Metals and Their Compounds Practice Questions — JAMB UTME — Page 4

Question 61Question →

When aqueous ammonia is added dropwise until in excess to an aqueous solution containing copper(II) ions, the pale blue precipitate that initially forms dissolves to produce a characteristic deep blue solution. Which complex ion is responsible for this deep blue color?

[\text{Cu}(\text{NH}_3)_4]^{2+}

[\text{Cu}(\text{NH}_3)_2]^+

[\text{Cu}(\text{OH})_4]^{2-}

[\text{Cu}(\text{H}_2\text{O})_4]^{2+}

Show answer & explanation

Answer: $[\text{Cu}(\text{NH}_3)_4]^{2+}$

Answer

The complex ion responsible for the deep blue solution is the tetraamminecopper(II) ion,

[\text{Cu}(\text{NH}_3)_4]^{2+}

.

When excess aqueous ammonia is added to a solution containing

\text{Cu}^{2+}

ions, ammonia molecules act as ligands to coordinate with the

\text{Cu}^{2+}

ion, forming the deep blue tetraamminecopper(II) complex ion,

[\text{Cu}(\text{NH}_3)_4]^{2+}

.

Step-by-Step Solution

1

Identify the initial reaction of copper(II) ions with aqueous ammonia.

A pale blue precipitate of copper(II) hydroxide,

\text{Cu(OH)}_2

, is formed when small amounts of aqueous ammonia are added:

\text{Cu}^{2+}_{(\text{aq})} + 2\text{NH}_{3(\text{aq})} + 2\text{H}_2\text{O}_{(\text{l})} \rightarrow \text{Cu(OH)}_{2(\text{s})} + 2\text{NH}_{4(\text{aq})}^+

.

Aqueous ammonia acts as a weak base, generating hydroxide ions.

2

Determine the effect of adding excess aqueous ammonia.

The copper(II) hydroxide precipitate dissolves due to ligand substitution, forming the soluble complex ion

[\text{Cu}(\text{NH}_3)_4]^{2+}_{(\text{aq})}

.

Ammonia molecules act as unidentate neutral ligands that bind strongly to copper(II) ions, displacing hydroxide ions and forming the deep blue tetraamminecopper(II) ion.

Key Concept

Complex ion formation and qualitative test for copper(II) ions using aqueous ammonia.

Estimated Time:1m 0s

Question 62Question →

When aluminium metal is dissolved in a concentrated aqueous solution of sodium hydroxide, a gas is evolved and a soluble complex ion is formed in the solution. What are the chemical formulas of the gas evolved and the complex ion formed, respectively?

H_2

and

[Al(OH)_4]^-

O_2

and

[Al(OH)_4]^-

H_2

and

Al^{3+}

NH_3

and

[Al(OH)_4]^-

Show answer & explanation

Answer: $H_2$ and $[Al(OH)_4]^-$

Answer

The gas evolved is hydrogen gas (

H_2

) and the complex ion formed is the tetrahydroxoaluminate(III) ion (

[Al(OH)_4]^-

).

Aluminium metal exhibits amphoteric behavior, dissolving in strong alkaline solutions such as aqueous sodium hydroxide. The reaction oxidizes aluminium metal to form the soluble aluminate complex ion,

[Al(OH)_4]^-

, while reducing hydrogen species to yield hydrogen gas,

H_2

.

Step-by-Step Solution

1

Identify the amphoteric property of aluminium metal in basic solutions.

Aluminium reacts with both acids and strong bases (alkalis) such as sodium hydroxide.

Aluminium forms protective oxide layers and has amphoteric chemical behavior.

2

Write the balanced ionic reaction between aluminium metal, hydroxide ions, and water.

2Al(s) + 2OH^-(aq) + 6H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq) + 3H_2(g)

Aluminium is oxidized to the soluble tetrahydroxoaluminate(III) complex ion, while water/hydrogen species are reduced to hydrogen gas.

3

Match the products to the question requirements.

Gas evolved:

H_2

; Soluble complex ion:

[Al(OH)_4]^-

The reaction produces hydrogen gas and tetrahydroxoaluminate(III).

Key Concept

Amphoteric Nature of Aluminium and Reaction with Strong Alkalis

Question 63Question →

What volume of nitrogen(IV) oxide gas ( $\text{NO}_2$ ), measured at s.t.p. in $\text{dm}^3$ , is evolved when $3.2\text{ g}$ of pure copper metal reacts completely with excess concentrated trioxonitrate(V) acid?

(Molar mass of $\text{Cu} = 64\text{ g mol}^{-1}$ ; Molar volume of gas at s.t.p. = $22.4\text{ dm}^3\text{ mol}^{-1}$ )

Show answer & explanation

Answer: 2.24

Answer

2.24 dm^3 of nitrogen(IV) oxide gas is produced at s.t.p.

