Oxidation-Reduction and Electrochemistry Practice Questions — JAMB UTME — Page 5

Question 81Question →

A solution containing trivalent ions $M^{3+}$ of an unknown metal $M$ is electrolyzed using inert electrodes. If a steady current of $2.50\text{ A}$ passed for $5790\text{ s}$ deposits $1.35\text{ g}$ of metal $M$ at the cathode, what is the relative atomic mass of metal $M$ ? [ $1\text{ Faraday} = 96500\text{ C/mol}$ ]

Show answer & explanation

Answer: 27

Answer

The relative atomic mass of metal

M

is

27\text{ g/mol}

.

Using Faraday's first law (

Q = I \times t

), the total charge passed is

2.50 \times 5790 = 14475\text{ C}

. Dividing by Faraday's constant (

96500\text{ C/mol}

) gives

0.15\text{ mol}

of electrons. Since the metal ion is trivalent (

M^{3+}

), the half-reaction is

M^{3+} + 3e^- \rightarrow M

, meaning

0.15\text{ mol}

of electrons deposits

0.05\text{ mol}

of metal. Dividing the mass (

1.35\text{ g}

) by the amount in moles (

0.05\text{ mol}

) yields a relative atomic mass of

27\text{ g/mol}

.

Step-by-Step Solution

1

Calculate the total electric charge passed (

Q

).

Q = 2.50\text{ A} \times 5790\text{ s} = 14475\text{ C}

.

Electric charge is the product of current in amperes and time in seconds.

2

Calculate the moles of electrons transferred.

n_e = \frac{14475\text{ C}}{96500\text{ C/mol}} = 0.15\text{ mol } e^-

.

One mole of electrons carries a charge of 1 Faraday (

96500\text{ C}

).

3

Determine the amount (in moles) of metal

M

deposited.

From

M^{3+} + 3e^- \rightarrow M

,

3\text{ moles of } e^-

deposit

1\text{ mole of } M

. Thus,

n_M = \frac{0.15\text{ mol}}{3} = 0.05\text{ mol}

.

The reduction of a trivalent ion requires 3 moles of electrons per mole of metal deposited.

4

Calculate the relative atomic mass (

M_r

).

M_r = \frac{1.35\text{ g}}{0.05\text{ mol}} = 27\text{ g/mol}

.

Molar mass is calculated by dividing the mass of the substance by the amount in moles.

Key Concept

Faraday's Laws of Electrolysis and Quantitative Calculations

Question 82Question →

An underground steel pipeline carrying natural gas is buried in moist soil. To protect the iron in the steel from corrosion, blocks of another metal are electrically connected to the pipeline at regular intervals. Given the standard reduction potentials below:

\text{Fe}^{2+}_{(\text{aq})} + 2e^- \rightarrow \text{Fe}_{(\text{s})}, \quad E^\circ = -0.44\text{ V}

\text{Cu}^{2+}_{(\text{aq})} + 2e^- \rightarrow \text{Cu}_{(\text{s})}, \quad E^\circ = +0.34\text{ V}

\text{Mg}^{2+}_{(\text{aq})} + 2e^- \rightarrow \text{Mg}_{(\text{s})}, \quad E^\circ = -2.37\text{ V}

\text{Ag}^{+}_{(\text{aq})} + e^- \rightarrow \text{Ag}_{(\text{s})}, \quad E^\circ = +0.80\text{ V}

Which metal can be attached to the pipeline to serve as an effective sacrificial anode, and what chemical change occurs to it during protection?

Magnesium, because it is oxidized preferentially due to its more negative reduction potential.

Copper, because its positive reduction potential allows it to supply electrons to the iron structure.

Silver, because metals with higher reduction potentials readily undergo sacrificial oxidation.

Magnesium, because it undergoes reduction at the cathode while keeping the iron oxidized.

Show answer & explanation

Answer: Magnesium, because it is oxidized preferentially due to its more negative reduction potential.

Answer

Magnesium serves as an effective sacrificial anode because it has a more negative standard reduction potential than iron, meaning it oxidizes more readily and supplies electrons to protect the iron pipeline.

Magnesium has a more negative standard reduction potential (-2.37 V) than iron (-0.44 V). In an electrochemical cell formed in moist soil, magnesium oxidizes preferentially (

\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-

), supplying electrons to the steel pipe. This makes the iron pipeline cathodic and prevents iron from oxidizing into rust.

Step-by-Step Solution

1

Compare the standard reduction potentials (

E^\circ

) of iron and the candidate metals.

E^\circ(\text{Mg}^{2+}/\text{Mg}) = -2.37\text{ V}

,

E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\text{ V}

,

E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}

, and

E^\circ(\text{Ag}^{+}/\text{Ag}) = +0.80\text{ V}

.

A metal acts as a sacrificial anode only if it is more easily oxidized (more electropositive) than iron, which corresponds to having a more negative standard reduction potential.

2

Identify which metal will oxidize preferentially over iron.

Magnesium has the most negative reduction potential (-2.37 V < -0.44 V), so it oxidizes preferentially:

\text{Mg}_{(\text{s})} \rightarrow \text{Mg}^{2+}_{(\text{aq})} + 2e^-

.

