Chemistry · Volume 1 · Chapter 1

Samacheer Class 12 Chemistry - Metallurgy

40 Book Back Q&AVerified AnswersFree Content

Complete Class 12 Chemistry book back solutions for Metallurgy with exam-ready answers.

Every answer on this page includes a verified and validated tag for study confidence.
What's on this page
Choose the correct answer 24Answer the following questions 16
Your Progress - Chapter 10% complete
Choose the correct answerChoose the correct answer24 questions
Q.1 Bauxite has the composition a)Al2O3 b)Al2O3·nH2O c)Fe2O3·H2O d)None of these
Answer: b

Bauxite is hydrated alumina; general formula is hydrated alumina: \(\mathrm{Al_2O_3\cdot nH_2O}\).

Q.2 Roasting of sulphide ore gives the gas (A). (A) is a colourless gas. Aqueous solution of (A) is acidic. The gas (A) is a)CO2 b)SO3 c)SO2 d)H2S
Answer: c

Roasting of metal sulfides (e.g. \(2ZnS+3O_2\to2ZnO+2SO_2\)) yields sulfur dioxide, \(\mathrm{SO_2}\), a colourless gas whose aqueous solution is acidic (forms sulfurous acid).

Q.3 Which one of the following reactions represents calcination?
Answer: c

Calcination is thermal decomposition of carbonate/hydroxide in absence of air to give the oxide and CO2. Example: \(\mathrm{MgCO_3\to MgO+CO_2}\). Option (a) is combustion/oxidation; (b) is roasting.

Q.4 The metal oxide which cannot be reduced to metal by carbon is a) PbO b)Al2O3 c) ZnO d) FeO
Answer: b

Aluminium oxide (\(\mathrm{Al_2O_3}\)) has very large negative free energy of formation and is too stable to be reduced by carbon; hence carbon reduction is not feasible for alumina.

Q.5 Which of the metal is extracted by Hall-Heroult process?
Answer: a

The Hall–Héroult process is the electrolytic extraction of aluminium from molten alumina dissolved in cryolite.

Q.6 Which of the following statements, about the advantage of roasting of sulphide ore before reduction is not true? a) ΔGf° of sulphide is greater than those for CS2 and H2S. b) ΔGr° is negative for roasting of sulphide ore to oxide c) Roasting of the sulphide to its oxide is thermodynamically feasible. d) Carbon and hydrogen are suitable reducing agents for metal sulphides.
Answer: d

Roasting converts sulphides to oxides which are easier to reduce; carbon and hydrogen are not generally suitable reducing agents for sulphides (they are used for oxides). Hence statement (d) is not true.

Q.7 Match column I with column II: A Cyanide process; B Froth flotation; C Electrolytic reduction; D Zone refining. Column-II: (i) Ultrapure Ge (ii) Dressing of ZnS (iii) Extraction of Al (iv) Extraction of Au (v) Purification of Ni. A B C D
Answer: (c)

A: Cyanide process → extraction of Au (iv). B: Froth flotation → concentration/dressing of sulphide ores like ZnS (ii). C: Electrolytic reduction → extraction of Al (iii). D: Zone refining → production of ultrapure semiconductors (e.g. Ge) (i). So mapping is (iv,ii,iii,i).

Q.8 Wolframite ore is separated from tinstone by the process of
Answer: d

Wolframite (iron–manganese tungstate) is magnetic while cassiterite (tinstone, SnO2) is non‑magnetic; electromagnetic separation exploits this magnetic difference.

Q.9 Which one of the following is not feasible? a) Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) b) Cu(s) + Zn2+(aq) → Zn(s) + Cu2+(aq) c) Cu(s) + 2Ag+(aq) → 2Ag(s) + Cu2+(aq) d) Fe(s) + Cu2+(aq) → Cu(s) + Fe2+(aq)
Answer: b

Feasible reactions require the metal to be above the ion in the activity series. Zn can reduce Cu2+ (a feasible). Cu cannot reduce Zn2+ (b not feasible). Cu can reduce Ag+ (c feasible). Fe reduces Cu2+ (d feasible).

