Chemistry Report

I. Title

Electrolysis reaction of Na2SO4.

II. Aim

- to know the color change in electrolysis reaction of Na2SO4 solution,
- to know the product in anode and cathode after electrolysis.

III. Basic theory

Electrolysis cell is opposite from Volta cell. In electrolysis, electric current use to redox reaction not spontanly. Electrolysis consist of electrode, electrolyte, and power supply. Electron insert to the electrolysis reaction through cathode. The definite space in solution absorb the electron from cathode and undergo reduction. Beside that, the other space loose the electron in anode and undergo oxidation.

The space that is involved in the anode and cathode reaction depend to the its potential electrode. The certainty are bellow:
- Space that experience reduction in cathode is the space that have higher potential reduction.
- Space that experience oxidation in anode is the space that have higher potential reduction.
Electrode in electrolysis reaction divide become two;

1. Inert electrode: C, PT, and Au.
2. Active electrode: not including C, Pt, and au. (such as; Zn, Fe, Ni, etc.)

IV. The tools and materials

• Na2SO4 solution 0,1 M



• Universal indicator



• Grafit electrode



• Drop pipette



• Statif and kelp



• Cable



• Power supply



• Beaker glasses



• U pipe



• Funnel



• spatula

V. Work steps
1. Combination the tools and materials like the picture bellow

2. Add 10 drops of universal indicator into 50 ml of Na2SO4 solution 0,1 M and mix it.
3. Put the solution into the “U” pipe.
4. Electrolysis the solution until it occur color change around both of the electrode.
5. Write down the change of the electrode.

VI. The table of the experiment

The color before electrolysis reaction The color after electrolysis reaction
in anode in cathode
green yellow purple

VI. Analyze the data

• Before electrolysis reaction, the color of the solution is green.


• After electrolysis reaction, the color on anode change become yellow and in cathode become purple.
  

VI. Questions
1. From the indicator color change, what that be formed in:
a) cathode?
b) anode?
2. If gas that happen in cathode is H2 (hydrogen) and in anode is O2, how the half reaction formula that happen in both electrode?
3. Give the explanation about the product of electrolysis reaction above!

Answer

1. a) OH-
b) H+
2. oxidation/anode : 2H2O (l) 4H+ (aq) + O2 (aq) +4e
reduction/reduces: 2H2O (l) + 2e 2OH- (aq) + H2 (g)
3. After electrolysis reaction have finished, in anode happen oxidation process. There are SO42- in anode but it don’t experience oxidation because the standard potential of SO42- more lower than water so in anode happen oxidation of water form H+ and O2 gas. Beside that, in cathode happen reduction process, but Na+ have potential standard more lower than water, it make water experience oxidation form OH- and H2 gas.

VI. Conclusion

From the experiment above, we know that the change in Na2SO4 solution after electrolysis reaction are
in anode form H+ and O2 gas and the color become yellow,
in cathode form OH- and H2 and the color become purple.