`MnO_(4)^(2-)+H^(+) to "Product"`
Product is formed

Options

A. `MnO_(4)^(-)`
B. `MnO_(2)`
C. Mn
D. Both(1) & (2)

Answer

To determine the product formed when \( \text{MnO}_4^{2-} \) reacts with \( \text{H}^+ \), we will analyze the reaction step by step.

### Step 1: Identify the Reaction Type
The reaction of \( \text{MnO}_4^{2-} \) with \( \text{H}^+ \) is an example of a disproportionation reaction. In this type of reaction, a single substance is both oxidized and reduced.

**Hint:** Look for changes in oxidation states to identify disproportionation.

### Step 2: Determine the Oxidation States
Let's assign oxidation states to manganese in \( \text{MnO}_4^{2-} \):
- Let the oxidation state of Mn be \( x \).
- The oxidation state of oxygen is -2, and there are 4 oxygen atoms.
- The overall charge of the ion is -2.

Setting up the equation:
\[ x + 4(-2) = -2 \]
\[ x - 8 = -2 \]
\[ x = +6 \]

So, the oxidation state of Mn in \( \text{MnO}_4^{2-} \) is +6.

**Hint:** Assign oxidation states carefully to understand the electron transfer.

### Step 3: Identify the Products
In a disproportionation reaction, the same element is transformed into two different oxidation states. Here, \( \text{Mn}^{+6} \) will be oxidized to \( \text{Mn}^{+7} \) and reduced to \( \text{Mn}^{+4} \).

- The product formed from oxidation is \( \text{MnO}_4^{-} \) (where Mn is +7).
- The product formed from reduction is \( \text{MnO}_2 \) (where Mn is +4).

**Hint:** Look for the products that correspond to the oxidation and reduction of the same element.

### Step 4: Write the Balanced Equation
The unbalanced equation can be written as:
\[ \text{MnO}_4^{2-} + \text{H}^+ \rightarrow \text{MnO}_4^{-} + \text{MnO}_2 + \text{H}_2\text{O} \]

To balance the equation, we need to ensure that both mass and charge are balanced. The balanced equation is:
\[ 3 \text{MnO}_4^{2-} + 4 \text{H}^+ \rightarrow 2 \text{MnO}_4^{-} + \text{MnO}_2 + 2 \text{H}_2\text{O} \]

**Hint:** Always check that both sides of the equation have the same number of atoms and the same total charge.

### Step 5: Conclusion
The products formed from the reaction of \( \text{MnO}_4^{2-} \) with \( \text{H}^+ \) are \( \text{MnO}_4^{-} \) and \( \text{MnO}_2 \).

**Final Answer:** The products are \( \text{MnO}_4^{-} \) and \( \text{MnO}_2 \).