MCQ Questions for Class 12 Engineering Chemistry General Principles And Processes Of Isolation Of Elements Quiz 10

Gravity separation method is used to concentrate which type of ores?

0%
Sulphide ores
0%
Phosphate ores
0%
Silicate ores
0%
Oxide ores

The impurities present with minerals are collectively known as

0%
Slag
0%
Flux
0%
Matrix
0%
Ores

Which of the following is used in froth floatation process for the benefication of ore?

0%
Coconut oil
0%
Olive oil
0%
Pine oil
0%
None of these

Which of the following statement is true?

0%
All ore are minerals
0%
All minerals are ore
0%
A mineral cannot be an ore
0%
An ore cannot be a mineral

Malachite and azurite are used respectively are :

0%
Blue and green pigment
0%
Red and green pigment
0%
Green and blue pigment
0%
Green and red pigment

For which of the following ores froth floatation method is used for concentration ?

0%
Haematite
0%
Zinc blende
0%
Magnetite
0%
Carnallite

Metal used for making joints in jewellery is

0%
Zn
0%
Cu
0%
Ag
0%
Cd

Which of the following metals is sometimes found native in nature ?

0%
Aluminium
0%
Copper
0%
Iron
0%
Magnesium

Explanation

Copper is very slowly attacked by air, moisture and

$CO_2$ and hence sometimes occurs in native state.

Froth flotation method cannot be employed for the concentration of:

0%
Zinc blende
0%
Pyrolusite
0%
Copper pyrites
0%
Cinnabar

Explanation

Pyrolusite

$(MnO_2)$ is an oxide ore, while zinc blende

$(ZnS)$ , copper pyrites

$(CuFeS_2)$ and cinnabar

$(HgS)$ are sulphide ores. Only sulphide ores are purified by froth flotatlon.

Hence, froth flotation method cannot be employed for the concentration of pyrolusite.

Option B is correct.

Observe the following statements regarding purification of bauxite:
I. During Hall's process, silica is removed as Si (vapour).
II. Bauxite ore contaminated with

$Fe_{2} O_{3}$ is purified in Baeyer's process.
III. During Serpeck's process,

$AlN$ is formed.

0%
I , II and III are correct
0%
Only I and II are correct
0%
Only I and III are correct
0%
Only II and III are correct

Explanation

Hall-Heroult's process is the electrolysis refining of bauxite ore. $CO_2$ is removed as impurities. Thus statement $I$ is incorrect.
The Bayer process is the principal industrial means of refining bauxite to produce alumina (aluminium oxide).

$TiO_2, Fe_2O_3, SiO_2$ exists as impurities in the bauxite ore. All the impurities are removed during this purification process.

Thus statement

$II$ is correct.

Serpeck's Process:

$Al_2O_3. 2H_2O + 3C + N_2 \rightarrow 2AlN + 3CO+ 2H_2O$

$AlN$ is formed in the Serpeck's process. Hence III is correct statement.

Option D is correct.

A solution of

$Na_{2}SO_{4}$ in water is electrolysed using inert electrodes. The products at the cathode and anode respectively, are:

0%
$O_{2}, \ H_{2}$
0%
$O_{2},\ Na$
0%
$H_{2},\ O_{2}$
0%
$O_{2},\ SO_{2}$

Explanation

When a solution of

$Na_2SO_4$ in water is electrolysed then reactions take place at cathode and anode are

At anode (oxidation):

$2H_2O \rightarrow O_2 + 4e^- + 4H^+$

At cathode ( reduction):

$2H^+ + 2e^- \rightarrow H_2$ .

So, the gas librated at cathode

$= H_2$

and gas librated at anode

$= O_2$ .

Option C is correct.

Metals which can be obtained by carbon reduction of their oxides are _________.

0%
Cu
0%
Fe
0%
Al
0%
Pb

Metals that are refined by liquation are:

0%
Si
0%
Pb
0%
Sn
0%
Bi

Explanation

Liquation is applicable for metals such as

$Sn, Pb$ and

$Bi$ , which have low melting points as compared to their impurities.