The metal with the lower (more negative) reduction potential readily loses electrons to the iron structure, making the iron the cathode in the electrochemical cell.

3

Determine the chemical process occurring at the sacrificial block.

Oxidation takes place at the sacrificial magnesium block (the anode), sacrificing the block while preserving the steel pipeline.

By definition, oxidation occurs at the anode (

\text{anode} = \text{oxidation}

).

Key Concept

Sacrificial Anodic Protection

Question 83Question →

The standard reduction potentials for two half-reactions at

25^\circ\text{C}

are given below:

\text{Co}^{2+}(aq) + 2e^- \rightarrow \text{Co}(s) \quad E^\circ = -0.28\text{ V}

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad E^\circ = +0.34\text{ V}

What is the standard electromotive force ( $E^\circ_{\text{cell}}$ ) for the spontaneous reaction that occurs when these two half-cells are connected under standard conditions?

+0.62\text{ V}

+0.06\text{ V}

-0.62\text{ V}

-0.06\text{ V}

Show answer & explanation

Answer: $+0.62\text{ V}$

Answer

The standard electromotive force for the spontaneous cell reaction is

+0.62\text{ V}

.

For a spontaneous electrochemical cell, reduction occurs at the electrode with the more positive reduction potential (cathode:

\text{Cu}^{2+}/\text{Cu}

at

+0.34\text{ V}

), and oxidation occurs at the electrode with the more negative reduction potential (anode:

\text{Co}^{2+}/\text{Co}

at

-0.28\text{ V}

). Subtracting the anode potential from the cathode potential yields

E^\circ_{\text{cell}} = +0.34\text{ V} - (-0.28\text{ V}) = +0.62\text{ V}

.

Step-by-Step Solution

1

Identify the cathode and anode based on standard reduction potentials.

Copper half-cell (

E^\circ = +0.34\text{ V}

) has the higher reduction potential and acts as the cathode (reduction). Cobalt half-cell (

E^\circ = -0.28\text{ V}

) has the lower reduction potential and acts as the anode (oxidation).

A species with a higher reduction potential is more easily reduced, driving spontaneous electron flow from anode to cathode.

2

Calculate the standard cell electromotive force (

E^\circ_{\text{cell}}

).

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34\text{ V} - (-0.28\text{ V}) = +0.34\text{ V} + 0.28\text{ V} = +0.62\text{ V}

.

The cell EMF for a spontaneous reaction must be positive.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Estimated Time:1m 0s

Question 84Question →

Fill in the missing terms to complete the statement describing the electrochemical roles of different regions during the rusting of iron.

Fill in the blanks below

In the electrochemical mechanism of rusting, the site on the iron surface where metallic iron is oxidized to iron(II) ions functions as the , while the region exposed to moisture and oxygen where reduction occurs acts as the .

Show answer & explanation

Answer

The site where metallic iron is oxidized functions as the anode, and the region where reduction occurs acts as the cathode.

During the rusting of iron, a micro-galvanic cell is established on the surface of the metal. The region where iron loses electrons to become

Fe^{2+}

ions undergoes oxidation and therefore functions as the anode. The oxygen-rich droplet region where electrons are consumed to reduce oxygen to

OH^-

ions functions as the cathode.

Step-by-Step Solution

1

Identify the region of oxidation during rusting.

Iron atoms lose electrons (

Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-

), which corresponds to oxidation.

By electrochemical definition, oxidation always occurs at the anode.

2

Identify the region of reduction during rusting.

Dissolved oxygen accepts electrons in the presence of water (

O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)

), which corresponds to reduction.

By electrochemical definition, reduction always takes place at the cathode.

Key Concept

Electrochemical mechanism of metallic corrosion

Question 85Question →

Aluminium resists atmospheric corrosion better than iron because it rapidly forms a tough, non-porous coating of aluminium oxide ( $Al_2O_3$ ) that prevents oxygen and moisture from reaching the underlying metal.

True

False

Show answer & explanation

Answer: True

Answer

The statement is TRUE. Aluminium forms an adherent, non-porous protective oxide film (

Al_2O_3

) upon exposure to air, which renders the metal passive and prevents further atmospheric corrosion.

The statement is true because aluminium undergoes passivation in atmospheric air, forming an impermeable, self-healing oxide film (

Al_2O_3

) that insulates the underlying metal from environmental oxygen and moisture.

Step-by-Step Solution

1

Analyze the surface reactivity of aluminium exposed to atmospheric air.

Aluminium rapidly reacts with oxygen to form a thin surface layer of aluminium oxide (

Al_2O_3

).

Aluminium has a high affinity for oxygen due to its electropositive nature.

2

Compare the physical properties of aluminium oxide with those of iron rust.

Unlike iron rust (

Fe_2O_3 \cdot xH_2O

), which is porous, brittle, and flakes off, the

Al_2O_3

coating is continuous, tough, and impermeable.

The non-porous nature of

Al_2O_3

prevents further diffusion of oxygen and moisture to the unreacted metal underneath.

