Oxidation-Reduction and Electrochemistry Practice Questions — JAMB UTME — Page 2

Match each redox concept on the left with its corresponding definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Classical oxidation corresponds to addition of oxygen or removal of hydrogen; classical reduction corresponds to removal of oxygen or addition of hydrogen; modern oxidation corresponds to loss of electrons or increase in oxidation number; modern reduction corresponds to gain of electrons or decrease in oxidation number.

Each concept correctly aligns with its historical or electronic definition: classical oxidation gains oxygen/loses hydrogen, classical reduction loses oxygen/gains hydrogen, modern oxidation loses electrons/increases oxidation state, and modern reduction gains electrons/decreases oxidation state.

Step-by-Step Solution

1

Identify the classical definitions of oxidation and reduction.

Classical oxidation involves adding oxygen or removing hydrogen. Classical reduction involves removing oxygen or adding hydrogen.

Classical chemistry based redox concepts strictly on the movement of oxygen and hydrogen atoms.

2

Identify the modern electronic definitions of oxidation and reduction.

Modern oxidation involves losing electrons or increasing oxidation number (OIL: Oxidation Is Loss). Modern reduction involves gaining electrons or decreasing oxidation number (RIG: Reduction Is Gain).

Modern chemistry expanded redox to include all electron transfer processes regardless of whether oxygen or hydrogen is present.

3

Match each left term to its exact definition on the right.

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

Aligning concepts with their foundational definitions ensures accurate distinction between classical and modern theories.

Key Concept

Classical vs Modern Concepts of Oxidation and Reduction

Question 22Question →

Complete the reduction half-reaction by identifying the correct coefficient for the electrons needed to balance the charge.

Fill in the blanks below

In the balanced reduction half-reaction in acidic medium: $\text{MnO}_4^- + 8\text{H}^+ + e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$, the missing coefficient for the electrons is .

Show answer & explanation

Answer

5

The total charge on the left side of the equation is

+7

(from one

\text{MnO}_4^-

ion carrying

-1

and eight

\text{H}^+

ions carrying

+8

). The total charge on the right side is

+2

(from one

\text{Mn}^{2+}

ion). To make both sides equal in charge, 5 electrons (each having a

-1

charge) must be added to the reactant side.

Step-by-Step Solution

1

Calculate the total ionic charge on the reactant side before adding electrons.

Net reactant charge =

(-1) + 8(+1) = +7

One permanganate ion contributes a charge of

-1

and eight hydrogen ions contribute

+8

.

2

Calculate the total ionic charge on the product side.

Net product charge =

+2

One manganese(II) ion contributes

+2

and four water molecules are neutral (

0

).

3

Determine the number of electrons required to balance the overall charge.

+7 + 5(-1) = +2

, so 5 electrons are required.

Electrons carry a

-1

charge, so adding

5e^-

to the reactant side lowers its charge from

+7

to

+2

to match the product side.

Key Concept

Balancing charge in half-reactions by adding electrons

Question 23Question →

Consider the following chemical reaction:

\text{H}_2\text{S}_{(g)} + \text{Cl}_{2(g)} \rightarrow 2\text{HCl}_{(g)} + \text{S}_{(s)}

Which of the following statements correctly interprets the chemical behavior of hydrogen sulfide (

\text{H}_2\text{S}

) using both classical and modern concepts of oxidation and reduction?

It is oxidized according to the classical definition because it loses hydrogen, and acts as a reducing agent according to the modern definition because sulfur increases its oxidation state from

-2

to

0

.

It is reduced according to the classical definition because it loses hydrogen, and acts as an oxidizing agent according to the modern definition because sulfur gains electrons.

It acts as an oxidizing agent according to the classical definition because it donates hydrogen to chlorine, and is reduced according to the modern definition because sulfur loses electrons.

It is oxidized according to the modern definition because the oxidation number of hydrogen increases, and acts as an oxidizing agent according to the classical definition by gaining chlorine.

Show answer & explanation

Answer: It is oxidized according to the classical definition because it loses hydrogen, and acts as a reducing agent according to the modern definition because sulfur increases its oxidation state from $-2$ to $0$ .

Answer

Hydrogen sulfide (

\text{H}_2\text{S}

) is oxidized according to the classical definition because it loses hydrogen, and acts as a reducing agent according to the modern definition because sulfur increases its oxidation state from

-2

to

0

.

The correct option correctly applies both definitions: under the classical concept, loss of hydrogen from hydrogen sulfide (converting it to elemental sulfur) is oxidation. Under the modern electronic concept, the sulfur atom in hydrogen sulfide changes its oxidation number from -2 to 0, which represents a loss of electrons (oxidation), thereby acting as a reducing agent.

Step-by-Step Solution

1

Analyze

\text{H}_2\text{S}

using the classical concept of redox (hydrogen transfer).

In the reaction,

\text{H}_2\text{S}

is converted into elemental sulfur (

\text{S}

). Since

\text{H}_2\text{S}

loses hydrogen atoms, it undergoes oxidation under the classical definition.

Classically, oxidation is defined as the addition of oxygen or the removal of hydrogen.

2

Determine the oxidation state of sulfur in reactants and products.

In

\text{H}_2\text{S}

, hydrogen has an oxidation number of

+1

, so sulfur has an oxidation number of

-2

. In elemental sulfur (

\text{S}

), the oxidation number is

0

.

Uncombined elements have an oxidation number of zero.

3

Analyze

\text{H}_2\text{S}

using the modern concept of redox (electron transfer and oxidation number change).

