Oxidation-Reduction and Electrochemistry Practice Questions — JAMB UTME

Question 1Question →

In the reaction between octasulfur, $S_8$ , and concentrated trioxonitrate(V) acid, sulfur is oxidized to tetraoxosulfate(VI) acid, $H_2SO_4$ . What are the oxidation numbers of sulfur in $S_8$ and $H_2SO_4$ respectively?

0

and

+6

-2

and

+6

+8

and

+6

0

and

+4

Show answer & explanation

Answer: $0$ and $+6$

Answer

The oxidation numbers of sulfur in

S_8

and

H_2SO_4

are

0

and

+6

respectively.

Free uncombined elements carry an oxidation state of zero, so sulfur in

S_8

is

0

. In tetraoxosulfate(VI) acid (

H_2SO_4

), setting the neutral molecule oxidation sum to zero yields

2(+1) + S + 4(-2) = 0

, giving

S = +6

.

Step-by-Step Solution

1

Determine the oxidation state of sulfur in free elemental form (

S_8

).

Oxidation state of sulfur in

S_8 = 0

.

By definition, an element in its free or uncombined state has an oxidation state of zero regardless of its atomicity.

2

Calculate the oxidation state of sulfur in

H_2SO_4

.

Oxidation state of sulfur in

H_2SO_4 = +6

.

Assign

+1

for each hydrogen atom and

-2

for each oxygen atom. Solving

2(+1) + S + 4(-2) = 0

gives

+2 + S - 8 = 0

, hence

S = +6

.

Key Concept

Assigning oxidation states to free elemental forms and central atoms in polyatomic oxoacids

Question 2Question →

Match each redox process description on the left with its corresponding definition type on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Addition of oxygen or removal of hydrogen corresponds to Classical definition of oxidation; Gain of electrons or decrease in oxidation number corresponds to Modern definition of reduction; Loss of electrons or increase in oxidation number corresponds to Modern definition of oxidation; Removal of oxygen or addition of hydrogen corresponds to Classical definition of reduction.

Classical concepts define oxidation as gaining oxygen or losing hydrogen, and reduction as losing oxygen or gaining hydrogen. Modern electronic concepts define oxidation as losing electrons (increasing oxidation state) and reduction as gaining electrons (decreasing oxidation state).

Step-by-Step Solution

1

Analyze classical definitions of redox reactions.

Classical oxidation involves adding oxygen or removing hydrogen, while classical reduction involves removing oxygen or adding hydrogen.

Classical redox concepts were based historically on element transfer, mainly oxygen and hydrogen.

2

Analyze modern electronic definitions of redox reactions.

Modern oxidation involves losing electrons (increase in oxidation number), while modern reduction involves gaining electrons (decrease in oxidation number).

Modern redox concepts expand the definitions to apply universally via electron transfer and oxidation state changes.

Key Concept

Classical (oxygen/hydrogen transfer) vs. Modern (electron transfer/oxidation state) definitions of oxidation and reduction.

Question 3Question →

Galvanizing protects iron from rusting by coating it with zinc, which acts as a sacrificial anode because zinc is more electropositive than iron.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. Zinc is more electropositive than iron and acts as a sacrificial anode to prevent rusting.

Zinc has a higher oxidation potential than iron. During exposure to atmospheric moisture and oxygen, zinc corrodes preferentially, sacrificing itself to prevent the oxidation of iron.

Step-by-Step Solution

1

Identify the relative positions of zinc and iron in the reactivity/electrochemical series.

Zinc is more electropositive (more reactive) than iron.

Metals higher in the reactivity series lose electrons more easily.

2

Analyze the mechanism of galvanization.

Zinc oxidizes preferentially to form zinc ions (

Zn^{2+}

) while protecting iron from oxidation (

Fe^{2+}

).

The more reactive metal serves as a sacrificial anode in an electrochemical corrosion cell.

Key Concept

Galvanization and Sacrificial Protection

Question 4Question →

What time, in seconds, is required to deposit $4.5\text{ g}$ of aluminium at the cathode during the electrolysis of molten aluminium oxide using a constant current of $5.0\text{ A}$ ? $[\text{Al} = 27,\, 1\text{ F} = 96,500\text{ C mol}^{-1}]$

9650\text{ s}

3217\text{ s}

6433\text{ s}

86850\text{ s}

Show answer & explanation

Answer: $9650\text{ s}$

Answer

The time required to deposit

4.5\text{ g}

of aluminium is

9650\text{ s}

.

The deposit of

4.5\text{ g}

of

\text{Al}

corresponds to

0.1667\text{ mol}

. Since

\text{Al}^{3+}

requires 3 electrons per atom deposited (

z = 3

), the total charge needed is

0.5\text{ F} = 48,250\text{ C}

. Dividing this charge by a current of

5.0\text{ A}

gives

9650\text{ seconds}

.

Step-by-Step Solution

1

Determine the moles of aluminium deposited and the moles of electrons transferred.

