Question

Balance the following redox reaction in basic solution. Cl ⒠(aq) + MnO ⒠4 (aq) → Cl 2 (g) + MnO 2 (s)

Answer 1

$6\; \text{Cl}^{-} \; (aq) + 2\; \text{MnO}_4^{-} \; (aq) + 4 \; \text{H}_2 \text{O} \; (l) \to 3\; \text{Cl}_2 \; (g) + 2\; \text{MnO}_2 \; (s) + 8\; \text{OH}^{-} \; (aq)$

Explanation

• The oxidation state of the manganese atom $\text{Mn}$ changes from +7 as in $\text{MnO}_4^{-}$ to +4 as in $\text{MnO}_2$.
• There are one $\text{Mn}$ atom in each $\text{MnO}_4$ ion.
• Reducing each $\text{MnO}_4^{-}$ ion would thus consume three electrons.

• Similarly, the oxidation state of the chlorine atom $\text{Cl}$ changes from -1 as in $\text{Cl}^{-}$ to 0 as in $\text{Cl}_2$.
• It takes two $\text{Cl}^{-}$ ions to produce one $\text{Cl}_2$ molecule.
• Oxidizing every two $\text{Cl}^{-}$ would thus produce one $\text{Cl}_2$ while releasing two electrons.

Three $\text{Cl}_2$ molecules contain six chlorine atoms. Three $\text{Cl}_2$ would thus correspond to six $\text{Cl}^{-}$ ions.

Combining six $\text{Cl}^{-}$, two $\text{MnO}_4^{-}$ ions, three $\text{Cl}_2$ ions, and two $\text{MnO}_2$ would balance the changes in oxidation state.

$6\; \text{Cl}^{-} \; (aq) + 2\; \text{MnO}_4^{-} \; (aq) \to 3\; \text{Cl}_2 \; (g) + 2\; \text{MnO}_2 \; (s)$ (NOT BALANCED)

Still, the product side lacks four oxygen atoms.

• In an acidic environment, oxygen atoms would combine with protons to produce water.
• In a basic environment (like this one,) there are nearly no protons for oxygen atoms to combine with. Oxygen atoms will likely combine with water to produce hydroxide ions.

Each oxygen atom combine with a water molecule to produce two hydroxide ions. Adding four water molecules and eight hydroxide ions would balance the equation.