Describe the electrolysis of potassium iodide in water. Include the overall balanced chemical reaction and the electrode reactions

The overall reaction is 2I^-_(aq) +2H₂O_(l) --> I_2(aq) + H_2(g) + 2OH^-_(aq), the anode reaction is 2I^-_(aq) --> I_2(aq)­­ + 2e^- and the cathode reaction is 2H⁺_(aq) + 2e^- --> H_2(g). In aqueous solution potassium iodide, KI, an ionic solid, dissociates into the ions K⁺ and I^- (K⁺ ions are not shown in the reaction schemes, because they do not participate in the reactions, though they are present). The water itself is also in equilibrium between H₂O and H⁺ and OH^-. The negative ions, I^- and OH^- are both attracted to the positive anode, but the iodide ion is in greater abundance, so the dominant anode reaction is the oxidation of iodide to iodine. This can be observed by the production of a brownish substance around the anode. If this is tested with a starch solution, it will go black, indicating the brown substance to be iodine. Similarly the negative ions K⁺ and H⁺ are both attracted to the negative cathode, but although the K⁺ ions are more abundant, the dominant cathode reaction is the reduction of protons to hydrogen gas. The reaction can be observed by effervescence at the cathode as the hydrogen gas bubbles off. Since the hydrogen gas is lost from the system, the solution becomes basic due to the relative overabundance of hydroxide ions in solution. This can be tested with a pH indicator such as phenolphthalein pink. This reaction happens in preference to the reduction of potassium partially because reduction of potassium ions would produce potassium metal, which would immediately react with the water, oxidising again to potassium hydroxide and hydrogen gas. The net reaction is the same.