$Si$ has a high melting point. So, liquation can not be used for its refining.

Options B,C and D are correct.

The principle of fractional crystallisation is involved in which one of the following processes?

0%
Van Arkel's method
0%
Distillation
0%
Cupellation
0%
Zone refining

Which of the following statements is/are correct?
a) In the decomposition of an oxide into oxygen and metal vapour, entropy increases.
b) Decomposition of an oxide is an endothermic change.
c) To make

$\Delta G^{0}$ negative, temperature should be high enough so that

$T\Delta S^{0}> \Delta H^{0}$ .

0%
(a), (b) only are correct
0%
(b), (c) only are correct
0%
(a), (c) only are correct
0%
(a), (b), (c) are correct

Explanation

Decomposition of metal oxide:

$MO_2\rightarrow M+O_2$

If we compare the reactant and product, we see that the number of products are more. Also among the product side even at room temperature, one product is gas so in this process randomness increases, i.e., entropy increases.
This is an endothermic reaction.

So, to increase free energy change, the temperature should be increased to make

$T\Delta S^{o}> \Delta H^{o}$ .

Hence option D is correct.

Choose the correct statement(s) for froth flotation process, which is used for the concentration of sulphide ore.

a) It is based on the difference in the ability of different minerals to become wet.

b) It uses sodium ethyl xanthate,

$C_{2}H_{5}OCS_{2}Na$ as collector.

c) It uses

$NaCN$ as a depressant in the mixture of

$ZnS$ and

$PbS$ , where

$ZnS$ forms soluble complex and

$PbS$ forms froth.

0%
Only (a) and (b) are correct
0%
Only (b) and (c) are correct
0%
Only (a) and (c) are correct
0%
All statements are correct

The mass of carbon anode consumed (giving the only carbon dioxide) in the production of

$270\ kg$ of aluminium metal from bauxite by the Hall process is:
[Atomic mass :

$Al = 27$ ]

0%
$90\ kg$
0%
$540\ kg$
0%
$180\ kg$
0%
$270\ kg$

Explanation

The reaction is :

$2Al_{2}O_{3} +3C \rightarrow 4Al+3CO_{2}$

Molar mass of aluminium is

$27\ gmol^{-1}$

Molar mass of carbon is

$12\ gmol^{-1}$

Carbon required is

$\dfrac{12 \times 3 \times 270}{4 \times 27}=90Kg$

Option A is correct.

The method of zone refining of metals is based on the principle of:

0%
greater mobility of the pure metal than that of the impurity
0%
higher melting point of the impurity than that of the pure metal
0%
greater noble character of the solid metal than that of the impurity
0%
greater solubility of the impurity in the molten state than in the solid

The main function of frothers is to _________________.

0%
stick to the ore and take it to the top
0%
convert the insoluble ore into soluble part
0%
make the ore hydrophobic
0%
none of the above

Select the correct statement for electrolysis of

$Al_2O_3$ by Hall-Heroult process.

0%
Cryolite $Na_3[AlF_6]$ lowers the melting point of $Al_2O_3$ and increases its electrical conductivity.
0%
Al is obtained at cathode and probably $CO_2$ at anode.
0%
Both of the above statements are correct.
0%
None of the above statements are correct.

Explanation

In the extraction of aluminium by electrolytic process, alumina

$Al_2O_3$ is used along with fluorspar and cryolite.
Cryolite reduces the melting temperature of the mixture and also acts as a solvent and helps dissolve

$Al_2O_3$ .
Cryolite:

$Na_3[AlF_6]$

$Na_{3}[AlF_{6}]\rightarrow 3NaF+AlF_{3}$

$NaF$ and

$AlF_3$ , both are ionic compounds, and ionise to give ions. This increases the electrical conductivity and lowers the melting point of

$Al_2O_3$ .
At cathode:

$Al^{3+} +3e^-\rightarrow Al$
At anode:

$C(s) + O^{2-} (melt) \rightarrow CO (g) +2e^-$

$C(s) + 2O^{2-} (melt) \rightarrow CO_2 (g) + 4e^-$

Froth floatation process is used for the concentration of sulphide ores. This process is an illustration of the practical application of:

0%
adsorption
0%
absorption
0%
sedimentation
0%
coagulation

Which of the following is not correctly matched with respect to the processes involved in the extractive metallurgy of the respective metal?

