Balance these redox reactions by the oxidation number method.
[tex]MnO₄⁻ + I⁻⁷ \rightarrow MnO₂ + I₂[/tex]
Answer (1)
To balance a redox reaction using the oxidation number method, you'll need to follow several steps. Let's balance the reaction MnO 4 − + I − → MnO 2 + I 2 :
Assign Oxidation Numbers:
For MnO 4 − , Mn has an oxidation state of +7.
For I − , I has an oxidation state of -1.
In MnO 2 , Mn has an oxidation state of +4.
In I 2 , I has an oxidation state of 0.
Identify the Changes in Oxidation Numbers:
Mn changes from +7 to +4, which is a decrease in oxidation number, indicating reduction.
I changes from -1 to 0, which is an increase in oxidation number, indicating oxidation.
Calculate the Change in Electrons (Half Reactions):
Reduction half-reaction: Mn 7 + + 3 e − → Mn 4 +
Oxidation half-reaction: 2 I − → I 2 + 2 e −
Balance the Electrons Transferred:
To balance the electrons in the half reactions, multiply the reduction half-reaction by 2, and the oxidation half-reaction by 3:
Reduction: 2 Mn 7 + + 6 e − → 2 Mn 4 +
Oxidation: 6 I − → 3 I 2 + 6 e −
Combine the Balanced Half-Reactions:
Add the two balanced half-reactions:
2 MnO 4 − + 6 I − → 2 MnO 2 + 3 I 2
Final Balancing (Check other atoms and charge):
Check that all other atoms and the charges are balanced. In this case, they are balanced, so the balanced redox reaction is:
2 MnO 4 − + 6 I − → 2 MnO 2 + 3 I 2
By following these steps, you have balanced the redox reaction using the oxidation number method.
