SPM Chemistry Form 5 Notes – Terminology and Concepts: Oxidation and Reduction (Part 7)

by BerryBerryTeacher

in Berry Reference (Notes)

The topic of Oxidation and Reduction is indeed a very long chapter in SPM Chemistry Form 5. Presented in this post is the Part 7 on the “Redox Reactions by the Transfer of Electrons as a Distance” from Berry Berry Easy. This is by far one of the most popular topics in exams for the written paper. Also one of the most hated topic as it is deemed too difficult by many students as there are a lot of things to be memorised. Berry Readers need not fear this topic as you should understand the concept of this ‘experiment’ and how electrons move across the salt bridge.

Six Easy Steps to Understand this SubTopic:

  • First, try to understand why the apparatus are set up this way. Is it for isolation? If so, why? (Answer: So that redox reaction can occur through the transfer of electrons without contact but through a conduit.
  • Secondly, try to understand the nature of both the reducing and oxidising agents. Why would certain combinations be chosen?
  • Thirdly, remember what is added onto the solution in the reducing agent arm of the U-tube (U-shaped glass, not the video site).
  • Fourthly, try to understand why is it tested on the reducing agent arm only.
  • Fifth, try to logically visualise the observation, visualising it is sometimes and important way to understand things.
  • Finally, use all your chemistry knowledge to infer your observations.

A summary of the most important part of this subtopic is summarised in the tables below.

SPM Chemistry Form 5 – Terminology and Concepts: Oxidation and Reduction (Part 7)

Rusting of Iron Nails

Rusting of Iron Nails

Redox Reactions by the Transfer of Electrons at a Distance

Set I

Reducing agent Oxidising agent Test on the solution in the reducing agent arm of U-tube
Iron(II) sulphate, FeSO4 solution Acidified potassium dichromate(VI), K2Cr2O7 solution Add a few drops of potassium thiocyanate, KSCN solution
Observation Inference
The electrode in the iron(II) sulphate, FeSO4 solution acts as the negative terminal, whereas the electrode in the acidified potassium dichromate(VI), K2Cr2O7 solution acts as the positive terminal. Electrons flow from iron(II) sulphate, FeSO4 solution to acidified potassium dichromate(VI), K2Cr2O7 solution
Iron(II) sulphate solution changes from pale green to yellow/brown. It gives blood-red colouration with potassium thiocyanate solution (KSCN) Iron(III) ions are present. Iron(II) ions are oxidised to  iron(III) ions.
Acidified potassium dichromate(VI), K2Cr2O7 solution changes colour from orange to green. Dichromate(VI) ions are reduced to chromium(III) ions.
  • Oxidation half-equation: Fe2+(aq) –> Fe3+(aq) + e
  • Reduction half-equation: Cr2O72-(aq) + 14H+(aq) + 6e –> 2Cr3+(aq) + 7H2O(l)
  • Overall reaction: Cr2O72-(aq) + 6Fe2+(aq) 14H+(aq) –> 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)

Set II

Reducing agent Oxidising agent Test on the solution in the reducing agent arm of U-tube
Iron(II) sulphate, FeSO4 solution Acidified manganate(VII), KMnO4 solution Add sodium hydroxide, NaOH solution
Observation Inference
The electrode in the iron(II) sulphate, FeSO4 solution acts as the negative terminal, whereas the electrode in the acidified potassium manganate(VII), KMnO4 solution acts as the positive terminal. Electrons flow from iron(II) sulphate, FeSO4 solution to acidified potassium manganate(VII), KMnO4 solution
Iron(II) sulphate solution changes from pale green to yellow/brown. It formed a brown precipitate when the brown solution is tested with sodium hydroxide solution (NaOH) Iron(III) ions are present. Iron(II) ions are oxidised to  iron(III) ions.
Purple acidified manganate(VII), KMnO4 solution turns colourless. Manganate(VII) ions are reduced to manganese(II) ions.
  • Oxidation half-equation: Fe2+(aq) –> Fe3+(aq) + e
  • Reduction half-equation: MnO4-(aq) + 8H+(aq) + 5e –> Mn2+(aq) + 4H2O(l)
  • Overall reaction: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) –> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

Set III

Reducing agent Oxidising agent Test on the solution in the reducing agent arm of U-tube
Potassium iodide, KI solution Bromine water, Br2 Add a few drops of starch solution
Observation Inference
The electrode in the potassium iodide, KI solution acts as the negative terminal, whereas the electrode in the bromine water acts as the positive terminal. Electrons flow from potassium iodide, KI solution to bromine water, Br2 (aq).
Colourless potassium iodide solution turns brown. It formed a dark blue colouration when the brown solution is tested with starch solution. Iodine is present.Iodide ions have oxidised to iodine.
Brown bromine water turns colourless. Bromines are reduced to bromide ion.
  • Oxidation half-equation: 2I-(aq) –> I2(aq) + 2e
  • Reduction half-equation: Br2(aq) + 2e –> 2Br-(aq)
  • Overall reaction: Br2(aq) + 2I-(aq) –> 2Br-(aq) + I2(aq)

Other pairs of reducing agent and oxidising agent

Reducing agent Oxidising agent
Potassium iodide,KI solution Iron(III) sulphate,Fe2(SO4)3 solution
Potassium iodide,KI solution Acidified potassium dichromate(VI),K2Cr2O7 solution
Potassium bromide,KBr solution Chlorine, Cl2 water

Do log in back to Berry Berry Easy for Part 8 in the series focusing on “rusting

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