0%
$Al_2O_3.2H_2O\rightarrow Al$ : Leaching, precipitation, calcination and electrolytic reduction (molten state)
0%
$Ag_2S\rightarrow Ag$ : Leaching and displacement method
0%
$PbS\rightarrow Pb$ : Froth flotation process, roasting and self reduction
0%
$KCl.MgCl_2.6H_2O\rightarrow Mg$ : Dehydration by simple heating and electrolytic reduction in aqueous phase

Explanation

A) Extraction of

$Al$ from its ore follows these steps:
Leaching, precipitation, calcination and electrolytic reduction (molten state)

$Al_2O_3(s)+3H_2O(l)+2NaOH(aq)\xrightarrow {Leaching} 2Na[Al(OH)_4](aq)$

$2Na[Al(OH)_4](aq)+CO_2(g)\rightarrow Al_2O_3.xH_2O(s)+2NaHCO_3(aq)$

$Al_2O_3.xH_2O(s)\xrightarrow [Calcination]{1470K} Al_2O_3(s)+xH_2O(g)$

Electrolytic reduction of molten pure

$Al_2O_3$ is mixed with

$Na_2AlF_6$ or

$CaF_2$

This method is known as Hall-Heroult process

$Cathode: Al^{3+}(melt)+3e^-\rightarrow Al(l)$

$Anode: C(s)+O^{2-}(melt)\rightarrow CO(g)+2e^-$

$C(s)+2O^{2-}(melt)\rightarrow CO_2(g)+4e^-$

B) In case of

$Ag$ , extraction follows leaching and displacement method as:
Leaching:

$Ag_2S+4NaCN\rightleftharpoons 2Na[Ag(CN)_2]+Na_2S$

$Na_2S+2O_2\rightarrow Na_2SO_4$

Displacement by zinc in aqueous solution:

$2Na[Ag(CN)_2]+Zn\rightarrow Na_2[Zn(CN)_4]+2Ag\downarrow$

C) Extraction of

$Cu$ follows these as:
Roasting:

$2PbS+3O_2\xrightarrow {\Delta} 2PbO+2SO_2$

Self reduction:

$2PbO+PbS\rightarrow 2Pb+SO_2$

D) Extraction of

$Mg$ follows:
Calcination:

$MgCl_2.6H_2O\xrightarrow [Dry HCl(g)]{\Delta (calcination)} MgCl_2+5H_2O$

It is not made anhydrous by simple heating because it gets hydrolysed

$MgCl_2.6H_2O\xrightarrow {\Delta} MgO.5H_2O+2HCl$

Electrolytic reduction:
Electrolytic reduction of molten anhydrous carnalite.

$MgCl_2\rightleftharpoons Mg^{2+}+2Cl^-$

At cathode:

$Mg^{2+}+2e^-\rightarrow Mg$ (99% pure);

At anode:

$2Cl^-\rightarrow Cl_2+2e^-$

Option D represents the incorrect statement.

Ans-option D

Which of the following reaction is a part of Hall's process?

0%
$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$
0%
$Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3$
0%
$AlN + 3H_2O \rightarrow Al(OH)_3 + NH_3$
0%
$Al_2O_3. 2H_2O + 2Na_2CO_3 \rightarrow 2NaAlO_2 + CO_2 + 2H_2O$

Explanation

Hall's process is used for the extraction of Aluminum from Alumina.

The reaction involved in the Hall's process is as follows :

$Al_2O_3. 2H_2O + 2Na_2CO_3 \rightarrow 2NaAlO_2 + CO_2 + 2H_2O$

Option D is correct.