Q.10 Electrochemical process is used to extract
Answer: c

Highly reactive metals like sodium are obtained by electrolysis (Downs cell) of molten salts; sodium is extracted electrolytically.

Q.11 Flux is a substance which is used to convert a) Mineral into silicate b) Infusible impurities to soluble impurities c) Soluble impurities to infusible impurities d) All of these
Answer: b

Flux combines with gangue to form a fusible (soluble in slag) compound (e.g. silicate) so that impurities are removed as slag. It converts infusible impurities into fusible/soluble slag.

Q.12 Which one of the following ores is best concentrated by froth - floatation method?
Answer: c

Froth flotation is ideal for concentrating sulphide ores such as galena (PbS).

Q.13 In the extraction of aluminium from alumina by electrolysis, cryolite is added to a) Lower the melting point of alumina b) Remove impurities from alumina c) Decrease the electrical conductivity d) Increase the rate of reduction
Answer: a

Cryolite (Na3AlF6) dissolves alumina and lowers its melting point and improves conductivity, making electrolysis of alumina practical.

Q.14 Zinc is obtained from ZnO by a) Carbon reduction b) Reduction using silver c) Electrochemical process d) Acid leaching
Answer: a

Zinc oxide is reduced by carbon (coke) at high temperature: \(\mathrm{ZnO + C\to Zn + CO}\).

Q.15 Extraction of gold and silver involves leaching with cyanide ion. Silver is later recovered by (NEET-2017)
Answer: c

After cyanide leaching, silver is commonly precipitated by displacement using zinc (Merrill–Crowe process): \(\mathrm{2[Ag(CN)_2^-] + Zn \to 2Ag + [Zn(CN)_4^{2-}] }\).

Q.16 Considering Ellingham diagram, which of the following metals can be used to reduce alumina? (NEET-2018)
Answer: c

From Ellingham diagram MgO line yields a more negative total free energy (for the stoichiometry required) than Al2O3; magnesium can reduce alumina thermodynamically whereas Fe, Cu and Zn cannot.

Q.17 The following set of reactions are used in refining Zirconium: Zr(impure) + 2I2 → ZrI4 at 523 K; ZrI4 → Zr(pure) + 2I2 at 1800 K. This method is known as
Answer: b

This is the Van Arkel (iodide) process (also called crystal bar process) used to purify Zr (and Ti) via volatile iodides.

Q.18 Which of the following is used for concentrating ore in metallurgy?
Answer: d

Concentration methods include leaching (dissolving unwanted soluble components) and froth flotation (separation of sulphide ores). Roasting is a chemical conversion, not a concentration method.

Q.19 The incorrect statement among the following is a) Nickel is refined by Mond's process b) Titanium is refined by Van Arkel's process c) Zinc blende is concentrated by froth floatation d) In the metallurgy of gold, the metal is leached with dilute sodium chloride solution
Answer: d

Gold is leached with dilute sodium cyanide (not sodium chloride). Statements (a),(b),(c) are correct.

Q.20 In the electrolytic refining of copper, which one of the following is used as anode?
Answer: b

In electrolytic refining the impure metal is used as the anode; pure copper is deposited at the cathode while impurities fall off as anode sludge.

Q.21 Which of the following plot gives Ellingham diagram a) ΔS Vs T b) Δ G Vs T c) ΔG Vs 1 T d) ΔG Vs T
Answer: b

Ellingham diagram is a plot of standard Gibbs free energy change of oxide formation against temperature: plot of ΔG° (y-axis) vs T (x-axis). The slope of each line equals −ΔS° for the reaction.