Moles of

\text{Al} = \frac{4.5\text{ g}}{27\text{ g mol}^{-1}} = 0.1667\text{ mol}

. Reduction half-reaction:

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}

. Therefore,

z = 3

moles of electrons are required per mole of

\text{Al}

.

Aluminium is a trivalent metal (

z = 3

), so depositing 1 mole of

\text{Al}

requires 3 Faradays of charge.

2

Calculate the total quantity of electricity (

Q

) required in coulombs.

Q = n_{e^-} \times F = (0.1667\text{ mol} \times 3) \times 96,500\text{ C mol}^{-1} = 0.5 \times 96,500 = 48,250\text{ C}

.

Total charge is the product of moles of electrons transferred and Faraday's constant.

3

Calculate the required time (

t

) using

Q = I \times t

.

t = \frac{Q}{I} = \frac{48,250\text{ C}}{5.0\text{ A}} = 9650\text{ s}

.

Dividing total charge by current yields time in seconds.

Key Concept

Quantitative application of Faraday's first and second laws of electrolysis for multivalent ions.

Estimated Time:1m 30s

Question 5Question →

In ammonium trioxonitrate(V), $NH_4NO_3$ , nitrogen exists in two distinct ionic environments. What are the respective oxidation numbers of the nitrogen atom in the cation and the nitrogen atom in the anion, and what is the systematic IUPAC name of the nitrogen-containing oxoanion formed when the anion is reduced by gaining two electrons?

-3

and

+5

; dioxonitrate(III) ion

+1

and

+5

; trioxonitrate(III) ion

0

and

+5

; dioxonitrate(III) ion

-3

and

+3

; trioxonitrate(V) ion

Show answer & explanation

Answer: $-3$ and $+5$ ; dioxonitrate(III) ion

Answer

The respective oxidation numbers are

-3

for nitrogen in the ammonium cation and

+5

for nitrogen in the trioxonitrate(V) anion. The reduced species formed upon gaining two electrons is the dioxonitrate(III) ion.

In the ionic compound

NH_4NO_3

, the nitrogen atom in the cation

NH_4^+

has an oxidation state of

-3

because

x + 4(+1) = +1

. The nitrogen atom in the anion

NO_3^-

has an oxidation state of

+5

because

y + 3(-2) = -1

. When

NO_3^-

is reduced by two electrons, the oxidation number of nitrogen decreases from

+5

to

+3

, converting

NO_3^-

to

NO_2^-

. In IUPAC nomenclature,

NO_2^-

is named the dioxonitrate(III) ion due to its two oxo ligands and nitrogen oxidation state of

+3

.

Step-by-Step Solution

1

Determine the oxidation state of nitrogen in the ammonium cation (

NH_4^+

).

Let the oxidation state of N be

x

. Since hydrogen is

+1

and the net charge is

+1

:

x + 4(+1) = +1 \implies x = -3

.

Ammonium is a polyatomic cation where the sum of oxidation numbers equals the net ionic charge.

2

Determine the oxidation state of nitrogen in the trioxonitrate(V) anion (

NO_3^-

).

Let the oxidation state of N be

y

. Since oxygen is

-2

and the net charge is

-1

:

y + 3(-2) = -1 \implies y = +5

.

Nitrate is a polyatomic anion where oxygen exhibits an oxidation number of

-2

.

3

Calculate the oxidation state of nitrogen after two-electron reduction of the anion and determine its IUPAC name.

Reduction by gaining 2 electrons decreases the oxidation number of nitrogen:

+5 - 2 = +3

. The resulting species containing nitrogen in the

+3

state with two oxygen atoms (

NO_2^-

) is systematically named the dioxonitrate(III) ion.

Reduction corresponds to a gain of electrons (decrease in oxidation number), and IUPAC rules for oxoanions specify the prefix for oxygen count ('dioxo-') followed by the central element and its oxidation state in Roman numerals.

Key Concept

Polyatomic Ion Oxidation Numbers and IUPAC Nomenclature of Oxoanions

Question 6Question →

A current of $2.0\text{ A}$ is passed through an aqueous solution of copper(II) tetraoxosulfate(VI) for $965\text{ seconds}$ . What is the mass of copper deposited at the cathode? $[F = 96,500\text{ C mol}^{-1},\text{ Cu} = 64]$

0.64\text{ g}

1.28\text{ g}

2.56\text{ g}

0.32\text{ g}

Show answer & explanation

Answer: $0.64\text{ g}$

Answer

The mass of copper deposited at the cathode is

0.64\text{ g}

.

The total charge passed is

Q = 2.0\text{ A} \times 965\text{ s} = 1930\text{ C}

. Dividing by Faraday's constant yields

0.02\text{ mol}

of electrons. Because copper(II) ions require

2\text{ moles}

of electrons per mole of copper metal (

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

), the amount of copper deposited is

0.01\text{ mol}

, which corresponds to

0.01 \times 64 = 0.64\text{ g}

.

