Prove that the reaction between fluorine and ice is a disproportionation reaction:
$H_{2}O_{(s)} + F_{2(g)} \to HF_{(g)} + HOF_{(g)}$
$H_{2}O_{(s)} + F_{2(g)} \to HF_{(g)} + HOF_{(g)}$
(N/A) Assigning oxidation states to the atoms in the reaction:
$H_{2}^{+1}O^{-2} + F_{2}^{0} \to H^{+1}F^{-1} + H^{+1}O^{-2}F^{+1}$
In this reaction,the oxidation state of fluorine $(F)$ changes from $0$ in $F_{2}$ to $-1$ in $HF$ (reduction) and to $+1$ in $HOF$ (oxidation).
Since the same element $(F)$ is simultaneously oxidized and reduced,the reaction is a disproportionation reaction.
$H_{2}^{+1}O^{-2} + F_{2}^{0} \to H^{+1}F^{-1} + H^{+1}O^{-2}F^{+1}$
In this reaction,the oxidation state of fluorine $(F)$ changes from $0$ in $F_{2}$ to $-1$ in $HF$ (reduction) and to $+1$ in $HOF$ (oxidation).
Since the same element $(F)$ is simultaneously oxidized and reduced,the reaction is a disproportionation reaction.
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Similar Questions
$3NaClO \xrightarrow{\Delta} NaClO_3 + 2NaCl$
Which of the following is true for the given reaction?
Which of the following is true for the given reaction?
Medium
View SolutionRefer to the periodic table and answer the following questions:
$(a)$ Select the possible non-metals that can show disproportionation reactions.
$(b)$ Select three metals that can show disproportionation reactions.
$(a)$ Select the possible non-metals that can show disproportionation reactions.
$(b)$ Select three metals that can show disproportionation reactions.
Medium
View SolutionWhich of the following are disproportionation reactions?
$(A)$ $Cl_2 + 2 NaOH \rightarrow NaCl + NaOCl + H_2O$
$(B)$ $2 H_2O_2 \rightarrow 2 H_2O + O_2$
$(C)$ $2 KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$
$(D)$ $3 MnO_4^{2-} + 4 H^+ \rightarrow 2 MnO_4^- + MnO_2 + 2 H_2O$
$(A)$ $Cl_2 + 2 NaOH \rightarrow NaCl + NaOCl + H_2O$
$(B)$ $2 H_2O_2 \rightarrow 2 H_2O + O_2$
$(C)$ $2 KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$
$(D)$ $3 MnO_4^{2-} + 4 H^+ \rightarrow 2 MnO_4^- + MnO_2 + 2 H_2O$
The species that undergo disproportionation in alkaline medium are
$(a) \, Cl_2 \, \, \, (b) \, MnO_4^{2-} \, \, \, (c) \, NO_2 \, \, \, (d) \, ClO_4^{-} $
$(a) \, Cl_2 \, \, \, (b) \, MnO_4^{2-} \, \, \, (c) \, NO_2 \, \, \, (d) \, ClO_4^{-} $
Difficult
View SolutionObserve the following reaction:
$Cl_{2(g)} + 2OH^{-}_{(aq)} \rightarrow ClO^{-}_{(aq)} + Cl^{-}_{(aq)} + H_2O_{(l)}$
Identify the correct statements about this reaction:
$I$. $Cl_{2(g)}$ is oxidized to $ClO^{-}_{(aq)}$
$II$. $Cl_{2(g)}$ is oxidized to $Cl^{-}_{(aq)}$
$III$. $Cl_{2(g)}$ is reduced to $ClO^{-}_{(aq)}$
$IV$. $Cl_{2(g)}$ is reduced to $Cl^{-}_{(aq)}$
The correct answer is:
$Cl_{2(g)} + 2OH^{-}_{(aq)} \rightarrow ClO^{-}_{(aq)} + Cl^{-}_{(aq)} + H_2O_{(l)}$
Identify the correct statements about this reaction:
$I$. $Cl_{2(g)}$ is oxidized to $ClO^{-}_{(aq)}$
$II$. $Cl_{2(g)}$ is oxidized to $Cl^{-}_{(aq)}$
$III$. $Cl_{2(g)}$ is reduced to $ClO^{-}_{(aq)}$
$IV$. $Cl_{2(g)}$ is reduced to $Cl^{-}_{(aq)}$
The correct answer is:
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