Key Concept

Metal Passivation via Protective Oxide Layer

Estimated Time:45s

Question 86Question →

In an industrial chlor-alkali membrane cell, concentrated sodium chloride solution (brine) is electrolyzed using a constant current of $19.3\text{ A}$ for $50\text{ minutes}$ . What is the mass, in grams, of sodium hydroxide ( $\text{NaOH}$ ) produced in the solution? [Molar mass of $\text{NaOH} = 40.0\text{ g mol}^{-1}$ , Faraday constant $F = 96,500\text{ C mol}^{-1}$ ]

Show answer & explanation

Answer: 24

Answer

The mass of sodium hydroxide produced is

24.0\text{ g}

.

Converting the electrolysis time to seconds (

3000\text{ s}

) yields a total charge of

Q = 19.3\text{ A} \times 3000\text{ s} = 57,900\text{ C}

. Dividing by Faraday's constant (

96,500\text{ C mol}^{-1}

) gives

0.6\text{ mol}

of electrons. In the chlor-alkali process, reduction of water produces

1\text{ mol}

of

\text{OH}^-

ions per mole of electrons, producing

0.6\text{ mol}

of

\text{NaOH}

. Multiplying by the molar mass (

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

) gives a mass of

24.0\text{ g}

.

Step-by-Step Solution

1

Convert electrolysis time into seconds and calculate total charge.

Q = 19.3\text{ A} \times (50 \times 60\text{ s}) = 57,900\text{ C}

.

Faraday's equations require time in seconds (

Q = I \times t

).

2

Calculate the moles of electrons transferred using the Faraday constant.

n(e^-) = \frac{57,900\text{ C}}{96,500\text{ C mol}^{-1}} = 0.6\text{ mol}

.

One Faraday (

96,500\text{ C}

) corresponds to one mole of electrons.

3

Determine the stoichiometry of the cathode reaction.

Cathode reaction:

2\text{H}_2\text{O}_{(l)} + 2e^- \rightarrow \text{H}_{2(g)} + 2\text{OH}^-_{(aq)}

. Thus,

1\text{ mol } e^-

forms

1\text{ mol } \text{OH}^-

, giving

0.6\text{ mol}

of

\text{NaOH}

.

During brine electrolysis, water is preferentially reduced at the cathode, generating hydroxide ions that pair with sodium ions.

4

Calculate the mass of sodium hydroxide produced.

\text{Mass} = 0.6\text{ mol} \times 40.0\text{ g mol}^{-1} = 24.0\text{ g}

.

Mass is obtained by multiplying the number of moles by the molar mass.

Key Concept

Industrial Electrolysis Stoichiometry (Chlor-Alkali Process) and Faraday's First Law

Estimated Time:1m 30s

Question 87Question →

Match each industrial electrolytic process on the left with its corresponding chemical characteristic or operating condition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hall-Héroult Process matches with 'Electrolysis of molten alumina dissolved in molten cryolite using carbon electrodes'. Electrorefining of Copper matches with 'Impure metal serves as the dissolving anode while pure metal deposits at the cathode'. Electroplating of Iron with Silver matches with 'The article to be coated serves as the cathode in an electrolyte containing silver ions'. Chlor-Alkali Process matches with 'Electrolysis of concentrated brine yielding chlorine gas, hydrogen gas, and sodium hydroxide'.

Each industrial process relies on specific redox reactions and cell configurations: Hall-Héroult process uses molten cryolite as a solvent for alumina; copper refining uses impure copper as the dissolving anode; electroplating places the article to be coated at the cathode; chlor-alkali electrolyzes brine to yield chlorine, hydrogen, and sodium hydroxide.

Step-by-Step Solution

1

Identify the key components and conditions of aluminium extraction.

Aluminium is extracted by electrolyzing molten alumina (

Al_2O_3

) dissolved in cryolite (

Na_3AlF_6

) to lower its melting point.

This corresponds to the Hall-Héroult process operating condition.

2

Identify the electrode roles in metal purification.

Impure metal dissolves at the anode and deposits as pure metal at the cathode.

This defines electrorefining of copper.

3

Identify the arrangement for electroplating.

The target object to be coated forms the cathode where metal cations are reduced.

This describes the electroplating of iron with silver.

4

Identify the reactants and products of brine electrolysis.

Concentrated

NaCl

solution electrolyzed gives

Cl_2

,

H_2

, and

NaOH

.

This defines the chlor-alkali industrial process.

Key Concept

Industrial applications of electrolysis including metal extraction, refining, electroplating, and chlor-alkali synthesis

Question 88Question →

Four identical iron nails are placed into separate test tubes filled with aerated sodium chloride solution. Nail 1 is wrapped with a zinc wire, Nail 2 is wrapped with a copper wire, Nail 3 is completely covered with petroleum jelly, and Nail 4 is left bare. After three days, Nail 2 exhibits significantly more severe rusting than the untreated Nail 4. Which of the following best explains why coupling iron with copper accelerates the corrosion of iron?

Iron is more electropositive than copper, causing iron to serve as the anode and undergo rapid oxidation.