Q.22 In the Ellingham diagram, for the formation of carbon monoxide a) ΔS(T) is negative b) ΔG(T) is positive c) ΔG(T) is negative d) initially ΔG(T) is positive, after 700 °C ΔG(T) is negative
Answer: d

For C + 1/2 O2 → CO the entropy change ΔS° is positive (gas formed), so the slope (−ΔS°) on the Ellingham diagram is negative. Hence ΔG° decreases with T; it may be positive at low T and becomes negative above a certain temperature (~700 °C), so statement (d) is correct.

Q.23 Which of the following reduction is not thermodynamically feasible? a) Cr2O3 + 2Al → Al2O3 + 2Cr b) Al2O3 + 2Cr → Cr2O3 + 2Al c) 3TiO2 + 4Al → 2Al2O3 + 3Ti d) none of these
Answer: b

Reaction (b) is the reverse of (a). Al2O3 is more stable (more negative ΔG°) than Cr2O3, so chromium cannot reduce Al2O3 to Al. Thus (b) is not thermodynamically feasible by Ellingham criteria.

Q.24 Which of the following is not true with respect to Ellingham diagram? a) Free energy changes follow a straight line. Deviation occurs when there is a phase change. b) The graph for the formation of CO2 is a straight line almost parallel to free energy axis. c) Negative slope of CO shows that it becomes more stable with increase in temperature. d) Positive slope of metal oxides shows that their stabilities decrease with increase in temperature.
Answer: b

The CO2 formation line is nearly horizontal (i.e., almost parallel to the temperature axis), not parallel to the free energy (vertical) axis. Statements (a), (c) and (d) are correct descriptions of Ellingham behaviour.

Answer the following questionsAnswer the following questions16 questions
Q.1What are the differences between minerals and ores?v
Solution

Key differences: (1) Definition: mineral = naturally occurring substance (e.g. quartz, hematite); ore = mineral containing metal in sufficient concentration (e.g. hematite Fe2O3 as an iron ore). (2) Economic value: only ores have economic value for metal extraction. (3) Composition: minerals may be pure compounds; ores contain metal plus gangue (impurities). (4) Examples: mineral—quartz (SiO2); ore—bauxite (Al2O3·nH2O) for Al.

Answer:

Mineral: naturally occurring chemical compound or element; may or may not be of economic value. Ore: a mineral (or mixture) from which a metal can be profitably extracted.

Q.2What are the various steps involved in the extraction of pure metals from their ores?v
Solution

Briefly: (1) Concentration (beneficiation) to remove gangue (e.g. froth flotation, magnetic separation). (2) Calcination/roasting to remove volatile impurities and convert ores to oxides/sulfates. (3) Reduction (chemical or electrolytic) of oxide to metal (e.g. carbon reduction, aluminothermy, electrolysis). (4) Refining to obtain high purity metal (electrolytic refining, Mond process, zone refining etc.).

Answer:

Concentration → Calcination/Roasting → Reduction/Smelting → Refining.

Q.3What is the role of quick lime in the extraction of Iron from its oxide Fe2O3?v
Solution

In the blast furnace CaO reacts with silica (gangue) to form calcium silicate (slag): CaO + SiO2 → CaSiO3. Slag is molten and less dense than iron and is removed, thus separating impurities from the metal.

Answer:

Quicklime (CaO) acts as a flux to form a removable slag with acidic impurities (silica).

Q.4Which type of ores can be concentrated by froth floatation method? Give two examples for such ores.v
Solution

Froth flotation separates hydrophobic sulfide minerals from hydrophilic gangue using collectors (e.g. xanthates) and frothers. Examples: lead ore (galena, PbS) and zinc ore (sphalerite, ZnS).

Answer:

Sulfide (and other hydrophobic) ores, e.g. galena (PbS), sphalerite (ZnS), chalcopyrite (CuFeS2).

Q.5Describe a method for refining nickel.v
Solution

Impure nickel is treated with CO at 50–60 °C to form volatile nickel tetracarbonyl: Ni + 4CO → Ni(CO)4. The gas is separated and decomposed at 200–250 °C on a heated surface to give pure nickel and CO: Ni(CO)4 → Ni + 4CO. This yields high-purity nickel; CO is recycled.

