Which of the following statements applies during the electroysis of sodium hydroxide sodium hydroxide solution using platinum electrodes?
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Correct Answer: Option D
Explanation:
The correct answer is D. The concentration of sodium hydroxide increases at the cathode only.
Here's why:
- During the electrolysis of sodium hydroxide solution (NaOH) using platinum electrodes:
- At the cathode (negative electrode), reduction occurs. The hydrogen ions (H⁺) are reduced to hydrogen gas (H₂) through the reaction:
\[
2H^+ + 2e^- \rightarrow H_2(g)
\]
This decreases the concentration of H⁺ ions and increases the concentration of OH⁻ ions in the solution, raising the concentration of sodium hydroxide (NaOH).
- At the anode (positive electrode), oxidation occurs, where hydroxide ions (OH⁻) are oxidized to oxygen gas (O₂) and water:
\[
4OH^- \rightarrow O_2(g) + 2H_2O + 4e^-
\]
This reaction decreases the concentration of OH⁻ ions, which can decrease the concentration of NaOH in the compartment near the anode.
Thus, the concentration of sodium hydroxide increases near the cathode due to the increase in OH⁻ ions, while it decreases near the anode. Therefore, option D is correct.
The correct answer is D. The concentration of sodium hydroxide increases at the cathode only.
Here's why:
- During the electrolysis of sodium hydroxide solution (NaOH) using platinum electrodes:
- At the cathode (negative electrode), reduction occurs. The hydrogen ions (H⁺) are reduced to hydrogen gas (H₂) through the reaction:
\[
2H^+ + 2e^- \rightarrow H_2(g)
\]
This decreases the concentration of H⁺ ions and increases the concentration of OH⁻ ions in the solution, raising the concentration of sodium hydroxide (NaOH).
- At the anode (positive electrode), oxidation occurs, where hydroxide ions (OH⁻) are oxidized to oxygen gas (O₂) and water:
\[
4OH^- \rightarrow O_2(g) + 2H_2O + 4e^-
\]
This reaction decreases the concentration of OH⁻ ions, which can decrease the concentration of NaOH in the compartment near the anode.
Thus, the concentration of sodium hydroxide increases near the cathode due to the increase in OH⁻ ions, while it decreases near the anode. Therefore, option D is correct.