Copper acts as a sacrificial anode by transferring positive ions to iron, increasing rust accumulation.

Copper reacts with water to release hydrogen ions, which directly oxidize the iron surface.

Iron acts as the cathode and undergoes cathodic reduction due to the higher electrical conductivity of copper.

Show answer & explanation

Answer: Iron is more electropositive than copper, causing iron to serve as the anode and undergo rapid oxidation.

Answer

Iron is more electropositive than copper, causing iron to serve as the anode and undergo rapid oxidation.

When iron is in electrical contact with a less reactive metal like copper in the presence of an electrolyte, a galvanic cell is formed. Because iron has a higher oxidation potential than copper, iron acts as the anode and is oxidized to iron(II) ions much faster than if it were un-coupled. Copper acts as the cathode where atmospheric oxygen is reduced.

Step-by-Step Solution

1

Identify the relative positions of iron and copper in the electrochemical series.

Iron (

Fe

) is more electropositive (higher tendency to oxidize) than copper (

Cu

).

Metals higher in the reactivity series lose electrons more readily.

2

Determine the galvanic cell configuration formed when iron and copper are in contact in an electrolyte.

Iron becomes the anode (site of oxidation:

Fe \rightarrow Fe^{2+} + 2e^-

) while copper acts as the cathode (site of reduction:

O_2 + 2H_2O + 4e^- \rightarrow 4OH^-

).

Electrons flow from the more reactive metal (anode) to the less reactive metal (cathode).

3

Evaluate the effect of this galvanic coupling on the corrosion rate of iron.

The oxidation of iron is significantly accelerated compared to un-coupled iron.

The presence of a cathodic metal (copper) facilitates the removal of electrons from iron, speeding up rust formation.

Key Concept

Galvanic Corrosion and Electrochemical Series

Estimated Time:1m 0s

Question 89Question →

During the electrolysis of a very dilute solution of sodium chloride using inert platinum electrodes, which gas is evolved at the anode and what primary factor determines its discharge?

Oxygen gas, because the position of

OH^-

in the electrochemical series is lower than that of

Cl^-

Chlorine gas, because chloride ions are present in higher concentration than hydroxide ions

Chlorine gas, because chloride ions are lower than hydroxide ions in the electrochemical series

Hydrogen gas, because the nature of the platinum electrode promotes cation oxidation

Show answer & explanation

Answer: Oxygen gas, because the position of $OH^-$ in the electrochemical series is lower than that of $Cl^-$

Answer

Oxygen gas is evolved at the anode because the position of hydroxide ions (

OH^-

) in the electrochemical series is lower than that of chloride ions (

Cl^-

).

In a very dilute solution of sodium chloride, the position of ions in the electrochemical series is the primary factor determining preferential discharge. Because hydroxide ions (

OH^-

) lie below chloride ions (

Cl^-

) in the electrochemical series,

OH^-

ions are preferentially discharged at the anode to produce oxygen gas.

Step-by-Step Solution

1

Identify all anions migrating to the positive electrode (anode).

The anions present in aqueous

NaCl

are

Cl^-

from sodium chloride and

OH^-

from the auto-ionization of water.

Negatively charged ions migrate to the anode during electrolysis.

2

Determine which factor governs preferential discharge in this setup.

Because the solution is explicitly stated as very dilute, position in the electrochemical series predominates over concentration.

Ion concentration only overrides standard discharge potential when the concentration of a competing ion is significantly high.

3

Compare the electrochemical series positions of the anions.

OH^-

lies below

Cl^-

in the electrochemical series, so

OH^-

is oxidized preferentially to produce oxygen gas:

4OH^- \rightarrow 2H_2O + O_2 + 4e^-

.

Anions positioned lower in the series require less energy to give up electrons.

Key Concept

Factors affecting preferential discharge of ions in electrolysis: position in the electrochemical series vs. ion concentration.

Question 90Question →

In an industrial electroplating plant, a steel component is coated with silver in an electrolytic bath. If a constant electric current of $2.0\text{ A}$ is passed through the bath for $48.25\text{ minutes}$ , what is the mass of silver, in grams, deposited on the cathode? [Molar mass of $\text{Ag} = 108\text{ g mol}^{-1}$ , $1\text{ F} = 96500\text{ C mol}^{-1}$ ]

Show answer & explanation

Answer: 6.48

Answer

The mass of silver deposited on the cathode during electroplating is 6.48 g.

According to Faraday's first law of electrolysis, charge

Q = 2.0\text{ A} \times (48.25 \times 60\text{ s}) = 5790\text{ C}

. The number of moles of electrons passed is

5790 / 96500 = 0.06\text{ mol}

. Since silver reduction (

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

) requires

1\text{ mole}

of electrons per mole of silver,

0.06\text{ mol}

of

\text{Ag}

is formed. The mass of silver deposited is

0.06\text{ mol} \times 108\text{ g mol}^{-1} = 6.48\text{ g}

.

Step-by-Step Solution

1

Convert time to seconds

t = 48.25 \text{ min} \times 60 \text{ s/min} = 2895 \text{ s}

Time must be expressed in seconds to calculate electric charge in coulombs.

