In an experiment,50 mL of 0.1 M solution of | vedclass

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(C) The reaction involves the reduction of the metal ion and the oxidation of the sulphite ion.
For the metal ion: $M^{3+} + (3-n)e^- 	o M^{n+}$ (Re

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$2$

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(C) The reaction involves the reduction of the metal ion and the oxidation of the sulphite ion.
For the metal ion: $M^{3+} + (3-n)e^- 	o M^{n+}$ (Reduction).
The $n$-factor for the metal is $(3-n)$.
For the sulphite ion: $SO_3^{2-} 	o SO_4^{2-} + 2e^-$.
The $n$-factor for the sulphite ion is $2$.
Using the principle of equivalence: $M_1 V_1 n_1 = M_2 V_2 n_2$.
$(0.1 \ M) 	imes (50 \ mL) 	imes (3-n) = (0.1 \ M) 	imes (25 \ mL) 	imes (2)$.
$5 	imes (3-n) = 5$.
$3-n = 1$.
$n = 2$.
Thus,the new oxidation number of the metal is $2$.

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