When copper reacts with concentrated trioxonitrate(V) acid, the reaction follows the stoichiometry

\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}

. Thus,

1\text{ mole}

of copper metal (

64\text{ g}

) yields

2\text{ moles}

of

\text{NO}_2

gas (

44.8\text{ dm}^3

at s.t.p.). For

3.2\text{ g}

(

0.05\text{ moles}

) of copper, the volume of

\text{NO}_2

formed is

0.10\text{ moles} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

.

Step-by-Step Solution

1

Write the balanced equation for the reaction of copper with concentrated trioxonitrate(V) acid

\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}

Concentrated trioxonitrate(V) acid acts as a strong oxidizing agent, converting copper to copper(II) ions and reducing itself to brown nitrogen(IV) oxide gas.

2

Calculate the moles of copper metal present in 3.2 g

3.2 / 64 = 0.05 mol

Number of moles is mass divided by relative molar mass.

3

Determine the amount of nitrogen(IV) oxide gas produced in moles

2 * 0.05 = 0.10 mol

The stoichiometric mole ratio between Cu and NO2 is 1:2.

4

Convert moles of nitrogen(IV) oxide gas into volume at s.t.p.

0.10 * 22.4 = 2.24 dm^3

One mole of any ideal gas occupies 22.4 dm^3 at standard temperature and pressure.

Key Concept

Stoichiometry of copper redox reaction with concentrated trioxonitrate(V) acid producing nitrogen(IV) oxide gas.

Question 64Question →

A laboratory technician reacts separate samples of two copper-based alloys, $X$ and $Y$ , with excess concentrated trioxonitrate(V) acid ( $HNO_3$ ). Sample $X$ dissolves completely to yield a clear blue solution with no residue, whereas Sample $Y$ yields a blue solution containing an insoluble white solid. Which of the following correctly identifies Alloys $X$ and $Y$ and the constituent element responsible for the insoluble white solid?

Alloy

X

is brass, Alloy

Y

is bronze, and tin forms the insoluble solid.

Alloy

X

is bronze, Alloy

Y

is brass, and zinc forms the insoluble solid.

Alloy

X

is brass, Alloy

Y

is bronze, and copper forms the insoluble solid.

Alloy

X

is duralumin, Alloy

Y

is brass, and aluminium forms the insoluble solid.

Show answer & explanation

Answer: Alloy $X$ is brass, Alloy $Y$ is bronze, and tin forms the insoluble solid.

Answer

Alloy

X

is brass, Alloy

Y

is bronze, and tin forms the insoluble solid.

Brass (

Cu + Zn

) dissolves completely in concentrated trioxonitrate(V) acid to produce soluble copper(II) trioxonitrate(V) and zinc trioxonitrate(V). Bronze (

Cu + Sn

) contains tin, which is oxidized by concentrated trioxonitrate(V) acid to form insoluble metastannic acid (

H_2SnO_3

), creating a white precipitate in the blue copper solution.

Step-by-Step Solution

1

Identify the elemental compositions of the copper alloys brass and bronze.

Brass is an alloy of copper (

Cu

) and zinc (

Zn

). Bronze is an alloy of copper (

Cu

) and tin (

Sn

).

Knowing constituent elements is essential to predict chemical reactions with concentrated acid.

2

Analyze the chemical reaction of Brass (Alloy

X

) with concentrated

HNO_3

.

Both

Cu

and

Zn

react to form soluble metal trioxonitrate(V) salts:

Cu(NO_3)_2

(blue) and

Zn(NO_3)_2

(colorless). The mixture forms a completely clear blue solution.

All nitrate salts of copper and zinc are soluble in aqueous media.

3

Analyze the chemical reaction of Bronze (Alloy

Y

) with concentrated

HNO_3

.

Copper dissolves to form soluble blue

Cu(NO_3)_2

, while tin (

Sn

) is oxidized by concentrated

HNO_3

to form hydrated tin(IV) oxide (

SnO_2 \cdot xH_2O

or metastannic acid,

H_2SnO_3

), which precipitates as an insoluble white solid.

Tin exhibits anomalous behavior with concentrated nitric acid compared to zinc, forming an insoluble oxide rather than a soluble nitrate salt.

Key Concept

Chemical differentiation of brass and bronze based on constituent metal reactivity with concentrated trioxonitrate(V) acid

Estimated Time:1m 30s

Question 65Question →

Alloying pure copper with zinc to produce brass enhances its mechanical hardness and tensile strength, but leads to a decrease in its electrical conductivity compared to pure copper.

True

False

Show answer & explanation

Answer: True

Answer

True. Alloying copper with zinc to form brass increases mechanical strength and hardness via lattice distortion, but decreases electrical conductivity due to increased electron scattering.

