mixture of 1 mole each of FeC2O4, Fe2(C2O4)3, FeSO4 and Fe2(SO4)3 in acidic medium, the number of moles of KMnO4 required is [2026]

Q 7 : 200 cc of x×10-3 M potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here x=_______. [2026]

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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]

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 4 :

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

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

Q 5 :

In order to oxidise a mixture of 1 mole each of ${FeC}_{2} O_{4}$ , ${Fe}_{2} {(C_{2} O_{4})}_{3}$ , ${FeSO}_{4}$ and ${Fe}_{2} {({SO}_{4})}_{3}$ in acidic medium, the number of moles of ${KMnO}_{4}$ required is [2026]

Q 7 :

200 cc of $x \times 10^{- 3} M$ potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium. Here $x =$ _______. [2026]