Sulfur goes from

-2

to

0

, which is an increase in oxidation state (loss of 2 electrons per sulfur atom).

An increase in oxidation state (loss of electrons) is oxidation. The substance that undergoes oxidation causes reduction in the other reactant, making

\text{H}_2\text{S}

the reducing agent.

Key Concept

Classical vs. Modern Redox Concepts

Estimated Time:2m 0s

Question 24Question →

Match each chemical species containing a central transition metal or halogen atom with its corresponding systematic IUPAC name and central element oxidation number.

Click a left item, then click its matching right item

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Answer

\text{K}_3[\text{Fe}(\text{CN})_6]

matches Potassium hexacyanoferrate(III), +3;

\text{K}_2\text{Cr}_2\text{O}_7

matches Potassium heptaoxodichromate(VI), +6;

\text{NaIO}_4

matches Sodium tetraoxoiodate(VII), +7;

[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2

matches Pentaamminechlorocobalt(III) chloride, +3.

Each complex chemical formula correctly pairs with its systematic IUPAC designation based on assigning oxidation numbers to central atoms according to standard IUPAC rules.

Step-by-Step Solution

1

Determine the oxidation state of Fe in

\text{K}_3[\text{Fe}(\text{CN})_6]

and match its IUPAC name

Fe oxidation state is +3; IUPAC name is Potassium hexacyanoferrate(III)

Potassium is +1, cyanide ion

\text{CN}^-

is -1. Solving

3(1) + x + 6(-1) = 0

gives

x = +3

.

2

Determine the oxidation state of Cr in

\text{K}_2\text{Cr}_2\text{O}_7

and match its IUPAC name

Cr oxidation state is +6; IUPAC name is Potassium heptaoxodichromate(VI)

Potassium is +1, oxygen is -2. Solving

2(1) + 2x + 7(-2) = 0

gives

2x = 12

, so

x = +6

.

3

Determine the oxidation state of I in

\text{NaIO}_4

and match its IUPAC name

I oxidation state is +7; IUPAC name is Sodium tetraoxoiodate(VII)

Sodium is +1, oxygen is -2. Solving

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

gives

x = +7

.

4

Determine the oxidation state of Co in

[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2

and match its IUPAC name

Co oxidation state is +3; IUPAC name is Pentaamminechlorocobalt(III) chloride

Ammonia

\text{NH}_3

is a neutral molecule (0 charge), while chloride

\text{Cl}^-

has -1 charge. Solving

x + 5(0) + 3(-1) = 0

gives

x = +3

.

Key Concept

Calculation of oxidation numbers in complex salts, coordination complexes, and polyatomic oxo-compounds, and applying systematic IUPAC nomenclature rules.

Question 25Question →

The transformation of iron(II) ions to iron(III) ions in aqueous solution is represented by the ionic half-equation:

\text{Fe}^{2+}_{(aq)} \rightarrow \text{Fe}^{3+}_{(aq)} + e^-

Which of the following statements correctly explains this chemical change in terms of classical and modern redox concepts?

It is an oxidation process because iron(II) loses an electron, resulting in an increase in oxidation state from

+2

to

+3

, which expands beyond classical oxygen-addition definitions.

It is a reduction process because iron(II) loses a negatively charged electron, which decreases its net oxidation state.

It is an oxidation process because uncombined elemental iron must be formed with a non-zero oxidation state before becoming iron(III).

It is a reduction process because classical redox definitions require the addition of oxygen atoms for any oxidation to take place.

Show answer & explanation

Answer: It is an oxidation process because iron(II) loses an electron, resulting in an increase in oxidation state from $+2$ to $+3$ , which expands beyond classical oxygen-addition definitions.

Answer

The transformation is an oxidation process because iron(II) loses an electron, resulting in an increase in its oxidation number from

+2

to

+3

.

The correct option accurately applies the modern definition of redox. Loss of electrons (

e^-

) increases the charge/oxidation state of iron from

+2

to

+3

, which defines oxidation regardless of whether oxygen is present.

Step-by-Step Solution

1

Analyze the given ionic half-equation

Iron(II) ion (

\text{Fe}^{2+}

) releases one electron (

e^-

) to become iron(III) ion (

\text{Fe}^{3+}

).

Determining electron movement is key to applying modern redox definitions.

2

Apply modern redox definitions (electron transfer and oxidation number)

Loss of electrons corresponds to oxidation. The oxidation number increases from

+2

to

+3

.

By definition, oxidation is loss of electrons (OIL) and increase in oxidation state.

3

Compare with classical redox definitions

Classical definitions restricted oxidation to oxygen gain or hydrogen loss, whereas modern electron theory accounts for ion conversions without oxygen.

Modern definitions generalize redox behavior to non-oxygen reactions.

Key Concept

Modern vs Classical Redox Concepts: Oxidation as Electron Loss and Oxidation State Increase

Question 26Question →

When a gas $X$ is bubbled into an acidified solution of potassium heptaoxodichromate(VI), $K_2Cr_2O_7$ , the solution changes color from orange to green, accompanied by the formation of a yellow precipitate. Which of the following correctly identifies gas $X$ and describes its role in the reaction?

Gas

X

is hydrogen sulfide (

H_2S

), and it acts as a reducing agent because the oxidation state of sulfur increases from

-2

to

0

.

Gas

X

is sulfur(IV) oxide (

SO_2

), and it acts as an oxidizing agent because it causes chromium to be reduced.

