In the reaction
`C_(2) O_(4)^(-2) + MnO_(4)^(-) + H^(+) rarr Mn^(+2) +CO_(2) ` the reductants is -

Options

A. `C_(2)O_(4)^(-2)`
B. `H^(+)`
C. `MnO_(4)^(-)`
D. None of these

Explanation

To determine the reductant in the reaction:

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

we need to analyze the oxidation states of the elements involved in the reaction.

### Step 1: Identify the oxidation states of carbon in \(\text{C}_2\text{O}_4^{2-}\)

1. The formula for oxalate ion is \(\text{C}_2\text{O}_4^{2-}\).
2. Let the oxidation state of carbon be \(x\).
3. There are 2 carbon atoms and 4 oxygen atoms (each with an oxidation state of -2).
4. The overall charge of the ion is -2.

The equation for the oxidation state can be set up as:

\[ 2x + 4(-2) = -2 \]

This simplifies to:

\[ 2x - 8 = -2 \implies 2x = 6 \implies x = +3 \]

Thus, the oxidation state of carbon in \(\text{C}_2\text{O}_4^{2-}\) is +3.

### Step 2: Identify the oxidation state of manganese in \(\text{MnO}_4^{-}\)

1. The formula for permanganate ion is \(\text{MnO}_4^{-}\).
2. Let the oxidation state of manganese be \(y\).
3. There are 4 oxygen atoms (each with an oxidation state of -2).

The equation for the oxidation state can be set up as:

\[ y + 4(-2) = -1 \]

This simplifies to:

\[ y - 8 = -1 \implies y = +7 \]

Thus, the oxidation state of manganese in \(\text{MnO}_4^{-}\) is +7.

### Step 3: Determine the oxidation state changes

1. In the products, manganese is reduced to \(\text{Mn}^{2+}\) with an oxidation state of +2.
2. The change in oxidation state for manganese is:

\[ +7 \rightarrow +2 \quad (\text{Reduction}) \]

3. For carbon in \(\text{C}_2\text{O}_4^{2-}\), it is converted to \(\text{CO}_2\) where the oxidation state of carbon is +4.
4. The change in oxidation state for carbon is:

\[ +3 \rightarrow +4 \quad (\text{Oxidation}) \]

### Step 4: Identify the reductant

- The species that undergoes oxidation (increases in oxidation state) acts as the reductant.
- Here, \(\text{C}_2\text{O}_4^{2-}\) is oxidized from +3 to +4, which means it is losing electrons.
- Therefore, \(\text{C}_2\text{O}_4^{2-}\) is the reductant in this reaction.

### Conclusion

The reductant in the reaction is \(\text{C}_2\text{O}_4^{2-}\).

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