Chemical reactions - Redox (3)
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1.
Describe a practical method for identifying a reducing agent using acidified potassium permanganate(VII) or acidified potassium iodide. Include a list of reagents and observations, and explain the reasoning behind the observations in terms of oxidation and reduction.
Method:
- Prepare two test tubes.
- In the first test tube, add a small amount of the unknown substance.
- In the second test tube, add a small amount of acidified potassium permanganate(VII) (e.g., 0.5M) or acidified potassium iodide (e.g., 0.5M).
- Observe any colour changes.
Reagents:
- Unknown substance (reducing agent)
- Acidified potassium permanganate(VII) (KMnO4 in H2SO4)
- Acidified potassium iodide (KI in H2SO4)
Observations and Reasoning:
- If the unknown substance is a reducing agent:
- When acidified potassium permanganate(VII) is added, the purple colour will fade to colourless. This indicates that the reducing agent is oxidizing the permanganate ion (MnO4-) to manganate(II) (MnO22-).
- When acidified potassium iodide is added, the blue colour of copper(II) ions (if present) will change to a pale green. This indicates that the reducing agent is reducing the copper(II) ions (Cu2+) to copper(I) ions (Cu+). The reducing agent is being oxidized in the process.
- If the unknown substance is not a reducing agent:
- No colour change will be observed with either acidified potassium permanganate(VII) or acidified potassium iodide.
Table summarizing the observations:
| Test | Reagent | Observation |
| 1 | KMnO4 (acidified) | Purple fades to colourless |
| 2 | KI (acidified) | Blue changes to pale green (if Cu2+ present) |
2.
Consider the following unbalanced equation for the combustion of methane:
CH4(g) + O2(g) → CO2(g) + ?
Balance the equation and state whether the reaction is redox, explaining your answer in terms of oxygen gain and loss.
The balanced equation is: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
This is a redox reaction.
In methane (CH4), the carbon has an oxidation state of -4 and the hydrogen has an oxidation state of +1. In carbon dioxide (CO2), the carbon has an oxidation state of +4 and the oxygen has an oxidation state of -2. This shows that carbon is gaining oxygen (its oxidation state increases from -4 to +4).
In oxygen (O2), the oxygen has an oxidation state of 0. This shows that oxygen is losing electrons (and effectively gaining oxygen atoms).
The change in oxidation states confirms that oxygen is transferred during the reaction, making it a redox process.
3.
The following table shows the oxidation states of various elements in different compounds. Use this information to determine which element is the oxidizing agent and which is the reducing agent in the reaction:
| KMnO4(aq) + Fe2+(aq) → Mn2+(aq) + Fe3+(aq) |
Explain your answer.
Oxidizing Agent: Manganese(VII) ions (Mn7+) in KMnO4 are the oxidizing agent. The oxidation state of manganese changes from +7 in MnO4- to +2 in Mn2+. This represents a loss of 5 electrons.
Reducing Agent: Iron(II) ions (Fe2+) are the reducing agent. The oxidation state of iron changes from +2 in Fe2+ to +3 in Fe3+. This represents a gain of 1 electron.
Explanation: The oxidizing agent is the species that causes oxidation to occur (it accepts electrons), and the reducing agent is the species that causes reduction to occur (it donates electrons). In this reaction, KMnO4 oxidizes Fe2+, causing it to lose electrons and become Fe3+. Therefore, MnO4- is reduced and acts as the oxidizing agent.