Number of electrons (x) involved in the following reaction is :
\(N \equiv N+ 8H^{+}+x \rightarrow 2N H^{3}+ H_{2}\)
Nitrogen Reaction Electron Transfer Analysis
The question asks us to determine the number of electrons, represented by 'x', involved in the following chemical reaction:
$ N \equiv N + 8H^{+} + x \rightarrow 2NH^{3} + H_{2} $
This reaction involves a change in the oxidation state of nitrogen, indicating a redox process. To find 'x', we need to analyze the oxidation states of nitrogen before and after the reaction.
Nitrogen Oxidation State Analysis
Initial Oxidation State of Nitrogen:
In the reactant side, we have diatomic nitrogen, \( N \equiv N \). In its elemental form, the oxidation state of nitrogen is 0.
Final Oxidation State of Nitrogen:
On the product side, nitrogen is present in ammonia, \( NH^3 \). In ammonia, hydrogen typically has an oxidation state of +1. Let the oxidation state of nitrogen be \( N_{ox} \).
$ N_{ox} + 3 \times (+1) = 0 $
Solving for \( N_{ox} \):
$ N_{ox} = -3 $
So, the oxidation state of nitrogen in \( NH^3 \) is -3.
Calculating Electrons Transferred (x)
The reaction shows \( N \equiv N \) converting to \( 2NH^3 \). This means we are looking at the transformation of one molecule of \( N_2 \) into two molecules of \( NH_3 \).
The change in oxidation state for one nitrogen atom is from 0 to -3. This represents a gain of 3 electrons per nitrogen atom.
$ N(0) \rightarrow N(-3) \quad (\text{Gain of } 3e^-) $
Since the reactant molecule is \( N_2 \) (containing two nitrogen atoms), and it forms \( 2NH_3 \) (also containing two nitrogen atoms in total), the total number of electrons gained during this reduction process is:
$ \text{Total electrons gained} = (\text{Number of N atoms}) \times (\text{Electrons gained per N atom}) $
$ \text{Total electrons gained} = 2 \times 3 $
$ \text{Total electrons gained} = 6 $
The variable 'x' in the reaction equation represents the total number of electrons transferred. Therefore, \( x = 6 \).
Conclusion
The number of electrons involved in the reduction of \( N_2 \) to \( NH_3 \) in this reaction is 6. Thus, \( x = 6 \).
Checking the Options:
- Option 1: 8
- Option 2: 6
- Option 3: 4
- Option 4: 10
Based on our calculation, the correct value for 'x' is 6.
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
- In the following redox reaction, which compound is acting as a reducing agent ?
$N_2H_4 + 2H_2O_2 \rightarrow N_2 + 4H_2O$ - 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?
