In balancing the half reaction
`CN^(ө)rarrCNO^(ө)`(skeltan)
The number of electron that must be added is

Options

A. `1` on the right
B. `0`
C. `1` on the left
D. `2` on the right

Explanation

To balance the half-reaction \( \text{CN}^- \rightarrow \text{CNO}^- \), we need to determine the change in oxidation state of carbon and find out how many electrons must be added to balance the reaction. Here’s a step-by-step solution:

### Step 1: Determine the oxidation state of carbon in CN^-
Let the oxidation state of carbon in \( \text{CN}^- \) be \( x \).
The oxidation state of nitrogen (N) is -3 (since it is in the -1 charge state).
The overall charge of the ion is -1.

So, we can set up the equation:
\[ x + (-3) = -1 \]
Solving for \( x \):
\[ x - 3 = -1 \\ x = +2 \]
Thus, the oxidation state of carbon in \( \text{CN}^- \) is +2.

### Step 2: Determine the oxidation state of carbon in CNO^-
Let the oxidation state of carbon in \( \text{CNO}^- \) be \( y \).
The oxidation state of nitrogen is -3 and that of oxygen is -2.
The overall charge of the ion is -1.

So, we can set up the equation:
\[ y + (-3) + (-2) = -1 \]
Solving for \( y \):
\[ y - 5 = -1 \\ y = +4 \]
Thus, the oxidation state of carbon in \( \text{CNO}^- \) is +4.

### Step 3: Calculate the change in oxidation state
Now we can find the change in oxidation state of carbon:
\[ \text{Change} = \text{Oxidation state in CNO}^- - \text{Oxidation state in CN}^- \\ \text{Change} = +4 - (+2) = +2 \]
This indicates that carbon is oxidized by 2 units.

### Step 4: Balancing the half-reaction
To balance the half-reaction, we need to add electrons to the left side (reactants) to account for the increase in oxidation state:
\[ \text{CN}^- \rightarrow \text{CNO}^- + 2e^- \]
This shows that 2 electrons must be added to balance the oxidation process.

### Conclusion
The number of electrons that must be added is **2**.

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