Unit 6: Qualitative Analysis (Redox)
Using redox reactions to identify the presence of specific oxidising and reducing agents.
6.11 Test for Oxidising Agents (e.g., Nitrate ions)
To test for the presence of an unknown oxidising agent, you must add a known reducing agent and look for a characteristic sign of a reaction.
General Test for Oxidising Agents:
A common test involves adding potassium iodide ($KI$) solution. The iodide ion ($I^-$) is a good reducing agent and is easily oxidised to iodine ($I_2$).
Half-Equation: $2I^-(aq) \rightarrow I_2(aq) + 2e^-$
A positive result is the formation of a brown solution due to the presence of aqueous iodine. The presence of iodine can be confirmed by adding starch solution, which will turn an intense blue-black colour.
Specific Test for Nitrate Ions ($NO_3^-$):
The nitrate ion is a mild oxidising agent. A specific and more sensitive test is required for its identification.
- Add sodium hydroxide solution to the sample (to make it alkaline).
- Add a small piece of aluminium foil or Devarda's alloy (a reducing agent).
- Gently warm the mixture.
- If nitrate ions are present, they will be reduced by the aluminium to form ammonia gas ($NH_3$).
Equation: $8Al(s) + 3NO_3^-(aq) + 5OH^-(aq) + 2H_2O(l) \rightarrow 8AlO_2^-(aq) + 3NH_3(g)$
The ammonia gas is identified by its pungent smell and by its ability to turn damp red litmus paper blue.
Solved Examples:
- What is a general chemical test for an oxidising agent?
Solution: Add potassium iodide solution. A positive test is the formation of a brown colour (iodine), which turns blue-black with starch. - A student adds KI solution to an unknown solution, and it turns brown. What does this indicate?
Solution: The unknown solution contains an oxidising agent. - What is the reducing agent in the test for nitrate ions?
Solution: Aluminium foil (or Devarda's alloy). - What is the final observable product in the test for nitrate ions?
Solution: Ammonia gas ($NH_3$). - How is the ammonia gas confirmed in the nitrate ion test?
Solution: By its pungent smell and by testing with damp red litmus paper, which turns blue. - Why is the mixture made alkaline with NaOH in the nitrate test?
Solution: The reduction of nitrate by aluminium works under alkaline conditions. - Write the oxidation half-equation for aluminium in the nitrate test.
Solution: $Al(s) + 4OH^-(aq) \rightarrow AlO_2^-(aq) + 2H_2O(l) + 3e^-$. - What is the change in oxidation number of nitrogen when $NO_3^-$ is reduced to $NH_3$?
Solution: It decreases from +5 in $NO_3^-$ to -3 in $NH_3$. - Would iron(III) chloride give a positive test with potassium iodide?
Solution: Yes. $Fe^{3+}$ is an oxidising agent and will oxidise $I^-$ to $I_2$, forming a brown solution. ($2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$). - Why is it necessary to warm the mixture in the nitrate test?
Solution: To increase the rate of reaction and to help the soluble ammonia gas to evolve from the solution.
6.12 Test for Reducing Agents (e.g., Sulphate(IV) ions)
To test for the presence of an unknown reducing agent, you must add a known oxidising agent and look for a characteristic colour change. Two common oxidising agents used for this are acidified potassium permanganate and acidified potassium dichromate.
Test with Acidified Potassium Manganate(VII) ($KMnO_4$):
- Reagent: Potassium manganate(VII) solution, acidified with dilute sulphuric acid.
- Observation: The intense purple colour of the manganate(VII) ion ($MnO_4^-$) is discharged, leaving a colourless solution. This is because the $MnO_4^-$ ion (Mn is +7) is reduced to the nearly colourless $Mn^{2+}$ ion.
- Equation: $MnO_4^-(aq) + 8H^+(aq) + 5e^- \rightarrow Mn^{2+}(aq) + 4H_2O(l)$
Test with Acidified Potassium Dichromate(VI) ($K_2Cr_2O_7$):
- Reagent: Potassium dichromate(VI) solution, acidified with dilute sulphuric acid.
- Observation: The orange colour of the dichromate(VI) ion ($Cr_2O_7^{2-}$) changes to a green colour. This is because the $Cr_2O_7^{2-}$ ion (Cr is +6) is reduced to the green chromium(III) ion ($Cr^{3+}$).
- Equation: $Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$
These tests are used to identify reducing agents like the sulphite (sulphate(IV)) ion, $SO_3^{2-}$, which is oxidised to the sulphate (sulphate(VI)) ion, $SO_4^{2-}$.
Solved Examples:
- What is the positive test result for a reducing agent using acidified potassium permanganate?
Solution: The purple solution turns colourless. - What is the colour change observed when a reducing agent is added to acidified potassium dichromate(VI)?
Solution: Orange to green. - Why must the oxidising agents be acidified for these tests?
Solution: The reduction of $MnO_4^-$ and $Cr_2O_7^{2-}$ requires hydrogen ions ($H^+$) as shown in their half-equations. - A student adds an unknown solution to acidified potassium dichromate(VI) and the solution turns green. What can be concluded?
Solution: The unknown solution contains a reducing agent. - What is the sulphite ion ($SO_3^{2-}$) converted into when it acts as a reducing agent?
Solution: The sulphate ion ($SO_4^{2-}$). - What is the oxidation state of chromium in the green $Cr^{3+}$ ion?
Solution: +3. - What is the oxidation state of manganese in the purple $MnO_4^-$ ion?
Solution: +7. - Would iron(II) ions, $Fe^{2+}$, give a positive test with acidified potassium permanganate?
Solution: Yes. $Fe^{2+}$ is a reducing agent (it can be oxidised to $Fe^{3+}$), so it would decolourise the purple permanganate solution. - How could you distinguish between a solution containing carbonate ions ($CO_3^{2-}$) and one containing sulphite ions ($SO_3^{2-}$)?
Solution: Add acidified potassium dichromate(VI). The sulphite solution would turn it from orange to green, while the carbonate solution would show no colour change (though it might fizz if the solution is strongly acidic). - Sulphur dioxide gas is bubbled through an orange solution of acidified potassium dichromate(VI). What is the final colour of the solution?
Solution: Green, because $SO_2$ is a reducing agent.