Question

\[ 2 \text{MnO}_4^{-} + b \text{I}^{-} + c \text{H}_2\text{O} \rightarrow x \text{I}_2 + y \text{MnO}_2 + z \text{OH}^{-} \]
If the above equation is balanced with integer coefficients, the value of \( z \) is \( \_\_\_\_\_\_ \).

Answer 1

Correct Answer: 8

Reduction Half Reaction:

\[ 2\text{MnO}_4^- \rightarrow 2\text{MnO}_2 \] \[ 2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^- \]

Oxidation Half Reaction:

\[ 2\text{I}^- \rightarrow \text{I}_2 + 2e^- \] \[ 6\text{I}^- \rightarrow 3\text{I}_2 + 6e^- \]

Adding the oxidation half and reduction half, we get the net reaction as:

\[ 2\text{MnO}_4^- + 6\text{I}^- + 4\text{H}_2\text{O} \rightarrow 3\text{I}_2 + 2\text{MnO}_2 + 8\text{OH}^- \]

Thus, \( z = 8 \).

Answer 2

The problem requires balancing the given redox reaction in a basic or neutral medium and determining the value of the stoichiometric coefficient \( z \) for \( \text{OH}^{-} \).

Concept Used:

The balancing of the redox reaction will be performed using the ion-electron method (half-reaction method). This method involves the following steps:

1. Separate the overall reaction into two half-reactions: one for oxidation and one for reduction.
2. Balance the atoms other than oxygen and hydrogen in each half-reaction.
3. Balance the oxygen atoms by adding \( \text{H}_2\text{O} \) molecules.
4. Balance the hydrogen atoms by adding \( \text{H}^+ \) ions.
5. Balance the charge by adding electrons (\( e^- \)).
6. Multiply the half-reactions by appropriate integers to make the number of electrons equal in both.
7. Add the balanced half-reactions and cancel any common species.
8. If the reaction is in a basic medium, neutralize any \( \text{H}^+ \) ions by adding an equal number of \( \text{OH}^- \) ions to both sides of the equation. The \( \text{H}^+ \) and \( \text{OH}^- \) on one side combine to form \( \text{H}_2\text{O} \).

Step-by-Step Solution:

Step 1: Identify and separate the oxidation and reduction half-reactions.

The oxidation state of Mn in \( \text{MnO}_4^{-} \) is +7, and in \( \text{MnO}_2 \) it is +4. This is a reduction.

The oxidation state of I in \( \text{I}^{-} \) is -1, and in \( \text{I}_2 \) it is 0. This is an oxidation.

Reduction half-reaction:

\[ \text{MnO}_4^{-} \rightarrow \text{MnO}_2 \]

Oxidation half-reaction:

\[ \text{I}^{-} \rightarrow \text{I}_2 \]

Step 2: Balance the atoms in each half-reaction.

For the reduction half-reaction, Mn is already balanced. To balance the four oxygen atoms on the left, we add two water molecules on the right:

\[ \text{MnO}_4^{-} \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \]

Now, balance the four hydrogen atoms on the right by adding four H⁺ ions on the left:

\[ \text{MnO}_4^{-} + 4\text{H}^{+} \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \]

For the oxidation half-reaction, balance the iodine atoms:

\[ 2\text{I}^{-} \rightarrow \text{I}_2 \]

Step 3: Balance the charges in each half-reaction by adding electrons.

In the reduction half-reaction, the net charge on the left is \( (-1) + 4(+1) = +3 \), and on the right is 0. Add 3 electrons to the left side:

\[ \text{MnO}_4^{-} + 4\text{H}^{+} + 3e^{-} \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O} \]

In the oxidation half-reaction, the net charge on the left is -2, and on the right is 0. Add 2 electrons to the right side:

\[ 2\text{I}^{-} \rightarrow \text{I}_2 + 2e^{-} \]

Step 4: Equalize the number of electrons in both half-reactions.

To make the number of electrons equal, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3.

Reduction:

\[ 2(\text{MnO}_4^{-} + 4\text{H}^{+} + 3e^{-} \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}) \] \[ \Rightarrow 2\text{MnO}_4^{-} + 8\text{H}^{+} + 6e^{-} \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O} \]

Oxidation:

\[ 3(2\text{I}^{-} \rightarrow \text{I}_2 + 2e^{-}) \] \[ \Rightarrow 6\text{I}^{-} \rightarrow 3\text{I}_2 + 6e^{-} \]

Step 5: Add the two half-reactions and cancel the electrons.

\[ 2\text{MnO}_4^{-} + 8\text{H}^{+} + 6\text{I}^{-} \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O} + 3\text{I}_2 \]

This is the balanced equation in an acidic medium.

Step 6: Convert the equation to a basic medium.

Since the original equation produces \( \text{OH}^{-} \), we must convert this equation to a basic medium. To neutralize the \( 8\text{H}^{+} \) ions, add \( 8\text{OH}^{-} \) ions to both sides of the equation.

\[ 2\text{MnO}_4^{-} + 8\text{H}^{+} + 8\text{OH}^{-} + 6\text{I}^{-} \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O} + 3\text{I}_2 + 8\text{OH}^{-} \]

Combine \( \text{H}^{+} \) and \( \text{OH}^{-} \) to form water:

\[ 2\text{MnO}_4^{-} + 8\text{H}_2\text{O} + 6\text{I}^{-} \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O} + 3\text{I}_2 + 8\text{OH}^{-} \]

Cancel the common \( \text{H}_2\text{O} \) molecules from both sides (subtract 4\( \text{H}_2\text{O} \) from each side):

\[ 2\text{MnO}_4^{-} + 4\text{H}_2\text{O} + 6\text{I}^{-} \rightarrow 2\text{MnO}_2 + 3\text{I}_2 + 8\text{OH}^{-} \]

Final Computation & Result:

The final balanced equation is:

\[ 2\text{MnO}_4^{-} + 6\text{I}^{-} + 4\text{H}_2\text{O} \rightarrow 3\text{I}_2 + 2\text{MnO}_2 + 8\text{OH}^{-} \]

Comparing this with the given equation format:

\[ 2 \text{MnO}_4^{-} + b \text{I}^{-} + c \text{H}_2\text{O} \rightarrow x \text{I}_2 + y \text{MnO}_2 + z \text{OH}^{-} \]

We can determine the coefficients: \( b=6, c=4, x=3, y=2, \) and \( z=8 \).

The value of \( z \) is 8.