The statement accurately reflects physical metallurgy principles. Foreign zinc atoms create strain fields in the copper lattice that block dislocation movement (increasing hardness), while simultaneously disrupting the periodic potential required for efficient electron transport (decreasing electrical conductivity).

Step-by-Step Solution

1

Analyze the composition and structure of brass.

Brass is a substitutional alloy formed by dissolving zinc atoms into the copper metallic lattice.

Understanding the atomic mixture establishes how foreign atoms fit into the primary metal matrix.

2

Evaluate the effect of foreign zinc atoms on mechanical properties.

Zinc atoms differ in size from copper atoms, causing localized lattice distortion that impedes dislocation motion.

Restricting the slipping of crystal planes increases the alloy's hardness and tensile strength compared to pure copper.

3

Evaluate the effect of lattice distortion on electrical conductivity.

The loss of periodic lattice symmetry increases the scattering of free conduction electrons.

Increased electron scattering reduces the mean free path of charge carriers, lowering overall electrical conductivity relative to pure copper.

Key Concept

Influence of alloying on mechanical hardness and electrical conductivity

Question 66Question →

Match each alloy listed on the left with its characteristic elemental composition and primary application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Duralumin pairs with

\text{Al}

,

\text{Cu}

,

\text{Mg}

,

\text{Mn}

(aircraft construction); Brass pairs with

\text{Cu}

,

\text{Zn}

(musical instruments/fittings); Stainless Steel pairs with

\text{Fe}

,

\text{Cr}

,

\text{Ni}

,

\text{C}

(cutlery/surgical tools); Soft Solder pairs with

\text{Pb}

,

\text{Sn}

(joining electrical connections).

Each alloy is correctly matched according to its primary constituent elements and application: Duralumin (

\text{Al}

,

\text{Cu}

,

\text{Mg}

,

\text{Mn}

) for aircraft bodywork; Brass (

\text{Cu}

,

\text{Zn}

) for musical instruments and fittings; Stainless Steel (

\text{Fe}

,

\text{Cr}

,

\text{Ni}

,

\text{C}

) for corrosion-resistant cutlery and medical tools; Soft Solder (

\text{Pb}

,

\text{Sn}

) for low-temperature electrical joint soldering.

Step-by-Step Solution

1

Identify the base metals and secondary additives for light-engineering alloys.

Duralumin is an aluminium-based alloy with

\text{Cu}

,

\text{Mg}

, and

\text{Mn}

engineered for aerospace applications due to low density and high mechanical strength.

Aluminium provides low mass while added metals induce lattice distortion to increase hardness.

2

Distinguish between copper-zinc and copper-tin alloys.

Brass is made of copper and zinc, which differs from Bronze (copper and tin). Brass is widely used for decorative fittings and musical instruments.

Zinc substitution in the copper matrix enhances workability and corrosion resistance.

3

Identify steel variations based on anti-corrosion alloying elements.

Stainless steel contains iron, chromium, nickel, and carbon. Chromium imparts a self-healing passive oxide coating.

Chromium content (typically >10.5%) resists oxidative rusting in moist atmospheric conditions.

4

Recall low-melting-point joining alloys.

Soft solder is an alloy of lead and tin engineered to melt at relatively low temperatures (

< 250^\circ\text{C}

).

The eutectic composition of lead and tin depresses the melting point below that of either constituent element.

Key Concept

Elemental compositions, structural properties, and functional uses of key industrial alloys

Question 67Question →

Match each characteristic property of transition metals on the left with its fundamental atomic or electronic explanation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Formation of colored ions matches excitation of electrons between split d-orbital energy levels; Variable oxidation states matches small energy difference between (n-1)d and ns subshells; Paramagnetism matches presence of unpaired d-electrons; High catalytic efficiency matches ability to adopt multiple oxidation states and provide active surface sites.

Transition elements owe their distinct properties to incompletely filled

d

-subshells. Colored compounds are created by

d

-

d

electron transitions when visible light is absorbed. Variable oxidation states arise because

3d

and

4s

energy levels are very close, so electrons from both subshells participate in reaction pathways. Paramagnetism originates from unpaired

d

-electrons, while catalytic behavior is driven by vacant/partially filled

d

-orbitals that adsorb reactants and facilitate intermediate oxidation states.

Step-by-Step Solution

1

Analyze the cause of color in transition metal complexes.

Ligands split the degenerate

d

-orbitals into different energy levels. Absorption of visible light promotes an electron (

d

-

d

transition), imparting color.

Relates the macroscopic color property to internal crystal field splitting.

2

Analyze why transition metals exhibit multiple oxidation states.

The energy gap between

(n-1)d

and

ns

subshells (such as

3d

and

4s

) is minimal, enabling electrons from both subshells to participate in bonding.

Explains why metals like Iron can exist as

Fe^{2+}

and

Fe^{3+}

.

