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MnO(4)^(-)+H^(+)+Br^(-) to Mn^(3+)(aq.)+...

`MnO_(4)^(-)+H^(+)+Br^(-) to Mn^(3+)(aq.)+Br_(2)uarr`

A

For disproportionation reaction.

B

For comproportionation reaction.

C

For either intermolecular redox reaction or displacement reaction

D

For either thermal combination redox reaction or thermal decomposition redox reaction.

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The correct Answer is:
To solve the given question regarding the reaction: \[ \text{MnO}_4^{-} + \text{H}^{+} + \text{Br}^{-} \rightarrow \text{Mn}^{3+} + \text{Br}_2 \] we will follow these steps: ### Step 1: Identify Oxidation States First, we need to determine the oxidation states of all the elements involved in the reaction. - For \(\text{MnO}_4^{-}\): - Let the oxidation state of Mn be \(X\). - The oxidation state of O is \(-2\). - Therefore, the equation becomes: \[ X + 4(-2) = -1 \implies X - 8 = -1 \implies X = +7 \] - For \(\text{H}^{+}\): - The oxidation state is \(+1\). - For \(\text{Br}^{-}\): - The oxidation state is \(-1\). - For \(\text{Mn}^{3+}\): - The oxidation state is \(+3\). - For \(\text{Br}_2\): - The oxidation state is \(0\) (as it is in its elemental form). ### Step 2: Determine Changes in Oxidation States Next, we analyze the changes in oxidation states: - Manganese changes from \(+7\) in \(\text{MnO}_4^{-}\) to \(+3\) in \(\text{Mn}^{3+}\). This is a reduction (gain of electrons). - Bromine changes from \(-1\) in \(\text{Br}^{-}\) to \(0\) in \(\text{Br}_2\). This is an oxidation (loss of electrons). ### Step 3: Classify the Reaction Since we have both oxidation and reduction occurring simultaneously, this reaction is classified as a redox reaction. ### Step 4: Identify the Type of Redox Reaction In this case, the reaction is an intermolecular redox reaction because it involves different species (MnO4- and Br-). It is not a disproportionation reaction (where a single species undergoes both oxidation and reduction) or a comproportionation reaction (where a single species is present in two different oxidation states). ### Conclusion Thus, the correct classification of the reaction is that it is an intermolecular redox reaction. ### Final Answer The answer to the question is that the reaction is an **intermolecular redox reaction**. ---
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Balance the equation by ion electron method MnO_(4)^(-) + Br^(-) to Mn^(2+) + Br_(2) (acidic medium)

Consider the following reaction occuring in basic medium 2MnO_(4)^(-) + Br^(-)(aq) to 2MnO_(2) (s) + BrO_(3)^(-) (aq) How the above reaction can be balanced further ?

For the following balanced redox reaction, 2MnO_(4)^(-) + 4H^(+) + Br_(2) hArr 2Mn^(2+) + 2BrO_(3)^(-) + 2H_(2)O . If the molecular weight of MnO_(4)^(-) and Br_(2) are x & y respectively then

Identify the oxidant and the reductant respectively in the following reaction. Cl_(2)(g) + 2Br^(-)(aq) to 2Cl^(-)(aq) + Br_(2)(aq)

I^(-)(aq.)+MnO_(4)^(-)(aq.)overset(H^(+))toX+Mn^(2+)(aq.) I^(-)(aq.)+MnO_(4)^(-)(aq.) underset("weakly "OH^(-))overset("Neutral or")to Y+MnO_(2) MnO_(4)^(-)(aq.)+Mn^(2+)(aq.) overset(ZnSO_(4)) to Z+4H^(+) Product X, Y and Z are respectively.

I^(-)(aq.)+MnO_(4)^(-)(aq.)overset(H^(+))toX+Mn^(2+)(aq.) I^(-)(aq.)+MnO_(4)^(-)(aq.) underset("weakly "OH^(-))overset("Neutral or")to Y+MnO_(2) MnO_(4)^(-)(aq.)+Mn^(2+)(aq.) overset(ZnSO_(4)) to Z+4H^(+) Product X, Y and Z are respectively.

NaBr+MnO_(2)+H_(2)SO_(4)(Conc.) to MnSO_(4)+Br_(2)uarr

NaBr+MnO_(2)+H_(2)SO_(4)(Conc.) to MnSO_(4)+Br_(2)uarr

Assertion :- Bromide ion is serving as a reducing agent in the reaction 2MnO_(4)^(-)(aq.)+Br^(-)(aq.)+H_(2)Orarr2MnO_(2)(aq.)+BrO_(3)^(-)(aq.)+2OH^(-)(aq.) Reason :- Oxidation number of Br increases from -1 to +5

MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H_(2)O , E^(@) = 1.51V MnO_(2) + 4H^(+) + 2e^(-) righarrow Mn^(2+) + 2H_(2)O E^(@) = 1.23V E_(MnO_(4)^(-)|MnO_(2)

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