Step-by-Step Solution

1

Calculate total quantity of electricity (

Q

) passed

Q = I \times t = 2.0\text{ A} \times 965\text{ s} = 1930\text{ C}

Faraday's first law relates charge to current and time.

2

Calculate the moles of electrons transferred

\text{Moles of } e^- = \frac{Q}{F} = \frac{1930\text{ C}}{96,500\text{ C mol}^{-1}} = 0.02\text{ mol } e^-

One mole of electrons carries one Faraday (

96,500\text{ C}

) of charge.

3

Determine moles of copper deposited using electrode reduction half-equation

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \implies n(\text{Cu}) = \frac{0.02\text{ mol } e^-}{2} = 0.01\text{ mol}

Reduction of one

\text{Cu}^{2+}

ion requires

2

moles of electrons per mole of copper deposited.

4

Calculate mass of copper deposited

\text{Mass} = n \times \text{Molar Mass} = 0.01\text{ mol} \times 64\text{ g mol}^{-1} = 0.64\text{ g}

Multiplying moles of substance by its molar mass yields mass.

Key Concept

Faraday's First Law of Electrolysis and Quantitative Stoichiometry of Charge

Estimated Time:1m 0s

Question 7Question →

The standard reduction potentials for iron and copper half-cells are given as follows:

\text{Fe}^{2+}(aq) + 2e^- \rightarrow \text{Fe}(s) \quad E^\circ = -0.44\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 overall reaction $\text{Fe}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Cu}(s)$ , and is the reaction spontaneous under standard conditions?

+0.78\text{ V}

, spontaneous

-0.78\text{ V}

, non-spontaneous

-0.10\text{ V}

, non-spontaneous

+0.10\text{ V}

, spontaneous

Show answer & explanation

Answer: $+0.78\text{ V}$ , spontaneous

Answer

The standard electromotive force (

E^\circ_{\text{cell}}

) is

+0.78\text{ V}

, and the reaction is spontaneous.

In the given reaction,

\text{Cu}^{2+}

ions are reduced to copper metal at the cathode (

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

), while

\text{Fe}

metal is oxidized to

\text{Fe}^{2+}

ions at the anode (

E^\circ = -0.44\text{ V}

). Using the standard formula

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

, we calculate

E^\circ_{\text{cell}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}

. Because the cell potential is positive, the reaction is spontaneous under standard conditions.

Step-by-Step Solution

1

Identify the reduction and oxidation half-reactions

Cathode (reduction):

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

with

E^\circ_{\text{cathode}} = +0.34\text{ V}

. Anode (oxidation):

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

with

E^\circ_{\text{anode}} = -0.44\text{ V}

.

Copper ions gain electrons (reduction at cathode) while iron metal loses electrons (oxidation at anode).

2

Calculate the standard cell potential (

E^\circ_{\text{cell}}

)

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}

.

The cell potential is the difference between the reduction potential of the cathode species and that of the anode species.

3

Determine reaction spontaneity

Since

E^\circ_{\text{cell}} = +0.78\text{ V} > 0

, the reaction is spontaneous.

A positive standard cell potential indicates a thermodynamically feasible (spontaneous) redox reaction under standard conditions.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Estimated Time:1m 15s

Question 8Question →

Match each standard reduction half-reaction on the left with its corresponding property regarding reducing/oxidizing strength or reaction spontaneity on the right. Which pairs correctly match each half-reaction with its chemical behavior?

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match

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

with being a stronger reducing agent than hydrogen that displaces

\text{H}_2

from acid,

\text{Ag}^+/\text{Ag}

with a metal that cannot displace hydrogen,

\text{2H}^+/\text{H}_2

with the standard reference electrode, and

\text{F}_2/\text{F}^-

with the strongest oxidizing agent.

Zinc has a negative reduction potential and displaces hydrogen from acid; silver has a positive reduction potential and cannot displace hydrogen; hydrogen serves as the reference potential at zero; fluorine gas possesses the highest positive reduction potential, functioning as the strongest oxidizing agent.

Step-by-Step Solution

1

Examine standard reduction potential (

E^\circ

) values

Higher positive values indicate a stronger tendency to gain electrons (stronger oxidizing agent). Negative values indicate that the reduced form easily loses electrons (stronger reducing agent).

Standard reduction potentials dictate relative oxidizing/reducing strength and reaction feasibility.

2

Relate

E^\circ

values to hydrogen displacement and spontaneity

Metals with

E^\circ < 0.00\text{ V}

(like

\text{Zn}

) spontaneously displace

\text{H}_2

from acids. Metals with

E^\circ > 0.00\text{ V}

(like

\text{Ag}

) do not.

A reaction is spontaneous when the overall standard cell potential

E^\circ_{\text{cell}}

is positive.

