In basic medium,CrO_{4}^{2-} oxidises S_{2}O_ | vedclass

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(B) The balanced redox reaction is:$8CrO_{4}^{2-} + 3S_{2}O_{3}^{2-} + 10H_{2}O \rightarrow 8Cr(OH)_{4}^{-} + 6SO_{4}^{2-} + 2OH^{-}$From the stoichio

Vedclass · Accepted Answer

$173$

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(B) The balanced redox reaction is:$8CrO_{4}^{2-} + 3S_{2}O_{3}^{2-} + 10H_{2}O ightarrow 8Cr(OH)_{4}^{-} + 6SO_{4}^{2-} + 2OH^{-}$From the stoichiometry,the $n$-factor for $CrO_{4}^{2-}$ ($Cr^{+6}$ to $Cr^{+3}$) is $3$,and for $S_{2}O_{3}^{2-}$ ($S^{+2}$ to $S^{+6}$),the total change in oxidation state is $2 	imes (6-2) = 8$.Using the law of equivalence: $n_{1}M_{1}V_{1} = n_{2}M_{2}V_{2}$$3 	imes 0.154 	imes V = 8 	imes 0.25 	imes 40$$0.462 	imes V = 80$$V = \frac{80}{0.462} \approx 173.16 \ mL$Rounding to the nearest integer,we get $173 \ mL$.

In basic medium,$CrO_{4}^{2-}$ oxidises $S_{2}O_{3}^{2-}$ to form $SO_{4}^{2-}$ and itself changes into $Cr(OH)_{4}^{-}$. The volume of $0.154 \ M \ CrO_{4}^{2-}$ required to react with $40 \ mL$ of $0.25 \ M \ S_{2}O_{3}^{2-}$ is ........... $mL$ (Rounded-off to the nearest integer).

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