HYPOTHESIS:
Changing the oxidation state of magnesium contained in a solution, the color of the solution will change as well.
MATERIAL AND EQUIPMENT:
50ml beaker
200ml conical flask
3ml pipette
Test tubes
1M HCl
3M NaOH
0.05M KMnO4
0.1M Na2SO3
METHOD:
We first prepare 500ml solutions of the following substances: 1M HCl, 3M NaOH, 0.05M KMnO4 and 0.1M Na2SO3.
In a test tube, we place 3ml of KMnO4 solution and then add to it 3ml of NaOH solution. Then we add some drops of Na2SO3.
In the same solution, we add a considerable excess of sodium sulfite. To the solution that solution, we add a few drops of HCl, until we observe a color change.
We should be very careful when handling HCl and NaOH because they are very corrosive. Furthermore, HCl is fuming so we use it under the fume cupboard. KMnO4 is an oxidizing agent and very harmful.
CALCULATIONS:
MM(NaOH) = 40
n = c x v → n = 3 x 0.05 = 0.15 moles
m = 0.15 x 40 = 6g
MM (KMnO4) = 158
n = 0.05 x 0.05 = 2.5 x 10-3 moles
m = 2.5 x 10-3 x 158 = 0.139g
MM(Na2SO3) = 125.8
n = 0.1 x 0.05 = 5 x 10-3 moles
m = 5 x 10-3 x 125.5 = 0.627g
DATA AND OBSERVATIONS:
-
MnO4- + SO3-2 → MnO4-2 + SO4-2
Oxidation states: Mn: 7→ 6 S: 4→ 6
Oxidising agent: SO3-2
Reducing agent: MnO4-
-
MnO4-2 + SO3-2 → MnO2 + SO4-2
Oxidation states: Mn: 6 → 4 S: 4 → 6
Oxidising agent: SO3-2
Reducing agent: MnO4-2
-
MnO2 + HCl → Mn+2 + Cl2
Oxidation states: Mn: 4 → 2 Cl: -1 → 0
Oxidising agent: HCl
Reducing agent: MnO2
In the second reaction, we may observe that the surface of the solution is transparent with some brown particles, whereas at the bottom of the beaker the solution is very dark brown/ black and a precipitate has formed.
DATA ANALYSIS AND CONCLUSION:
As it was state in the hypothesis, we expected to see different colors as the oxidation state of manganese changes. Manganese may gain or lose electrons when undergoes such reactions and thus its oxidation state changes and therefore we are able to see the difference in color.