Gas

X

is chlorine (

Cl_2

), and it acts as a reducing agent because it is oxidized to chloride ions (

Cl^-

).

Gas

X

is carbon(II) oxide (

CO

), and it acts as an oxidizing agent by precipitating elemental sulfur from the acid solution.

Show answer & explanation

Answer: Gas $X$ is hydrogen sulfide ( $H_2S$ ), and it acts as a reducing agent because the oxidation state of sulfur increases from $-2$ to $0$ .

Answer

Gas

X

is hydrogen sulfide (

H_2S

), acting as a reducing agent as sulfur is oxidized from an oxidation state of

-2

to

0

.

Hydrogen sulfide (

H_2S

) acts as a reducing agent by donating electrons to dichromate ions (

Cr_2O_7^{2-}

), reducing chromium from oxidation state

+6

(orange) to

+3

(green). Simultaneously, sulfide ions (

S^{2-}

) in

H_2S

are oxidized to elemental sulfur (

S_{(s)}

), which appears as a characteristic yellow precipitate.

Step-by-Step Solution

1

Analyze the color change of acidified potassium heptaoxodichromate(VI),

K_2Cr_2O_7

.

The orange color is due to

Cr_2O_7^{2-}

ions (where Cr has an oxidation state of

+6

). The green color indicates reduction to

Cr^{3+}

ions (oxidation state

+3

).

Potassium heptaoxodichromate(VI) acts as an oxidizing agent when reduced from

+6

to

+3

.

2

Identify the role of gas

X

and the origin of the yellow precipitate.

Gas

X

causes the reduction of dichromate, so gas

X

must be a reducing agent. The yellow precipitate formed is insoluble elemental sulfur (

S_{(s)}

).

A yellow colloidal precipitate in wet redox tests of gases is characteristic of sulfide oxidation to sulfur.

3

Determine the oxidation state change for hydrogen sulfide (

H_2S

).

In

H_2S

, sulfur has an oxidation state of

-2

. In elemental sulfur (

S

), its oxidation state is

0

.

An increase in oxidation state (from

-2

to

0

) represents oxidation; the species undergoing oxidation is the reducing agent.

Key Concept

Laboratory identification of reducing agents using acidified potassium heptaoxodichromate(VI)

Question 27Question →

When acidified potassium heptaoxodichromate(VI), $K_2Cr_2O_7$ , oxidizes sulfur(IV) oxide gas, $SO_2$ , the sulfur species is converted into a polyatomic oxoanion. What is the systematic IUPAC name of the resulting oxoanion, and what is the net change in the oxidation state of the sulfur atom per atom during this redox reaction?

Tetraoxosulfate(VI) ion, with an increase of

+2

Tetraoxosulfate(VI) ion, with an increase of

+6

Trioxosulfate(IV) ion, with a decrease of

-2

Heptaoxodisulfate(VI) ion, with an increase of

+4

Show answer & explanation

Answer: Tetraoxosulfate(VI) ion, with an increase of $+2$

Answer

The resulting oxoanion is the tetraoxosulfate(VI) ion, and the net change in the oxidation state of sulfur is an increase of

+2

.

In the reaction between

SO_2

and

K_2Cr_2O_7

in acid, sulfur is oxidized from

+4

in

SO_2

to

+6

in the sulfate ion (

SO_4^{2-}

). The oxidation state increases by

+2

. The oxoanion

SO_4^{2-}

contains four oxygen atoms bonded to sulfur in the

+6

state, which systematically names it tetraoxosulfate(VI) ion.

Step-by-Step Solution

1

Determine the oxidation state of sulfur in the reactant sulfur(IV) oxide (

SO_2

).

In

SO_2

, oxygen has an oxidation state of

-2

. Let

x

be the oxidation state of sulfur:

x + 2(-2) = 0 \Rightarrow x = +4

.

Establishing the initial oxidation state is essential to calculate the oxidation state change.

2

Identify the chemical formula of the product oxoanion formed when

SO_2

is oxidized by acidified

K_2Cr_2O_7

.

Acidified dichromate oxidizes

SO_2

to sulfate ions,

SO_4^{2-}

.

Oxidation of sulfur(IV) oxide in aqueous acid yields the sulfate oxoanion.

3

Determine the oxidation state of sulfur in the product oxoanion (

SO_4^{2-}

) and calculate the net change.

In

SO_4^{2-}

, let

y

be the oxidation state of sulfur:

y + 4(-2) = -2 \Rightarrow y = +6

. The change in oxidation state is

+6 - (+4) = +2

(an increase of 2).

Determining the final state gives the quantitative change required by the problem.

4

Apply systematic IUPAC nomenclature rules to name the oxoanion

SO_4^{2-}

.

The ion has four oxygen atoms attached to central sulfur with oxidation state

+6

, giving the name tetraoxosulfate(VI) ion.

IUPAC oxoanion rules require specifying the number of oxygen atoms (tetraoxo-), the central element root (sulfate), and its Roman numeral oxidation state ((VI)).

Key Concept

Calculation of oxidation numbers in oxoanions and systematic IUPAC nomenclature of inorganic redox species.

Estimated Time:2m 0s

Question 28Question →

Consider the conversion of iron(II) chloride to iron(III) chloride in aqueous solution:

2\text{FeCl}_{2(aq)} + \text{Cl}_{2(g)} \rightarrow 2\text{FeCl}_{3(aq)}

Which of the following statements correctly explains why the conversion of

\text{FeCl}_2

to

\text{FeCl}_3

is classified as an oxidation process under both classical and modern concepts of redox?