Answer:

Mond process (carbonyl refining) is commonly used for nickel.

Q.6Explain zone refining process with an example.v
Solution

Principle: impurity distribution coefficient k (= impurity in solid/impurity in melt) is ≪1 for many impurities. A narrow molten zone is passed along a crystalline rod; impurities preferentially enter the melt and are transported to one end. Repeating the process yields extremely pure metal/semiconductor (e.g. Si, Ge).

Answer:

Zone refining removes impurities by moving a molten zone along a bar; impurities concentrate in the melt and are carried to one end. Example: production of ultrapure silicon for semiconductors.

Q.7Using the Ellingham diagram, (A) Predict the conditions under which (i) Aluminium might be expected to reduce magnesia. (ii) Magnesium could reduce alumina. (B) is it possible to reduce Fe2O3 by coke at a temperature around 1200K?v
Solution

Ellingham criterion: a metal A will reduce oxide BX if ΔG°(formation of A oxide) is more negative than ΔG°(formation of B oxide) (i.e. A2O3 line lies below BxOy). (A)(i,ii) On standard Ellingham diagrams MgO and Al2O3 lines lie very low; neither Al nor Mg can thermodynamically reduce the other oxide under ordinary temperatures (their oxide lines do not lie above the other). (B) Carbon (as CO) line crosses the Fe-oxide line at temperatures used in blast furnaces, so reduction of Fe2O3 by coke/CO is thermodynamically feasible around ~1200 K.

Answer:

(A)(i) Aluminium cannot reduce MgO under normal conditions; (ii) Magnesium cannot reduce Al2O3. (B) Yes — Fe2O3 can be reduced by coke (via CO/CO2) near 1200 K.

Q.8Give the uses of zinc.v

Uses of zinc include:

  • Galvanising iron/steel to prevent corrosion.
  • Alloys: brass (Cu–Zn), some bronzes and die-casting alloys.
  • Batteries: zinc–carbon and alkaline cells (zinc acts as the anode).
  • Sacrificial anodes for corrosion protection of ships and pipelines.
  • Zinc oxide (ZnO) in the rubber industry, paints, cosmetics and medicines (e.g., skin ointments).
  • Electroplating and as a white pigment (zinc white).
Q.9Explain the electrometallurgy of aluminium.v
Solution

Purified alumina (from Bayer process) is dissolved in molten cryolite (Na3AlF6) to lower melting point and increase conductivity. Electrolysis at ~940–980 °C: at cathode 4e− + Al3+ → Al (liquid), at carbon anode C + O2− → CO/CO2 + electrons. Overall (approx.): 2Al2O3 + 3C → 4Al + 3CO2. Cryolite reduces melting point, increases solubility and conductivity; carbon anodes are consumed and must be replaced.

Answer:

Hall–Héroult electrolytic process: Al2O3 dissolved in molten cryolite is electrolysed to produce Al.

Q.10Explain the following terms with suitable examples. (i) Gangue (ii) slagv
Solution

Gangue: non-metallic impurities associated with an ore (e.g. SiO2, clay). Slag: product formed by reaction of flux (e.g. CaO) with gangue (e.g. SiO2) during smelting; it is molten, less dense than metal and removed. Example in iron extraction: CaO (flux) + SiO2 (gangue) → CaSiO3 (slag).

Answer:

(i) Gangue: unwanted earthy impurities in an ore (e.g. silica). (ii) Slag: molten flux–gangue compound removed during smelting (e.g. CaSiO3).

Q.11Give the basic requirement for vapour phase refining.v
Solution

Requirements: (1) Metal forms a volatile, reversible compound (e.g. metal halide, carbonyl) at moderate temperature. (2) Impurities either do not form volatile compounds or form ones with different volatility. (3) Volatile compound can be transported and decomposed under different conditions to yield pure metal. Example: Mond process (Ni + CO ⇌ Ni(CO)4).

