Redox equations are balanced either by ion-electron method or by oxidation number method. Both methods lead to the correct from of the balanced equation. The ion electron methodd has two advantages. So some chemists prefer to use the ion-electron method for redox reactions carried out in dilute aqueous solutions, where free ions have more or less independent existance.
The oxidation state method for redox reactions is mostly used for solid chemicals or for reactions in concentrated acid media.
For the reaction
`K_(4)[Fe(CN)_(6)]rarrFe^(3+)+CO_(2)+NO_(3)^(ө)`
the `n`-factor is

Options

A. `1`
B. `11`
C. `(5)/(3)`
D. `61`

Explanation

To find the n-factor for the reaction \( K_4[Fe(CN)_6] \rightarrow Fe^{3+} + CO_2 + NO_3^- \), we will follow these steps:

### Step 1: Identify the oxidation states
- In \( K_4[Fe(CN)_6] \):
- Potassium (K) has an oxidation state of +1.
- The cyanide ion (CN) has an oxidation state of -1.
- Let the oxidation state of iron (Fe) be \( x \).

The overall charge of the complex ion \( [Fe(CN)_6]^{4-} \) is -4. Therefore, we can set up the equation:
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]
Thus, the oxidation state of Fe in \( K_4[Fe(CN)_6] \) is +2.

### Step 2: Determine the changes in oxidation states
- In the reaction, \( Fe^{2+} \) is oxidized to \( Fe^{3+} \):
- Change in oxidation state for Fe: \( +2 \rightarrow +3 \) (1 electron lost).

- The cyanide ions \( CN^- \) are oxidized to \( CO_2 \):
- Each \( CN^- \) (with an oxidation state of -1) is converted to \( CO_2 \) (with an oxidation state of 0).
- The change in oxidation state for each \( CN^- \) is \( -1 \rightarrow 0 \) (1 electron lost per cyanide).
- Since there are 6 cyanide ions, the total change for cyanide is \( 6 \times 1 = 6 \) electrons lost.

### Step 3: Determine the reduction process
- The cyanide ions are also converted to nitrate ions \( NO_3^- \):
- Each \( CN^- \) is converted to \( NO_3^- \) (with an oxidation state of +5).
- The change in oxidation state for each \( CN^- \) is \( -1 \rightarrow +5 \) (6 electrons lost per cyanide).
- Thus, for 6 cyanide ions, the total change is \( 6 \times 6 = 36 \) electrons lost.

### Step 4: Calculate the total n-factor
- The total number of electrons transferred in the reaction is the sum of the electrons lost in the oxidation of \( Fe^{2+} \) and the cyanide ions:
\[ n = 1 \text{ (from Fe)} + 36 \text{ (from CN to NO}_3^-) + 6 \text{ (from CN to CO}_2) = 1 + 36 + 6 = 43 \]

### Conclusion
Thus, the n-factor for the reaction \( K_4[Fe(CN)_6] \rightarrow Fe^{3+} + CO_2 + NO_3^- \) is **43**.

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