2

Calculate total quantity of electricity

Q

Q = I \times t = 2.0 \text{ A} \times 2895 \text{ s} = 5790 \text{ C}

Electric charge is the product of current and duration of electrolysis.

3

Determine moles of electrons transferred

n(e^-) = \frac{5790 \text{ C}}{96500 \text{ C mol}^{-1}} = 0.06 \text{ mol}

Faraday's constant indicates that

96500\text{ C}

corresponds to

1\text{ mole}

of electrons.

4

Relate electron flow to silver discharge at cathode

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

, so

0.06 \text{ mol of } e^- \text{ yields } 0.06 \text{ mol of Ag}

Silver ion discharge requires one electron per silver atom deposited.

5

Calculate mass of silver deposited

\text{Mass} = 0.06 \text{ mol} \times 108 \text{ g mol}^{-1} = 6.48 \text{ g}

Multiplying the molar quantity by relative atomic mass yields the total mass deposited.

Key Concept

Quantitative application of Faraday's laws of electrolysis in industrial electroplating.

Question 91Question →

In the industrial Chlor-Alkali process, concentrated sodium chloride solution (brine) is electrolyzed using inert carbon electrodes. Which substance is liberated at the cathode, and what is the reason for its discharge?

Hydrogen gas, because

H^+

ions are lower than

Na^+

ions in the electrochemical series and require less energy to gain electrons.

Sodium metal, because

Na^+

ions are present in a significantly higher concentration in brine than

H^+

ions.

Chlorine gas, because

Cl^-

ions migrate to the cathode to undergo reduction.

Oxygen gas, because

OH^-

ions are preferentially discharged over

Cl^-

ions at the cathode.

Show answer & explanation

Answer: Hydrogen gas, because $H^+$ ions are lower than $Na^+$ ions in the electrochemical series and require less energy to gain electrons.

Answer

Hydrogen gas is liberated at the cathode because hydrogen ions (

H^+

) lie lower in the electrochemical series than sodium ions (

Na^+

) and are thus preferentially reduced.

In the electrolysis of concentrated aqueous sodium chloride (brine), water provides

H^+

and

OH^-

ions alongside

Na^+

and

Cl^-

ions. At the cathode (the negative electrode), cations

Na^+

and

H^+

compete for discharge. Hydrogen (

H^+

) is much lower than sodium (

Na^+

) in the electrochemical series, meaning it accepts electrons far more easily. Therefore, hydrogen ions are preferentially reduced to yield hydrogen gas (

H_2

).

Step-by-Step Solution

1

Identify the ions present in the electrolyte

In aqueous

NaCl

(brine), the ions present are

Na^+

and

Cl^-

from sodium chloride, and

H^+

and

OH^-

from the auto-ionization of water.

Electrolysis of aqueous solutions involves ions from both the solute and the solvent.

2

Determine which ions migrate to the cathode

Cations (

Na^+

and

H^+

) migrate to the negatively charged cathode.

Oppositely charged ions are attracted to the electrodes.

3

Apply the principles of preferential discharge at the cathode

H^+

is placed much lower than

Na^+

in the reactivity/electrochemical series, so

2H^+ + 2e^- \rightarrow H_2(g)

occurs.

Ions lower in the electrochemical series gain electrons (are reduced) more readily than ions higher up.

Key Concept

Preferential discharge of cations during electrolysis of aqueous brine in the Chlor-Alkali industry

Question 92Question →

An offshore oil platform made of steel is submerged in seawater. To protect the submerged steel structure from corrosion, an impressed current cathodic protection system is installed using an external direct current source and an inert anode. Which of the following best describes the electrochemical role and behavior of the steel structure in this system?

It acts as the cathode, receiving electrons from the external power source to prevent iron from oxidizing.

It acts as the anode, releasing electrons into the DC power source to corrode preferentially.

It acts as the cathode, undergoing continuous oxidation to form a protective metal oxide layer.

It acts as the anode, attracting hydroxide ions to passivate the steel surface.

Show answer & explanation

Answer: It acts as the cathode, receiving electrons from the external power source to prevent iron from oxidizing.

Answer

The steel structure acts as the cathode, receiving electrons from the external DC power source to prevent iron from oxidizing.

In impressed current cathodic protection (ICCP), an external DC power source supplies electrons to the steel structure. By forcing electrons into the steel, the structure becomes the cathode of the electrochemical cell, preventing the oxidation of iron (

Fe \rightarrow Fe^{2+} + 2e^-

) and effectively stopping corrosion.

Step-by-Step Solution

1

Identify the electrochemical mechanism of Cathodic Protection via Impressed Current (ICCP).

An external direct current (DC) power source forces electrons onto the metal structure to be protected.

Corrosion of iron occurs via an anodic oxidation process (

Fe \rightarrow Fe^{2+} + 2e^-

). Supplying electrons reverses/prevents this oxidation.

2

Determine the electrode polarities in the cathodic protection circuit.

