The electrolysis of Sodium Sulfate (Na2SO4)

Cathode →          2H2O + 2e-     →     2OH- + H2

                                   

Anode   →           2H2O   →  4H+ + O2 + 4e-

        In cathode, the product the reduction is OH-, which is base becomes purple after being added by phenolphthalein and in anode, the product of oxidation is H+ is acid. If acid is added by phenolphthalein, the solution will stay the same, which is colorless. In both sides, cathode and anode, there is bubble, which is produced by hydrogen.

        In conclusion,

II. The electrolysis of Potassium Iodide (KI)   

        Data collection

        Data analysis

        From the experiment, we can see that reduction happens in cathode and oxidation happens in anode. As a result of electrolysis, OH- and H2 are produced in cathode, therefore the color turns into brownish and bubble is formed. Also in cathode, the color changes into purple after being added phenolphthalein to show that the solution is base. It is because OH- + PP becomes purple. However, after being added starch the color turns into colorless and precipitates.

        In anode, oxidation occurs after electrolysis and Iodine is being oxidized so that the solution the color changes into brownish. However, after being added phenolphthalein or starch the color remains the same, colorless. It is because H+ + PP becomes colorless.

 The reaction is in the following:

Cathode        →         2H2O + 2e-         →         2OH- + H2

Anode                →          2I-                 →         I2 + 2e-

In conclusion,