1. Oxidation and Reduction Reactions:
We are given the following overall reaction for the electrochemical process:
$ \text{N}_2\text{H}_4 + \text{O}_2 \longrightarrow \text{N}_2 + \text{H}_2\text{O}^{2-} $
2. At the Anode (Oxidation):
The oxidation reaction at the anode is:
$ \text{N}_2\text{H}_4 + 4 \text{OH}^- \longrightarrow \text{N}_2 + 4 \text{H}_2\text{O} + 4 e^- $
3. At the Cathode (Reduction):
The reduction reaction at the cathode is:
$ \text{O}_2 + 2 \text{H}_2\text{O} + 4 e^- \longrightarrow 4 \text{OH}^- $
4. Complete Reaction:
The complete reaction combining both oxidation and reduction is:
$ \text{N}_2\text{H}_4 + \text{O}_2 \longrightarrow \text{N}_2 + 2 \text{H}_2\text{O} $
5. Conclusion:
The correct statements are (A) and (C).
To solve the problem, analyze the electrochemical oxidation of hydrazine (\(N_2H_4\)) by oxygen, focusing on reactions at the anode and cathode and products formed.
1. Anode reaction:
In alkaline medium, hydrazine is oxidized at the anode:
\[
N_2H_4 + 4OH^- \rightarrow N_2(g) + 4H_2O + 4e^-
\]
Hydroxide ions react with hydrazine to form nitrogen gas, water, and release electrons.
Option 1 is correct.
2. Cathode reaction:
Oxygen is reduced at the cathode:
\[
O_2 + 2H_2O + 4e^- \rightarrow 4OH^-
\]
Molecular oxygen gets converted to hydroxide ions.
Option 3 is correct.
3. Option 2 analysis:
Hydrazine does not undergo reduction at the cathode in this process; instead, oxygen is reduced. Nascent hydrogen is not formed.
Option 2 is incorrect.
4. By-products:
Electrochemical oxidation of hydrazine mainly produces nitrogen and water; oxides of nitrogen are not major by-products.
Option 4 is incorrect.
Final Answer:
Options 1 and 3 are correct.
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm$^{-1}$ mol$^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm$^{-1}$ mol$^{-1}$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm$^{-1}$ mol$^{-1}$, its degree of dissociation is given by x $\times$ 10$^{-1}$. The value of x is ______. (Nearest integer)
Consider the following half cell reaction $ \text{Cr}_2\text{O}_7^{2-} (\text{aq}) + 6\text{e}^- + 14\text{H}^+ (\text{aq}) \longrightarrow 2\text{Cr}^{3+} (\text{aq}) + 7\text{H}_2\text{O}(1) $
The reaction was conducted with the ratio of $\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$
The pH value at which the EMF of the half cell will become zero is ____ (nearest integer value)
[Given : standard half cell reduction potential $\text{E}^\circ_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\text{V}, \quad \frac{2.303\text{RT}}{\text{F}} = 0.059\text{V}$
| Concentration of KCl solution (mol/L) | Conductivity at 298.15 K (S cm-1) | Molar Conductivity at 298.15 K (S cm2 mol-1) |
|---|---|---|
| 1.000 | 0.1113 | 111.3 |
| 0.100 | 0.0129 | 129.0 |
| 0.010 | 0.00141 | 141.0 |
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.