The negative terminal of the DC power source is connected to the steel structure, making it the cathode. The positive terminal is connected to an inert material, making it the anode.

Reduction or electron supply occurs at the cathode, ensuring the protected metal remains in its unoxidized elemental state.

Key Concept

Impressed Current Cathodic Protection (ICCP)

Question 93Question →

A steady electric current of $5.0\text{ A}$ is passed through molten lead(II) bromide ( $PbBr_2$ ) for $32\text{ minutes}$ and $10\text{ seconds}$ . What mass of lead, in grams, is deposited at the cathode? [ $\text{Pb} = 207$ , $1\text{ F} = 96\text{ }500\text{ C mol}^{-1}$ ]

Show answer & explanation

Answer: 10.35

Answer

The mass of lead deposited at the cathode is

10.35\text{ g}

.

Passing a steady current of

5.0\text{ A}

for

1930\text{ s}

transfers

9650\text{ C}

of charge, corresponding to

0.10\text{ mol}

of electrons (

0.10\text{ F}

). Since reduction of lead(II) ions (

Pb^{2+} + 2e^- \rightarrow Pb

) requires 2 moles of electrons per mole of lead metal,

0.05\text{ mol}

of lead is deposited. Multiplying by the relative atomic mass of lead (

207\text{ g/mol}

) gives

10.35\text{ g}

.

Step-by-Step Solution

1

Convert time from minutes and seconds into total seconds

t = (32 \times 60\text{ s}) + 10\text{ s} = 1930\text{ s}

Time must be expressed in seconds to calculate electric charge in coulombs.

2

Calculate total quantity of electricity (

Q

) passed

Q = I \times t = 5.0\text{ A} \times 1930\text{ s} = 9650\text{ C}

Electric charge is the product of current in amperes and duration in seconds.

3

Calculate the moles of electrons transferred

\text{Moles of } e^- = \frac{9650\text{ C}}{96500\text{ C mol}^{-1}} = 0.10\text{ mol e}^-

One Faraday (

96500\text{ C}

) corresponds to one mole of electrons.

4

Relate moles of electrons to moles of lead metal using the cathode half-reaction

Pb^{2+} + 2e^- \rightarrow Pb(s) \implies \text{Moles of } Pb = \frac{0.10\text{ mol e}^-}{2} = 0.05\text{ mol}

Lead has a valency of 2 in

PbBr_2

, requiring 2 moles of electrons per mole of lead deposited.

5

Calculate the mass of deposited lead

\text{Mass} = 0.05\text{ mol} \times 207\text{ g mol}^{-1} = 10.35\text{ g}

Mass is obtained by multiplying the amount of substance in moles by its relative atomic mass.

Key Concept

Faraday's Laws of Electrolysis and Quantitative Calculations

Question 94Question →

Arrange the following cations in order of INCREASING ease of preferential discharge at an inert cathode during electrolysis, starting from the least easily discharged to the most easily discharged:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from least easily discharged to most easily discharged is Sodium ion (

Na^+

), Zinc ion (

Zn^{2+}

), Hydrogen ion (

H^+

), and Copper(II) ion (

Cu^{2+}

).

During electrolysis with inert electrodes, cations migrate to the cathode to undergo reduction. The ease with which a cation accepts electrons depends on its position in the electrochemical series; cations situated lower in the series gain electrons more readily than those above them. Since the relative order from top to bottom is

Na^+

,

Zn^{2+}

,

H^+

,

Cu^{2+}

, the ease of preferential discharge increases in the order:

Na^+ < Zn^{2+} < H^+ < Cu^{2+}

.

Step-by-Step Solution

1

Identify the relative positions of the cations (

Na^+

,

Zn^{2+}

,

H^+

,

Cu^{2+}

) in the electrochemical series.

The order from top (most electropositive metal) to bottom is

Na^+

>

Zn^{2+}

>

H^+

>

Cu^{2+}

.

Metals higher up lose electrons more readily, whereas ions lower down accept electrons more readily.

2

Apply the principle of preferential discharge for cations at the cathode.

Cations lower down in the electrochemical series are preferentially reduced over those higher up.

Ions lower in the series have more positive standard reduction potentials, making reduction thermodynamically more favorable.

3

Arrange the cations in increasing order of ease of discharge.

Sequence: Sodium ion (

Na^+

), Zinc ion (

Zn^{2+}

), Hydrogen ion (

H^+

), Copper(II) ion (

Cu^{2+}

).

The ease of preferential discharge increases progressively down the electrochemical series.

Key Concept

Position of cations in the electrochemical series determines their relative ease of preferential discharge during electrolysis.

Estimated Time:1m 30s

Question 95Question →

During the electrolysis of concentrated hydrochloric acid ( $HCl_{(aq)}$ ) using inert graphite electrodes, a gaseous product $X$ is evolved at the anode while gas $Y$ is evolved at the cathode. Which option correctly identifies gas $X$ , gas $Y$ , and the primary factor responsible for the preferential discharge that yields gas $X$ ?

Gas

X

is

Cl_2

, Gas

Y

is

H_2

, driven by the high concentration of chloride ions.

