Redox is a reaction in which both oxidation and reduction will take place simultaneously . It is obvious that if one substance gives electron there must be another substance to accept these electrons . In some reactions, same substance is reduced as well as oxidised, these reactions are termed as disproportionation reactions.
For calculating equivalent mass in redox reaction change in oxidtaion number is realted to n-factor which is reciprocal of molar ratio.
Which of the following is an example of redox reaction ?

Options

A. `2NO_(2) to N_(2)O_(4)`
B. `NH_(4)OH to NH_(4)^(+) + OH^(-)`
C. `2NO_(2) + H_(2)O to HNO_(3) + HNO_(2)`
D. `N_(2)O_(5) +H_(2)O to 2HNO_(3)`

Answer

To determine which of the given options is an example of a redox reaction, we need to analyze the oxidation states of the elements involved in each reaction. A redox reaction involves both oxidation (loss of electrons, increase in oxidation state) and reduction (gain of electrons, decrease in oxidation state).

### Step-by-step Solution:

1. **Identify the Reactions**: We have four options to analyze. We need to check the oxidation states of the relevant elements in each reaction.

2. **Option 1: \(2NO_2 \rightarrow N_2O_4\)**
- Calculate the oxidation state of nitrogen in \(NO_2\):
\[ \text{Let oxidation state of N be } x. \]
\[ x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4 \]
- Calculate the oxidation state of nitrogen in \(N_2O_4\):
\[ 2x + 4(-2) = 0 \implies 2x - 8 = 0 \implies 2x = 8 \implies x = +4 \]
- **Conclusion**: No change in oxidation state (both +4). Therefore, this is **not a redox reaction**.

3. **Option 2: \(NH_4OH \rightarrow NH_4^+ + OH^-\)**
- Calculate the oxidation state of nitrogen in \(NH_4OH\):
\[ \text{Let oxidation state of N be } x. \]
\[ x + 4(-1) + (-1) = 0 \implies x - 4 - 1 = 0 \implies x = -3 \]
- Calculate the oxidation state of nitrogen in \(NH_4^+\):
\[ x + 4(-1) = +1 \implies x - 4 = 1 \implies x = +3 \]
- **Conclusion**: No change in oxidation state (both -3). Therefore, this is **not a redox reaction**.

4. **Option 3: \(2NO_2 + H_2O \rightarrow HNO_3 + HNO_2\)**
- Calculate the oxidation state of nitrogen in \(NO_2\):
\[ x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4 \]
- Calculate the oxidation state of nitrogen in \(HNO_3\):
\[ 1 + x - 6 = 0 \implies x - 5 = 0 \implies x = +5 \]
- Calculate the oxidation state of nitrogen in \(HNO_2\):
\[ x + 1 - 4 = 0 \implies x - 3 = 0 \implies x = +3 \]
- **Conclusion**: Oxidation state changes: \(NO_2\) (from +4 to +5 in \(HNO_3\) - oxidation) and from +4 to +3 in \(HNO_2\) (reduction). Therefore, this is a **redox reaction**.

5. **Option 4: \(N_2O_5 + H_2O \rightarrow 2HNO_3\)**
- Calculate the oxidation state of nitrogen in \(N_2O_5\):
\[ 2x + 5(-2) = 0 \implies 2x - 10 = 0 \implies 2x = 10 \implies x = +5 \]
- Calculate the oxidation state of nitrogen in \(HNO_3\):
\[ 1 + x - 6 = 0 \implies x - 5 = 0 \implies x = +5 \]
- **Conclusion**: No change in oxidation state (both +5). Therefore, this is **not a redox reaction**.

### Final Answer:
The correct answer is **Option 3: \(2NO_2 + H_2O \rightarrow HNO_3 + HNO_2\)**, which is a redox reaction.