The product , which is not obtained during the electrolysis of brine solution is ____ .
- \(cl_2\)
- H2
- HCl
- NaOH
The Correct Option is B
Solution and Explanation
Electrolysis of Brine Products
Brine is a concentrated aqueous solution of \( \text{NaCl} \). During electrolysis of brine, the following reactions occur:
At the cathode (reduction):
Water molecules are preferentially reduced over sodium ions because the reduction potential of water is higher than that of sodium ions.
\( 2\text{H}_2\text{O(l)} + 2e^- \rightarrow \text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)} \)
At the anode (oxidation):
Chloride ions are oxidized to chlorine gas.
\( 2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2e^- \)
Overall Reaction:
Combining the cathode and anode reactions:
\( 2\text{NaCl(aq)} + 2\text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{(g)} + \text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \)
Conclusion:
The products obtained during the electrolysis of brine are hydrogen gas (\( \text{H}_2 \)), chlorine gas (\( \text{Cl}_2 \)), and sodium hydroxide (\( \text{NaOH} \)). \( \text{HCl} \) is not formed.
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Electrolysis Questions
- Find out charge (in C) required for electrolysis of \(1\) mole of \(H_2O\) to produce \(O_2\) on one of the electrodes. \((F = 96500 \;C)\)
- $\text{A constant current was passed through a solution of AuCl}_4^- \text{ ion between gold electrodes. After a} \\\text{period of 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed} \\$$\text{through the solution is \_\_\_\_\_\_\_\_} \times 10^{-2} \, \text{F.} \\$$\text{(Given atomic mass of Au = 197)}$
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ….. (Nearest integer).
A solution of aluminium chloride is electrolyzed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is _________
O\(_2\) gas will be evolved as a product of electrolysis of:
(A) an aqueous solution of AgNO3 using silver electrodes.
(B) an aqueous solution of AgNO3 using platinum electrodes.
(C) a dilute solution of H2SO4 using platinum electrodes.
(D) a high concentration solution of H2SO4 using platinum electrodes.
Choose the correct answer from the options given below :
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,