Which of the following statements is correct during the electrolysis of a caustic soda solution using platinum electrodes?
Oxygen gas is given off at the cathode
Hydrogen gas is given of at the anode
Sodium metals is deposited at the anode
Sodium metal is deposited at the cathode
AAlkalinity at the cathode increases
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dis my school sef who is the person answering all dis questions given stud wrong answer 😠🤬😡
The correct answer is D
First during the electrolysis of NaOH using platinum(passive) electrode
Here is the reaction
Na2+ + OH- ----> Na(s) + 4OH(g)
Now cation should migrate to the cathode and be deposited the
While anion should migrate at the anode and be evolve there
From the above reaction Na metal should be deposited at the cathode while Oxygen gas should be evolve at the anode
The answer can't be E alkalinity deals with solutions not electrode
Also not that the question didn't say dilute caustic soda so we can compare it with H2O
They are correct, the answer is E
A 
Oxygen is formed at the anode, not the cathode.
B Hydrogen is formed at the cathode, not the anode.
C Sodium metal is not deposited in aqueous solution.
D Sodium cannot be deposited from aqueous NaOH (water is reduced instead)
E .is the correct answer .the alkalinity increase at the cathode due to the formation of OH- ions
The correct answer is B.
During the electrolysis of a caustic soda (sodium hydroxide) solution using platinum electrodes, hydrogen gas is given off at the cathode, and oxygen gas is given off at the anode. This is because when an electric current is passed through the solution, the sodium and hydroxide ions in the solution are attracted to opposite electrodes. At the cathode, the positively charged hydrogen ions (H+) are reduced to hydrogen gas (H2), while at the anode, the negatively charged hydroxide ions (OH-) are oxidized to oxygen gas (O2).
Option C is incorrect because sodium metal cannot be deposited at the anode as it will be oxidized to sodium ions. Option D is also incorrect because sodium metal cannot be deposited at the cathode as it will be reduced to sodium ions. Option E is incorrect because alkalinity at the cathode does not increase during the electrolysis of a caustic soda solution using platinum electrodes.
During the electrolysis of a caustic soda (sodium hydroxide) solution using platinum electrodes, option B is correct: hydrogen gas is given off at the anode.
The electrolysis of a sodium hydroxide solution involves the following half-reactions at the electrodes:
At the cathode: 2H2O + 2e- → H2 + 2OH-
At the anode: 4OH- → O2 + 2H2O + 4e-
As we can see, hydrogen gas is produced at the cathode while oxygen gas is produced at the anode. Sodium metal is not deposited at either electrode since platinum electrodes are used instead of reactive electrodes such as sodium or aluminum. Additionally, the alkalinity at the cathode (i.e., the concentration of hydroxide ions) does increase as hydroxide ions are consumed in the reduction of water to hydrogen gas.
the answer is E. D cannot be the answer because sodium metal is not deposited in anode of electrolysis of caustic soda but only in the electrolysis of molten sodium chloride through the down cell.
A. Oxygen gas is given off at the cathode.
This statement is incorrect. Oxygen gas is typically given off at the anode during the electrolysis of water.
B. Hydrogen gas is given off at the anode.
This statement is incorrect. Hydrogen gas is usually given off at the cathode during the electrolysis of water.
C. Sodium metal is deposited at the anode.
This statement is incorrect. During the electrolysis of aqueous sodium chloride (NaCl) solution, sodium ions are discharged at the cathode, not the anode.
D. Sodium metal is deposited at the cathode.
This statement is incorrect. During the electrolysis of aqueous sodium chloride (NaCl) solution, sodium ions (Na⁺) are discharged at the cathode, but sodium metal is not deposited. Instead, sodium ions are reduced to form sodium metal, which then reacts with water to produce sodium hydroxide (NaOH) and hydrogen gas (H2).
E. Alkalinity at the cathode increases.
This statement is incorrect. The alkalinity typically decreases at the cathode during the electrolysis of water because hydrogen gas is given off, leading to a decrease in hydroxide ion concentration and a decrease in alkalinity.
The answer is D which is sodium. Sodium is deposit at the cathode because it is a Cation. On the other hand Hydrogen (H+) is deposited at the cathode too. But when you look closely on the electrochemical series to check which one will be preferentially deposit...NA+ is higher than H+ . So Na+ will be deposited preferentially..
Assuming the said alkalinity will increase at the alkalinity, it will be the best answer. But 2 OH- is being deposited in the anode because anions(-ve) are deposited in the anode(+ve)
USING THE ELECTROLYTIC CELL FOR EXPLANATION
simce its NaOH solution, it means NaOH and water. so the ions in competition are Na+,H+. Na+ is higher in the ECS. Therefore it is discharged prefencially. and since reduction occurs at cathode, we get sodium metal.
