`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+)+ne^(-)`
What is the number `(n)` of electrons involved in the process, divided by `10` ?

Explanation

To solve the problem of determining the number of electrons involved in the oxidation of the cyanide ion (CN⁻) to nitrate (NO₃⁻) and carbon dioxide (CO₂), we will follow these steps:

### Step 1: Write the half-reaction
The half-reaction for the oxidation of CN⁻ can be expressed as:
\[ \text{CN}^- \rightarrow \text{CO}_2 + \text{NO}_3^- + \text{H}^+ + n e^- \]

### Step 2: Determine oxidation states
- In CN⁻, carbon (C) has an oxidation state of -1 and nitrogen (N) has an oxidation state of -3.
- In CO₂, carbon has an oxidation state of +4.
- In NO₃⁻, nitrogen has an oxidation state of +5.

### Step 3: Calculate the change in oxidation states
- For carbon:
- Change from -1 (in CN⁻) to +4 (in CO₂) = +5
- For nitrogen:
- Change from -3 (in CN⁻) to +5 (in NO₃⁻) = +8

### Step 4: Total change in oxidation states
The total change in oxidation states is the sum of the changes for carbon and nitrogen:
\[ \text{Total change} = 5 + 8 = 13 \]

### Step 5: Determine the number of electrons transferred
Since each oxidation corresponds to the loss of electrons, the total number of electrons (n) involved in the reaction is equal to the total change in oxidation states:
\[ n = 13 \]

### Step 6: Divide by 10
The problem asks for the number of electrons involved in the process divided by 10:
\[ \frac{n}{10} = \frac{13}{10} = 1.3 \]

### Final Answer
The final answer is:
\[ \frac{n}{10} = 1.3 \]