3

Determine the electronic basis of paramagnetism.

Unpaired electrons possess a net magnetic spin moment, causing the ion or atom to be attracted into an external magnetic field.

Distinguishes paramagnetism (unpaired electrons) from diamagnetism (all paired electrons).

4

Examine how transition elements act as catalysts.

Partially filled

d

-orbitals adsorb reactants onto active sites, and variable oxidation states allow the metal to lower activation energy by forming intermediate species.

Connects surface adsorption and redox cycles to catalytic mechanism.

Key Concept

Electronic Configurations and Characteristic Properties of Transition Elements

Question 68Question →

Match each copper-containing ore with its corresponding chemical formula.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Malachite matches

\text{CuCO}_3\cdot\text{Cu(OH)}_2

, Copper pyrites matches

\text{CuFeS}_2

, Cuprite matches

\text{Cu}_2\text{O}

, and Chalcocite matches

\text{Cu}_2\text{S}

.

Each copper ore correctly maps to its characteristic chemical formula: Malachite is basic copper carbonate

\text{CuCO}_3\cdot\text{Cu(OH)}_2

; Copper pyrites is copper iron disulfide

\text{CuFeS}_2

; Cuprite is copper(I) oxide

\text{Cu}_2\text{O}

; and Chalcocite is copper(I) sulfide

\text{Cu}_2\text{S}

.

Step-by-Step Solution

1

Identify the chemical nature of Malachite

Malachite is a basic carbonate mineral of copper with the formula

\text{CuCO}_3\cdot\text{Cu(OH)}_2

.

Basic copper carbonate consists of copper carbonate and copper hydroxide in a 1:1 mole ratio.

2

Identify the chemical nature of Copper pyrites

Copper pyrites (chalcopyrite) has the formula

\text{CuFeS}_2

.

It is the chief ore of copper containing both iron and copper sulfides.

3

Identify the chemical nature of Cuprite

Cuprite is copper(I) oxide,

\text{Cu}_2\text{O}

.

Cuprite is a red oxide ore containing copper in the +1 oxidation state.

4

Identify the chemical nature of Chalcocite

Chalcocite is copper(I) sulfide,

\text{Cu}_2\text{S}

.

Chalcocite is a dark sulfide ore of copper.

Key Concept

Chemical composition and formulas of primary copper ores

Estimated Time:1m 0s

Question 69Question →

Stainless steel exhibits high resistance to atmospheric corrosion primarily because incorporated chromium reacts with oxygen to form a micro-thin, passive oxide surface layer that prevents further oxidation of the underlying iron.

True

False

Show answer & explanation

Answer: True

Answer

The statement is TRUE. The addition of chromium to iron in stainless steel leads to the formation of an adherent, passive chromium(III) oxide (

Cr_2O_3

) film on the surface, protecting the alloy from ongoing oxidation.

Chromium in stainless steel reacts with ambient oxygen to generate a thin, self-repairing surface film of

Cr_2O_3

. This passive layer serves as a barrier preventing moisture and oxygen from coming into direct contact with the iron matrix, stopping corrosion.

Step-by-Step Solution

1

Identify the primary alloying constituent responsible for corrosion resistance in stainless steel.

Stainless steel is an alloy composed of iron (

Fe

), carbon (

C

), chromium (

Cr

), and often nickel (

Ni

). Chromium is the element specifically added to impart corrosion resistance.

Understanding the chemical roles of individual constituent metals in an alloy is necessary to evaluate physical and chemical property modifications.

2

Analyze the surface chemical reaction between chromium in the alloy and atmospheric oxygen.

Chromium oxidizes preferentially over iron to form a continuous, insoluble layer of chromium oxide (

Cr_2O_3

).

This process, known as passivation, prevents reactive agents like water and oxygen from penetrating to the core iron atoms.

3

Evaluate the correctness of the statement based on passivation principles.

The statement accurately describes the passivation process of stainless steel.

Because chromium passivation is the accepted scientific mechanism for stainless steel's rust prevention, the statement is true.

Key Concept

Passivation and Corrosion Resistance of Chromium in Steel Alloys

Question 70Question →

Cobalt is a first-row transition element with an atomic number of 27. What is the ground-state electronic configuration of the cobalt(II) ion, $Co^{2+}$ ?

[Ar] 3d^7

[Ar] 3d^5 4s^2

[Ar] 3d^6 4s^1

[Ar] 3d^9

Show answer & explanation

Answer: $[Ar] 3d^7$

Answer

The electronic configuration of the cobalt(II) ion,

Co^{2+}

, is

[Ar] 3d^7

.

Neutral cobalt (

Z = 27

) has the electronic configuration

[Ar] 3d^7 4s^2

. When transition metals ionize, electrons are lost first from the outermost

4s

orbital because

n=4

electrons experience lower electrostatic attraction from the nucleus than

n=3

electrons. Removing two electrons yields

[Ar] 3d^7

.