At what temperature do zinc and carbon have equal affinity for oxygen?

0%
$1000^oC$
0%
$1500^oC$
0%
$500^oC$
0%
$1200^oC$

Explanation

As both oxides of zinc and carbon, i.e.,

$ZnO\: \& \: CO$ met at

$1000^oC$ in graph, so at this temperature, both carbon and zinc have equal affinity with oxygen.

Concentration of which of the following ore is carried out by froth floatation process?

0%
Bauxite
0%
Chalcocite
0%
Haematite
0%
Calamine

Explanation

Froth flotation process is selectively used for separating hydrophobic and hydrophilic substances. Chalcocite (

$Cu_2S$ ) is a sulphide ore and sulphide ores are concentrated by froth floatation.

When copper rod is dipped in

$CuSO_4$ solution, the potential developed at the interface is called as __________ .

0%
electrode potential
0%
standard potential
0%
cell potential
0%
none of the above

Explanation

Copper rod is dipped into

$CuSO_4$ solution. Since, copper is a less active metal,

$Cu^{+2}$ ions present in the solution accept an electron from the metal rod and copper is deposited on the metal rod. Hence, copper rod becomes positively charged, and attracts the negatively charged

$SO_4^{-2}$ ions from the solution. Thus, an electrical double layer is created and a potential at the interface of copper rod and

$CuSO_4$ solution. The electrical potential developed at the interface of the metal and its solution due to Helmholtz electrical double layer is called single electrode potential. Thus, when copper rod is dipped in

$CuSO_4$ solution, the potential developed at the interface is called electrode potential.

At temperature

$1000^oC$ ,

$\Delta G$ of the reaction

$ZnO+C\rightarrow Zn+CO$ is _________.

0%
-ve
0%
+ve
0%
zero
0%
nothing can be said

Explanation

Both oxides of zinc and carbon, i.e.,

$ZnO\: \& \: CO$ met at

$1000^oC$ in graph. So, at this temperature,

$\Delta G$ of the given reaction will be zero.

Which of the above given curves is wrongly presented?

0%
$C\rightarrow CO_2$
0%
$Pb\rightarrow PbO$
0%
$Zr\rightarrow ZrO_2$
0%
$Mg\rightarrow MgO$

Explanation

For the transformation

$Pb\rightarrow PbO$ , the curve is wrongly represented.

Consider the reaction

$2Pb(s) + O_2(g) \rightarrow 2PbO(s)$

This is the combustion reaction. It is exothermic in nature.

Hence, for this reaction,

$\Delta H$ is negative.

For the above reaction,

$\Delta n < 0$ . Hence,

$\Delta S$ is negative.

At low temperature,

$\Delta G$ will be negative, as the temperature increases,

$\Delta G$ value increases and becomes positive.

Which occurs to minimum extent in nature?

0%
$HgS$
0%
$H_2S$
0%
$Bi_2S_3$
0%
$CS_2$

Explanation

As at natural temperature

$\triangle G^{\small\circ}$ of

$CS_2$ is least negative so it is most unstable and has a minimum extent in nature.

At what approximate temperature, zinc and carbon have equal affinity for oxygen?

0%
$1000^{\small\circ}C$
0%
$1500^{\small\circ}C$
0%
$500^{\small\circ}C$
0%
$1200^{\small\circ}C$

Explanation

As both oxides met at

$1000^oC$ in the graph so at this temperature both

$C$ and

$Zn$ have an equal affinity with oxygen.

Which of the above metal oxide has the minimum thermal decomposition temperature?

0%
$CaO$
0%
$FeO$
0%
$ZrO_2$
0%
$MgO$

Explanation

$ZrO_2$ has the minimum thermal decomposition temperature, as it crosses zero

$\Delta G$ line at minimum temperature as shown in the Ellingham diagram.

Option C is correct.