3

Match each half-reaction to its appropriate description

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

matches with displacing

\text{H}_2

;

\text{Ag}^+/\text{Ag}

matches with inability to displace

\text{H}_2

;

\text{2H}^+/\text{H}_2

matches with the zero reference electrode;

\text{F}_2/\text{F}^-

matches with the strongest oxidizing agent.

Each standard reduction potential maps directly to these electrochemical behaviors.

Key Concept

Electrochemical Series and Reaction Spontaneity

Question 9Question →

An electric current of $5.00\text{ A}$ is passed through an aqueous solution of a metal chloride using inert electrodes for $3860\text{ seconds}$ . Calculate the volume of chlorine gas, in $\text{dm}^3$ , liberated at standard temperature and pressure (STP). (Take $1\text{ Faraday} = 96,500\text{ C mol}^{-1}$ , Molar volume of gas at $\text{STP} = 22.4\text{ dm}^3\text{ mol}^{-1}$ )

Show answer & explanation

Answer: 2.24

Answer

The volume of chlorine gas liberated at STP is

2.24\text{ dm}^3

.

Passing

5.00\text{ A}

for

3860\text{ s}

transfers

19,300\text{ C}

of charge, corresponding to

0.200\text{ mol}

of electrons. Because oxidation of chloride ions (

2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-

) requires

2\text{ moles}

of electrons per mole of diatomic chlorine gas,

0.100\text{ mol}

of

\text{Cl}_2

gas is produced. At STP, this occupies

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

.

Step-by-Step Solution

1

Calculate the total electric charge passed during electrolysis.

Q = 19,300\text{ C}

Using

Q = I \times t

, where current

I = 5.00\text{ A}

and time

t = 3860\text{ s}

.

2

Determine the amount of substance of electrons transferred in moles.

n(e^-) = 0.200\text{ mol}

Dividing total charge by Faraday's constant (

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

).

3

Apply stoichiometric ratio from the anode half-reaction to find moles of chlorine gas.

n(\text{Cl}_2) = 0.100\text{ mol}

The reaction

2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-

indicates that

2\text{ moles}

of electrons liberate

1\text{ mole}

of

\text{Cl}_2

.

4

Calculate the volume of chlorine gas produced at STP.

V = 2.24\text{ dm}^3

Multiplying the moles of

\text{Cl}_2

by the molar volume of gas at STP (

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

).

Key Concept

Faraday's laws of electrolysis applied to gas volume calculations at STP

Question 10Question →

Calculate the volume of oxygen gas, in $\text{cm}^3$ measured at STP, liberated at the anode during the electrolysis of dilute tetraoxosulfate(VI) acid when a steady current of $1.93\text{ A}$ is passed through the electrolyte for $50\text{ minutes}$ .

[Take Faraday's constant $F = 96500\text{ C mol}^{-1}$ , Molar volume of gas at STP $= 22.4\text{ dm}^3\text{ mol}^{-1}$ ]

Show answer & explanation

Answer: 336

Answer

The volume of oxygen gas liberated at STP is

336\text{ cm}^3

.

Passing a current of

1.93\text{ A}

for

3000\text{ s}

delivers

5790\text{ C}

of charge, equivalent to

0.06\text{ moles}

of electrons. Because the anodic discharge of hydroxide ions (

4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^-

) requires

4\text{ moles}

of electrons per mole of

\text{O}_2

,

0.015\text{ moles}

of

\text{O}_2

are generated. Multiplying by the molar volume at STP (

22400\text{ cm}^3\text{mol}^{-1}

) gives

336\text{ cm}^3

.

Step-by-Step Solution

1

Convert the duration of electrolysis from minutes to seconds

t = 50\text{ min} \times 60\text{ s/min} = 3000\text{ s}

Current calculations require time in SI units (seconds).

2

Calculate the total electric charge passed through the electrolyte

Q = I \times t = 1.93\text{ A} \times 3000\text{ s} = 5790\text{ C}

Charge passed is the product of electric current and time.

3

Calculate the quantity of electrons passed in moles

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

One mole of electrons corresponds to 1 Faraday (

96500\text{ C}

).

4

Use the anodic half-reaction equation to determine the molar ratio of electrons to oxygen gas

4\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g) + 4e^-

;

n(\text{O}_2) = \frac{0.06\text{ mol}}{4} = 0.015\text{ mol}

The discharge of hydroxide ions requires 4 moles of electrons per mole of oxygen gas evolved.

5

Calculate the volume of liberated oxygen gas at STP in

\text{cm}^3

V = 0.015\text{ mol} \times 22400\text{ cm}^3\text{ mol}^{-1} = 336\text{ cm}^3

One mole of gas occupies

22.4\text{ dm}^3 = 22400\text{ cm}^3

at standard temperature and pressure.

Key Concept

Quantitative electrochemistry using Faraday's laws and stoichiometric electron-to-gas relationships at electrodes.