Answer:

Metal must form a volatile compound selectively which can be decomposed to give pure metal; impurity must not form the volatile species or decomposes differently.

Q.12Describe the role of the following in the process mentioned. (i) Silica in the extraction of copper. (ii) Cryolite in the extraction of aluminium. (iii) Iodine in the refining of Zirconium. (iv) Sodium cyanide in froth floatation.v
Solution

(i) In copper smelting silica (SiO2) reacts with iron oxide to form fayalite slag (2FeO·SiO2), removing iron impurities. (ii) Cryolite (Na3AlF6) dissolves Al2O3, lowers melting point and increases electrical conductivity in Hall–Héroult cell. (iii) Iodine reacts with impure Zr to form volatile ZrI4 which decomposes on a hot filament to give pure Zr (van Arkel–de Boer). (iv) NaCN is used in froth flotation to depress pyrite and other unwanted sulfides by forming complexes with Fe, improving selectivity for desired sulfide minerals.

Answer:

(i) Flux to form slag with FeO. (ii) Solvent and flux to lower melting point & increase conductivity. (iii) Forms volatile ZrI4 for zone purification (van Arkel–de Boer). (iv) Depressant/activator to control flotation of sulfide minerals.

Q.13Explain the principle of electrolytic refining with an example.v
Solution

In copper refining: anode = impure Cu, cathode = pure Cu starter sheet, electrolyte = CuSO4 + H2SO4. At anode: Cu → Cu2+ + 2e− (impurities fall as sludge). At cathode: Cu2+ + 2e− → Cu (pure). Result: copper dissolves from anode and plates pure at cathode; precious impurities (Au, Ag) collect as anode slime.

Answer:

Impure metal anode dissolves; pure metal plates at cathode; impurities either remain in solution or collect as anode sludge. Example: electrolytic refining of copper.

Q.14The selection of reducing agent depends on the thermodynamic factor: Explain with an example.v
Solution

Using Ellingham diagram: metal A reduces oxide BxOy if ΔG°(formation of A oxide) < ΔG°(formation of B oxide). Example: carbon (as CO) reduces Fe2O3 to Fe at high T because ΔG° for formation of CO/CO2 makes the overall reaction thermodynamically favorable; carbon cannot reduce Al2O3 because Al2O3 is thermodynamically more stable than CO/CO2 at accessible temperatures.

Answer:

A reducing agent is chosen such that its oxide is more stable (more negative ΔG°) than the oxide of the metal to be reduced; thus the reduction is thermodynamically downhill.

Q.15Give the limitations of Ellingham diagram.v
Solution

Limitations: (1) Only standard free energies (ΔG°) — real activities and non‑ideal solutions can change outcomes. (2) Kinetics not considered; a thermodynamically feasible reaction may be slow. (3) Assumes standard pressure (1 bar) and pure phases; partial pressures (e.g. of O2, CO) alter lines. (4) Does not include formation of intermediate species or complex oxides, chlorides, etc. (5) Phase changes cause line breaks and some curves may shift with composition/pressure.

Answer:

Ellingham diagrams give only thermodynamic feasibility (ΔG°) not kinetics; assume pure substances and standard states; ignore activities/non‑ideal behaviour, partial pressures, and complex formation.

Q.16Write a short note on electrochemical principles of metallurgy.v
Solution

Principles: metal reduction/oxidation are viewed as electrode processes with characteristic potentials. The feasibility of electrolysis is given by the cell EMF (related to ΔG° by ΔG° = −nFE°). Electrolytic refining/electrowinning apply an external voltage greater than decomposition potential to reduce metal ions at the cathode. Comparison of standard potentials helps choose conditions and predict whether a metal can be extracted electrolytically. Examples: copper electrorefining, Hall–Héroult aluminium production, and electrolysis for metals less reactive than hydrogen under appropriate conditions.

Answer:

Electrochemical metallurgy uses electrode potentials and cell EMFs to predict and carry out extraction/refining of metals by electrolysis or electrowinning.