To make the following reduction process spontaneous, temperature should be:

$ZnO + C \longrightarrow Zn + CO$

0%
$1000^{\small\circ}C$
0%
$>1000^{\small\circ}C$
0%
$<500^{\small\circ}C$
0%
$<1000^{\small\circ}C$

Explanation

Above

$1000^oC$ ,

$ZnO$ in the graph is above

$CO$ graph. So, above this temperature, the difference between the free energy of both will be negative and hence the reaction will be spontaneous.

$1100^{\small\circ}C$ , which reaction is spontaneous to a maximum extent?

0%
$MgO + C \rightarrow Mg + CO$
0%
$ZnO + C \rightarrow Zn + CO$
0%
$MgO + Zn \rightarrow Mg + ZnO$
0%
$ZnO + Mg \rightarrow MgO + Zn$

To make the following reduction process spontaneous, temperature should be

$ZnO+C\xrightarrow {\Delta} Zn+CO$

0%
$< 1000^oC$
0%
$> 1000^oC$
0%
$< 500^oC$
0%
$> 500^oC$ but $< 1000^oC$

Explanation

In the Ellingham diagram, the curve which lies below act as a reducing agent.
Above

$1000^oC$ , the curve of

$ZnO$ lies above

$CO$ as shown in graph. So, above this temperature, the difference between free energies of both will be negative and reaction will be spontaneous.

Which of the above given metal oxides can be reduced by

$Fe$ (reducing agent) to its respective metal at temperature

$T_1$ ?

0%
$ZrO_2$
0%
$CaO$
0%
$MgO$
0%
None of the above

$1000^oC$ ,

$\Delta G^{\small\circ}$ of the reaction is:

$ZnO + C \longrightarrow Zn + CO$

0%
$-ve$
0%
$+ve$
0%
zero
0%
None of these

Explanation

As they both oxides met at

$1000^oC$ in graph so at this temperature

$\Delta G$ of the reaction will be zero.

To make the following reduction process spontaneous, temperature should be

$ZnO + C \longrightarrow ZN + CO$

0%
$<1000^{\small\circ}C$
0%
$>1100^{\small\circ}C$
0%
$<500^{\small\circ}C$
0%
$>500^{\small\circ}C$ but $<1000^{\small\circ}C$

Explanation

above

$1000^oC$ ZnO in graph is above than CO so above this temp., difference between free energy of both will be negative and reaction will be spontaneous.
So option B is correct.

Which sulphide occurs to minimum extent in nature?

0%
$HgS$
0%
$H_2S$
0%
$Bi_2S_3$
0%
$CS_2$

Explanation

As at natural temperature

$\triangle G^{\small\circ}$ of

$CS_2$ is least negative so it is most unstable and has a minimum extent in nature.

Option D is correct.

Formation of which of the sulphides is most spontaneous?

0%
$HgS$
0%
$Bi_2S_3$
0%
$PbS$
0%
$CS_2$

Explanation

As at all temperature

$\triangle G^{\small\circ}$ of

$PbS$ is highest negative so it is the most spontaneous.

As per the Ellingham diagram of oxides which of the following conclusion is true?

0%
Al reduces $Fe_{2}O_{3}$ , whereas $MgO$ cannot be reduced by Al at $1500^{o}C$
0%
Fe reduces $Al_{2}O_{3}$ , whereas $MgO$ cannot be reduced by Al at $1500^{o}C$
0%
Al reduces $Fe_{2}O_{3}$ , whereas $MgO$ cannot be reduced by Ca at $1500^{o}C$
0%
Al can reduce both $Fe_{2}O_{3}$ and $MgO$ to the corresponding metal at $1500^{0}C$

When

$FeCr_2O_4$ (chromite) is reduced with Carbon in an electric-arc furnace:

0%
Cr and $Fe_2O_3$ are formed
0%
Fe and $Cr_2O_3$ are formed
0%
Fe and Cr (ferrochrome) are formed
0%
$FeCrO_4$ is formed

Explanation

When chromite

$(FeCr_2O_4)$ is reduced with carbon in an electric-arc furnace, ferrochrome (Fe and Cr) is obtained.

