Reaction Between Iron(II) Sulphate and Bromine Water


Step 1: Identifying the Oxidising Agent and Reducing Agent
Oxidising Agent: Bromine water
Reducing Agent: Iron(II) sulphate

Step 2: Determining the Oxidation and Reduction Process and Predicting the Observation
Oxidation
The reducing agent undergoes oxidation
Fe2+ + 2e --> Fe3+

Observation: The green colour of iron(II) sulphate solution turn brown.

Note: Iron(II) ion is green in colour whereas iron(III) ion is brown in colour.

Reduction
The oxidising agent undergoes reduction
Br2 + 2e --> 2Br-

Observation: The yellow/orange colour of bromine water become colourless.

Note: Bromine is yellow/orange in water whereas bromide is colourless.

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.

1 comment:

  1. correction (Fe2+ + 2e --> Fe3+) to (Fe2+ -> Fe3+ + e)

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