Question 11Question →

Calculate the quantity of electricity, in Coulombs, required to deposit $0.108\text{ g}$ of silver at the cathode during the electrolysis of silver trioxonitrate(V) solution. ( $\text{Ag} = 108\text{ g/mol}$ , $1\text{ F} = 96,500\text{ C/mol}$ )

Show answer & explanation

Answer: 96.5

Answer

The quantity of electricity required is

96.5\text{ C}

.

Depositing

0.108\text{ g}

of silver (molar mass

108\text{ g/mol}

) requires

0.001\text{ moles}

of silver atoms. According to the cathodic reaction

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

,

1\text{ mole}

of electrons (

96,500\text{ C}

) is required to deposit

1\text{ mole}

of

\text{Ag}

. Therefore, the total charge required is

0.001 \times 96,500\text{ C} = 96.5\text{ C}

.

Step-by-Step Solution

1

Calculate the number of moles of silver deposited

\text{Moles of Ag} = \frac{0.108\text{ g}}{108\text{ g/mol}} = 0.001\text{ mol}

Number of moles is calculated by dividing mass by molar mass.

2

Determine the quantity of electricity (charge) required

Q = 0.001\text{ mol} \times 96,500\text{ C/mol} = 96.5\text{ C}

The reduction reaction

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

shows

1\text{ mole}

of electrons (

1\text{ F} = 96,500\text{ C}

) deposits

1\text{ mole}

of silver.

Key Concept

Faraday's First Law of Electrolysis and Quantitative Mass-Charge Relationship

Question 12Question →

During the electrolysis of dilute tetraoxosulfate(VI) acid, a steady current of $2.50\text{ A}$ is passed through the electrolyte for $1930\text{ seconds}$ . What volume of hydrogen gas, measured at STP, is liberated at the cathode?

$[1\text{ Faraday} = 96500\text{ C mol}^{-1}, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]$

0.56\text{ dm}^3

0.60\text{ dm}^3

1.12\text{ dm}^3

1.20\text{ dm}^3

Show answer & explanation

Answer: $0.56\text{ dm}^3$

Answer

The volume of hydrogen gas liberated at STP is

0.56\text{ dm}^3

.

The correct answer is derived by calculating the electric charge passed (

Q = 2.50 \times 1930 = 4825\text{ C}

), converting to moles of electrons (

0.050\text{ mol}

), applying the electrode reduction stoichiometry (

2\text{ mol } e^-

per

1\text{ mol } H_2

), and converting the resulting

0.025\text{ mol } H_2

to volume at STP using

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

, which yields

0.56\text{ dm}^3

.

Step-by-Step Solution

1

Calculate total quantity of electricity (

Q

) passed.

Q = I \times t = 2.50\text{ A} \times 1930\text{ s} = 4825\text{ C}

Faraday's first law states quantity of charge is the product of current and time in seconds.

2

Calculate the amount of electrons in moles transferred.

n(e^-) = \frac{Q}{F} = \frac{4825\text{ C}}{96500\text{ C mol}^{-1}} = 0.050\text{ mol}

One Faraday (

96500\text{ C}

) represents one mole of electrons.

3

Determine moles of

H_2

gas produced using the cathode half-reaction stoichiometry.

Cathode reaction:

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

. Moles of

H_2 = \frac{0.050\text{ mol } e^-}{2} = 0.025\text{ mol } H_2

Two moles of electrons are required to reduce hydrogen ions to produce one mole of hydrogen gas.

4

Calculate the volume of hydrogen gas at STP.

V = 0.025\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 0.56\text{ dm}^3

One mole of any ideal gas occupies

22.4\text{ dm}^3

at STP.

Key Concept

Quantitative application of Faraday's Laws of Electrolysis for gas evolution at electrodes.

Estimated Time:1m 30s

Question 13Question →

The standard reduction potentials for three half-cells are given below:

\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s) \quad E^\circ = -0.74\text{ V}

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

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

A chemistry student intends to store an aqueous solution of iron(II) tetraoxonitrate(V), $\text{Fe(NO}_3)_2$ , in metal containers. Based on standard electrode potentials, which container choice is suitable for safely storing the solution without undergoing a spontaneous redox reaction?

Only the Silver container

Only the Chromium container

Both the Chromium and Silver containers

Neither container

Show answer & explanation

Answer: Only the Silver container

Answer

Only the Silver container can safely store the iron(II) tetraoxonitrate(V) solution.

A redox reaction is spontaneous if the standard cell potential is positive (E°cell > 0). When storing aqueous iron(II) ions in a Silver container, the potential reaction involves oxidation of Silver metal and reduction of iron(II) ions: E°cell = E°(reduction) - E°(oxidation) = -0.44 V - (+0.80 V) = -1.24 V. Because E°cell is negative, the reaction cannot occur spontaneously, making the Silver container suitable. Conversely, for Chromium, E°cell = -0.44 V - (-0.74 V) = +0.30 V, which is positive and causes a spontaneous reaction that destroys the Chromium container.