Gas

X

is

O_2

, Gas

Y

is

H_2

, driven by the position of hydroxide ions in the electrochemical series.

Gas

X

is

H_2

, Gas

Y

is

Cl_2

, driven by the high concentration of hydrogen ions.

Gas

X

is

O_2

, Gas

Y

is

Cl_2

, driven by the chemical nature of the graphite electrodes.

Show answer & explanation

Answer: Gas $X$ is $Cl_2$ , Gas $Y$ is $H_2$ , driven by the high concentration of chloride ions.

Answer

Gas

X

is

Cl_2

, Gas

Y

is

H_2

, driven by the high concentration of chloride ions.

In concentrated hydrochloric acid, the ions migrating to the anode are

Cl^-

and

OH^-

. Because the solution is concentrated, the concentration of

Cl^-

ions is far higher than that of

OH^-

. Consequently, the concentration factor predominates over position in the electrochemical series, resulting in

Cl^-

being preferentially oxidized to chlorine gas (

Cl_2

). At the cathode,

H^+

ions gain electrons to form hydrogen gas (

H_2

).

Step-by-Step Solution

1

Identify the ions present in concentrated hydrochloric acid (

HCl_{(aq)}

).

The cations present are

H^+

(from

HCl

and

H_2O

) and the anions present are

Cl^-

and

OH^-

.

Hydrochloric acid ionizes completely in water into

H^+

and

Cl^-

, alongside minor autoionization of water.

2

Determine the reaction taking place at the anode (positive electrode).

Chloride ions (

Cl^-

) are discharged preferentially to form chlorine gas (

Cl_2

).

Although

OH^-

is higher in the electrochemical series than

Cl^-

, the significantly higher concentration of

Cl^-

in concentrated

HCl

overrides position in the series.

3

Determine the reaction taking place at the cathode (negative electrode).

Hydrogen ions (

H^+

) are discharged to form hydrogen gas (

H_2

).

H^+

is the only cation present in solution, gaining electrons at the cathode.

Key Concept

Concentration effect on preferential discharge of anions during electrolysis

Question 96Question →

Arrange the following anions in order of INCREASING ease of preferential discharge at an inert platinum anode during the electrolysis of dilute solutions, starting from the least easily discharged to the most easily discharged.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of anions in order of increasing ease of discharge at an inert anode is:

NO_3^-

followed by

Cl^-

, then

Br^-

, and finally

OH^-

.

In dilute aqueous solutions using inert electrodes, preferential discharge of anions at the anode is determined by their relative positions in the electrochemical series. The order of increasing ease of discharge (from hardest to easiest) is nitrate (

NO_3^-

), chloride (

Cl^-

), bromide (

Br^-

), and hydroxide (

OH^-

).

Step-by-Step Solution

1

Identify the factor governing preferential discharge

Since the solutions are dilute and electrodes are inert (platinum), preferential discharge depends entirely on the positions of the anions in the electrochemical series.

Concentration effects and electrode nature do not alter the standard relative discharge order in dilute solutions with inert electrodes.

2

Recall the electrochemical series position for anions

The relative positions from highest (hardest to discharge) to lowest (easiest to discharge) are

F^- < SO_4^{2-} < NO_3^- < Cl^- < Br^- < I^- < OH^-

.

Anions lower in the series lose electrons (get oxidized) more readily due to lower standard oxidation potentials.

3

Sequence the given anions from least easily discharged to most easily discharged

The correct sequence is

NO_3^- \rightarrow Cl^- \rightarrow Br^- \rightarrow OH^-

.

Nitrate is positioned highest among the listed ions, followed by chloride, then bromide, with hydroxide positioned lowest.

Key Concept

Position of anions in the electrochemical series determines preferential discharge at the anode in dilute solutions.

Question 97Question →

Match each chemical formula of the inorganic redox species on the left with its correct IUPAC name on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match

KMnO_4

to Potassium tetraoxomanganate(VII),

K_2Cr_2O_7

to Potassium heptaoxodichromate(VI),

NaClO_3

to Sodium trioxochlorate(V), and

Fe_2O_3

to Iron(III) oxide.

Each formula is correctly paired by evaluating the algebraic sum of oxidation numbers to find the oxidation state of the central transition metal or non-metal, then matching the corresponding oxo prefix and Roman numeral according to standard IUPAC conventions.

Step-by-Step Solution

1

Determine the oxidation state of Mn in

KMnO_4

Let oxidation state of Mn be

x

:

(+1) + x + 4(-2) = 0 \implies x - 7 = 0 \implies x = +7

.

The sum of oxidation numbers in a neutral compound is equal to zero. This identifies the Roman numeral for manganese as (VII).

2

Determine the oxidation state of Cr in

K_2Cr_2O_7

Let oxidation state of Cr be

x

:

2(+1) + 2x + 7(-2) = 0 \implies 2x - 12 = 0 \implies x = +6

.

The compound contains seven oxygen atoms (heptaoxo) and two chromium atoms (dichromate), giving chromium an oxidation state of (VI).

