chemistry experiment

CELL ELECTROLYSIS

 

I.              THE AIM

The aim of this experiment is to know the changing of salt solution (Na_sSO₄ and KI)  after electrolysis process.

 

II.            BASIC THEORY

Electrolysis is one of electrochemist reaction. In electrolysis process, the electrics are used to do unspontaneous redox reaction. So, it changes electric energy into chemist energy.

Electrolysis doesn’t need the salt bridge. The main components are tube, electrode, electrolyte solution, and power supply. Electron (electric charges) will enter to the solution through the catode. The reduction will be happened at catode. Electron will be released at anode. So, the oxydation will be happened at anode. Catode’s charge is negative, while anode’s charge is positive. Kation will be reducted, and anion will be oxydated.

Electrolysis is accorded to the electrode. There are two types of electrode, inert and active. Inert electrode (such as C, Pt, Au) will not be reacted. Active electrode (every logam except C, Pt, Au) will be reacted.

Reaction in catode is accorded to the kation. If the kation solution belongs to active logam (IA, IIA, Al, Mn), so water will be reducted. But, if the active logam is in solid condition, it will be reducted. If the solution is acid (contains H⁺ ion), it will produce H_2­ gas. Every ion except the ion above will be reducted into it’s logam compound.

If the electrode is inert, reaction in anode is accorded to the anion. Anion belongs to oxy-acid (such as SO₄^2-, NO₃^-, PO₄^3-) will not be oxydated. So, the water will be oxydated. The solution contains Halida ion (such as F^-, Cl^-, Br^-, I^-), will be oxydated into their compounds. Base solution (contains OH^- ion) will produce O₂ gas and H₂O. But, if the electrode is active, the electrode will be oydated into the compound.

 

 

III.          INSTRUMENTS AND MATERIALS

The instruments are

1.      U-tube

2.      Beaker Glass

3.      Measurer Cylinder

4.      Drop Pipette

5.      Test tubes

6.      Power Supply

The materials are

1.      Carbon Electrode

2.      Na₂SO₄ 0,1 M 50 mL

3.      KI 0,1 M

4.      Universal Indicator 10 mL

5.      Fenolftalein 10 mL

6.      Amylum 10 mL

 

IV.         WORK STEPS

1.      Na₂SO₄ Electrolysis

a.      Adding 10 drops of universal indicator to 50 mL Na₂SO₄ 0,1 M_.

b.      Pouring the solution into U-tube.

c.       Electrolyzing the solution until the color changes around the electrodes.

                                                   

2.      KI Electrolysis

a.      Electrolyzing KI 0,1 M until the color changes around the electrodes.

b.      Moving the solution from catode space into 2 test tubes (each tubes ±20 mL).

c.       Adding 2 drops fenolftalein into the first tube, and adding 2 drops of amylum into the second tube.

d.      Repeating the steps for solution from anode space.

 

V.           THE RESULT

1.      Na₂SO₄ Electrolysis

The color before electrolysis process was dark green.

The color after electrolysis process were purple in catode space and red in anode space.

2.      KI Electrolysis

Space	The Color after
	Electrolysis	Adding of PP	Adding of Amylum
Anode	Yellow	Yellow	Purplish blue
Catode	Transparent/No color	Pink	Muddy white

 

VI.         THE ANALYSIS

1.      Na₂SO₄ Electrolysis

§         The Formula of Electrolysis

                  2Na₂SO_4(aq)       →     4Na⁺_(aq)      +   2SO₄^2-_(aq)

Catode   : 4H₂O_(l)  + 4e   →     2H_2(g)          +   4OH^-_(aq)

Anode    : 2H₂O_(l)             →     4H⁺_(aq)        +   O_2(g)        + 4e

   2 Na₂SO_{4(aq) +} 6H₂O_(l)      →     4NaOH_(aq) + 2H₂SO_4(aq) + 2H_2(g)  + O_2(g)  

§         OH^- will be formed in catode space.

§         H⁺ will be formed in anode space.

§         In anode space, the color changes from dark green into red. It means that the solution become acid, and form H₂SO₄.

§         In catode space, the color changes from dark green into purple. It means that the solution become base, and form NaOH.

2.      KI Electrolysis

§         The Formula of Electrolysis

                  2KI_(aq)                   →     2K⁺_(aq)        +   2I^-_(aq)

Catode   : 2H₂O_(l)  + 2e   →     H_2(g)            +   2OH^-_(aq)

Anode    : 2I^-_(aq)                →     I_2(g)             +   2e          

               2 KI_{(aq) +} 2H₂O_(l)  →     2KOH_(aq)     +  H_2(g)    + I_2(g)  

§         In anode space, there will be I₂. It’s proved by the changing color into purplish blue after the adding of amylum.

§         In catode space, there will be H₂ gas because of water reduction. The color will be change into pink after the adding of PP. So, it becomes base (KOH).

 

VII.       THE CONCLUSION

1.      After electrolysis process, Na₂SO₄ will form acid solution (H₂SO₄) in anode room and base solution (NaOH) in catode room.

2.      After electrolysis process, KI will form I₂ gas in anode room and H₂ gas in catode room. KI also become base solution (KOH) in catode room.

 

VIII.     THE REFFERENTION

Purba, Michael. 2004. Kimia 3A untuk SMA Kelas XII. Jakarta: Penerbit Erlangga.

Purba, Michael. 2007. Kimia 3 untuk SMA Kelas XII. Jakarta: Penerbit Erlangga.