The ion `A^(n+)` is oxidised to `AO_(3)^(-)` by `MnO_(4)^(-)` changing to `Mn^(2+)` in acid medium. Given that `2.68 xx10^(-3)` mole of `A^(n+)` required `1.61 xx10^(-3)` mole of `MnO_(4)^(-)` . What is the value of n.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Determine the mole ratio of the reactants
We are given that 2.68 x 10^(-3) moles of A^(n+) reacts with 1.61 x 10^(-3) moles of MnO4^(-).

To find the mole ratio, we can divide the moles of A^(n+) by the moles of MnO4^(-):

\[ \text{Mole ratio} = \frac{2.68 \times 10^{-3}}{1.61 \times 10^{-3}} = \frac{2.68}{1.61} \approx 1.66 \]

This ratio simplifies to approximately 5:3 when multiplied to eliminate decimals.

### Step 2: Write the half-reactions
1. **Reduction half-reaction for MnO4^(-):**
\[ \text{MnO}_4^{-} + 8\text{H}^+ + 5e^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]
Here, Mn changes from +7 in MnO4^(-) to +2 in Mn^(2+).

2. **Oxidation half-reaction for A^(n+):**
\[ \text{A}^{n+} \rightarrow \text{AO}_3^{-} + 3e^{-} \]
Here, we need to determine the oxidation state of A.

### Step 3: Balance the charges
From the mole ratio, we know that 5 moles of A^(n+) reacts with 3 moles of MnO4^(-). Therefore, we can equate the number of electrons transferred:

- From the MnO4^(-) half-reaction, 5 electrons are involved.
- From the A^(n+) half-reaction, 3 electrons are involved.

### Step 4: Set up the equation for charge balance
The total charge on the left side must equal the total charge on the right side.

For the A^(n+) half-reaction:
\[ n - 3 = \text{charge of AO}_3^{-} = -1 \]
Thus, we can set up the equation:
\[ n - 3 = -1 \implies n = 2 \]

### Step 5: Conclusion
The value of n is 2, which means A is in the +2 oxidation state.

### Final Answer:
The value of n is **2**.

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