Step-by-Step Solution

1

Determine the ground-state electronic configuration of neutral cobalt (

Co

,

Z = 27

).

The neutral cobalt atom has 27 electrons, giving an electronic configuration of

[Ar] 3d^7 4s^2

.

According to the Aufbau principle, orbitals are filled in increasing order of energy, placing two electrons in the

4s

orbital and seven in the

3d

orbitals.

2

Apply cation formation rules to remove two electrons for the

Co^{2+}

ion.

Two electrons are removed from the outermost principal energy level (

n = 4

), which is the

4s

orbital.

Although

4s

is filled before

3d

, electrons in the principal shell with the highest principal quantum number (

n = 4

) are held least tightly and are lost first upon ionization.

3

Write the resulting electronic configuration for

Co^{2+}

.

[Ar] 3d^7

Removing the two

4s

electrons leaves seven electrons in the

3d

subshell.

Key Concept

Electronic Configuration of Transition Metal Cations

Question 71Question →

Aircraft structural components require materials that combine low density with high tensile strength. Which of the following alloys, composed predominantly of aluminium along with copper, magnesium, and manganese, is extensively used in aircraft construction because it is significantly stronger than pure aluminium?

Duralumin

Magnalium

Alnico

Solder

Show answer & explanation

Answer: Duralumin

Answer

Duralumin is the aluminium alloy composed of aluminium, copper, magnesium, and manganese used in aircraft construction.

Duralumin is composed of aluminium (

95\%

), copper (

4\%

), magnesium (

0.5\%

), and manganese (

0.5\%

). Adding these elements to aluminium alters the crystal lattice and enhances its hardness and tensile strength without significantly increasing its density, making it ideal for aircraft bodies.

Step-by-Step Solution

1

Identify the required material properties from the question prompt.

The target material must be a light, high-tensile-strength aluminium alloy used in aircraft construction.

Aircraft construction requires low-density materials to minimize weight while maintaining structural integrity under stress.

2

Analyze the elemental composition specified in the prompt (

Al + Cu + Mg + Mn

).

Aluminium combined with small percentages of copper (

4\%

), magnesium (

0.5\%

), and manganese (

0.5\%

) forms Duralumin.

Alloying pure aluminium with these transition and main-group elements creates lattice distortions that prevent dislocation movement, dramatically increasing tensile strength.

3

Match the composition and applications to the correct alloy name.

Duralumin is the correct choice.

Magnalium lacks copper/manganese reinforcement for heavy structural framing, Alnico is a magnetic alloy, and Solder is a tin-lead joining alloy.

Key Concept

Composition, properties, and applications of aluminium alloys (Duralumin)

Question 72Question →

Match each metallurgical process or extraction stage on the left with its corresponding chemical principle or operational method on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Concentration of low-grade sulfide ores matches Froth flotation using oil collectors; Extraction of highly electropositive metals matches Electrolysis of fused salts; Reduction of haematite matches Chemical reduction by carbon monoxide gas; Refining of crude blister copper matches Electrolytic dissolution of impure anode and deposition of pure metal on cathode.

The paired principles directly correspond to standard industrial metallurgy: froth flotation uses oil wettability to concentrate sulfide ores; fused salt electrolysis extracts top-series electropositive metals; carbon monoxide reduces iron oxide in the blast furnace; and electro-refining purifies crude metal using crude anode oxidation and pure cathode deposition.

Step-by-Step Solution

1

Determine the physical concentration technique suitable for sulfide minerals.

Sulfide ores like galena and chalcopyrite selectively attach to air bubbles created by oil collectors and froth up, separating from waste gangue.

Difference in surface wettability between sulfide ore and siliceous gangue.

2

Evaluate the extraction strategy based on the reactivity series of metals.

Metals positioned near the top of the reactivity series (

Al

,

Na

,

Ca

) require electrical energy for reduction because carbon cannot displace them from their oxides.

High electropositivity means these metals have higher affinity for oxygen than carbon has.

3

Identify the primary chemical reducing agent in the blast furnace.

Coke burns to form

CO_2

, which reacts further with coke to give

CO

.

CO

gas then reduces

Fe_2O_3

step-by-step to molten iron.

CO

is a effective gaseous reducing agent that penetrates porous ore charges.

4

Analyze the electrolytic purification mechanism for crude metals.

Impurities stay in solution or form anode sludge while metal ions migrate and plate onto the cathode as pure copper metal.

Anodic oxidation releases

Cu^{2+}

ions while cathode reduction ensures selective plating of pure copper.