Classically, iron(II) chloride gains an electronegative element (chlorine); modernly, the oxidation state of iron increases from +2 to +3 due to the loss of an electron.

Classically, elemental chlorine undergoes oxidation because its oxidation number is +1; modernly, iron(II) ions gain electrons to form iron(III) ions.

Classically, iron(II) chloride loses oxygen to chlorine gas; modernly, chlorine gas is oxidized because it accepts two moles of electrons per mole of molecules.

Classically, the process represents an acid-base neutralization reaction; modernly, iron(II) chloride acts as a dehydrating agent by eliminating net positive charge.

Show answer & explanation

Answer: Classically, iron(II) chloride gains an electronegative element (chlorine); modernly, the oxidation state of iron increases from +2 to +3 due to the loss of an electron.

Answer

The conversion of iron(II) chloride to iron(III) chloride is an oxidation process because classically, it involves the addition of an electronegative element (chlorine) to iron(II) chloride, and modernly, iron undergoes an increase in oxidation number from +2 to +3 via electron loss.

The correct answer accurately contrasts classical and modern definitions. Under classical rules, adding an electronegative element like chlorine to a compound constitutes oxidation. Under modern rules, oxidation is defined by electron loss and an increase in oxidation number, which occurs when iron(II) with an oxidation state of +2 loses an electron to become iron(III) with an oxidation state of +3.

Step-by-Step Solution

1

Analyze the reaction using classical redox definitions.

Classical redox extends beyond oxygen/hydrogen transfer: oxidation is defined as the addition of an electronegative element (such as chlorine) or removal of an electropositive element. In

2\text{FeCl}_2 + \text{Cl}_2 \rightarrow 2\text{FeCl}_3

,

\text{FeCl}_2

gains a chlorine atom (electronegative element), making it an oxidation process.

Classical definitions account for reactions not involving oxygen or hydrogen by tracking electronegative and electropositive additions/removals.

2

Analyze the reaction using modern electronic and oxidation state concepts.

The ionic half-reaction for iron is

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

. The oxidation state of iron increases from

+2

in

\text{FeCl}_2

to

+3

in

\text{FeCl}_3

.

Modern theory defines oxidation as the loss of electrons (LEO) resulting in an increase in oxidation state.

3

Synthesize classical and modern findings to identify the correct statement.

The statement identifying chlorine addition as classical oxidation and

\text{Fe}^{2+}

to

\text{Fe}^{3+}

electron loss as modern oxidation is correct.

Both classical and modern frameworks independently confirm that

\text{FeCl}_2

undergoes oxidation.

Key Concept

Classical vs Modern Definitions of Oxidation and Reduction

Estimated Time:2m 0s

Question 29Question →

Consider the redox reaction represented by the following equation:

\text{CuO}_{(s)} + \text{H}_{2(g)} \rightarrow \text{Cu}_{(s)} + \text{H}_2\text{O}_{(g)}

Which of the following statements correctly describes hydrogen (

\text{H}_2

) from both classical and modern perspectives of redox?

Classically, hydrogen is oxidized because it gains oxygen; modernly, it is oxidized because its oxidation state increases from 0 to +1.

Classically, hydrogen is reduced because it gains oxygen; modernly, it is oxidized because it gains electrons.

Classically, hydrogen is oxidized because it accepts electrons; modernly, its oxidation state decreases from +1 to 0.

Classically, hydrogen is reduced because it acts as an oxygen donor; modernly, it is reduced because its oxidation state increases from 0 to +1.

Show answer & explanation

Answer: Classically, hydrogen is oxidized because it gains oxygen; modernly, it is oxidized because its oxidation state increases from 0 to +1.

Answer

Classically, hydrogen is oxidized because it gains oxygen; modernly, it is oxidized because its oxidation state increases from 0 to +1.

In the given reaction, hydrogen gas (

\text{H}_2

) combines with oxygen to form water. Classically, gaining oxygen is defined as oxidation. Modernly, the oxidation number of hydrogen increases from 0 (in

\text{H}_2

) to +1 (in

\text{H}_2\text{O}

), which also represents oxidation (loss of electrons). Thus, both concepts agree that hydrogen undergoes oxidation.

Step-by-Step Solution

1

Analyze hydrogen under the classical concept of redox.

In

\text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O}

, elemental hydrogen (

\text{H}_2

) gains oxygen to form water (

\text{H}_2\text{O}

). Under the classical definition, oxidation is defined as the addition or gain of oxygen.

Classical redox concepts define oxidation as the addition of oxygen (or removal of hydrogen) to a substance.

2

Analyze hydrogen under the modern concept of redox.

The oxidation state of free elemental hydrogen (

\text{H}_2

) is 0. In water (

\text{H}_2\text{O}

), hydrogen has an oxidation state of +1. The oxidation state increases from 0 to +1, which corresponds to a loss of electrons.

Modern redox concepts define oxidation as a loss of electrons or an increase in oxidation state.

3

Synthesize classical and modern findings to select the correct description.

Hydrogen is oxidized under both classical (gain of oxygen) and modern (increase in oxidation number from 0 to +1) definitions.

Both definitions consistently classify hydrogen as undergoing oxidation in this reaction.

