`CN^(-)` is oxidised by `NO_(3)^(-)` in presence of acid :
`can^(-)+bNO_(3)^(-)+cH^(+)rarr(a+b)NO+aCO_(2)+(c)/(2)H_(2)O`
`What are the values of a, b, c in that order:

Options

A. 3, 7, 7
B. 3, 10, 7
C. 3, 10, 10
D. 3, 7, 10

Answer

To solve the problem of balancing the redox reaction where CN⁻ is oxidized by NO₃⁻ in the presence of acid, we will follow a systematic approach using the half-reaction method. Here’s a step-by-step solution:

### Step 1: Identify the oxidation and reduction half-reactions
- **Oxidation half-reaction**: CN⁻ is oxidized to CO₂.
- **Reduction half-reaction**: NO₃⁻ is reduced to NO.

### Step 2: Write the oxidation half-reaction
1. The oxidation of CN⁻ to CO₂ can be represented as:
\[ CN^- \rightarrow CO_2 \]
2. Determine the oxidation states:
- In CN⁻, C is at +2 and N is at -3.
- In CO₂, C is at +4.
3. The change in oxidation state for carbon is from +2 to +4, indicating a loss of 2 electrons.

### Step 3: Balance the oxidation half-reaction
1. Balance the carbon atoms:
\[ CN^- \rightarrow CO_2 \]
2. Balance the oxygen by adding water:
\[ CN^- \rightarrow CO_2 + 3H_2O \]
3. Balance the hydrogen by adding H⁺:
\[ CN^- + 6H^+ \rightarrow CO_2 + 3H_2O \]
4. Balance the charge by adding electrons:
- Left side: 1 (from CN⁻) + 6 (from H⁺) = +7
- Right side: 0 (neutral CO₂) = 0
- Thus, we need to add 6 electrons to the left side:
\[ CN^- + 6H^+ \rightarrow CO_2 + 3H_2O + 2e^- \]

### Step 4: Write the reduction half-reaction
1. The reduction of NO₃⁻ to NO can be represented as:
\[ NO_3^- \rightarrow NO \]
2. Determine the oxidation states:
- In NO₃⁻, N is at +5.
- In NO, N is at +2.
3. The change in oxidation state for nitrogen is from +5 to +2, indicating a gain of 3 electrons.

### Step 5: Balance the reduction half-reaction
1. Balance the nitrogen atoms:
\[ NO_3^- \rightarrow NO \]
2. Balance the oxygen by adding water:
\[ NO_3^- \rightarrow NO + 2H_2O \]
3. Balance the hydrogen by adding H⁺:
\[ NO_3^- + 4H^+ \rightarrow NO + 2H_2O \]
4. Balance the charge by adding electrons:
- Left side: 1 (from NO₃⁻) + 4 (from H⁺) = +5
- Right side: 0 (neutral NO) = 0
- Thus, we need to add 3 electrons to the left side:
\[ NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O \]

### Step 6: Equalize the number of electrons
To combine the half-reactions, we need to equalize the number of electrons lost and gained. The oxidation half-reaction loses 2 electrons, and the reduction half-reaction gains 3 electrons. We can multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the electrons:

- Oxidation half-reaction (multiplied by 3):
\[ 3CN^- + 18H^+ \rightarrow 3CO_2 + 9H_2O + 6e^- \]

- Reduction half-reaction (multiplied by 2):
\[ 2NO_3^- + 8H^+ + 6e^- \rightarrow 2NO + 4H_2O \]

### Step 7: Combine the half-reactions
Now we can add the two half-reactions together:
\[ 3CN^- + 2NO_3^- + 10H^+ \rightarrow 3CO_2 + 2NO + 5H_2O \]

### Step 8: Identify the coefficients a, b, c
From the balanced equation:
- \( a = 3 \) (for CO₂)
- \( b = 2 \) (for NO₃⁻)
- \( c = 5 \) (for H⁺)

### Final Answer
Thus, the values of \( a, b, c \) are:
\[ \text{a = 3, b = 2, c = 5} \]