MnO_4^{-} oxidises (i) oxalate ion in acidic | vedclass

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Question

(B) The balanced chemical equation for the oxidation of oxalate ion by $MnO_4^{-}$ in acidic medium is:$2 MnO_4^{-} + 5 C_2O_4^{2-} + 16 H^{+} \longri

Vedclass · Accepted Answer

$2: 5$ and $1: 8$

Vedclass · Answer

(B) The balanced chemical equation for the oxidation of oxalate ion by $MnO_4^{-}$ in acidic medium is:$2 MnO_4^{-} + 5 C_2O_4^{2-} + 16 H^{+} \longrightarrow 2 Mn^{2+} + 10 CO_2 + 8 H_2O$From the stoichiometry,$2$ moles of $MnO_4^{-}$ react with $5$ moles of $C_2O_4^{2-}$.Thus,the ratio $[MnO_4^{-} : C_2O_4^{2-}] = 2 : 5$.The balanced chemical equation for the oxidation of $HCl$ by $MnO_4^{-}$ is:$2 MnO_4^{-} + 16 HCl \longrightarrow 2 MnCl_2 + 5 Cl_2 + 8 H_2O$From the stoichiometry,$2$ moles of $MnO_4^{-}$ react with $16$ moles of $HCl$.Thus,the ratio $[MnO_4^{-} : HCl] = 2 : 16 = 1 : 8$.Therefore,the required ratios are $2:5$ and $1:8$.

$MnO_4^{-}$ oxidises $(i)$ oxalate ion in acidic medium at $333 \ K$ and $(ii)$ $HCl$. For balanced chemical equations,the ratios $[MnO_4^{-} : C_2O_4^{2-}]$ in $(i)$ and $[MnO_4^{-} : HCl]$ in $(ii)$ respectively are

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