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Question

Which of the following processes is required for extracting metal from cinnabar ore?

The correct answer is Roasting

Extracting Metal from Cinnabar Ore

The question asks to identify the specific process required for extracting metal from cinnabar ore. Cinnabar is the primary ore of mercury, and its chemical composition is mercury(II) sulfide ($HgS$). The extraction of metals from their ores involves various metallurgical processes, chosen based on the nature of the ore and the reactivity of the metal.

Understanding Cinnabar and Mercury Extraction

Cinnabar ($HgS$) is a sulfide ore. Mercury is a relatively low-reactivity metal. The extraction process aims to convert the sulfide ore into the pure metal.

Analysis of the Roasting Process

Roasting is a process where an ore is heated strongly in the presence of excess air, typically below its melting point. This process is commonly used for sulfide ores.

  • For cinnabar ($HgS$), roasting involves heating it in the presence of air. This converts mercury(II) sulfide into mercury(II) oxide ($HgO$) and sulfur dioxide ($SO_2$). The relevant chemical equation is: $$2HgS_{(s)} + 3O_{2(g)} \xrightarrow{Heat} 2HgO_{(s)} + 2SO_{2(g)}$$
  • The mercury(II) oxide ($HgO$) formed is then further heated, causing it to decompose into mercury vapor ($Hg_{(g)}$) and oxygen ($O_{2(g)}$). The equation for this decomposition is: $$2HgO_{(s)} \xrightarrow{Heat} 2Hg_{(l)} + O_{2(g)}$$
  • Alternatively, direct heating of cinnabar in excess air can directly yield mercury vapor and sulfur dioxide: $$HgS_{(s)} + O_{2(g)} \xrightarrow{Heat} Hg_{(g)} + SO_{2(g)}$$
  • The mercury vapor is then condensed to obtain liquid mercury. Roasting is effective because it converts the sulfide into an oxide, which can be easily decomposed by heat to yield the metal.

Evaluating Other Options

  • Electrolytic reduction: This method is generally used for extracting highly reactive metals (like sodium, aluminum) or for refining metals. It involves using electrical energy to reduce the metal ions, typically from molten salts or solutions. Mercury is not reactive enough to require this method for extraction from its ore.
  • Thermit process: The thermit process uses aluminum powder as a reducing agent to reduce metal oxides, especially those of metals that are difficult to reduce by common methods (e.g., $Fe_2O_3$, $Cr_2O_3$). It is not suitable for extracting mercury from cinnabar.
  • Calcination: Calcination involves heating ores in the absence or limited supply of air. It is used to remove volatile impurities, moisture, or carbon dioxide from ores, or to change the ore's physical state (e.g., converting carbonates to oxides). It is not the primary method for extracting metals from sulfide ores like cinnabar.

Conclusion

Based on the chemical nature of cinnabar ($HgS$) and the properties of mercury, roasting is the appropriate process for its extraction. This method effectively converts the sulfide ore into a form that can be readily reduced to the metal, often through simple heating.

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Important Questions from Electrochemistry

  1. At $298 \, K$, given the standard electrode potentials: $E^\circ_{Cu^{2+}/Cu} = 0.34 \, V$, $E^\circ_{Zn^{2+}/Zn} = -0.76 \, V$, $E^\circ_{Fe^{2+}/Fe} = -0.44 \, V$, and $E^\circ_{Ag^{+}/Ag} = 0.80 \, V$.
    Based on these values, which of the following reactions is NOT expected to occur spontaneously under standard conditions?
  2. You are given three metals 'X', 'Y' and 'Z'. Metal 'X' is found to react with an aqueous solution of both YSO 4and ZSO 4whereas metal 'Z' is found to react only with aqueous solution of YSO 4. Based on these observations, select the correct statement from the following.

  3. The mobility of a divalent cation in water is 8 × 10-8 m2 V-1 s-1. The effective radius of the ion is (viscosity of water = 1 cP : c = 1.6 × 10-19 C)

  4. The electrical double layer model among the following that consists of both fixed and diffuse layers is

  5. If the overpotential of an electrolysis process is increased from 0.5 V to 0.6 V, then the ratio of current densities (In \(\frac{\int0.6 }{\int0.5}\)) of the electrolysis will be equal to (given transfer co - efficient = 0.5)

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