Define oxidation number - Leaving Cert Chemistry - Question d - 2001

In the given reaction:

1. Assign oxidation numbers:

   • For $H^+$: +1
   • For $MnO_4^-$: +7 (for Mn)
   • For $C_2O_4^{2-}$: +3 (for C)
   • For $CO_2$: +4 (for C)
   • For $Mn^{2+}$: +2

2. Identify oxidation and reduction:

   • Oxidation: C in $C_2O_4^{2-}$ is oxidized from +3 to +4 in $CO_2$, indicating loss of electrons.
   • Reduction: Mn in $MnO_4^-$ is reduced from +7 to +2 in $Mn^{2+}$, indicating gain of electrons.

3. Balance the half-reactions:

   • Oxidation half-reaction: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
   • Reduction half-reaction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

4. Combine and balance the overall equation: $16H^+ + 2MnO_4^- + 5C_2O_4^{2-} \rightarrow 10CO_2 + 2Mn^{2+} + 8H_2O$

The reduction in decolorising time upon the addition of the second portion of potassium manganate(VII) solution can be explained by the concept of catalyst and reaction kinetics. The initial addition of potassium manganate(VII) produces a substantial amount of the $Mn^{2+}$ catalyst. When the second portion is added, it provides more reactants that react rapidly with the remaining oxalic acid, increasing the reaction rate. Therefore, the reaction proceeds more quickly, resulting in a shorter decolorising time as more of the purple manganese species is reduced to the colorless manganese ions.