Electrolysis of Aqueous Potassium Iodide PDF

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Electrolysis of Aqueous Potassium Iodide Purpose: To understand the process of electrolysis and being able to identify the products that are formed through this process. Hypothesis: Hydrogen Gas will form on the cathode (negative end) while iodine will form on the anode (positive end). Pre-Lab Questions: The reaction will produce hydrogen gas on the cathode and iodine on the anode. Materials: graphite pencil lead (2.0 cm) 2 wire leads with alligator clips U-tube 500.0 mL beaker 3 toothpicks 2 droppers 70 mL 1 mol/L KI

a piece of copper graphite rod Cotton wool 10 mL of 1% starch solution 10 mL of 1% phenolphthalein 9-V battery

Procedure: 1. Place the U-tube on a 500.0 mL beaker to hold it in place. 2. Add a small piece of cotton wool on the U-tube. Add 1 mol/L KI solution to the U-tube until the level of the solution is 2 cm from the opening. 3. Poke the opening with toothpick to remove air bubbles if necessary. 4. Attach the black alligator clip onto the lead and place it on one end of the U-tube. 5. Clamp the red alligator clip onto the graphite and place the graphite on the other end of the U-tube. 6. Attach the black end to the negative terminal of the 9-V battery. 7. Let the reaction proceed for three minute. Examine and record any changes occuring in the U-tube. 8. Remove the electrodes from the U-tube and the alligator clips from the battery. 9. Add one drop of 1% starch solution onto U-tube end that contains the anode. 10. Add one drop of phenolphthalein on the cathode end of the U-tube. 11. Dispose the reactants and products and rinse out both the U-tube and electrode. Observations: The black lead started to bubble almost immediately once it is connected to the power supply. Yellow substances form on the graphite side of the utube. The reaction occur for around 3 to 4 minutes. The yellow substance diffuses until it reach the other side of the tube (not entirely).

Analysis: 1. Diagram 1: Electrolytic Cell

2. Iodine and hydroxide ion is attracted to the anode while potassium and hydrogen ion is attracted to the cathode. It can be observed through the yellow substance forming around the anode during the experiment. Since the solution used is aqueous and does not contain pure KI, other products such as hydroxide and hydrogen ion also forms from the dissociation of water. 3. 2 H2O(l) + 2 e- → H2 (g) + 2 OH- (aq) E o  = - 0.83 V I2 (s) + 2 e- → 2 I- (aq) Eo  = 0.54 V * H+  and K +  is attracted to the cathode (-) although a reduction reaction occur. 4. 2 H2O(l) → O2 (g) + 4 H+ (aq) + 4 e- K+(aq) + e- → K(s) E o  = - 2.92 V 

E o  = 1.23 V

* OH-  and I -  is attracted to the anode (+) although an oxidation reaction occur. 5. Since the water used is not under standard conditions, the electrode potentials differ for those at [H+ ] = 10-7  and [OH- ] = 10-7  . Oxidation (Anode) :

2 H2O(l) → O2 (g) + 4 H+ (aq) + 4 e- K(s) → K+ (aq) + e- E o  = 2.92 V 

E o  = 0.82 V ( [H+] = 10-7  )

The oxidation of water is more spontaneous than those of K since the electrode potential is more negative. Which means that the oxidation of water is favourable at the anode. Reduction (Cathode): 2 H2O(l) + 2 e- → H2 (g) + 2 OH- (aq) Eo  = - 0.41 V ( [OH-] = 10-7  ) - - o I2 (s) + 2 e → 2 I (aq) E = 0.54 V

The reduction of I2 is more spontaneous than those of water since the electrode potential is more positive. Which means that it is easier to reduce at the cathode. E cell = E cation − E anion = 0.54 V − 0.82 V = − 0.28 V 6. It is due to the kinetic factors which increases the need of energy for the reaction to occur. Overvoltage is an additional voltage that must be applied in order for nonspontaneous reactions to occur. Some energy might be loss in the process. Either the anode is to positive or the cathode is to negative, it will result in overvoltage in an electrolysis reaction. 7. Iodine is formed on the anode and hydrogen gas is formed on the cathode. The prediction is true for this electrolysis reaction. Iodine tend to form more than hydroxide ion since the electrode potential is more positive. While hydrogen ion, which forms into hydrogen gas is more favorable than the oxidation of potassium since its electrode potential is more negative. Conclusion: The hypothesis is correct since it was observed that iodine does form on the anode, proven with the drop of starch, and hydrogen gas is produced. Although during the experiment, hydroxide ion is also formed since the phenolphthalein indicator changes colour. Which means that one side of the u-tube is basic and the other acidic. References: Tro, N. (2017). Chemistry: A Molecular Approach. Pearson Education....