A.Electrolysis of Water. Use your electrolysis unit to split water (01-0) I. Obt

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A.Electrolysis of Water. Use your electrolysis unit to split water (01-0) I. Obtain a Petri dish. Fill it about half fuall with distilled water. Add 2 drops of phenolphthalein indicator and swirl gently to mix. This indicator is colorless in neutral or acidic solutions and pink in basic solutions. 2. Immerse the tips of the graphite rods and observe carefully. Do you see any evidence of a chemical reaction? No chemicol cacHn 3. Add about 1 gram of solid potassium nitrate, KNO. Swirl the dish gently to aid in dissolving the KNO crystals. The job of the KNO, is to be the electrolyte, i.e, is to help the solution conduct electricity by providing K and NO ions, without directly participating in any reactions. Again immerse the graphite rods in the solution. Observe carefully for a couple of minutes. You should see evidence of chemical change. (Set the dish on a light-colored background to he observations clearer.) Record what is happening at the electrode attached to the red wires and what is happening at the electrode attached to the black wires. Which one produces more gas bubbles? Which causes the indicator 4. KNo3-ca n irolcokr to change color? lok wire buobles pink more gas bubbe to dreroe B. Electrolysis of a Water Solution of Potassium lodide, K(ag) I.Rinse the solution from the previous part down the drain with excess water. Then add about 1 gram of solid potassium iodide, KI, to the dish. L1.o03g 2. Fill your Petri dish at least half full with tap water. Swirl the dish gently to aid in dissolving the KI crystals. The KI does two jobs here. It serves as the electrolyte, helping the water conduct electricity, but it also will be a reactant. 3.Immerse the graphite electrodes in the solution. Record your observations, noting what is happening at or near each of the two electrodes. Block-gas bubbies more bubbes Red Vellow lgreen 4.While the electrolysis is running, add 2 drops of phenolphthalein indicator directly next to each electrode. Record any color changes you see and note whether this was at the red or black wire electrode. Red wire- remauned sama

Explanation / Answer

4. 2H2O (l) --> 2H2 (g) + O2 (g)

5. Phenolphthalein indicator turns pink in basic solution . From the given half reactions, OH- (basic) is being formed at black wire. Also from the overall reaction, less bubbles of O2 will be formed compared to H2 per mole of water. Both confirm formation of H2 at black wire and O2 at red wire.

6. Black wire is supplying electrons as from given half reactions, electrons ae being supplied to cathode to reduce water . Red wire is receiving electrons at anode to complete circuit.

7. a.At black wire, phenolphthalein indicator turned pink confirming presence of basic solution being formed hence at cathode,

4H2O (l) + 4e- --> 2H2(g) + 4OH- (aq).

b. At red wire, yellow product was formed confirming presence of I2. Hence at Red wire,

2 I- (aq) --> I2 (aq) + 2e-

c.In both cases, H2 (g) was being formed.

d. O2 (g) was not formed instead I2 (s) was formed. It is easier to remove electrons from KI solution as compared to water as yellow product was formed.

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