Chlorine undergoes disproportionation in alka | vedclass

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(A) The given reaction is a disproportionation reaction of chlorine in an alkaline medium:$Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$Step $1$: Assi

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$1, 2, 1$ and $1$

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(A) The given reaction is a disproportionation reaction of chlorine in an alkaline medium:$Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$Step $1$: Assign oxidation states.$Cl_2$ $(0)$ $\rightarrow Cl^-$ $(-1)$ and $ClO^-$ $(+1)$.Step $2$: Write half-reactions.Reduction: $Cl_2 + 2e^- \rightarrow 2Cl^-$Oxidation: $Cl_2 + 4OH^- \rightarrow 2ClO^- + 2H_2O + 2e^-$Step $3$: Add the half-reactions.$2Cl_2 + 4OH^- \rightarrow 2Cl^- + 2ClO^- + 2H_2O$Step $4$: Simplify by dividing by $2$.$Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O$Comparing with $a Cl_2 + b OH^- \rightarrow c ClO^- + d Cl^- + e H_2O$,we get $a=1, b=2, c=1, d=1$.

Chlorine undergoes disproportionation in alkaline medium as shown below:
$a Cl_{2(g)} + b OH^{-}_{(aq)} \rightarrow c ClO^{-}_{(aq)} + d Cl^{-}_{(aq)} + e H_2O_{(l)}$
The values of $a, b, c$ and $d$ in a balanced redox reaction are respectively:

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Chlorine undergoes disproportionation in alkaline medium as shown below:$a Cl_{2(g)} + b OH^{-}_{(aq)} \rightarrow c ClO^{-}_{(aq)} + d Cl^{-}_{(aq)} + e H_2O_{(l)}$The values of $a, b, c$ and $d$ in a balanced redox reaction are respectively: