Extraction of metals as per the activity series - Future Study Point

Extraction of metals as per the activity series

Extraction of metals as per the label in the activity series

Extraction of metals as per the activity series

extraction of the metals

 

Metals are extracted as per their level in the activity series. Some of the metals are found in the earth’s crust in free states and some of them are found in the form of their compound known as ore. The metals are differentiated on the basis of their reactivity known as activity series. The metals at the bottom of the activity series are the least reactive found in a free state for example gold, silver, platinum, and copper are found in the free state. The metals in the middle of the activity series are moderately reactive as example zinc, iron, lead, etc. These metals are found in the earth”s crust in the form of their oxides, carbonates, and sulfides The metals at the top of the activity series are so reactive that they never found in nature as free elements as for example potassium, sodium, calcium, magnesium, and aluminum.

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Conclusively, the metals are classified into three groups on the basis of reactivity.(i) Metals of low reactivity (ii) Metals of medium reactivity (iii) Metals of high reactivity. Different techniques are to be used for obtaining the metals falling in each category.

Extraction of metals low in the activity series

The oxides of these metals can be reduced to metals by heating alone. For example, cinnabar (HgS)  is an ore of mercury. When it is heated in air, it is first converted into mercuric oxide (HgO). Mercuric oxide is then reduced to mercury on further heating.

2HgS (s) + 3O2(g) →2HgO(s) + 2SO2(g)

2HgO(s) →2Hg(l) + O2((g)

Similarly, copper which is found as Cu2S in nature can be obtained from its ore by just heating in air.

2CuS(s) + 3O2g) → 2CuO(s) + 2SO2(g)

2CuO + Cu2S →6Cu(s) + SO2(g)

extraction of metals

Extraction of metals in the middle of the activity series

The metals in the middle of the activity series such as Fe, Zn, Pb, Cu, etc, are moderately reactive. These are usually available in the earth’s crust as sulfides or carbonates. Since it is easier to extract metal from their oxides by the method of reduction, so the sulfides and carbonates of metals must be converted to their metal oxides. Sulfide ores of metals are converted into oxides by heating strongly in presence of excess air, this process is known as roasting.

Carbonate ores are converted into oxides of metal by heating strongly in limited air. This process is known as calculation.

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As example zinc is available in the earth as zinc sulfide and zinc carbonate, so ZnO is derived from these ores by the method of roasting and calcination as follows.

Roasting

2ZnS(s) + 3O2(g) →2ZnO(s) +2SO2(g)

Calcination

ZnCO3(s) → ZnO(s) + CO2(g)

Zinc oxide is then heated with carbon and thus  reduced to Zinc.

ZnO(s) + C(s) →Zn(s) + CO(g)

Oxides of the metals can also be reduced to the metal by using the displacement reaction. In this way the metal of higher activity series displaces the required metal from its oxide.

As an example when magnesium oxide is heated with aluminum powder, it gives magnese.

3MnO(s) + 4Al(s) →3Mn(l) + 2Al2O3(s) +Heat

These displacement reactions are highly exothermic. The amount of heat is so large that metal is produced in the molten state, such reaction is also known as thermit reaction.

Fe2O3(s) + 2Al(s) → 2Fe(l) – Al2O3(s) + Heat

This reaction is also used to join railway tracks and cracked machine parts.

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Extraction of metals in the top of the activity series

The metals high up in the reactivity series are very reactive, they can’t be obtained from their compound by heating with carbon. Carbon is unable to reduce the oxides of  Na K Mg and Ca, this is because these metals have more affinity for oxygen than carbon. These metals are extracted by the electrolysis of their molten chlorides. The metal is deposited at the cathode (-ve charged electrode) and chlorine is liberated at the anode (+ve charged electrode). The reactions are

Na(+) → e + Na

2Cl( – )  + 2e →Cl2

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