Key Concept

General Principles of Metallurgy and Metal Extraction

Estimated Time:1m 30s

Question 73Question →

During the industrial extraction of sodium metal by the electrolysis of molten sodium chloride in a Downs cell, a steady current of $9.65\text{ A}$ is passed through the electrolytic cell for $50\text{ minutes}$ . What mass of pure sodium metal is collected at the cathode? [ $\text{Na} = 23$ , $1\text{ F} = 96,500\text{ C mol}^{-1}$ ]

6.9\text{ g}

3.45\text{ g}

13.8\text{ g}

6.72\text{ g}

Show answer & explanation

Answer: $6.9\text{ g}$

Answer

The mass of pure sodium metal collected at the cathode is

6.9\text{ g}

.

At the cathode of the Downs cell, sodium ions undergo single-electron reduction (

\text{Na}^+ + e^- \rightarrow \text{Na}

). Passing

28,950\text{ C}

of charge transfers

0.3\text{ mol}

of electrons. Multiplying

0.3\text{ mol}

by the atomic mass of sodium (

23\text{ g mol}^{-1}

) yields exactly

6.9\text{ g}

.

Step-by-Step Solution

1

Calculate the total electric charge (

Q

) passed through the cell in seconds.

Q = I \times t = 9.65\text{ A} \times (50 \times 60\text{ s}) = 28,950\text{ C}

.

Electric charge is determined by multiplying current in amperes by time in seconds.

2

Calculate the amount of substance (in moles) of electrons transferred.

Moles of

e^- = \frac{Q}{F} = \frac{28,950\text{ C}}{96,500\text{ C mol}^{-1}} = 0.3\text{ mol}

.

One Faraday (

96,500\text{ C}

) corresponds to the electric charge of one mole of electrons.

3

Relate the moles of electrons to the moles of sodium metal deposited using the cathode reduction half-equation.

Cathode reaction:

\text{Na}^+ + e^- \rightarrow \text{Na}

. Thus,

1\text{ mol}

of

e^-

produces

1\text{ mol}

of

\text{Na}

, yielding

0.3\text{ mol}

of

\text{Na}

.

Sodium is a univalent alkali metal cation requiring 1 electron per discharged ion.

4

Calculate the mass of sodium metal deposited.

Mass of

\text{Na} = 0.3\text{ mol} \times 23\text{ g mol}^{-1} = 6.9\text{ g}

.

Mass is obtained by multiplying the number of moles by the relative atomic mass.

Key Concept

Quantitative Electrolysis of Molten Salts in Downs Cell Extraction

Estimated Time:2m 0s

Question 74Question →

During the industrial smelting stage of copper extraction from chalcopyrite ( $\text{CuFeS}_2$ ), silicon(IV) oxide ( $\text{SiO}_2$ ) is added to the furnace charge. What is the primary chemical role of silicon(IV) oxide in this process?

To act as an acidic flux that combines with iron(II) oxide impurity to form a fusible slag of iron(II) trioxosilicate(IV)

To serve as the primary reducing agent that converts copper(I) sulfide directly to metallic copper

To lower the melting point of the copper electrolyte during subsequent electro-refining

To react with acidic sulfur impurities and neutralize them into non-volatile salts

Show answer & explanation

Answer: To act as an acidic flux that combines with iron(II) oxide impurity to form a fusible slag of iron(II) trioxosilicate(IV)

Answer

Silicon(IV) oxide acts as an acidic flux that reacts with basic iron(II) oxide to form molten iron(II) trioxosilicate(IV) slag (

\text{FeSiO}_3

).

In the extraction of copper from sulfide ores, the ore contains significant iron impurities. Smelting oxidized iron into basic iron(II) oxide (

\text{FeO}

). Adding sand or silicon(IV) oxide (

\text{SiO}_2

), which acts as an acidic flux, causes a chemical reaction forming molten iron(II) trioxosilicate(IV) (

\text{FeSiO}_3

). This slag is less dense than the copper matte, allowing easy separation by skimming or tapping off.

Step-by-Step Solution

1

Identify the nature of impurities in copper ore roasting/smelting

Partial roasting of copper pyrites (

\text{CuFeS}_2

) produces iron(II) oxide (

\text{FeO}

), which is a basic oxide impurity.

Iron impurities must be separated from copper matte (

\text{Cu}_2\text{S} \cdot \text{FeS}

) prior to copper reduction.

2

Determine the role of the added flux

An acidic flux, silicon(IV) oxide (

\text{SiO}_2

), reacts with basic

\text{FeO}

according to the equation:

\text{FeO}(s) + \text{SiO}_2(s) \rightarrow \text{FeSiO}_3(l)

.

An acid-base reaction between flux and gangue forms a light, molten silicate layer known as slag.

3

Conclude the function of slag in pyrometallurgy

The molten slag (

\text{FeSiO}_3

) floats on top of the heavier copper matte layer and is tapped off.

Slag formation allows continuous mechanical removal of iron impurities.

