General Principles and Processes of Isolation of Elements Test

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General Principles and Processes of Isolation of Elements Test - 56

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Weekly Quiz Competition

Question 1
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Carbon reduction process is not commercially applicable for which of the following set of oxides to extract the respective metal?

(I) $$ZnO$$ (II) $$Fe_2O_3$$ (III) $$Al_2O_3$$ (IV) $$SnO_2$$ (V) $$MgO$$

A
$$ZnO, Fe_2O_3, SnO_2$$

B
$$ZnO, SnO_2, MgO$$

C
$$MgO, Al_2O_3$$

D
$$MgO, SnO_2, Al_2O_3$$,

Solution

Carbon reduction process is used to reduce less electropositive metals. $$Mg$$ and $$Al$$ have high electropositivity.

$$MgO + C \xrightarrow[2000^0C]{at \ high \ temp} Mg + CO$$

It is possible at high temperatures, so it is not used for commercial purposes.

Hence, cannot be extracted from oxides through the carbon reduction process.
Question 2
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Find out A :

A
$$CuFeS_2$$

B
$$MgCl_2.6H_2O$$

C
$$Al_2O_3.2H_2O$$

D
$$Fe_2O_3$$

Solution

Compound A is bauxite $$\displaystyle (Al_2O_3.2H_2O)$$
Described below is the Hall's process. Finely powdered bauxite is fused with sodium carbonate and extracted with water to obtain a solution containing aluminate and a residue of $$\displaystyle Fe_2O_3 $$
$$\displaystyle Al_2O_3.2H_2O + Na_2CO_3 \rightarrow 2NaAlO_2 + CO_2 + 2H_2O$$
Carbon dioxide gas is passed through solution and filtered. The filtrate contains sodium carbonate. Aluminium hydroxide is precipitated. $$\displaystyle 2NaAlO_2 + 3H_2O +CO_2 \xrightarrow {323 K - 333K}2Al(OH)_3 \downarrow + Na_2CO_3$$
Aluminium hydroxide is heated to form pure alumina
$$\displaystyle Al(OH)_3 \xrightarrow {\Delta} Al_2O_3$$
Pure alumina is then mixed with cryolite $$\displaystyle (Na_3AlF_6 )$$ and fluorspar $$\displaystyle CaF_2$$ and melted at high temperature. The molten material is then electrolysed with carbon electrodes to obtain pure Al at cathode.
Question 3
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Directions For Questions

Lead obtained from galena ore (PbS) by air reduction or carbon reduction process contains base metal (Cu, Bi, Sn, As) as impurities; it is due to the presence of these impurities that lead becomes hard and brittle.

...view full instructions

Zn-Ag alloy formed in the upper layer of molten lead is skimmed off from the surface of the molten lead by perforated ladles. This alloy contains leed as impurity, This impurity of Pb is removed by :

A
distillation process

B
cupellation

C
liquation

D
Bett's electrolysis

Solution

Impurity of lead is removed by liquation.
Liquation is used for the refining of metals having low melting point and are associated with high melting impurities.
Ex. Pb, Sn, Sb, Bi and Hg.
The impure metal is heated on the sloping hearth of a furnace. The pure metal flows down leaving behind the non- fusible material on the hearth.
Question 4
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Which of the following is incorrect on the basis of the above Ellingham diagram for carbon?

A
Upto 710$$^{\circ}$$C, the reaction of forrnation of $$CO_2$$ is energetically more favorable, but above 710$$^{\circ}$$C, the formation of CO is preferred.

B
In principle, carbon can be used to reduce any metal oxide at a sufficiently high temperature.

C
$$\Delta S(C_{(s)} + 1/2 O_{2(g)} \longrightarrow CO_{(g)}) < \Delta S (C_{(S)} + O_{2(g)} \longrightarrow CO_{2(g)}$$

D
Carbon reduces many oxides at elevated temperature because $$\Delta G$$ vs temperature line has a negative slope.

