Understanding the Balanced Redox Reaction for Permanganate and Chloride Ions

The reaction between permanganate (MnO4-) and chloride (Cl-) ions is a classic example of redox chemistry. The goal is to balance the reaction under acidic conditions to produce manganese(II) ions (Mn2 ), chlorine gas (Cl2), and water (H2O).

Step-by-Step Guide to Balancing the Redox Reaction

Balancing redox equations can be a bit tricky, but with a systematic approach, it can be done easily. Here are the steps to balance the redox reaction involving permanganate and chloride ions:

Step 1: Identify the Half-Reactions

The first step is to break down the overall reaction into their respective half-reactions. One half-reaction involves the reduction of permanganate to manganese ions, while the other involves the oxidation of chloride to chlorine gas.

Reduction Half-Reaction

The reduction of permanganate to manganese ions is as follows:

MnO4- 8 H 5 e- $rightarrow$ Mn2 4 H2O

Oxidation Half-Reaction

The oxidation of chloride ions to chlorine gas is as follows:

2 Cl- $rightarrow$ Cl2 2 e-

Step 2: Balance the Electrons

To balance the electrons in the half-reactions, we need to find a common multiple of the electrons involved in both half-reactions. The first half-reaction involves 5 electrons, while the second half-reaction involves 2 electrons. The least common multiple of 5 and 2 is 10.

To balance the electrons in the oxidation half-reaction, multiply it by 5:

10 Cl- $rightarrow$ 5 Cl2 10 e-

Step 3: Combine the Half-Reactions

Now, combine the balanced half-reactions to form the overall equation:

MnO4- 8 H 10 Cl- $rightarrow$ Mn2 4 H2O 5 Cl2

Step 4: Balance Charges and Atoms

Finally, ensure that both the charge and the number of atoms are balanced in the equation. The final balanced equation is:

MnO4- 8 H 10 Cl- $rightarrow$ Mn2 4 H2O 5 Cl2

The Final Balanced Redox Equation

If we write the equation in terms of molar amounts and with spectator ions included, we get:

K MnO4- 8 K HClaq $rightarrow$ K2MnCl2aq $frac{5}{2}$ Cl2g 4 K H2O

The reaction changes from a deep purple color due to the permanganate ion to a colorless solution when manganese(II) ions are formed. This color change is due to the fact that manganese(II) forms a high-spin d5 complex, which does not absorb light in the visible region, hence it is colorless.

Summary:

Reduction half-reaction: MnO4- 8 H 5 e- $rightarrow$ Mn2 4 H2O Oxidation half-reaction: 10 Cl- $rightarrow$ 5 Cl2 10 e- Final balanced equation: MnO4- 8 H 10 Cl- $rightarrow$ Mn2 4 H2O 5 Cl2