Key Concept

Comparison of Classical (Oxygen/Hydrogen Transfer) and Modern (Electron Transfer/Oxidation Number) Concepts of Redox

Question 30Question →

Match each inorganic chemical formula on the left with its corresponding systematic IUPAC name on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

\text{KMnO}_4

matches Potassium tetraoxomanganate(VII);

\text{KClO}_3

matches Potassium trioxochlorate(V);

\text{K}_2\text{CrO}_4

matches Potassium tetraoxochromate(VI);

\text{KNO}_2

matches Potassium dioxonitrate(III).

Each chemical formula is correctly matched to its systematic IUPAC name by evaluating the oxidation state of the central atom:

+7

for

\text{Mn}

in

\text{KMnO}_4

(Potassium tetraoxomanganate(VII)),

+5

for

\text{Cl}

in

\text{KClO}_3

(Potassium trioxochlorate(V)),

+6

for

\text{Cr}

in

\text{K}_2\text{CrO}_4

(Potassium tetraoxochromate(VI)), and

+3

for

\text{N}

in

\text{KNO}_2

(Potassium dioxonitrate(III)).

Step-by-Step Solution

1

Determine the oxidation state of the central atom in each chemical species using standard rules (Group 1 alkali metals

= +1

, Oxygen

= -2

, total neutral compound charge

= 0

).

Manganese in

\text{KMnO}_4

is

+7

; Chlorine in

\text{KClO}_3

is

+5

; Chromium in

\text{K}_2\text{CrO}_4

is

+6

; Nitrogen in

\text{KNO}_2

is

+3

.

The Roman numeral in IUPAC nomenclature for oxo-compounds directly corresponds to the calculated oxidation state of the central non-metal or transition metal.

2

Map each chemical formula to its IUPAC name based on the oxygen ligand count prefix and the central atom's Roman numeral oxidation state.

\text{KMnO}_4 \rightarrow

Potassium tetraoxomanganate(VII);

\text{KClO}_3 \rightarrow

Potassium trioxochlorate(V);

\text{K}_2\text{CrO}_4 \rightarrow

Potassium tetraoxochromate(VI);

\text{KNO}_2 \rightarrow

Potassium dioxonitrate(III).

IUPAC convention requires specifying oxygen count (dioxo-, trioxo-, tetraoxo-) followed by the central element suffix (-ate) and its oxidation state in parentheses.

Key Concept

Oxidation number determination and IUPAC nomenclature of inorganic redox species.

Question 31Question →

When the following redox reaction is balanced in an acidic medium using the smallest whole-number coefficients:

\text{ClO}_3^-(\text{aq}) + a\text{Fe}^{2+}(\text{aq}) + b\text{H}^+(\text{aq}) \rightarrow \text{Cl}^-(\text{aq}) + c\text{Fe}^{3+}(\text{aq}) + d\text{H}_2\text{O}(\text{l})

What is the value of the coefficient

b

for

\text{H}^+

?

3

5

6

12

Show answer & explanation

Answer: 6

Answer

The stoichiometric coefficient for hydrogen ions is 6.

In the reduction half-reaction,

\text{ClO}_3^-

is reduced to

\text{Cl}^-

. Balancing the 3 oxygen atoms requires

3\text{ H}_2\text{O}

on the product side. Consequently,

6\text{ H}^+

ions are required on the reactant side to balance the 6 hydrogen atoms, giving

b = 6

.

Step-by-Step Solution

1

Determine the oxidation state change for chlorine

In

\text{ClO}_3^-

, chlorine has an oxidation state of

+5

. In

\text{Cl}^-

, chlorine has an oxidation state of

-1

. The total change is a gain of

6\text{ e}^-

.

Knowing the electron transfer per mole of chlorate ion establishes the electron requirement for the reduction half-reaction.

2

Balance oxygen atoms using water

The chlorate ion

\text{ClO}_3^-

contains 3 oxygen atoms, requiring

3\text{ H}_2\text{O}

on the product side.

In acidic redox balancing, oxygen atoms are balanced by adding water molecules to the side deficient in oxygen.

3

Balance hydrogen atoms using hydrogen ions

To balance the 6 hydrogen atoms in

3\text{ H}_2\text{O}

, add

6\text{ H}^+

to the reactant side.

Hydrogen atoms from the water molecules on the product side must originate from hydrogen ions in the acidic medium.

4

Verify overall mass and charge balance

The reduction half-reaction is

\text{ClO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}

. Adding the oxidation half-reaction

6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-

yields the overall balanced equation with

b = 6

.

Combining half-reactions confirms that electrons cancel out and mass and charge are conserved.

Key Concept

Balancing Redox Equations via Half-Reactions in Acidic Medium

Question 32Question →

Complete the following statement regarding the chemical test for an oxidizing agent using moist starch-iodide paper.

Fill in the blanks below

When chlorine gas (

Cl_2

) is brought into contact with moist starch-iodide paper, the paper turns because iodide ions (

I^-

) undergo to form free iodine (

I_2

).

Show answer & explanation

Answer

The starch-iodide paper turns blue-black (or blue) because iodide ions (

I^-

) undergo oxidation to form elemental iodine (

I_2

).

Moist starch-iodide paper is used to test for oxidizing agents such as chlorine (

Cl_2

). Chlorine oxidizes iodide ions (

I^-

) to elemental iodine (

I_2

) through the loss of electrons (oxidation). The liberated iodine subsequently reacts with starch on the test paper to form a characteristic blue-black complex.

Step-by-Step Solution

1

Identify the chemical role of chlorine gas (

Cl_2

) reacting with iodide ions (

I^-

)

Chlorine acts as an oxidizing agent by accepting electrons from iodide ions:

Cl_2 + 2I^- \rightarrow 2Cl^- + I_2

.

