Step 1: Identifying the Oxidising Agent and Reducing Agent
Oxidising Agent: Potassium dichromate(VI)
Reducing Agent: Potassium Iodide
Step 2: Determining the Oxidation and Reduction Process and Predicting the Observation
The reducing agent (Potassium Iodide) undergoes oxidation
2I- --> I2 + 2e
Observation: The colourless solution turn yellow/orange.
Note: Potassium iodide is colourless whereas iodine is yellow or orange in colour when dissolve in water.
The oxidising agent undergoes reduction
Cr2O72- + 14H+ + 6e → 2Cr3+ + 7H2O
Observation: The orange colour of the solution turn green.
Note: Dichromate(VI) ion is orange in colour whereas chromium(III) ion is green in colour
Step 3: Identifying the Anode and Cathode
Electrode P: Anode
Electrode Q: Cathode
Note: Oxidation occurs at anode whereas reduction occurs at cathode.
Step 4: Determine the positive and Negative Terminal
Positive Terminal: Electrode Q
Negative Terminal: Electrode P
Note: Anode is the negative terminal whereas cathode is the positive terminal.
Step 5: Determine the Direction of Flow of Electrons.
From electrode P to electrode Q.
Note: Electrons flow from the negative terminal to the positive terminal through the wire.