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.
Standard electrode potential for \( \text{Sn}^{4+}/\text{Sn}^{2+} \) couple is +0.15 V and that for the \( \text{Cr}^{3+}/\text{Cr} \) couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be:
To calculate the cell potential (\( E^\circ_{\text{cell}} \)), we use the standard electrode potentials of the given redox couples.
Given data:
\( E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15V \)
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74V \)
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____. 
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
As shown in the figures, a uniform rod $ OO' $ of length $ l $ is hinged at the point $ O $ and held in place vertically between two walls using two massless springs of the same spring constant. The springs are connected at the midpoint and at the top-end $ (O') $ of the rod, as shown in Fig. 1, and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $ f_1 $. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2, and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $ f_2 $. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:
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 ________.