Extraction of Metals – Long Answer Questions
Medium Level (Application & Explanation)
Q1. Explain why the method of extraction used for aluminium is different from that for iron.
Answer:
- Aluminium is a highly reactive metal and is placed high in the reactivity series.
- Its compounds, such as Al₂O₃, have a very strong hold over oxygen and cannot be reduced by common reducing agents like carbon.
- Therefore, aluminium is extracted by electrolysis of its molten ore (alumina), using electricity to separate the metal from oxygen.
- Iron is less reactive and lies below aluminium in the reactivity series.
- Iron ore (Fe₂O₃) can be reduced by carbon (coke) in a blast furnace to obtain iron metal.
- Thus, the difference in reactivity decides the extraction process: electrolysis for aluminium and reduction by carbon for iron.
Q2. Describe the steps involved in the extraction of zinc from zinc sulphide ore (ZnS).
Answer:
- Zinc is extracted from zinc blende (ZnS), a sulphide ore, using two main steps.
- First, the ore is roasted in the presence of air. This converts ZnS to zinc oxide (ZnO) and releases SO₂ gas.
- The equation is: 2ZnS + 3O₂ → 2ZnO + 2SO₂.
- Second, the zinc oxide obtained is reduced using carbon (coke), which gives zinc metal and carbon monoxide: ZnO + C → Zn + CO.
- This method is used because zinc is in the middle of the reactivity series and can be displaced by carbon.
- Finally, zinc is purified if needed.
Q3. Why is electrolysis considered an expensive method for metal extraction? Give an example.
Answer:
- Electrolysis needs a large amount of electricity to break strong chemical bonds in highly reactive metals.
- The process also demands special equipment to operate at high temperatures and handle molten salts.
- All this raises the cost of extraction significantly compared to reduction by carbon or just heating.
- For example, extracting aluminium from bauxite via electrolysis uses huge energy, making it costly.
- Cheaper methods can't be used, since aluminium can't be reduced by ordinary chemicals.
- As a result, electrolysis is reserved for highly reactive and valuable metals.
Q4. Explain the terms “roasting” and “calcination” with suitable chemical equations.
Answer:
- Roasting is the process of heating a sulphide ore in excess air to convert it into an oxide.
- For example: 2ZnS + 3O₂ → 2ZnO + 2SO₂ (zinc sulphide is roasted to zinc oxide).
- Calcination is heating a carbonate ore in absence or limited supply of air. This removes volatile impurities and converts it into an oxide.
- Example: ZnCO₃ → ZnO + CO₂ (zinc carbonate is calcined to zinc oxide).
- Both are preliminary steps before reduction and help in converting the ore into a more easily reducible form.
- These are important for extracting moderately reactive metals.
Q5. Why are gold and platinum found in free (native) state in nature?
Answer:
- Gold and platinum are among the least reactive metals; they do not react easily with air, water, or most chemicals.
- Because of their low reactivity, they do not form compounds or combine with other elements naturally.
- This is why they are found as pure metals, also called native state, in the Earth’s crust.
- They do not require extraction or reduction from ores.
- At most, they need physical separation from surrounding rocks.
- Their presence in free state makes them rare and valuable.
High Complexity (Analysis & Scenario-Based)
Q6. Suppose you are given a sulphide ore and a carbonate ore of the same metal. How would you convert each to the oxide before reduction? Explain with reasons.
Answer:
- A sulphide ore should be converted to oxide by roasting—heating it in excess air. The oxygen reacts with the sulphur part, removing it as SO₂ and leaving the metal oxide.
- Example: 2MS + 3O₂ → 2MO + 2SO₂ (M is a metal).
- A carbonate ore is converted by calcination—heating it strongly without air. This decomposes the carbonate, releasing CO₂ and leaving the metal oxide.
- Example: MCO₃ → MO + CO₂.
- These methods are chosen because sulphide ores react well with oxygen, and carbonate ores decompose with heat.
- Both final products, metal oxides, are then easier to reduce to the metal.
Q7. Imagine iron was above aluminium in the reactivity series. How would this affect its extraction process?
Answer:
- If iron was more reactive than aluminium, it would have a greater affinity for oxygen and other non-metals.
- This means carbon could not reduce iron oxide; iron would not be extracted using the blast furnace method.
- Instead, electrolysis of iron’s molten ore (like Fe₂O₃) would become necessary, just like for aluminium.
- This would make iron extraction much more expensive and energy-intensive.
- Availability of iron might decrease, making it as precious as current-day aluminium or even more.
- Hence, the position in the reactivity series is crucial for its practical extraction and daily use.
Q8. A factory owner wants to extract copper using carbon as a reducing agent. Under what conditions is this possible, and why might it sometimes be unnecessary?
Answer:
- Copper is moderately reactive and can be extracted using carbon if present as an oxide.
- The reaction: 2CuO + C → 2Cu + CO₂. Thus, copper(II) oxide can be reduced by carbon to pure copper.
- However, copper is also found in the native (free) state or in relatively pure form; here, extraction is unnecessary since it can simply be separated physically.
- If copper exists as a sulphide or carbonate, ore must first be converted to oxide (by roasting or calcination) before reduction.
- Using carbon is not possible for extraction from high-reactivity ores.
- Thus, the specific condition depends on the copper ore’s chemical form and reactivity.
Q9. Compare the extraction of mercury from cinnabar (HgS) and zinc from zinc blende (ZnS). Highlight similarities and differences.
Highlight
Answer:
- Cinnabar (HgS) and zinc blende (ZnS) are both sulphide ores and both need conversion to oxides before reduction.
- The conversion step is similar: both ores are roasted in air to form their respective oxides.
- HgS + O₂ → HgO + SO₂
- 2ZnS + 3O₂ → 2ZnO + 2SO₂
- Main difference:
- Mercury oxide (HgO) is reduced by simple heating (thermal decomposition): 2HgO → 2Hg + O₂.
- Zinc oxide (ZnO) requires reduction by carbon: ZnO + C → Zn + CO.
- This is because mercury is less reactive than zinc and comes out easily on heating.
- Zinc needs a stronger reducing agent due to higher reactivity.
Q10. Predict and explain what might happen if sodium was tried to be extracted from NaCl by heating with carbon.
Answer:
- Sodium is a highly reactive metal and is placed above carbon in the reactivity series.
- When NaCl is heated with carbon, no reaction occurs because sodium's affinity for chlorine is much stronger than carbon's ability to reduce it.
- Carbon cannot displace sodium from NaCl.
- Thus, heating alone or reduction by carbon is useless for sodium extraction.
- The only effective method is electrolysis of molten NaCl, which uses electricity to break the bond and release sodium.
- This highlights how the reactivity series determines the choice of extraction process.