If Cl_2 is passed through hot aqueous NaOH, t | vedclass

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Question

(B) When $Cl_2$ is passed through hot aqueous $NaOH$, a disproportionation reaction occurs, resulting in the formation of sodium chloride $(NaCl)$ and

Vedclass · Accepted Answer

$-1$ and $+5$

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(B) When $Cl_2$ is passed through hot aqueous $NaOH$, a disproportionation reaction occurs, resulting in the formation of sodium chloride $(NaCl)$ and sodium chlorate $(NaClO_3)$: $6NaOH + 3Cl_2 \longrightarrow 5NaCl + NaClO_3 + 3H_2O$ In $NaCl$, the oxidation state of $Cl$ is $x$: $(+1) + x = 0 \implies x = -1$ In $NaClO_3$, the oxidation state of $Cl$ is $y$: $(+1) + y + 3(-2) = 0 \implies y - 5 = 0 \implies y = +5$ Thus, the $Cl$ atoms in the products exist in $-1$ and $+5$ oxidation states.

If $Cl_2$ is passed through hot aqueous $NaOH$, the products formed have $Cl$ in different oxidation states. These are indicated as

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