50 mL of 0.1 M solution of a salt reacted with 25 mL of 0.1 M solution of sodium sulphite. The half reaction for the oxidation of sulphite ion is :

${\mathrm{SO}}_3^{2-} \left(\mathrm{aq}\right) + {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \stackrel{ }{\to } \left(\mathrm{aq}\right) + 2{\mathrm{H}}^{+} \left(\mathrm{aq}\right) + 2{\mathrm{e}}^{-}$

If the oxidation number of metal in the salt was 3, what would be the new oxidation number of metal :

No. of equivalents = mole x n-factor

${\mathrm{SO}}_3^{2-} + {\mathrm{H}}_2\mathrm{O} \stackrel{ }{\to } {\mathrm{SO}}_4^{-2} + 2{\mathrm{H}}^{+} + 2{\mathrm{e}}^{-}$                                                                                                                                               For Sulphite n factor is 2.       

                 (VN)_salt = (VN)_Sulphite                                                                                                                                                          $$\begin{gathered} \Rightarrow 50 \times .1 \times \mathrm{n} = 25 \times .1 \times 2 (\because \mathrm{N}=\mathrm{M}\times \mathrm{n}-\mathrm{factor}) \\ \mathrm{n} = \frac{2.5 \times 2}{5} \\ \mathrm{n} = 1 \\ \Rightarrow \mathrm{Final} \mathrm{oxidation} \mathrm{state} \mathrm{will} \mathrm{be} \left(3-1\right)=2 \end{gathered}$$