The coefficients x, y, p, q, and r in the fol | vedclass

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(D) The given reaction is a disproportionation reaction where $Mn$ in $MnO_4^{2-}$ is oxidized and reduced.
Step $1$: Write the half-reactions.
Oxid

Vedclass · Accepted Answer

$3, 2, 1, 2, 4$

Vedclass · Answer

(D) The given reaction is a disproportionation reaction where $Mn$ in $MnO_4^{2-}$ is oxidized and reduced.
Step $1$: Write the half-reactions.
Oxidation: $MnO_4^{2-} \rightarrow MnO_4^{-} + e^-$
Reduction: $MnO_4^{2-} + 2H_2O + 2e^- \rightarrow MnO_2 + 4OH^{-}$
Step $2$: Balance the electrons by multiplying the oxidation half-reaction by $2$.
$2MnO_4^{2-} \rightarrow 2MnO_4^{-} + 2e^-$
$MnO_4^{2-} + 2H_2O + 2e^- \rightarrow MnO_2 + 4OH^{-}$
Step $3$: Add the two half-reactions:
$3MnO_4^{2-} + 2H_2O \rightarrow MnO_2 + 2MnO_4^{-} + 4OH^{-}$
Comparing this with the given equation, we get $x=3, y=2, p=1, q=2, r=4$.
Thus, the correct option is $D$.

The coefficients $x, y, p, q,$ and $r$ in the following balanced equation are respectively:
$xMnO_4^{2-}{_{\text{(aq)}}} + yH_2O_{\text{(l)}} \rightarrow pMnO_2{_{\text{(s)}}} + qMnO_4^{-}{_{\text{(aq)}}} + rOH^{-}{_{\text{(aq)}}}$

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The coefficients $x, y, p, q,$ and $r$ in the following balanced equation are respectively: $xMnO_4^{2-}{_{\text{(aq)}}} + yH_2O_{\text{(l)}} \rightarrow pMnO_2{_{\text{(s)}}} + qMnO_4^{-}{_{\text{(aq)}}} + rOH^{-}{_{\text{(aq)}}}$