Key Concept

Role of fluxes and slag formation in metallurgy

Estimated Time:1m 0s

Question 75Question →

When

16.8\text{ g}

of sodium hydrogentrioxocarbonate(IV) (

\text{NaHCO}_3

) is strongly heated in a closed system until decomposition is complete according to the equation:

2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)

What is the total volume of gaseous products liberated at standard temperature and pressure (STP)?
[

M_r\text{ of NaHCO}_3 = 84\text{ g mol}^{-1}

; Molar volume of gas at STP

= 22.4\text{ dm}^3\text{ mol}^{-1}

]

2.24\text{ dm}^3

4.48\text{ dm}^3

4.80\text{ dm}^3

8.96\text{ dm}^3

Show answer & explanation

Answer: $4.48\text{ dm}^3$

Answer

The total volume of gaseous products liberated at STP is

4.48\text{ dm}^3

.

Decomposing

16.8\text{ g}

(

0.20\text{ mol}

) of

\text{NaHCO}_3

yields

0.10\text{ mol}

of

\text{H}_2\text{O}(g)

and

0.10\text{ mol}

of

\text{CO}_2(g)

, totaling

0.20\text{ mol}

of gaseous products. At STP,

0.20\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 4.48\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the number of moles of

\text{NaHCO}_3

decomposed.

\text{Moles of NaHCO}_3 = \frac{16.8\text{ g}}{84\text{ g mol}^{-1}} = 0.20\text{ mol}

.

Converting the given mass into moles using molar mass.

2

Determine the mole ratio between

\text{NaHCO}_3

and total gaseous products.

From

2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)

,

2\text{ moles}

of

\text{NaHCO}_3

produce

1\text{ mole}

of

\text{H}_2\text{O}(g)

and

1\text{ mole}

of

\text{CO}_2(g)

, giving

2\text{ moles}

of total gaseous products.

Both water vapor (at high decomposition temperature) and carbon(IV) oxide exist in the gaseous state.

3

Calculate total moles and total volume of gas at STP.

\text{Total moles of gas} = 0.20\text{ mol}

.

\text{Total volume} = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 4.48\text{ dm}^3

.

Multiplying total gaseous moles by the molar volume at STP.

Key Concept

Thermal decomposition stoichiometry of sodium hydrogentrioxocarbonate(IV)

Estimated Time:1m 30s

Question 76Question →

Match each sodium compound listed on the left with its corresponding characteristic property, industrial preparation process, or commercial application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Sodium peroxide matches with acting as an oxygen generator in breathing apparatus; Sodium hydroxide matches with industrial manufacture via brine electrolysis in a mercury cathode cell; Sodium hydrogentrioxocarbonate(IV) matches with thermal decomposition releasing carbon dioxide gas; Sodium trioxocarbonate(IV) decahydrate matches with efflorescence in dry atmosphere to form a monohydrate powder.

Each sodium compound is paired correctly according to its fundamental chemical behavior or industrial utility: Sodium peroxide reacts with carbon dioxide to regenerate oxygen; Sodium hydroxide is manufactured via the chlor-alkali process in Castner-Kellner mercury cells; Sodium hydrogentrioxocarbonate(IV) undergoes thermal decomposition to release carbon dioxide; and Sodium trioxocarbonate(IV) decahydrate loses hydration water through efflorescence.

Step-by-Step Solution

1

Analyze the chemical property and application of sodium peroxide (

\text{Na}_2\text{O}_2

).

Sodium peroxide reacts with carbon dioxide according to the equation

2\text{Na}_2\text{O}_2 + 2\text{CO}_2 \rightarrow 2\text{Na}_2\text{CO}_3 + \text{O}_2

, releasing oxygen gas.

This property makes it effective as an oxygen rebreather component.

2

Identify the industrial preparation method for sodium hydroxide (

\text{NaOH}

).

Sodium hydroxide is produced by electrolyzing concentrated brine (

\text{NaCl}_{(aq)}

) in a Castner-Kellner cell containing a mercury cathode.

Sodium ions discharge at the mercury cathode forming an amalgam, which reacts with water to yield pure

\text{NaOH}

.

3

Examine the thermal stability of sodium hydrogentrioxocarbonate(IV) (

\text{NaHCO}_3

).

Heating

\text{NaHCO}_3

causes decomposition:

2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2

.

Alkali metal hydrogentrioxocarbonates decompose upon gentle heating to liberate carbon dioxide gas.

4

Determine the effect of air exposure on hydrated sodium trioxocarbonate(IV) (

\text{Na}_2\text{CO}_3\cdot 10\text{H}_2\text{O}

).

Exposure to dry air leads to the loss of nine water molecules:

\text{Na}_2\text{CO}_3\cdot 10\text{H}_2\text{O} \rightarrow \text{Na}_2\text{CO}_3\cdot \text{H}_2\text{O} + 9\text{H}_2\text{O}

.

