Balancing the Chemical Equation

The given unbalanced chemical reaction is:

\( \text{xMnO}_4^- + \text{yFe}_3\text{O}_4 + \text{H}_2\text{O} \rightarrow \text{xMnO}_2 + \text{zFe}_2\text{O}_3 + \text{xOH}^- \)

We need to find the values of the coefficients x, y, and z that balance this equation. Notice that the coefficient 'x' appears before \( \text{MnO}_4^- \), \( \text{MnO}_2 \), and \( \text{OH}^- \). The coefficient before \( \text{H}_2\text{O} \) is implicitly 1 as written.

Steps for Balancing Chemical Equations

We can balance the equation by ensuring that the number of atoms of each element and the total charge are the same on both sides of the reaction. We will use the algebraic method based on atom count.

Let's write down the balance equations for each element:

• Manganese (Mn): Number of Mn atoms on the left = x. Number of Mn atoms on the right = x. This is already balanced by the notation.
• Iron (Fe): Number of Fe atoms on the left = 3y (since each \( \text{Fe}_3\text{O}_4 \) molecule has 3 Fe atoms). Number of Fe atoms on the right = 2z (since each \( \text{Fe}_2\text{O}_3 \) molecule has 2 Fe atoms). \[ 3y = 2z \quad (1) \]
• Oxygen (O): Number of O atoms on the left = 4x (from \( \text{MnO}_4^- \)) + 4y (from \( \text{Fe}_3\text{O}_4 \)) + 1 (from \( \text{H}_2\text{O} \)). Total = \( 4x + 4y + 1 \). Number of O atoms on the right = 2x (from \( \text{MnO}_2 \)) + 3z (from \( \text{Fe}_2\text{O}_3 \)) + x (from \( \text{OH}^- \)). Total = \( 2x + 3z + x = 3x + 3z \). \[ 4x + 4y + 1 = 3x + 3z \] Simplifying this equation by subtracting 3x from both sides: \[ x + 4y + 1 = 3z \quad (2) \]
• Hydrogen (H): Number of H atoms on the left = 2 (from \( \text{H}_2\text{O} \)). Number of H atoms on the right = x (from \( \text{OH}^- \)). \[ 2 = x \quad (3) \]

Now, let's consider the total charge on both sides:

• Charge: Total charge on the left = x multiplied by the charge of \( \text{MnO}_4^- \) (-1) + y multiplied by the charge of \( \text{Fe}_3\text{O}_4 \) (0) + 1 multiplied by the charge of \( \text{H}_2\text{O} \) (0). Total = \( x(-1) + y(0) + 1(0) = -x \). Total charge on the right = x multiplied by the charge of \( \text{MnO}_2 \) (0) + z multiplied by the charge of \( \text{Fe}_2\text{O}_3 \) (0) + x multiplied by the charge of \( \text{OH}^- \) (-1). Total = \( x(0) + z(0) + x(-1) = -x \). \[ -x = -x \] This equation is always true and does not give us new information about x, y, or z, but it confirms the structure of the given reaction relative to charge.

Solving for Coefficients x, y, and z

We have a system of equations from the atom balance:

1. \( 3y = 2z \)
2. \( x + 4y + 1 = 3z \)
3. \( x = 2 \)

From equation (3), we directly find the value of x:

\( \mathbf{x = 2} \)

Now substitute the value of x into equation (2):

\( 2 + 4y + 1 = 3z \)

\( 3 + 4y = 3z \quad (4) \)

Now we have a system of two equations with y and z:

\( 3y = 2z \quad (1) \)

\( 3 + 4y = 3z \quad (4) \)

From equation (1), we can express z in terms of y:

\( z = \frac{3y}{2} \)

Substitute this expression for z into equation (4):

\( 3 + 4y = 3 \left( \frac{3y}{2} \right) \)

\( 3 + 4y = \frac{9y}{2} \)

To eliminate the fraction, multiply the entire equation by 2:

\( 2(3 + 4y) = 2 \left( \frac{9y}{2} \right) \)

\( 6 + 8y = 9y \)

Subtract 8y from both sides:

\( 6 = 9y - 8y \)

\( \mathbf{y = 6} \)

Now substitute the value of y into the expression for z:

\( z = \frac{3y}{2} = \frac{3(6)}{2} = \frac{18}{2} = 9 \)

\( \mathbf{z = 9} \)

So, the values of the coefficients are x=2, y=6, and z=9.

Verification

Let's substitute x=2, y=6, z=9 back into the original reaction:

\( 2\text{MnO}_4^- + 6\text{Fe}_3\text{O}_4 + 1\text{H}_2\text{O} \rightarrow 2\text{MnO}_2 + 9\text{Fe}_2\text{O}_3 + 2\text{OH}^- \)

Check atom counts:

Check charge:

All atoms and the charge are balanced. The coefficients are x=2, y=6, and z=9.

Final Coefficients

The balanced coefficients are x=2, y=6, and z=9.

Revision Table: Balancing Chemical Equations

Additional Information: Redox Reactions and Balancing

The reaction \( \text{MnO}_4^- + \text{Fe}_3\text{O}_4 \rightarrow \text{MnO}_2 + \text{Fe}_2\text{O}_3 \) is a redox reaction. In this reaction, Manganese changes oxidation state from +7 in \( \text{MnO}_4^- \) to +4 in \( \text{MnO}_2 \). This is a reduction.

\( \text{Mn}^{+7} \rightarrow \text{Mn}^{+4} \)

Iron in \( \text{Fe}_3\text{O}_4 \) (a mixed oxide containing \( \text{Fe}^{2+} \) and \( \text{Fe}^{3+} \)) is oxidized. Specifically, the \( \text{Fe}^{2+} \) portion of \( \text{Fe}_3\text{O}_4 \) is oxidized to \( \text{Fe}^{3+} \) in \( \text{Fe}_2\text{O}_3 \).

\( \text{Fe}^{+2} \rightarrow \text{Fe}^{+3} \)

\( \text{Fe}_3\text{O}_4 \) can be written as \( \text{FeO} \cdot \text{Fe}_2\text{O}_3 \). The Fe in \( \text{FeO} \) is +2, and the Fe in \( \text{Fe}_2\text{O}_3 \) is +3. In the product \( \text{Fe}_2\text{O}_3 \), all Fe is +3. The species being oxidized is the \( \text{Fe}^{2+} \) from \( \text{Fe}_3\text{O}_4 \). This is why balancing redox reactions involving complex oxides like \( \text{Fe}_3\text{O}_4 \) requires careful consideration of the oxidation states or using the overall change method as shown in the algebraic balance.

The presence of \( \text{OH}^- \) in the products and \( \text{H}_2\text{O} \) suggests the reaction occurs in a basic medium.