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(C) The reaction involves the oxidation of sulphite $(SO_{3}^{2-})$ to sulphate $(SO_{4}^{2-})$,which releases $2$ electrons per mole of sulphite.Numb

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$2$

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(C) The reaction involves the oxidation of sulphite $(SO_{3}^{2-})$ to sulphate $(SO_{4}^{2-})$,which releases $2$ electrons per mole of sulphite.Number of milliequivalents $(Meq)$ of sodium sulphite = $25 \ mL 	imes 0.1 \ M 	imes 2 = 5 \ Meq$.Since $Meq$ of salt = $Meq$ of sodium sulphite,we have $50 \ mL 	imes 0.1 \ M 	imes n = 5$,where $n$ is the change in oxidation state of the metal.$5 	imes n = 5 \implies n = 1$.Since the metal is reduced,the new oxidation number = $3 - 1 = 2$.

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