Chlorine has a stronger electron affinity and higher standard reduction potential than iodine.

2

Determine the type of redox process occurring at the iodide ion

The oxidation number of iodine increases from

-1

in

I^-

to

0

in

I_2

, which represents oxidation (loss of electrons).

Oxidation is defined as the loss of electrons or an increase in oxidation state.

3

Identify the visual observation produced on the indicator paper

The liberated elemental iodine (

I_2

) reacts with the starch embedded in the paper to form a deep blue-black starch-iodine complex.

Starch serves as a specific indicator that yields a blue-black coloration in the presence of free molecular iodine.

Key Concept

Laboratory test for oxidizing agents using moist starch-iodide paper

Estimated Time:1m 0s

Question 33Question →

Consider the redox reaction between dichromate ions (

\text{Cr}_2\text{O}_7^{2-}

) and iron(II) ions (

\text{Fe}^{2+}

) in an acidic medium:

\text{Cr}_2\text{O}_7^{2-} + x\text{Fe}^{2+} + y\text{H}^+ \rightarrow 2\text{Cr}^{3+} + x\text{Fe}^{3+} + z\text{H}_2\text{O}

What is the stoichiometric coefficient

x

of

\text{Fe}^{2+}

when the ionic equation is completely balanced?

Show answer & explanation

Answer: 6

Answer

The stoichiometric coefficient x of Fe²⁺ in the balanced redox reaction is 6.

In the reduction half-reaction, dichromate (

\text{Cr}_2\text{O}_7^{2-}

) contains two Cr atoms in the +6 oxidation state converting to two

\text{Cr}^{3+}

ions in the +3 state, which consumes 6 electrons. In the oxidation half-reaction, each

\text{Fe}^{2+}

ion loses 1 electron to form

\text{Fe}^{3+}

. To balance charge transfer, 6

\text{Fe}^{2+}

ions are needed for every 1

\text{Cr}_2\text{O}_7^{2-}

ion, making the stoichiometric coefficient

x

equal to 6.

Step-by-Step Solution

1

Determine the oxidation state changes for Chromium and Iron.

Chromium changes from +6 in

\text{Cr}_2\text{O}_7^{2-}

to +3 in

\text{Cr}^{3+}

, requiring 3 electrons per Chromium atom (

6e^-

total for two Cr atoms). Iron changes from +2 in

\text{Fe}^{2+}

to +3 in

\text{Fe}^{3+}

, releasing

1e^-

per Iron atom.

Identifying the number of electrons transferred in each half-reaction is required to balance the overall redox equation.

2

Balance the electron gain and loss.

The oxidation half-reaction (

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

) must be multiplied by 6 to balance the 6 electrons required by the dichromate ion.

The total number of electrons lost by the reducing agent must equal the total number of electrons gained by the oxidizing agent.

3

Combine the half-reactions and read the coefficient

x

.

The balanced chemical equation is

\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}

, giving

x = 6

.

The coefficient

x

corresponds directly to the stoichiometric multiplier applied to

\text{Fe}^{2+}

.

Key Concept

Balancing Redox Equations via Half-Reactions in Acidic Medium

Question 34Question →

In the redox reaction represented by the equation $H_2S_{(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)} + S_{(s)}$ , which substance acts as the reducing agent?

H_2S

, because sulfur increases its oxidation state from

-2

to

0

Cl_2

, because chlorine decreases its oxidation state from

0

to

-1

H_2S

, because hydrogen is reduced from

+1

to

0

Cl_2

, because it loses electrons to form

HCl

Show answer & explanation

Answer: $H_2S$ , because sulfur increases its oxidation state from $-2$ to $0$

Answer

Hydrogen sulfide (

H_2S

) is the reducing agent because sulfur is oxidized, increasing its oxidation number from

-2

in

H_2S

to

0

in elemental

S

.

Hydrogen sulfide (

H_2S

) acts as the reducing agent because sulfur undergoes oxidation. Its oxidation number increases from

-2

in

H_2S

to

0

in elemental sulfur (

S

), indicating that it loses electrons to reduce chlorine.

Step-by-Step Solution

1

Assign oxidation numbers to each element in the given reaction.

In

H_2S

:

H = +1, S = -2

. In

Cl_2

:

Cl = 0

. In

HCl

:

H = +1, Cl = -1

. In elemental

S

:

S = 0

.

Determining oxidation states allows tracking of electron loss and gain.

2

Identify which species loses electrons (is oxidized).

Sulfur changes from

-2

to

0

, representing an increase in oxidation number (loss of electrons).

The reactant containing the element that undergoes oxidation is the reducing agent.

Key Concept

Oxidizing and Reducing Agents via Oxidation State Tracking

Question 35Question →

A galvanic cell is constructed under standard conditions using the following two reduction half-reactions:

\text{Fe}^{3+}_{\text{(aq)}} + \text{e}^- \rightarrow \text{Fe}^{2+}_{\text{(aq)}}, \quad E^\circ = +0.77\text{ V}

\text{Sn}^{4+}_{\text{(aq)}} + 2\text{e}^- \rightarrow \text{Sn}^{2+}_{\text{(aq)}}, \quad E^\circ = +0.15\text{ V}

What is the standard cell potential ( $E^\circ_{\text{cell}}$ ) in volts for the spontaneous overall reaction?

