H_2S gas is not evolved when SO_3^{2-} ion re | vedclass

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(B) The reaction of $SO_3^{2-}$ with $Al$ in the presence of $NaOH$ is a reduction reaction where $SO_3^{2-}$ is reduced to $S^{2-}$.The chemical equa

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$Al +$ Conc. $NaOH$

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(B) The reaction of $SO_3^{2-}$ with $Al$ in the presence of $NaOH$ is a reduction reaction where $SO_3^{2-}$ is reduced to $S^{2-}$.The chemical equation is: $SO_3^{2-} + 2Al + 2OH^- + 3H_2O 	o S^{2-} + 2[Al(OH)_4]^-$.In this reaction,$H_2S$ is not evolved because the sulfide ion $(S^{2-})$ formed is immediately trapped in the alkaline medium as $Na_2S$ (or $S^{2-}$ ions),and no acidic conditions are present to release $H_2S$ gas.In contrast,reactions with dilute acids like $HCl$ or $H_2SO_4$ typically provide the $H^+$ ions necessary to convert $S^{2-}$ into $H_2S$ gas.

$H_2S$ gas is not evolved when $SO_3^{2-}$ ion reacts with the following reagents.

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