Reaction Between Potassium Dichromate(VI) and Potassium Iodide

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.

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