A student has three colourless solutions labelled 1, 2 and 3, shown in Fig. 11.1. One is a reducing agent, one is an oxidising agent and one is distilled wa...
Question 1 Report
A student has three colourless solutions labelled 1, 2 and 3, shown in Fig. 11.1. One is a reducing agent, one is an oxidising agent and one is distilled water. The student tested each solution by adding a drop of acidified potassium manganate(VII), and separately by adding potassium iodide with starch solution.
Fig. 11.1
(a) Complete the table of expected observations for the three solutions.
| solution | with acidified potassium manganate(VII) | with potassium iodide and starch |
|---|---|---|
| reducing agent | ||
| oxidising agent |
[4]
(b) Deduce how the results identify which solution is the reducing agent and which is the oxidising agent, giving reasons from the colour changes. [3]
(c) Plan how you would show that solution 3 is neither an oxidising nor a reducing agent. [3]
(d) Name the coloured substance that gives the blue-black colour with starch. [1]
(e) State which ion is oxidised when the oxidising agent reacts with potassium iodide. [1]
(f) Suggest two ways to make the tests fair and reliable. [2]
(g) Give one hazard of acidified potassium manganate(VII) and a suitable precaution. [2]
Answer Details
This question uses two contrasting reagents to tell an oxidising agent, a reducing agent and water apart. Acidified potassium manganate(VII) is decolourised by a reducing agent, while potassium iodide with starch turns blue-black in the presence of an oxidising agent.
(a) Expected observations [4]:
| solution | with acidified manganate(VII) | with potassium iodide and starch |
|---|---|---|
| reducing agent | purple decolourises / goes colourless [1] | no change [1] |
| oxidising agent | no change / stays purple [1] | turns blue-black [1] |
(b) Interpreting the results [3]: the reducing agent decolourises the purple manganate(VII) because it reduces \(MnO_4^{-}\) to \(Mn^{2+}\) [1]; the oxidising agent turns potassium iodide and starch blue-black because it oxidises iodide to iodine [1]; so the solution that decolourises the manganate(VII) is the reducing agent and the solution that makes the blue-black colour is the oxidising agent [1].
(c) To show solution 3 is neither (it is distilled water) [3]: add solution 3 to acidified potassium manganate(VII) and the purple colour does not change [1]; add solution 3 to potassium iodide and starch and no blue-black forms [1]; use the same volume/portion and clean apparatus as a control so the comparison is valid [1]. Two negative results confirm it is neither an oxidising nor a reducing agent.
(d) The coloured substance that gives the blue-black colour with starch is iodine [1].
(e) When the oxidising agent reacts with potassium iodide, the iodide ion (\(I^{-}\)) is oxidised [1] to iodine (\(2I^{-} \rightarrow I_2 + 2e^{-}\)).
(f) To make the tests fair and reliable (any two) [2]: use the same volume / portion of each solution; use clean apparatus for each test; repeat the tests; view against a white background.
(g) Hazard and precaution [2]: acidified potassium manganate(VII) is an irritant / oxidising / stains the skin [1]; a suitable precaution is to wear eye protection and handle it with care [1].
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