JAMB 2024 Chemistry · Question 7
When Sulphur(IV)oxide is passed into solution of acidified tetraoxomanganate(VII), the colour changes from
Options
- A.blue to green
- B.pink to yellow
- C.purple to colourless
- D.Orange to green
Answer: C. purple to colourless
Explanation written with AI and checked against the answer key
Analysis
The question asks for the color change observed when Sulphur(IV) oxide () reacts with an acidified solution of tetraoxomanganate(VII) (). This is a classic redox reaction frequently tested in JAMB Chemistry.
Concept Explanation
- The Reactants: • Acidified potassium tetraoxomanganate(VII) () is a powerful oxidizing agent. In its acidic state, the manganese ion is in the oxidation state, which gives the solution a deep purple color. • Sulphur(IV) oxide () acts as a reducing agent.
- The Reaction: When is bubbled through the acidified solution, the is oxidized to sulphate ions (), and the manganese(VII) ions are reduced to manganese(II) ions ().
The balanced ionic equation is: 2\text{MnO}_4^-_{(aq)} + 5\text{SO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} \rightarrow 2\text{Mn}^{2+}_{(aq)} + 5\text{SO}_4^{2-}_{(aq)} + 4\text{H}^+_{(aq)}
- The Color Change: • The ion in aqueous solution is virtually colourless (or very pale pink, which is often considered colourless in the context of this specific reaction). • Therefore, the intense purple color of the permanganate solution disappears as the reaction proceeds.
Step-by-Step Solution
- Step 1: Identify the initial state. Acidified is purple.
- Step 2: Identify the reaction type. This is a redox reaction where is reduced.
- Step 3: Determine the final state. The reduction of to results in a loss of color.
- Step 4: Match with options. The change is from purple to colourless.
Conclusion
The reaction causes the purple color of the permanganate to be discharged.
Exam Tip: Always remember that acidified is a "color-killer" for reducing agents. If you see it in a question, look for the option that involves the disappearance of the purple color!
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