Show answer & explanation

Answer: 0.62

Answer

The standard cell potential for the spontaneous reaction is +0.62 V.

In a spontaneous galvanic cell, reduction takes place at the cathode, which is the electrode with the higher standard reduction potential (+0.77 V for

\text{Fe}^{3+}/\text{Fe}^{2+}

). Oxidation occurs at the anode (+0.15 V for

\text{Sn}^{4+}/\text{Sn}^{2+}

). Substituting these values into

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

yields

+0.77\text{ V} - (+0.15\text{ V}) = +0.62\text{ V}

.

Step-by-Step Solution

1

Identify the cathode and anode based on standard reduction potentials.

Cathode: Fe3+/Fe2+ half-cell (E° = +0.77 V); Anode: Sn4+/Sn2+ half-cell (E° = +0.15 V).

In a spontaneous galvanic cell, reduction occurs at the electrode with the more positive standard reduction potential.

2

Apply the standard electromotive force equation.

E°cell = E°cathode - E°anode = +0.77 V - (+0.15 V) = +0.62 V

The cell potential measures the overall potential difference between the reduction half-cell and oxidation half-cell.

Key Concept

Standard Cell Potential Calculation

Estimated Time:1m 30s

Question 36Question →

Match each chemical species on the left with the correct oxidation state of its central atom on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potassium trioxovanadate(V) (

KVO_3

) matches +5; Sodium thiosulfate (

Na_2S_2O_3

) matches +2; Potassium hexachloroplatinate(IV) (

K_2PtCl_6

) matches +4; Trioxophosphate(III) acid (

H_3PO_3

) matches +3.

Each chemical species is correctly paired with the oxidation state of its central atom calculated by setting the sum of all oxidation numbers equal to zero.

Step-by-Step Solution

1

Calculate oxidation state of Vanadium in

KVO_3

Potassium is +1, Oxygen is -2.

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

.

The sum of oxidation numbers in a neutral compound equals zero.

2

Calculate oxidation state of Sulfur in

Na_2S_2O_3

Sodium is +1, Oxygen is -2.

2(1) + 2S + 3(-2) = 0 \implies 2S = 4 \implies S = +2

.

Average oxidation state calculation per sulfur atom.

3

Calculate oxidation state of Platinum in

K_2PtCl_6

Potassium is +1, Chlorine is -1.

2(1) + Pt + 6(-1) = 0 \implies Pt = +4

.

Halogens in binary complex salts take an oxidation number of -1.

4

Calculate oxidation state of Phosphorus in

H_3PO_3

Hydrogen is +1, Oxygen is -2.

3(1) + P + 3(-2) = 0 \implies P = +3

.

Hydrogen in covalent oxoacids takes an oxidation state of +1.

Key Concept

Oxidation number calculation using standard rules for alkali metals, halogens, hydrogen, and oxygen in neutral inorganic compounds.

Estimated Time:1m 30s

Question 37Question →

Consider the following ionic equation for the redox reaction between dichromate(VI) ions and iodide ions in an acidic medium:

\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + x\text{I}^-(\text{aq}) + y\text{H}^+(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + z\text{I}_2(\text{aq}) + w\text{H}_2\text{O}(\text{l})

What are the correct values of the stoichiometric coefficients

x

,

y

, and

z

respectively when the equation is completely balanced?

6, 14, 3

2, 14, 1

6, 7, 3

3, 14, 6

Show answer & explanation

Answer: 6, 14, 3

Answer

The correct stoichiometric coefficients for x, y, and z are 6, 14, and 3 respectively.

The reduction of one dichromate ion (

\text{Cr}_2\text{O}_7^{2-}

) requires 6 electrons and 14 hydrogen ions to yield two

\text{Cr}^{3+}

ions and 7 water molecules. Oxidation of iodide ions (

\text{I}^-

) to iodine (

\text{I}_2

) releases 2 electrons per molecule formed. To equalize electron exchange at 6 electrons, 6 moles of iodide ions (

x = 6

) produce 3 moles of iodine molecules (

z = 3

), requiring 14 moles of

\text{H}^+

(

y = 14

).

Step-by-Step Solution

1

Write and balance the reduction half-reaction for dichromate(VI) ions

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

Chromium changes oxidation state from +6 to +3 (total 6 electrons gained for two Cr atoms), and 7 oxygen atoms require

14\text{H}^+

to form

7\text{H}_2\text{O}

.

2

Write and balance the oxidation half-reaction for iodide ions

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

Iodine changes oxidation state from -1 to 0, losing 1 electron per iodide ion.

3

Equalize the number of electrons transferred in both half-reactions

Multiply the oxidation half-reaction by 3:

6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-

6 electrons lost must equal 6 electrons gained.

4

Combine the half-reactions and determine coefficients

\text{Cr}_2\text{O}_7^{2-} + 6\text{I}^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}

Matching coefficients gives

x = 6

,

y = 14

, and

z = 3

.

Key Concept

Balancing Redox Equations using Half-Reactions in Acidic Medium

Question 38Question →

Complete the balanced reduction half-reaction equation for the conversion of nitrate ions to nitrogen monoxide gas in an acidic medium by providing the missing stoichiometric coefficients. What are the values of the coefficients for hydrogen ions and electrons?

Fill in the blanks below

$$\text{NO}_3^-(\text{aq}) + \text{H}^+(\text{aq}) + e^- \rightarrow \text{NO}(\text{g}) + 2\text{H}_2\text{O}(\text{l})$$

Show answer & explanation

Answer

The balanced reduction half-reaction requires 4 hydrogen ions (H⁺) and 3 electrons (e⁻).