Which of the following sulphides can not be reduced to metal by

$H_2$ at about

$1000^{\small\circ}C$ ?

0%
$HgS$
0%
$PbS$
0%
$Bi_2S_3$
0%
All of these

Explanation

Hydrogen will reduce sulphide of other metals which lie above it in the Ellingham diagram because free energy change will become more negative and more the difference & more energy will be released and reaction will be more spontaneous.

So it can reduce others but can not reduce

$PbS$ as it lies below in the graph.

Si of high purity to be used in semiconductor can be prepared by following methods :

I. $SiO_2\, +\, 2C\, \rightarrow\, Si\, +\, 2CO$

II. $Si\, +\, 2Cl_2\, \rightarrow\, SiCl_4$

$SiCl_4\, +\, 2Mg\, \rightarrow\, Si\, +\, 2MgCl_2$

Better method is :

0%
I
0%
II
0%
both (A) and (B)
0%
None of these

Explanation

Si of high purity to be used in semiconductor can be prepared by the reaction of Si with chlorine to form

$SiCl_4$ followed by fractional distillation (to purify

$SiCl_4$ ) and reduction with hydrogen. Zone refining is used for the further purification.

$Si\, +\, 2Cl_2\, \rightarrow\, SiCl_4(l) \xrightarrow {2H_2(g)} Si(s) +4HCl (g) \xrightarrow {Zone refining} Ultra pure silicon$

In electrtolysis of

$Al_2O_3$ by Hall-Heroult process:

0%
Cryolite $Na_3[AIF_6]$ lowers the melting point of $Al_2O_3$ and increases its electrical conductivity.
0%
$Al$ is obtained at cathode and probably $CO_2$ at anode
0%
both (a) and (b) are correct
0%
none of the above is correct

Explanation

The electrolysis of alumina by Hall and Heroult's process is carried by using a fused mixture of alumina and cryolite along with minor quantities of aluminum fluoride and fluorspar. The addition of cryolite and fluorspar increases the electrical conductivity of alumina and lowers the fusion temperature. Thus, the option A is correct. Also at anode, alumina reacts with fluorine to give oxygen. The liberated oxygen reacts with carbon to form

$CO$ and

$CO_2$ . These gases are liberated at anode.

Which of the following elements can be prepared by heating the oxide above

$400^{o}C$ ?

0%
Hg
0%
Mg
0%
Fe
0%
Al

Explanation

$\Delta G$ is least negative for Hg among all metals above

$400^0C$ temp. So, Hg can be prepared by heating the oxide above

$400^0C$ .

$FeCr_2O_4$ (Chromite) is a good source of chromium and its compound like

$Na_2CrO_4$ and

$Na_2Cr_2O_7$

$FeCr_2O_4 + NaOH + air \longrightarrow (A) + Fe_2O_3$

II.

$(A) + (B) \longrightarrow Na_2Cr_2O_7$

III.

$Na_2Cr_2O_7 + X \overset{\triangle}{\longrightarrow} Cr_2O_3$

IV.

$Cr_2O_3 + Y \overset{\triangle}{\longrightarrow} Cr$

Compounds (A) and (B) are:

0%
$Na_2CrO_4, H_2SO_4$
0%
$Na_2Cr_2O_7, HCl$
0%
$Na_2CrO_6, H_2SO_4$
0%
$Na_4[Fe(OH)_6], H_2SO_4$

Explanation

Compounds (A) and (B) are

$Na_2CrO_4 \ and \ H_2SO_4$ .

$FeCr_2O_4\, +\, NaOH\, +\, air\, \rightarrow\, {(A)} {Na_2CrO_4}\, +\, Fe_2O_3$

$\underset {(A)} {Na_2CrO_4}\, +\, \underset {(B)} {H_2SO_4}\, \rightarrow\, Na_2Cr_2O_7$

From the Ellingham diagram, identify the oxide which is reduced by the carbon at highest temperature among the following oxides.

