`CN^(ө)` ion is oxidised by a powerful oxidising agent to `NO_(3)^(ө)` and `CO_(2)` or `CO_(3)^(2-)` depending on the acidity of the reaction mixture.
`CN^(ө)rarrCO_(2)+NO_(3)^(ө)+H^(o+)+e^(-)`
What is the number `(n)` of electrons involved in the process, divided by `10` ?

Explanation

To solve the problem, we need to determine the number of electrons involved in the oxidation of the cyanide ion (CN⁻) to nitrate (NO₃⁻) and carbon dioxide (CO₂), and then divide that number by 10.

### Step-by-Step Solution:

1. **Write the Half-Reaction**:
The oxidation half-reaction for the cyanide ion (CN⁻) can be represented as:
\[ CN^{-} \rightarrow CO_{2} + NO_{3}^{-} + H^{+} + e^{-} \]

2. **Balance the Half-Reaction**:
To balance the reaction, we need to ensure that the number of atoms of each element and the charge are balanced. The balanced equation for the oxidation of cyanide can be written as:
\[ 3 CN^{-} + 15 H_{2}O \rightarrow 3 CO_{2} + 3 NO_{3}^{-} + 30 H^{+} + 30 e^{-} \]

3. **Count the Electrons**:
From the balanced equation, we can see that a total of 30 electrons are involved in the reaction.

4. **Divide by 10**:
The problem asks for the number of electrons divided by 10:
\[ n = \frac{30}{10} = 3 \]

5. **Final Answer**:
Therefore, the value of \( n \) is 3.