How to Solve a Redox Reaction: A Case Study of KMnO? NH? MnO? N? KOH H?O

Understanding and solving redox reactions is a crucial skill in chemistry. This article will guide you through the process of balancing a complex redox reaction using the example of the reaction between potassium permanganate (KMnO?) and ammonia (NH?) to produce manganese dioxide (MnO?), nitrogen gas (N?), potassium hydroxide (KOH), and water (H?O).

Step 1: Identify Oxidation States

To begin, we need to determine the oxidation states of the elements involved in the reaction.

KMnO?

K: 1 Mn: 7 O: -2 (four Os contribute -8) Net charge: 1 - 7 - 8 0 (correct)

NH?

N: -3 H: 1 (three Hs contribute 3) Net charge: -3 3 0 (correct)

MnO?

Mn: 4 O: -2 (two Os contribute -4) Net charge: 4 - 4 0 (correct)

N?

N: 0 (elemental form)

KOH

K: 1 O: -2 H: 1 Net charge: 1 - 2 1 0 (correct)

H?O

H: 1 (two Hs contribute 2) O: -2 Net charge: 2 - 2 0 (correct)

Step 2: Write the Half-Reactions

Next, we identify and write the oxidation and reduction half-reactions.

Oxidation Half-Reaction

NH? to N?:
( text{n } text{NH}_3 rightarrow text{N}_2 6text{H}^ 6e^- )

Reduction Half-Reaction

MnO?? to MnO?:
( text{MnO}_4^- 8text{H}^ 5e^- rightarrow text{MnO}_2 4text{H}_2text{O} )

Step 3: Balance Electrons

To balance the number of electrons transferred in both half-reactions, we multiply the half-reactions by appropriate factors.

Oxidation half-reaction produces 6 electrons. Reduction half-reaction consumes 5 electrons. Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 6.

Oxidation Half-Reaction

( 5 text{NH}_3 rightarrow 5text{N}_2 30text{H}^ 30e^- )

Reduction Half-Reaction

( 6 text{MnO}_4^- 48text{H}^ 30e^- rightarrow 6text{MnO}_2 24text{H}_2text{O} )

Step 4: Combine the Half-Reactions

Combine the two half-reactions, ensuring that the electrons cancel out.

( 5text{NH}_3 6text{MnO}_4^- 48text{H}^ rightarrow 5text{N}_2 6text{MnO}_2 24text{H}_2text{O} 30e^- 30e^- )

Step 5: Add KOH to Neutralize

Since KOH is a product, we can add hydroxide ions (OH?) to the reaction to neutralize the protons (H?).

( 48text{OH}^- 48text{H}^ rightarrow 24text{H}_2text{O} )

Final Balanced Reaction

Substitute back to get the final balanced equation.

( 5text{NH}_3 6text{KMnO}_4 24text{KOH} rightarrow 5text{N}_2 6text{MnO}_2 15text{H}_2text{O} 6text{KOH} )

Conclusion:

The balanced redox reaction is:

[ 5text{NH}_3 6text{KMnO}_4 24text{KOH} rightarrow 5text{N}_2 6text{MnO}_2 15text{H}_2text{O} 6text{KOH} ]

This process ensures both mass and charge are balanced. If you have further questions or need additional clarification, feel free to ask!