Step-by-Step Solution

1

Determine the cell potential E°cell for storing iron(II) solution in a Chromium container.

E°cell = E°cathode - E°anode = E°(Fe²⁺/Fe) - E°(Cr³⁺/Cr) = -0.44 V - (-0.74 V) = +0.30 V.

Chromium acts as the anode (oxidation) and iron(II) as the cathode (reduction).

2

Evaluate spontaneity for the Chromium container reaction.

Since E°cell = +0.30 V > 0, the reaction is spontaneous.

A positive standard cell potential indicates a feasible redox reaction, meaning the Chromium container will dissolve and react with the solution.

3

Determine the cell potential E°cell for storing iron(II) solution in a Silver container.

E°cell = E°(Fe²⁺/Fe) - E°(Ag⁺/Ag) = -0.44 V - (+0.80 V) = -1.24 V.

Silver acts as the anode (oxidation) and iron(II) as the cathode (reduction).

4

Evaluate spontaneity for the Silver container reaction and conclude safe storage.

Since E°cell = -1.24 V < 0, the reaction is non-spontaneous, so the Silver container safely holds the solution.

A negative cell potential means no spontaneous reaction occurs between Silver metal and aqueous iron(II) ions.

Key Concept

Spontaneity of Redox Reactions and Standard Cell Potential

Question 14Question →

Match each electrolytic setup to the primary factor or mechanism governing the electrode reaction during electrolysis.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair each electrolytic process with its underlying discharge factor: concentrated brine anode discharge is governed by ion concentration; active copper anode reaction is governed by electrode nature; dilute acid anode discharge depends on electrochemical series position; and molten salt electrolysis operates in the absence of competing ions.

Each electrolytic scenario is correctly matched to the principal factor governing its reaction: concentrated aqueous

NaCl

anode output is determined by ion concentration; copper electrode electrolysis depends on active electrode participation; dilute

H_2SO_4

anode output is dictated by relative positions in the electrochemical series; and molten

NaCl

involves discharge without competing aqueous ions.

Step-by-Step Solution

1

Examine the electrolysis of concentrated

NaCl_{(aq)}

at the anode

High concentration of

Cl^-

overrides the electrochemical series position of

OH^-

.

Concentration factor dominates when halide ion concentration is high.

2

Examine the electrolysis of

CuSO_{4(aq)}

with copper electrodes

The copper anode dissolves into

Cu^{2+}

ions.

An active electrode participates chemically in the reaction instead of inert anion discharge.

3

Examine the electrolysis of dilute

H_2SO_{4(aq)}

with inert electrodes

OH^-

is discharged in preference to

SO_4^{2-}

to give oxygen gas.

OH^-

is positioned higher in the electrochemical series than

SO_4^{2-}

.

4

Examine the electrolysis of molten

NaCl_{(l)}

Na^+

ions are discharged at the cathode.

Without water, there are no competing

H^+

ions present.

Key Concept

Factors governing the preferential discharge of ions during electrolysis (concentration of ions, nature of electrodes, and position in the electrochemical series)

Question 15Question →

In the industrial refining of copper by electrolysis, a block of impure copper is used as the anode in an aqueous copper(II) tetraoxosulfate(VI) electrolyte. Which of the following half-equations correctly represents the primary reaction taking place at the anode?

Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-

Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}

4OH^-_{(aq)} \rightarrow 2H_2O_{(l)} + O_{2(g)} + 4e^-

2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}

Show answer & explanation

Answer: $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-$

Answer

The primary reaction occurring at the anode is the oxidation of copper metal into copper(II) ions:

Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-

.

During the industrial electrorefining of copper, an active copper anode is oxidized by losing two electrons per atom, dissolving into the electrolyte as copper(II) ions (

Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-

). This maintains the concentration of copper(II) ions in solution while transferring pure copper onto the cathode.

Step-by-Step Solution

1

Identify the general electrode process at the positive terminal (anode).

Oxidation (loss of electrons) always occurs at the anode.

Electrons are removed from species at the anode during electrolysis.

2

Determine whether the anode electrode material is inert or active (reactive).

The anode is made of impure copper metal, which is an active electrode.

Active metal anodes dissolve into solution more readily than anions in the solution can discharge.

3

Write the oxidation half-reaction for the dissolution of the copper anode.

Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-

Neutral copper atoms in the impure block lose two electrons each to enter the solution as soluble copper(II) cations.

Key Concept

Electrorefining of copper and active anode oxidation

Question 16Question →

During the industrial extraction of aluminium using the Hall-Héroult process, a steady current of $96.5\text{ A}$ is passed through an electrolytic cell containing molten alumina ( $Al_2O_3$ ) dissolved in molten cryolite for $5.0\text{ hours}$ . What is the mass of pure aluminium, in grams, deposited at the cathode?

(Take $1\text{ Faraday} = 96,500\text{ C mol}^{-1}$ , Relative atomic mass: $\text{Al} = 27$ )

Show answer & explanation

Answer: 162

Answer

The mass of pure aluminium deposited at the cathode is

162\text{ g}

.

