Question

A 0.1 M solution of a metal salt (50 mL) reacts with 25 mL of 0.1 M sodium sulphite. Given the half-reaction

$SO_3^{2-}$ + $H_2O$ → $SO_4^{2-}$ + 2$H^+$ + 2$e^-$

If the metal starts with oxidation number +2, what is its oxidation number after reaction.

• A. 0
• B. +1
• C. +2
• D. +4

Answer 1

Answer: (B)

+1

Answer 2

1. Calculate moles of reactants:

   • Moles of metal ions ($M^{2+}$) = Molarity × Volume = $0.1 \, \text{M} \times 50 \, \text{mL} = 0.1 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{mol}$.
   • Moles of sulfite ions ($SO_3^{2-}$) = Molarity × Volume = $0.1 \, \text{M} \times 25 \, \text{mL} = 0.1 \, \text{mol/L} \times 0.025 \, \text{L} = 0.0025 \, \text{mol}$.

2. Calculate moles of electrons transferred:

   • The given half-reaction is $SO_3^{2-}$ + $H_2O$ → $SO_4^{2-}$ + 2$H^+$ + 2$e^-$.
   • This shows that 1 mole of $SO_3^{2-}$ produces 2 moles of electrons.
   • Therefore, 0.0025 mol of $SO_3^{2-}$ produces $0.0025 \, \text{mol} \times 2 = 0.005 \, \text{mol}$ of electrons.

3. Determine electron transfer per metal ion:

   • The metal starts with an oxidation number of +2 ($M^{2+}$).
   • The electrons produced by the oxidation of sulfite are accepted by the metal ions, causing them to be reduced.
   • We have 0.005 mol of $M^{2+}$ ions and 0.005 mol of electrons available for reduction.
   • Number of electrons accepted per $M^{2+}$ ion = $\frac{\text{Moles of } e^-}{\text{Moles of } M^{2+}} = \frac{0.005 \, \text{mol}}{0.005 \, \text{mol}} = 1 \, e^-$.

4. Calculate the final oxidation state of the metal:

   • The metal ion starts with an oxidation state of +2.
   • It accepts 1 electron during the reaction.
   • The reduction half-reaction for the metal is $M^{2+} + 1e^- \rightarrow M^{+}$.
   • The final oxidation state of the metal is +2 - 1 = +1.