3

Determine the oxidation state of Cl in

NaClO_3

Let oxidation state of Cl be

x

:

(+1) + x + 3(-2) = 0 \implies x - 5 = 0 \implies x = +5

.

Three oxygen atoms give the prefix 'trioxo', and chlorine has an oxidation state of (V).

4

Determine the oxidation state of Fe in

Fe_2O_3

Let oxidation state of Fe be

x

:

2x + 3(-2) = 0 \implies 2x - 6 = 0 \implies x = +3

.

Iron has an oxidation state of

+3

, naming the simple binary metal oxide as Iron(III) oxide.

Key Concept

Calculating central atom oxidation numbers and applying IUPAC nomenclature rules for oxoanions, oxoacids, and binary oxides.

Question 98Question →

Match each chemical formula of the inorganic redox species on the left with its corresponding IUPAC name on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are:

HClO_2

matches with Dioxochloric(III) acid;

H_2S_2O_7

matches with Heptaoxodisulfuric(VI) acid;

NaNO_2

matches with Sodium dioxonitrate(III); and

KIO_4

matches with Potassium tetraoxoiodate(VII).

Each chemical species is systematically named by calculating the oxidation state of its central non-metal atom and prefixing the number of oxygen atoms present (dioxo-, tetraoxo-, heptaoxo-).

HClO_2

has chlorine in

+3

state (Dioxochloric(III) acid),

H_2S_2O_7

has sulfur in

+6

state across two atoms (Heptaoxodisulfuric(VI) acid),

NaNO_2

has nitrogen in

+3

state (Sodium dioxonitrate(III)), and

KIO_4

has iodine in

+7

state (Potassium tetraoxoiodate(VII)).

Step-by-Step Solution

1

Determine the oxidation state of chlorine in

HClO_2

.

Assign

+1

to

H

and

-2

to

O

:

(+1) + Cl + 2(-2) = 0 ⇒ Cl = +3

. Combined with two oxo groups, the IUPAC name is Dioxochloric(III) acid.

Oxoacids are named by specifying the number of oxygen atoms with oxo prefixes followed by the central element and its Roman numeral oxidation state.

2

Calculate the oxidation state of sulfur in

H_2S_2O_7

.

Assign

+1

to

H

and

-2

to

O

:

2(+1) + 2(S) + 7(-2) = 0 ⇒ 2S = +12 ⇒ S = +6

. With seven oxygen atoms and two sulfur atoms, the IUPAC name is Heptaoxodisulfuric(VI) acid.

The prefix 'heptaoxo-' accounts for seven oxygens and 'disulfuric' indicates two sulfur atoms.

3

Calculate the oxidation state of nitrogen in

NaNO_2

.

Assign

+1

to

Na

and

-2

to

O

:

(+1) + N + 2(-2) = 0 ⇒ N = +3

. The anion is dioxonitrate(III), making the salt Sodium dioxonitrate(III).

Salts of oxoanions state the cation name first followed by the IUPAC name of the oxoanion.

4

Calculate the oxidation state of iodine in

KIO_4

.

Assign

+1

to

K

and

-2

to

O

:

(+1) + I + 4(-2) = 0 ⇒ I = +7

. The anion is tetraoxoiodate(VII), making the salt Potassium tetraoxoiodate(VII).

Four oxygen atoms dictate the prefix 'tetraoxo-' and the iodine state

+7

gives Roman numeral (VII).

Key Concept

IUPAC Nomenclature and Oxidation State Calculation of Oxoacids and Oxosalts

Question 99Question →

Match each chemical process on the left with the correct classical or modern redox concept on the right that specifically describes the change taking place.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Removal of oxygen from zinc oxide corresponds to classical reduction; addition of oxygen to carbon corresponds to classical oxidation; loss of electrons by sodium corresponds to modern electron-transfer oxidation; and the decrease in oxidation number of nitrogen corresponds to modern oxidation number reduction.

Each chemical transformation is matched according to whether it describes classical oxygen transfer or modern electron/oxidation state changes. Removal of oxygen represents classical reduction; addition of oxygen represents classical oxidation; loss of electrons represents modern electronic oxidation; and a decrease in oxidation number represents modern reduction.

Step-by-Step Solution

1

Analyze classical definitions of redox

Classical oxidation involves the addition of oxygen or removal of hydrogen, while classical reduction involves the removal of oxygen or addition of hydrogen.

This establishes the ground rules for the first two pairs.

2

Analyze modern electronic and oxidation number definitions

Modern oxidation is the loss of electrons (OIL) or an increase in oxidation state. Modern reduction is the gain of electrons (RIG) or a decrease in oxidation state.

This establishes the ground rules for the remaining two pairs.

3

Match each process to its corresponding definition

Zinc oxide losing oxygen is classical reduction. Carbon gaining oxygen is classical oxidation. Sodium losing electrons is modern electronic oxidation. Nitrogen decreasing in oxidation number from 0 to -3 is modern oxidation number reduction.

Applying the concepts directly aligns each process with its primary redox definition.

Key Concept

Classical vs Modern Concepts of Redox Reactions

Oxidation-Reduction and Electrochemistry

Answer

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