Using

Q = I \times t

, a current of

96.5\text{ A}

for

5.0\text{ hours}

(

18,000\text{ s}

) yields a total charge of

1,737,000\text{ C}

, which equals

18\text{ Faradays}

(

18\text{ moles of electrons}

). Since the reduction of

\text{Al}^{3+}

to

\text{Al}

requires 3 electrons per atom (

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}

),

18\text{ moles}

of electrons liberate

6\text{ moles}

of aluminium metal. Multiplying by the molar mass of aluminium (

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

) gives

162\text{ g}

.

Step-by-Step Solution

1

Convert time to seconds and calculate total electric charge transferred

Q = 96.5\text{ A} \times (5.0 \times 3600\text{ s}) = 1,737,000\text{ C}

Electric charge is defined as current multiplied by time in seconds (

Q = I \times t

).

2

Calculate the moles of electrons transferred using Faraday's constant

Moles of

e^- = \frac{1,737,000\text{ C}}{96,500\text{ C mol}^{-1}} = 18\text{ moles}

One Faraday (

96,500\text{ C}

) corresponds to the charge carried by one mole of electrons.

3

Relate moles of electrons to moles of aluminium deposited using the half-equation

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}_{(s)}

, so

3\text{ mol } e^-

produces

1\text{ mol Al}

. Moles of

\text{Al} = \frac{18}{3} = 6\text{ moles}

Aluminium ion

\text{Al}^{3+}

requires three electrons for reduction to metallic aluminium.

4

Multiply moles of aluminium by its relative atomic mass

\text{Mass} = 6\text{ mol} \times 27\text{ g mol}^{-1} = 162\text{ g}

Mass equals molar amount multiplied by molar mass.

Key Concept

Quantitative application of Faraday's laws of electrolysis in the extraction of metals

Question 17Question →

Food containers are commonly manufactured from iron sheets coated with a thin protective layer of tin. If the protective tin coating is deeply scratched to expose both metals to damp atmosphere, which of the following electrochemical outcomes will occur?

Iron will rust more rapidly than un-plated iron because iron is more electropositive than tin and acts as the anode.

Tin will act as a sacrificial anode and dissolve preferentially to protect the exposed iron core.

Rusting is completely prevented because tin forms an amphoteric oxide layer over the exposed region.

Iron will gain electrons from tin and undergo reduction, preventing oxidation at the metal junction.

Show answer & explanation

Answer: Iron will rust more rapidly than un-plated iron because iron is more electropositive than tin and acts as the anode.

Answer

Iron will rust more rapidly than un-plated iron because iron is more electropositive than tin and acts as the anode.

In the reactivity series, iron is placed above tin. When a tin-plated iron surface is scratched, an electrochemical cell is formed in the presence of atmospheric water and oxygen. Because iron is more electropositive than tin, iron preferentially loses electrons (undergoes oxidation) and acts as the anode. This leads to accelerated rusting of the iron substrate compared to un-plated iron.

Step-by-Step Solution

1

Compare the standard electrode potentials (reactivity) of iron (

Fe

) and tin (

Sn

).

Fe

has a more negative standard reduction potential (

E^\circ = -0.44\text{ V}

) than

Sn

(

E^\circ = -0.14\text{ V}

), meaning iron is more electropositive (more reactive).

The more electropositive metal in electrical contact acts as the anode in the presence of an electrolyte.

2

Identify the anodic and cathodic reactions taking place at the scratched junction.

Iron oxidizes at the anode (

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

) while oxygen and water are reduced at the tin cathode surface (

O_{2(g)} + 2H_2O_{(l)} + 4e^- \rightarrow 4OH^-_{(aq)}

).

Since iron is the anode, its oxidation rate is accelerated due to the large cathodic area of tin relative to the small exposed iron anode area.

Key Concept

Tinning vs Galvanizing in Corrosion Prevention

Estimated Time:1m 0s

Question 18Question →

An electrolytic cell containing concentrated sodium chloride solution initially utilizes inert platinum electrodes, resulting in gas evolution at both the cathode and the anode. If the platinum cathode is replaced with a liquid mercury cathode while maintaining all other experimental conditions, which product is preferentially formed at the cathode, and what primary factor governs this change?

Sodium metal (as sodium amalgam), governed by the nature of the mercury electrode

Hydrogen gas, governed by the position of hydrogen below sodium in the electrochemical series

Sodium metal (as sodium amalgam), governed by the high concentration of sodium ions in the electrolyte

Chlorine gas, governed by the high concentration of chloride ions migrating to the cathode

Show answer & explanation

Answer: Sodium metal (as sodium amalgam), governed by the nature of the mercury electrode

Answer

Sodium metal (as sodium amalgam), governed by the nature of the mercury electrode

When a mercury cathode is used in the electrolysis of concentrated sodium chloride solution, sodium ions (

Na^+

) are preferentially discharged over hydrogen ions (

H^+

) because sodium dissolves in mercury to form a stable amalgam (

Na/Hg

). This interaction lowers the discharge energy for

Na^+

, making the nature of the electrode the determining factor.

