You can carry out an electrolysis of a neutral solution containing sodium sulfate $\ce{Na2SO4}$ plus some ammonium acetate $\ce{CH3COONH4}$ as buffer, and some drops of bromthymol blue. The solution is initially green. A buffer is needed, as the "neutral" $\ce{Na2SO4}$ solution is always a bit acidic, probably due to the atmospheric $\ce{CO2}$. After some time, hydrogen and oxygen are produced on the electrodes. But the cathodic zone (around the minus pole) turns basic and the solution gets blue, because of the appearance of $\ce{OH-}$ ions in solution : $$\ce{H2O + 2 e- -> 2 OH- + H2}$$ Simultaneously the anodic region (around the positive pole) turns acidic as $\ce{H+}$ ions are produced and the solution turns yellow, according to the equation $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ You may separate the two regions for example with a filter paper, allowing the ions to cross it. After some time, you obtain two separated solutions, one blue, and one yellow. If you remove the filter paper, the solution turns green again by stirring. You can inverse the polarity if you do want it, and start again.

You can carry out an electrolysis of a neutral solution containing sodium sulfate $\ce{Na2SO4}$ plus some ammonium acetate $\ce{CH3COONH4}$ as buffer, and some drops of bromthymol blue. The solution is initially green. A buffer is needed, as the "neutral" $\ce{Na2SO4}$ solution is always a bit acidic, probably due to the atmospheric $\ce{CO2}$. After some time, hydrogen and oxygen are produced on the electrodes. But the cathodic zone (around the minus pole) turns basic and the solution gets blue, because of the appearance of $\ce{OH-}$ ions in solution : $$\ce{2 H2O + 2 e- -> 2 OH- + H2}$$ Simultaneously the anodic region (around the positive pole) turns acidic as $\ce{H+}$ ions are produced and the solution turns yellow, according to the equation $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ You may separate the two regions for example with a filter paper, allowing the ions to cross it. After some time, you obtain two separated solutions, one blue, and one yellow. If you remove the filter paper, the solution turns green again by stirring. You can inverse the polarity if you do want it, and start again.

You can carry out an electrolysis of a neutral solution containing sodium sulfate $\ce{Na2SO4}$ plus some ammonium acetate $\ce{CH3COONH4}$ as buffer, and some drops of bromothymol blue. The solution is initially green. A buffer is needed, as the "neutral" $\ce{Na2SO4}$ solution is always a bit acidic, probably due to the atmospheric $\ce{CO2}$. After some time, hydrogen and oxygen are produced on the electrodes. But the cathodic zone (around the minus pole) turns basic and the solution gets blue, because of the appearance of $\ce{OH-}$ ions in solution : $$\ce{H2O + 2 e- -> 2 OH- + H2}$$ Simultaneously the anodic region (around the positive pole) turns acidic as $\ce{H+}$ ions are produced and the solution turns yellow, according to the equation $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ You may separate the two regions for example with a filter paper, allowing the ions to cross it. After some time, you obtain two separated solutions, one blue, and one yellow. If you remove the filter paper, the solution turns green again by stirring. You can inverse the polarity if you do want it, and start again.

You can carry out an electrolysis of a neutral solution containing sodium sulfate $\ce{Na2SO4}$ plus some ammonium acetate $\ce{CH3COONH4}$ as buffer, and some drops of bromthymol blue. The solution is initially green. A buffer is needed, as the "neutral" $\ce{Na2SO4}$ solution is always a bit acidic, probably due to the atmospheric $\ce{CO2}$. After some time, hydrogen and oxygen are produced on the electrodes. But the cathodic zone (around the minus pole) turns basic and the solution gets blue, because of the appearance of $\ce{OH-}$ ions in solution : $$\ce{H2O + 2 e- -> 2 OH- + H2}$$ Simultaneously the anodic region (around the positive pole) turns acidic as $\ce{H+}$ ions are produced and the solution turns yellow, according to the equation $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ You may separate the two regions for example with a filter paper, allowing the ions to cross it. After some time, you obtain two separated solutions, one blue, and one yellow. If you remove the filter paper, the solution turns green again by stirring. You can inverse the polarity if you do want it, and start again.