This spontaneous loss of water of crystallization to the atmosphere is termed efflorescence.

Key Concept

Properties, Preparation, and Uses of Major Sodium Compounds

Question 77Question →

In the industrial extraction of sodium metal using the Downs process, anhydrous molten sodium chloride is electrolyzed instead of an aqueous sodium chloride solution. Why is an aqueous solution of sodium chloride unsuitable for extracting sodium metal?

Hydrogen ions are preferentially discharged at the cathode because they have a higher reduction potential than sodium ions.

Sodium ions migrate to the anode in aqueous solution and form sodium hydroxide.

Water acts as a reducing agent at the cathode, converting sodium metal into sodium oxide.

Water elevates the melting point of sodium chloride, preventing electrolysis from occurring efficiently.

Show answer & explanation

Answer: Hydrogen ions are preferentially discharged at the cathode because they have a higher reduction potential than sodium ions.

Answer

Hydrogen ions are preferentially discharged at the cathode because they have a higher reduction potential than sodium ions.

When an aqueous solution of sodium chloride is electrolyzed, both

Na^+

and

H^+

cations migrate to the cathode. Because

H^+

lies lower in the electrochemical series and has a much higher reduction potential (

0.00\text{ V}

) than

Na^+

(

-2.71\text{ V}

),

H^+

ions are preferentially reduced to form hydrogen gas (

H_2

). To isolate metallic sodium, water must be absent, which is why molten (fused) sodium chloride is used in the Downs process.

Step-by-Step Solution

1

Identify all cations present in an aqueous solution of sodium chloride

The solution contains

Na^+

ions from sodium chloride and

H^+

ions from the self-ionization of water.

Electrolysis of aqueous ionic solutions involves competing ions from both the dissolved salt and water.

2

Compare the standard reduction potentials of the competing cations at the cathode

H^+

has a reduction potential of

0.00\text{ V}

, whereas

Na^+

has a standard reduction potential of

-2.71\text{ V}

.

The cation with the higher (more positive) reduction potential gains electrons more readily at the cathode.

3

Determine the cathode product and deduce why molten NaCl is required

In aqueous solution,

H^+

ions are reduced to

H_2(g)

at the cathode instead of

Na^+

forming

Na(s)

. Therefore, molten

NaCl

must be used to exclude water entirely.

Extracting reactive metals like sodium requires an electrolyte free of water so that

Na^+

is the only cation available for reduction.

Key Concept

Preferential discharge of ions during electrolysis and extraction of reactive metals

Question 78Question →

What is the correct chronological sequence of steps involved in the industrial extraction of sodium metal using the Downs process, from raw material preparation to the final collection of the metallic product?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence begins with purifying and drying rock salt to yield anhydrous

NaCl

, followed by mixing in

CaCl_2

to lower the fusion temperature. The mixture is then melted and electrolyzed in the Downs cell, leading to the reduction of

Na^+

at the cathode, and finally collecting liquid sodium as it floats to the top of the molten electrolyte.

The industrial extraction of sodium metal via the Downs process requires a strict sequence: raw rock salt is first purified and dried to remove water, then mixed with

CaCl_2

flux to lower the melting point to about

600^\circ\text{C}

. This solid mixture is melted in the cell, where direct current electrolyzes the melt, reducing

Na^+

at the iron cathode. Finally, the buoyant molten sodium rises to the top of the cathode chamber and is harvested in a nitrogen environment.

Step-by-Step Solution

1

Identify raw material pretreatment

Anhydrous

NaCl

is prepared by removing moisture and impurities.

Water must be completely eliminated to prevent hydrogen discharge and hazardous reactions with sodium metal.

2

Identify flux addition step

CaCl_2

is added to form a binary eutectic mixture.

Lowering the melting point from

801^\circ\text{C}

to approximately

600^\circ\text{C}

reduces operating costs and sodium vaporization.

3

Identify electrolytic cell setup and power application

The mixture is fused and direct current is passed through molten

NaCl

/

CaCl_2

.

Electrolysis requires mobile ions in a molten state to conduct electricity.

4

Identify electrode reactions

Reduction:

Na^+ + e^- \rightarrow Na(l)

at cathode; Oxidation:

2Cl^- \rightarrow Cl_2(g) + 2e^-

at anode.

Na^+

ions migrate to the iron cathode and accept electrons, while

Cl^-

ions migrate to the graphite anode.

5

Identify product separation and harvest

Molten sodium metal floats above the dense molten electrolyte into a collection riser.

The lower density of molten sodium (

0.97\text{ g/cm}^3

) relative to molten electrolyte (

2.1\text{ g/cm}^3

) allows passive buoyancy separation.

Key Concept

Downs Process for Sodium Extraction

Estimated Time:1m 30s

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