Describe with the help of a well labelled diagram an experiment that could be done on electrolysis of dilute sodium iodide. Include relevant half equations at electrodes.

Here's how to describe the experiment for the electrolysis of dilute sodium iodide: 1. Experiment Description: Apparatus: Set up an electrolytic cell consisting of a beaker containing dilute sodium iodide solution. Immerse two inert electrodes (e.g., graphite or platinum) into the solution. Connect these electrodes to a DC power supply, ensuring one is the anode (positive terminal) and the other is the cathode* (negative terminal). Procedure: Pass a direct current through the dilute sodium iodide solution. Observe the changes occurring at each electrode. Observations: At the cathode* (negative electrode), hydrogen gas will be produced, seen as effervescence (bubbles). The solution around the cathode will become alkaline due to the formation of hydroxide ions. At the anode* (positive electrode), iodine will be produced. This will be observed as a brown coloration in the solution around the anode, or as a brown precipitate if the concentration of iodine becomes high enough. 2. Well-labelled Diagram (Description): Imagine a diagram showing: A beaker containing the dilute sodium iodide solution*. Two graphite electrodes* immersed in the solution. One electrode labeled Anode (+) and the other Cathode (-)*. The electrodes connected to a DC power supply*. Bubbles of gas* shown rising from the cathode. Brown coloration or precipitate* shown around the anode. 3. Relevant Half-Equations at Electrodes: At the Cathode (Negative Electrode - Reduction): Hydrogen ions (from water dissociation) are preferentially reduced over sodium ions because hydrogen has a higher reduction potential. 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) At the Anode (Positive Electrode - Oxidation): Iodide ions are preferentially oxidized over hydroxide ions (from water dissociation) because iodide ions are easier to oxidize. 2I⁻(aq) → I₂(s) + 2e⁻ What's next? 📸