Solution

In Ellingham diagram,
higher is the slope higher is the entropy change so
$$\Delta S(C_{(s)} + 1/2 O_{2(g)} \longrightarrow CO_{(g)}) > \Delta S (C_{(S)} + O_{2(g)} \longrightarrow CO_{2(g)}$$
as CO have higher slope.
Also,
Carbon monoxide is a more effective reducing agent than carbon below 983 K but above this temperature the reverse is true.
Question 5
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Directions For Questions

Lead obtained from galena ore (PbS) by air reduction or carbon reduction process contains base metal (Cu, Bi, Sn, As) as impurities; it is due to the presence of these impurities that lead becomes hard and brittle.

...view full instructions

Ag can be obtained from purified Zn-Ag alloy by :

A
distillation

B
poling

C
liquation

D
reduction

Solution

Lead contains impurities such as Cu, Ag, Bi, Sb and Sn.
Silver is removed by Parke's process where molten zinc is added to molten impure lead. The former is immiscible with the latter. Silver is more soluble in molten zinc than in molten lead. Zinc-silver alloy solidifies earlier then molten lead and thus can be separated.
Ag can be thus obtained from Zn-Ag alloy by distillation process.
Question 6
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Find out F in the given flowchart.

A
cathode

B
anode

C
electrolyte

D
none of these

Solution

Al is extracted by the electrolysis of fused mixture of alumina, cryolite and fluorspar.
The reaction involved in the whole process is given as :

$$Al_2O_3.2H_2O \xrightarrow {conc.\ Na_2CO_3} Na[Al(OH)_4] \xrightarrow {CO_2} Al_2O_3 \xrightarrow {Mix\ with\ Na_3AlF_6} Al$$
A B C D E F

At cathode, Al(III) ions are oxidized to Al and at anode, fluoride ions are reduced to fluorine.
At cathode $$4Al^{+} + 12 e^- \rightarrow 4Al$$
At anode $$12F^- \rightarrow 6F_2 +12e^-$$
Hence, F is $$cathode$$.
Question 7
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Directions For Questions

These equilibria can be discussed in terms of thermodynamic functions.
For the reaction :
(i) $$\displaystyle M_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow MO_(s)$$
(ii) $$\displaystyle \frac{1}{2} C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow \frac{1}{2} CO_{2(g)}$$
(iii) $$\displaystyle C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$$
(iii) $$\displaystyle CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$$
The temperature dependence of $$\Delta \mathring{G}$$ of reaction $$(i)$$ to $$(iv)$$ is shown in the given diagram. This is known as Ellingham diagram. With the help of Ellingham diagram, one can easily predict the most suitable reducing agent for the reduction of metal oxides.

...view full instructions

A metal oxide $$MO_{(s)}$$ can be reduced by carbon or carbon monoxide successfully when the line in Ellingham diagram for reaction (i):

A
lies above the line for one of the reactions (ii) to (iv)

B
intersects the line for one of the reactions (ii) to (iv)

C
lies below the line for one of the reactions (ii) to (iv)

D
may lie above or below the line for one of the reactions (ii) to (iv) depending upon the temperature

Solution

A metal oxide $$MO_{(s)}$$ can be reduced by carbon or carbon monoxide successfully when the line in Ellingham diagram for reaction (i):
lies below the line for one of the reactions (ii) to (iv).
This is because, at a given temperature, the free energy change for the oxidation of metal to metal oxide will be more positive (or less negative) than that for the reactions (ii) to (iv)
A reaction is spontaneous when the free energy change is negative.
Question 8
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Directions For Questions

These equilibria can be discussed in terms of thermodynamic functions.
For the reaction :
(i) $$\displaystyle M_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow MO_(s)$$
(ii) $$\displaystyle \frac{1}{2} C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow \frac{1}{2} CO_{2(g)}$$
(iii) $$\displaystyle C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$$
(iii) $$\displaystyle CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$$
The temperature dependence of $$\Delta \mathring{G}$$ of reaction $$(i)$$ to $$(iv)$$ is shown in the given diagram. This is known as Ellingham diagram. With the help of Ellingham diagram, one can easily predict the most suitable reducing agent for the reduction of metal oxides.

