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(C) disproportionation reaction is one in which the same element in a given oxidation state is simultaneously oxidized and reduced.$A$: $2 H_2O_2 \rig

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$MnO_4^- + 4 H^+ + 3 e^- \rightarrow MnO_2 + 2 H_2O$

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(C) disproportionation reaction is one in which the same element in a given oxidation state is simultaneously oxidized and reduced.$A$: $2 H_2O_2 \rightarrow 2 H_2O + O_2$: Oxygen in $H_2O_2$ is in $-1$ state. It goes to $-2$ in $H_2O$ (reduction) and $0$ in $O_2$ (oxidation). This is a disproportionation reaction.$B$: $2 NO_2 + H_2O \rightarrow HNO_3 + HNO_2$: Nitrogen in $NO_2$ is in $+4$ state. It goes to $+5$ in $HNO_3$ (oxidation) and $+3$ in $HNO_2$ (reduction). This is a disproportionation reaction.$C$: $MnO_4^- + 4 H^+ + 3 e^- \rightarrow MnO_2 + 2 H_2O$: Manganese in $MnO_4^-$ is in $+7$ state and goes to $+4$ in $MnO_2$. This is a simple reduction reaction,not disproportionation.$D$: $3 MnO_4^{2-} + 4 H^+ \rightarrow 2 MnO_4^- + MnO_2 + 2 H_2O$: Manganese in $MnO_4^{2-}$ is in $+6$ state. It goes to $+7$ in $MnO_4^-$ (oxidation) and $+4$ in $MnO_2$ (reduction). This is a disproportionation reaction.

Which of the given reactions is not an example of a disproportionation reaction?

Similar Questions

Prove that the reaction between fluorine and ice is a disproportionation reaction:
$H_{2}O_{(s)} + F_{2(g)} \to HF_{(g)} + HOF_{(g)}$

Statement-$1$ : $I_2 \rightarrow IO_3^- + I^-$; This is a disproportionation reaction.
Statement-$2$ : Oxidation number of $I$ can vary from $-1$ to $+7$.

In the following reaction,$3Br_2 + 6CO_3^{2-} + 3H_2O \rightarrow 5Br^{-} + BrO_3^- + 6HCO_3^-$,what happens to bromine?

Which of the following species can function both as an oxidizing as well as a reducing agent?

What will $1 \ mol$ of $MnO_4^{2-}$ produce upon disproportionation in a neutral aqueous medium?

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