$MgO,$

$CaO,$

$TiO_2$ and

$Al_2O_3$

0%
$MgO$
0%
$CaO$
0%
$TiO_2$
0%
$Al_2O_3$

Use the relationship $\Delta \,G^{\circ}\,=\,- nFE^{\circ}_{cell}$ to estimate the minimum voltage required to electrolyse $Al_2O_3$ in the Hall-Heroult process.

$\Delta G^{\circ}_{f}(Al_2O_3)\,=\,-1520\,kJ\,mol^{-1}$

$\Delta G^{\circ}_{f}(CO_2)\,=\,-394\,kJ\,mol^{-1}$

The oxidation of the graphite anode to $CO_2$ permits the electrolysis to occur at a lower voltage than if the electrolysis reactions were :

$Al_2O_3\,\rightarrow\,2Al\,+\,3O_2$ .

What is the approximate value of low voltage?

0%
2 V
0%
3 V
0%
4V
0%
5V

Explanation

Net reaction in Hall-Heroult process is

$3C\,+\,2Al_2O_3\,\rightarrow\,4Al\,+\,3CO_2$

Or $4Al^{3+}\,+\,12e^{-}\,\rightarrow \,4Al,$

Number of electrons (n) = 12

$\Delta\,G^{\circ }\,=\,3 \Delta\, G^{\circ }_f (CO_2)\,-\,2 \Delta\,G^{\circ }_f(Al_2O_3)$

$=-3\,\times\,394\,-\,2(-1520)$

$= 1858\ kJ$

$\Delta G^{\circ}\,=\, -nFE^{\circ}_{cell}$

$-E^{\circ}_{cell}\,=\, \displaystyle \frac {\Delta G^{\circ}}{nF}\, =\, \displaystyle\frac {1858\,\times\, 1000}{12\, \times\, 96500}$

$= 1.60\ V$

Thus, Hall-Heroult process takes place at lower voltage.

Match column A (process) with column B (electrolyte).

A (process)	B (electrolyte)
(I) Downs Cell	(W) Fused $MgCl_2$
(II) Dow sea water	(X) Fused $Al_2O_3 + Na_3AlF_6$
(III) Hall-Heroult	(Y) Fused $KHF_2$
(IV) Moissan	(Z) Fused 40% NaCl + 60% CaCl $_2$

Choose the correct alternative:

0%
I-Z, II-W, III-X, IV-Y
0%
I-X, II-Y, III-Z, IV-W
0%
I-W, II-Z, III-X, IV-Y
0%
I-X, II-Z, III-W, IV-Y

Explanation

(1) In Downs cell molten

$MgCl_2$ is electrolyzed, the electrolyte used in the cell is a fused mixture of 40% NaCl and 60%

$CaCl_2$ .

(II) In Dow process,

$Mg(OH)_2$ is precipitated from sea-water by slurrying with calcined dolomite and then converting it to

$MgCl_2$ by reacting with HCl. In Dow sea water process, the electrolyte used is fused

$MgCl_2$ .

(III) In the Hall-Heroult process, the electrolyte used is a fused mixture of

$Al_2O_3$ and

$Na_3AlF_6$ .

(IV) In Moissan process, the electrolyte used is fused

$KHF_2$ .

The correct match is:
I-Z, II-W, III-X, IV-Y

$\Delta G$ vs

$T$ plot in the Ellingham diagram slopes downward for the reaction.

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$Mg + \dfrac {1}{2} O_{2} \longrightarrow MgO$
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$2Ag + \dfrac {1}{2} O_{2} \longrightarrow Ag_2O$
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$C + \dfrac {1}{2} O_{2} \longrightarrow CO$
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$CO + \dfrac {1}{2} O_{2} \longrightarrow CO_{2}$

Explanation

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.

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CBSE Questions for Class 12 Engineering Chemistry General Principles And Processes Of Isolation Of Elements Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry General Principles And Processes Of Isolation Of Elements Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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