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pH of the NaCl solution after electrolys...

pH of the NaCl solution after electrolysis will

A

increase

B

decrease

C

remain constant

D

can't be determined

Text Solution

AI Generated Solution

The correct Answer is:
To determine the effect on the pH of a NaCl solution after electrolysis, we can follow these steps: ### Step 1: Understand the Electrolysis Process During the electrolysis of a sodium chloride (NaCl) solution, two main reactions occur: one at the cathode and one at the anode. ### Step 2: Identify the Reactions At the cathode (negative electrode), water is reduced, producing hydrogen gas (H₂) and hydroxide ions (OH⁻): \[ 2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \] At the anode (positive electrode), chloride ions (Cl⁻) are oxidized to produce chlorine gas (Cl₂): \[ 2Cl^- \rightarrow Cl_2(g) + 2e^- \] ### Step 3: Write the Overall Reaction Combining the two half-reactions, the overall reaction for the electrolysis of NaCl solution can be represented as: \[ NaCl(aq) + H_2O(l) \rightarrow Na^+(aq) + OH^-(aq) + H_2(g) + \frac{1}{2}Cl_2(g) \] ### Step 4: Analyze the Effect on pH The production of hydroxide ions (OH⁻) at the cathode increases the basicity of the solution. Since pH is a measure of the hydrogen ion concentration, an increase in hydroxide ions will result in a decrease in hydrogen ion concentration, leading to an increase in pH. ### Step 5: Conclusion As a result of the electrolysis of NaCl solution, the pH of the solution will increase due to the formation of hydroxide ions. ### Final Answer The pH of the NaCl solution after electrolysis will increase. ---

To determine the effect on the pH of a NaCl solution after electrolysis, we can follow these steps: ### Step 1: Understand the Electrolysis Process During the electrolysis of a sodium chloride (NaCl) solution, two main reactions occur: one at the cathode and one at the anode. ### Step 2: Identify the Reactions At the cathode (negative electrode), water is reduced, producing hydrogen gas (H₂) and hydroxide ions (OH⁻): \[ 2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \] ...
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