Mno-4 +I-=2Mn2++I2?

jedyolumilade

2 years ago

Let’s balance the redox reaction for the given equation:

[ \text{MnO}_4^- + \text{I}^- = 2\text{Mn}^{2+} + \text{I}_2 ]

Assign Oxidation Numbers:
Mn: +7 (from MnO₄⁻)
I: -1 (from I⁻)
Mn: +2 (from Mn²⁺)
I: 0 (from I₂)
Write Half-Reactions:
Reduction (Iodine, I): [ \text{I}^- \rightarrow \text{I}_2 + 2e^- ]
Oxidation (Manganese, Mn): [ \text{MnO}_4^- + 8e^- \rightarrow 2\text{Mn}^{2+} ]
Balance Atoms and Charges:
Balance oxygen atoms by adding water molecules: [ \text{MnO}_4^- + 8\text{H}_2\text{O} + 5\text{I}^- \rightarrow 2\text{Mn}^{2+} + 10\text{OH}^- + \text{I}_2 ]
Balance hydrogen atoms by adding H⁺ ions: [ \text{MnO}_4^- + 8\text{H}_2\text{O} + 5\text{I}^- + 8\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{OH}^- + \text{I}_2 ]
Balance Electrons:
Multiply the reduction half-reaction by 5 and the oxidation half-reaction by 2 to balance electrons: [ 5(\text{I}^- \rightarrow \text{I}_2 + 2e^-) + 2(\text{MnO}_4^- + 8e^- \rightarrow 2\text{Mn}^{2+}) ]
Combine Half-Reactions: [ 5\text{I}^- + 2\text{MnO}_4^- + 8\text{H}_2\text{O} + 8\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{OH}^- + \text{I}_2 ]
Therefore, the balanced redox equation is:

[ 5\text{I}^- + 2\text{MnO}_4^- + 8\text{H}_2\text{O} + 8\text{H}^+ = 2\text{Mn}^{2+} + 10\text{OH}^- + \text{I}_2 ]

Feel free to ask if you need further clarification