Step-by-Step Solution

1

Identify the ions present in concentrated

NaCl(aq)

Cations:

Na^+

and

H^+

; Anions:

Cl^-

and

OH^-

Water partially ionizes to provide

H^+

and

OH^-

, while dissolved

NaCl

provides

Na^+

and

Cl^-

.

2

Analyze the baseline reaction at an inert platinum cathode

H^+

is preferentially discharged over

Na^+

to produce

H_2(g)

According to the electrochemical series,

H^+

has a higher reduction potential (is easier to reduce) than

Na^+

at inert electrodes.

3

Evaluate the effect of replacing the platinum cathode with a liquid mercury cathode

Na^+

is preferentially discharged over

H^+

, forming sodium amalgam (

Na/Hg

)

The chemical affinity between sodium and mercury lowers the overpotential/discharge potential required to reduce

Na^+

. Thus, the nature of the electrode overrides the standard position in the electrochemical series.

Key Concept

Factors affecting preferential discharge of ions during electrolysis: Nature of electrode

Estimated Time:1m 30s

Question 19Question →

Consider the standard reduction potentials for the following two half-cell reactions:

\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s) \quad E^\circ = -0.76\text{ V}

\text{Fe}^{3+}(aq) + 3e^- \rightarrow \text{Fe}(s) \quad E^\circ = -0.04\text{ V}

What is the standard cell potential (

E^\circ_{\text{cell}}

), in volts, for the spontaneous redox reaction represented by the balanced chemical equation:

3\text{Zn}(s) + 2\text{Fe}^{3+}(aq) \rightarrow 3\text{Zn}^{2+}(aq) + 2\text{Fe}(s)

Show answer & explanation

Answer: 0.72

Answer

The standard cell potential for the spontaneous reaction is +0.72 V.

To calculate the standard cell potential (

E^\circ_{\text{cell}}

), identify the cathode (reduction) and anode (oxidation) processes from the balanced chemical equation. Iron(III) ions are reduced to iron metal at the cathode (

E^\circ = -0.04\text{ V}

), while zinc metal is oxidized to zinc ions at the anode (

E^\circ = -0.76\text{ V}

). Using

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

, we calculate

E^\circ_{\text{cell}} = -0.04\text{ V} - (-0.76\text{ V}) = +0.72\text{ V}

. Because standard electrode potential is an intensive property, the stoichiometric coefficients (3 for Zn and 2 for Fe³⁺) do not alter the half-cell potentials.

Step-by-Step Solution

1

Determine the oxidation and reduction species from the overall equation

Zinc is oxidized at the anode (

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

), and

\text{Fe}^{3+}

is reduced at the cathode (

\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}

).

The equation shows elemental Zn losing electrons to form

\text{Zn}^{2+}

and

\text{Fe}^{3+}

gaining electrons to form Fe.

2

Recall that standard electrode potential is an intensive property

The values

E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}

and

E^\circ(\text{Fe}^{3+}/\text{Fe}) = -0.04\text{ V}

remain unchanged regardless of stoichiometric coefficients.

Potential measures electrical potential energy per unit charge, which does not depend on the total amount of substance reacting.

3

Calculate the standard cell potential using

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

E^\circ_{\text{cell}} = -0.04\text{ V} - (-0.76\text{ V}) = +0.72\text{ V}

.

Subtracting the anode reduction potential from the cathode reduction potential yields the net electromotive force of the spontaneous cell.

Key Concept

Standard Cell Potential Calculation and Independence of E° from Stoichiometric Coefficients

Question 20Question →

Calculate the quantity of electricity, in Coulombs, transferred when a steady electric current of $5.0\text{ A}$ is passed through an electrolytic cell for $20\text{ minutes}$ .

Show answer & explanation

Answer: 6000

Answer

The quantity of electricity transferred is

6000\text{ C}

.

According to Faraday's laws of electrolysis, the total charge

Q

passed through an electrolyte is calculated by

Q = I \times t

. Converting

20\text{ minutes}

to seconds gives

20 \times 60 = 1200\text{ s}

. Multiplying by the current

5.0\text{ A}

gives

Q = 5.0 \times 1200 = 6000\text{ C}

.

Step-by-Step Solution

1

Convert time from minutes to seconds

t = 1200\text{ s}

Electric current in Amperes measures charge per second, so time must be converted to seconds.

2

Calculate electric charge using

Q = I \times t

Q = 6000\text{ C}

The quantity of electricity (

Q

) in Coulombs equals current (

I

) in Amperes multiplied by time (

t

) in seconds.

Key Concept

Calculation of Quantity of Electricity (

Q = I \times t

)

Oxidation-Reduction and Electrochemistry

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