NEET 2018 · ChemistryPrevious Year Question

For the redox reaction $MnO X_{4} X^{-} + C X_{2} O X_{4} X^{2 -} + H X^{+} Mn X^{2 +} + CO X_{2} + H X_{2} O$ , the correct coefficients of the reactants for the balanced equation are ( $MnO X_{4} X^{-}$ , $C X_{2} O X_{4} X^{2 -}$ , $H X^{+}$ respectively)

Answer: (B) (b) 2, 5, 16. \textbf{Answer:} (b) 2, 5, 16 \textbf{Solution:} $Mn$ goes from +7 (in $MnO X_{4} X^{-}$ ) to +2 (in $Mn X^{2 +}$ ), gaining 5 electrons, so the $n$ -factor of $MnO X_{4} X^{-}$ is 5.

A.(a) 2, 16, 5
B.(b) 2, 5, 16✓
C.(c) 16, 5, 2
D.(d) 5, 16, 2

Correct Answer

(B) (b) 2, 5, 16

Solution & Explanation

\textbf{Answer:} (b) 2, 5, 16 \textbf{Solution:} $Mn$ goes from +7 (in $MnO X_{4} X^{-}$ ) to +2 (in $Mn X^{2 +}$ ), gaining 5 electrons, so the $n$ -factor of $MnO X_{4} X^{-}$ is 5. Each $C$ goes from +3 (in $C X_{2} O X_{4} X^{2 -}$ ) to +4 (in $CO X_{2}$ ), so each oxalate loses 2 electrons; the $n$ -factor of $C X_{2} O X_{4} X^{2 -}$ is 2. To balance the electrons, multiply $MnO X_{4} X^{-}$ by 2 and $C X_{2} O X_{4} X^{2 -}$ by 5 (the molar ratio is the inverse of the $n$ -factors). Balancing oxygen and charge then requires 16 $H X^{+}$ : $2 MnO X_{4} X^{-} + 5 C X_{2} O X_{4} X^{2 -} + 16 H X^{+} 2 Mn X^{2 +} + 10 CO X_{2} + 8 H X_{2} O$ The coefficients of the reactants are 2, 5, 16, which is option (b).

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For the redox reaction MnOX4​X−+CX2​OX4​X2−+HX+​MnX2++COX2​+HX2​O, the correct coefficients of the reactants for the balanced equation are (MnOX4​X−, CX2​OX4​X2−, HX+ respectively)

Solution & Explanation

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For the redox reaction $MnO X_{4} X^{-} + C X_{2} O X_{4} X^{2 -} + H X^{+} Mn X^{2 +} + CO X_{2} + H X_{2} O$ , the correct coefficients of the reactants for the balanced equation are ( $MnO X_{4} X^{-}$ , $C X_{2} O X_{4} X^{2 -}$ , $H X^{+}$ respectively)