$N_2H_4 + 2H_2O_2 \rightarrow N_2 + 4H_2O$
Redox Reaction Analysis: Identifying the Reducing Agent
This question asks us to identify the reducing agent in the given redox reaction: $N_2H_4 + 2H_2O_2 \rightarrow N_2 + 4H_2O$
Understanding Redox Reactions and Agents
In a redox reaction, electrons are transferred between chemical species. This involves changes in oxidation states.
- Oxidation is the loss of electrons or an increase in oxidation state.
- Reduction is the gain of electrons or a decrease in oxidation state.
- A reducing agent is a substance that causes reduction in another substance and is itself oxidized.
- An oxidizing agent is a substance that causes oxidation in another substance and is itself reduced.
To find the reducing agent, we need to determine the oxidation states of the elements in the reactants and see which substance gets oxidized (its oxidation state increases).
Calculating Oxidation States
Let's calculate the oxidation states for each element in the reactants and products:
Reactants:
1. $N_2H_4$ (Hydrazine)
- Hydrogen (H) typically has an oxidation state of +1.
- Let the oxidation state of Nitrogen (N) be $x$.
- The sum of oxidation states in a neutral molecule is 0.
- Equation: $2x + 4(+1) = 0$
- $2x + 4 = 0$
- $2x = -4$
- $x = -2$. The oxidation state of N in $N_2H_4$ is -2.
2. $H_2O_2$ (Hydrogen Peroxide)
- Hydrogen (H) has an oxidation state of +1.
- Let the oxidation state of Oxygen (O) be $y$.
- Equation: $2(+1) + 2y = 0$
- $2 + 2y = 0$
- $2y = -2$
- $y = -1$. The oxidation state of O in $H_2O_2$ is -1.
Products:
1. $N_2$ (Nitrogen Gas)
- Nitrogen is in its elemental form.
- The oxidation state of an element in its elemental form is 0.
- The oxidation state of N in $N_2$ is 0.
2. $H_2O$ (Water)
- Hydrogen (H) has an oxidation state of +1.
- Let the oxidation state of Oxygen (O) be $z$.
- Equation: $2(+1) + z = 0$
- $2 + z = 0$
- $z = -2$. The oxidation state of O in $H_2O$ is -2.
Analyzing Oxidation State Changes
Now, let's compare the oxidation states from reactants to products:
- Nitrogen (N): Changes from -2 in $N_2H_4$ to 0 in $N_2$. The oxidation state increases, meaning Nitrogen is oxidized.
- Oxygen (O): Changes from -1 in $H_2O_2$ to -2 in $H_2O$. The oxidation state decreases, meaning Oxygen is reduced.
- Hydrogen (H): Remains at +1 in all species ($N_2H_4$, $H_2O_2$, $H_2O$).
Determining the Reducing Agent
Since $N_2H_4$ is the substance that gets oxidized (its Nitrogen atom's oxidation state increases from -2 to 0), it acts as the reducing agent. It causes the reduction of $H_2O_2$ (where Oxygen's oxidation state decreases from -1 to -2).
| Compound | Element | Initial Oxidation State | Final Oxidation State | Change | Role |
|---|---|---|---|---|---|
| $N_2H_4$ | N | -2 | 0 (in $N_2$) | Increase (+2) | Reducing Agent |
| $H_2O_2$ | O | -1 | -2 (in $H_2O$) | Decrease (-1) | Oxidizing Agent |
Therefore, $N_2H_4$ is the reducing agent in this reaction.
Important Questions from Redox Reactions
Nitric acid is reduced to nitrogen dioxide in which of the following reactions?
In reaction, 2Na2S2O3 + I2 → A + 2NaI; A will be _________.
The role of H3PO4 in the estimation of Fe(II) with K2Cr2O7 using diphenylamine sulphonate as indicator is to
Number of electrons (x) involved in the following reaction is :
\(N \equiv N+ 8H^{+}+x \rightarrow 2N H^{3}+ H_{2}\)- Consider the following pairs of reactions and properties :
Reactions Properties I. $\text{Fe} \, (s) + 2\text{HCl} \, (aq) \rightarrow \text{FeCl}_2 \, (aq) + \text{H}_2 \, (g)$ $\text{Fe}$ is reductant II. $\text{Zn} \, (s) + \text{CuSO}_4 \, (aq) \rightarrow \text{ZnSO}_4 \, (aq) + \text{Cu} \, (s)$ $\text{Cu}$ is being reduced III. $\text{Br}_2 \, (l) + 2\text{I}^- \, (aq) \rightarrow 2\text{Br}^- \, (aq) + \text{I}_2 \, (s)$ $\text{Br}_2$ is reductant IV. $\text{CH}_4 \, (g) + 2\text{O}_2 \, (g) \rightarrow \text{CO}_2 \, (g) + 2\text{H}_2\text{O} \, (l)$ $\text{C}$ atom in $\text{CH}_4$ is oxidized
Which of the above pairs are correct?