The balanced half-reaction is NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(l). Four hydrogen ions balance the four hydrogens present in the two water molecules on the right. Three electrons balance the net charge (+3 on the left reactant side versus 0 on the neutral product side), confirming that nitrogen is reduced from oxidation state +5 in NO₃⁻ to +2 in NO.

Step-by-Step Solution

1

Balance oxygen atoms using water molecules

One nitrate ion (NO₃⁻) has 3 oxygen atoms, and NO has 1 oxygen atom, requiring 2 H₂O molecules on the product side: NO₃⁻ → NO + 2 H₂O

In aqueous acid media, oxygen atoms are balanced by adding H₂O molecules to the oxygen-deficient side.

2

Balance hydrogen atoms using hydrogen ions

The right side has 4 hydrogen atoms in 2 H₂O, so 4 H⁺ ions must be added to the left side: NO₃⁻ + 4 H⁺ → NO + 2 H₂O

Hydrogen atoms in acidic media are balanced by adding H⁺ ions to the hydrogen-deficient side.

3

Balance electrical charge using electrons

Left side net charge = (-1) + 4(+1) = +3. Right side net charge = 0. Adding 3 electrons (3 e⁻) to the left side gives a net charge of 0 on both sides: NO₃⁻ + 4 H⁺ + 3 e⁻ → NO + 2 H₂O

Electrons are added to the side with the higher net positive charge to satisfy conservation of charge.

Key Concept

Half-Reaction Method for Balancing Redox Equations in Acidic Medium

Estimated Time:1m 30s

Question 39Question →

Match each redox description on the left with its corresponding classical or modern definition concept on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Addition of oxygen matches Classical oxidation; Loss of electrons matches Modern oxidation (electron transfer); Decrease in oxidation state matches Modern reduction (oxidation state change); Removal of oxygen matches Classical reduction.

Classical oxidation involves gaining oxygen, while classical reduction involves losing oxygen. In modern electronic terms, oxidation is the loss of electrons (OIL), and reduction is a decrease in oxidation state (reduction of oxidation number).

Step-by-Step Solution

1

Identify classical redox concepts based on oxygen transfer.

Addition of oxygen corresponds to classical oxidation, whereas removal of oxygen corresponds to classical reduction.

Classical definitions focused on the transfer of oxygen and hydrogen atoms.

2

Identify modern redox concepts based on electron transfer and oxidation numbers.

Loss of electrons corresponds to modern oxidation, and a decrease in oxidation state corresponds to modern reduction.

Modern concepts expand redox beyond oxygen/hydrogen to include electron movement and formal charge changes.

Key Concept

Distinguishing between classical (oxygen/hydrogen transfer) and modern (electron transfer and oxidation number) definitions of oxidation and reduction.

Estimated Time:1m 0s

Question 40Question →

Match each chemical transformation on the left with the correct classical or modern redox definition that specifically describes it on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Conversion of methane to methanol matches classical oxidation via direct addition of oxygen; conversion of mercury(II) chloride to mercury(I) chloride matches classical reduction via removal of an electronegative element; transformation of magnesium to magnesium fluoride matches modern oxidation via electron loss (

0

to

+2

); conversion of dichromate to chromium(III) ion matches modern reduction via decrease in oxidation number (

+6

to

+3

).

Each chemical transformation is matched to its corresponding classical or modern definition based on fundamental redox principles: oxygen addition represents classical oxidation; removal of an electronegative element represents classical reduction; electron loss and oxidation number increase represent modern oxidation; and oxidation number decrease represents modern reduction.

Step-by-Step Solution

1

Analyze the conversion of

\text{CH}_4

to

\text{CH}_3\text{OH}

using classical definitions.

An oxygen atom is directly added to methane without removing hydrogen. Under classical rules, addition of oxygen is defined as oxidation.

Classical redox concepts classify oxygen gain as oxidation.

2

Analyze

2\text{HgCl}_2 + \text{SnCl}_2 \rightarrow \text{Hg}_2\text{Cl}_2 + \text{SnCl}_4

for

\text{HgCl}_2

.

In

\text{HgCl}_2

, mercury is bound to two chlorines per atom. In

\text{Hg}_2\text{Cl}_2

, mercury is bound to one chlorine per atom. The removal/loss of an electronegative element (

\text{Cl}

) is classical reduction.

Classical definitions extend reduction to include the removal of electronegative elements or addition of electropositive elements.

3

Examine

\text{Mg}_{(s)} \rightarrow \text{MgF}_{2(s)}

under modern concepts.

Elemental magnesium (

\text{Mg}^0

) forms

\text{Mg}^{2+}

ions by losing

2

electrons. Loss of electrons and an increase in oxidation state from

0

to

+2

represents modern oxidation.

Modern redox theory defines oxidation as loss of electrons (OIL) or increase in oxidation number.

4

Examine

\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}

using oxidation states.

In

\text{Cr}_2\text{O}_7^{2-}

,

2x + 7(-2) = -2 \Rightarrow x = +6

. In

\text{Cr}^{3+}

, the oxidation state is

+3

. The decrease in oxidation number from

+6

to

+3

represents modern reduction.

A decrease in oxidation state is the defining feature of reduction under modern IUPAC guidelines.

Key Concept

Definitions and Classical vs Modern Concepts of Redox

Oxidation-Reduction and Electrochemistry

Answer

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