Define oxidation number - Leaving Cert Chemistry - Question 10 - 2002

For the reaction given:

MnO$_4^-$ + Cl$^- + H^+$ → Mn$^{2+}$ + Cl$_2$ + H$_2$O,

let's assign oxidation numbers:

• For Mn in MnO$_4^-$, the oxidation number is +7.
• For Cl$^-$, the oxidation number is -1.
• For Mn$^{2+}$, the oxidation number is +2.
• For Cl$_2$, the oxidation number is 0.

Now, let's analyze the changes:

• Mn changes from +7 (in MnO$_4^-$) to +2 (in Mn$^{2+}$), indicating it is reduced.
• Cl changes from -1 (in Cl$^-$) to 0 (in Cl$_2$), indicating it is oxidised.

Thus, MnO$_4^-$ is reduced and Cl$^-$ is oxidised.

To balance the equation:

1. Start with the half-reactions:

   • For Mn: MnO$_4^-$ → Mn$^{2+}$
   • For Cl: Cl$^-$ → Cl$_2$

2. Balance the Mn reaction:

   • 2MnO$_4^-$ → 2Mn$^{2+}$
   • Balance charges with H$^+$ and water (H$_2$O): 2MnO$_4^-$ + 16H$^+$ → 2Mn$^{2+}$ + 8H$_2$O

3. Balance the Cl reaction:

   • 2Cl$^-$ → Cl$_2$ + 2e$^-$

4. Combine half-reactions:

   • 2MnO$_4^-$ + 10Cl$^-$ + 16H$^+$ → 2Mn$^{2+}$ + 5Cl$_2$ + 8H$_2$O

Thus, the balanced equation is:

2MnO$_4^-$ + 10Cl$^-$ + 16H$^+$ → 2Mn$^{2+}$ + 5Cl$_2$ + 8H$_2$O.