...view full instructions

CuO can be reduced to Cu conveniently by using C or CO at any temperature :

A
approximately equal to room temperature

B
above room temperature

C
below room temperature

D
at all temperature

Solution

The Ellingham diagram for oxides shows several important features:
(a)The graph for metal oxide all slope upwards, because the free energy change increases with an increase of temperature as discussed above. (b)The free energy changes all follows a straight line unless the materials metal or vaporize.
(c)When the temperature is raised, a point will be reached where the graph crosses the $$\Delta G = 0$$ line.
Below this temperature the free energy of formation of the oxide is negative, so the oxide is stable. Above this temperature the free of formation of the oxide is positive, and the oxide becomes unstable, and should decompose into the metal and dioxygen.
(d)Any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram because the free energy will become more negative by an amount equal to the different between the two graphs at that particular temperature.
(e) Al metal can reduce the oxide of metal placed above in the diagram. Due to more negative free energy Al can reduce Zn but not Mg.
From figure it can be seen that,
CuO can be reduced to Cu conveniently by using C or CO above room temperature.
Question 9
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Find out E :

A
$$Na_3GeF_6$$

B
$$Na_3AlF_6$$

C
$$Al_2O_3$$

D
none of these

Solution

Compound A is bauxite $$\displaystyle (Al_2O_3.2H_2O)$$
Described below is the Hall's process. Finely powdered bauxite is fused with sodium carbonate and extracted with water to obtain a solution containing aluminate and a residue of $$\displaystyle Fe_2O_3 $$
$$\displaystyle Al_2O_3.2H_2O + Na_2CO_3 \rightarrow 2NaAlO_2 + CO_2 + 2H_2O$$
Carbon dioxide gas is passed through solution and filtered. The filtrate contains sodium carbonate. Aluminium hydroxide is precipitated. $$\displaystyle 2NaAlO_2 + 3H_2O +CO_2 \xrightarrow {323 K - 333K}2Al(OH)_3 \downarrow + Na_2CO_3$$
Aluminium hydroxide is heated to form pure alumina
$$\displaystyle Al(OH)_3 \xrightarrow {\Delta} Al_2O_3$$
Pure alumina is then mixed with cryolite $$\displaystyle (Na_3AlF_6 )$$ and fluorspar $$\displaystyle CaF_2$$ and melted at high temperature. The molten material is then electrolysed with carbon electrodes to obtain pure Al at cathode.
Question 10
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These equilibria can be discussed in terms of thermodynamic functions.
For the reaction :

(i) $$\displaystyle M_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow MO_(s)$$
(ii) $$\displaystyle \frac{1}{2} C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow \frac{1}{2} CO_{2(g)}$$
(iii) $$\displaystyle C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$$
(iii) $$\displaystyle CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$$

The temperature dependence of $$\Delta \mathring{G}$$ of reaction (i) to (iv) is shown in the given diagram. This is known as the Ellingham diagram. With the help of the Ellingham diagram, one can easily predict the most suitable reducing agent for the reduction of metal oxides.

$$AI_2O_3$$ can be reduced by carbon at a temperature of:

A
$$30^{\circ}$$C

B
$$500^{\circ}$$C

C
$$2000^{\circ}$$C

D
$$> 2500^{\circ}$$C

Solution

The Ellingham diagram for oxides shows several important features:
(a)The graph for metal oxide all slope upwards, because the free energy change increases with an increase of temperature as discussed above.
(b)The free energy changes all follows a straight line unless the materials metal or vaporize.
(c)When the temperature is raised, a point will be reached where the graph crosses the $$\Delta G = 0$$ line.
Below this temperature the free energy of formation of the oxide is negative, so the oxide is stable. Above this temperature the free of formation of the oxide is positive, and the oxide becomes unstable, and should decompose into the metal and dioxygen.
(d)Any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram because the free energy will become more negative by an amount equal to the different between the two graphs at that particular temperature.
(e) Al metal can reduce the oxide of metal placed above in the diagram. Due to more negative free energy Al can reduce Zn but not Mg.
From figure it can be seen that,
Alumina can be reduced by carbon at a temperature > 2500 C.