Oxidation-Reduction and Electrochemistry Practice Questions — JAMB UTME — Page 3

Question 41Question →

Pair each of the chemical transformation descriptions listed on the left with its corresponding classical or modern redox definition concept on the right.

Click a left item, then click its matching right item

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Answer

Loss of electrons corresponds to the modern electronic concept of oxidation; removal of hydrogen corresponds to the classical concept of oxidation via hydrogen transfer; decrease in oxidation number corresponds to the modern oxidation-number concept of reduction; addition of oxygen corresponds to the classical concept of oxidation via oxygen transfer.

The matching pairs correctly connect historical definitions (oxygen addition and hydrogen removal) and modern definitions (electron loss and oxidation number reduction) to their corresponding oxidation or reduction classifications.

Step-by-Step Solution

1

Differentiate classical definitions from modern redox concepts.

Classical concepts focus on oxygen and hydrogen transfer, whereas modern concepts focus on electron transfer and changes in oxidation state.

Historical definitions arose before the electron was discovered, while modern definitions generalize redox processes to all chemical species.

2

Match electron transfer and oxidation state statements.

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

Oxidation increases charge/oxidation state via electron loss; reduction decreases charge/oxidation state via electron gain.

3

Match hydrogen and oxygen transfer statements.

Removal of hydrogen is classical oxidation, and addition of oxygen is classical oxidation.

Classical oxidation is defined by adding oxygen or removing hydrogen from a substance.

Key Concept

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

Estimated Time:45s

Question 42Question →

Match each chemical species listed on the left with its correct systematic IUPAC name on the right.

Click a left item, then click its matching right item

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Answer

The correct pairings are:

MnO_4^{2-}

matches Tetraoxomanganate(VI) ion;

ClO_4^-

matches Tetraoxochlorate(VII) ion;

N_2O

matches Dinitrogen(I) oxide; and

Fe_2O_3

matches Iron(III) oxide.

Each chemical formula correctly corresponds to its IUPAC name based on the calculated oxidation number of the electropositive element and standard IUPAC nomenclature rules for oxoanions and oxides.

Step-by-Step Solution

1

Determine the oxidation number of manganese in

MnO_4^{2-}

Manganese has an oxidation state of

+6

.

Applying the algebraic rule for polyatomic ions:

x + 4(-2) = -2 \implies x = +6

. The four oxygen atoms prefix as 'tetraoxo-', giving tetraoxomanganate(VI) ion.

2

Determine the oxidation number of chlorine in

ClO_4^-

Chlorine has an oxidation state of

+7

.

Applying the algebraic rule:

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

. The species is named tetraoxochlorate(VII) ion.

3

Determine the oxidation number of nitrogen in

N_2O

Nitrogen has an oxidation state of

+1

.

Applying the neutrality rule:

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

. The IUPAC name is dinitrogen(I) oxide.

4

Determine the oxidation number of iron in

Fe_2O_3

Iron has an oxidation state of

+3

.

Applying the neutrality rule:

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

. The IUPAC name is iron(III) oxide.

Key Concept

Calculation of oxidation numbers for central atoms in oxoanions and binary oxides to deduce standard IUPAC names.

Question 43Question →

Match each oxoanion on the left with the correct oxidation number of its central element on the right.

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Answer

The correct matches pair

PO_4^{3-}

with

+5

,

NO_2^-

with

+3

,

SO_4^{2-}

with

+6

, and

CO_3^{2-}

with

+4

.

Each central atom's oxidation state is determined by setting the sum of the oxidation states equal to the net ionic charge, using

-2

for each oxygen atom.

Step-by-Step Solution

1

Assign the standard oxidation number of

-2

to oxygen in oxoanions.

Each oxygen atom contributes an oxidation state of

-2

.

Oxygen is more electronegative than phosphorus, nitrogen, sulfur, and carbon.

2

Set up an algebraic sum where the total oxidation numbers equal the ion's net charge.

For

PO_4^{3-}

:

P + 4(-2) = -3

; for

NO_2^-

:

N + 2(-2) = -1

; for

SO_4^{2-}

:

S + 4(-2) = -2

; for

CO_3^{2-}

:

C + 3(-2) = -2

.

The sum of oxidation states in a polyatomic species equals the charge on the species.

3

Solve each linear equation for the oxidation number of the central atom.

P = +5

,

N = +3

,

S = +6

, and

C = +4

.

Algebraic isolation of the unknown variable determines the oxidation state.

Key Concept

Assigning Oxidation Numbers in Polyatomic Oxoanions

Question 44Question →

In chemical analysis, oxidation-reduction reactions are interpreted using either classical transfer principles (oxygen/hydrogen) or modern electronic and oxidation-state theories. Match each chemical transformation on the left with the specific classical or modern redox definition concept on the right that governs it.

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Answer

The conversion of ammonia to nitrogen gas matches classical oxidation via hydrogen removal. The oxidation of magnesium metal to magnesium ions matches modern electronic oxidation via electron loss. The change of permanganate ion to manganese(II) ion matches modern reduction via a decrease in oxidation state from +7 to +2. The conversion of ethene to ethane matches classical reduction via hydrogen addition.

Each pair correctly matches a specific chemical transformation with its governing classical or modern redox rule. The removal of hydrogen from ammonia is classical oxidation. The loss of electrons from magnesium metal is modern electronic oxidation. The reduction in oxidation number of manganese from +7 to +2 is modern reduction. The addition of hydrogen to ethene is classical reduction.

Step-by-Step Solution

1

Analyze the conversion of

2\text{NH}_3

to

\text{N}_2

Ammonia loses hydrogen atoms during the reaction.

According to classical redox concepts, the removal of hydrogen from a compound constitutes oxidation.

2

Analyze the half-reaction

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

Magnesium loses two electrons.

Under modern electronic theory, oxidation is defined as electron loss (OIL - Oxidation Is Loss).

3

Analyze the reduction of

\text{MnO}_4^-

to

\text{Mn}^{2+}

Manganese's oxidation state drops from

+7

to

+2

.

According to modern oxidation number conventions, a drop/decrease in oxidation number represents reduction.

4

Analyze the hydrogenation of ethene

\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6

Hydrogen is added across the carbon-carbon double bond.

Classical redox principles define reduction as the gain or addition of hydrogen.

Key Concept

Classical vs Modern Concepts of Redox

Estimated Time:2m 0s

Question 45Question →

Complete the following statement regarding manganese redox species by filling in the correct numerical oxidation states.

Fill in the blanks below

In potassium manganate(VI),

K_2MnO_4

, the oxidation state of the central manganese atom is , whereas in potassium tetraoxomanganate(VII),

KMnO_4

, the oxidation state of manganese is .

Show answer & explanation

Answer

The oxidation state of manganese in

K_2MnO_4

is +6, and in

KMnO_4

it is +7.

In potassium manganate(VI),

K_2MnO_4

, two potassium ions yield a

+2

total charge and four oxide ions yield

-8

, requiring manganese to be

+6

to balance to zero. In potassium tetraoxomanganate(VII),

KMnO_4

, one potassium ion (

+1

) and four oxide ions (

-8

) leave manganese at

+7

.

Step-by-Step Solution

1

Calculate the oxidation state of manganese in

K_2MnO_4

.

Assign standard oxidation numbers:

K = +1

and

O = -2

. For the neutral molecule

K_2MnO_4

,

2(+1) + \text{Mn} + 4(-2) = 0 \implies +2 + \text{Mn} - 8 = 0 \implies \text{Mn} = +6

.

The sum of oxidation states in a neutral chemical compound must equal zero.

2

Calculate the oxidation state of manganese in

KMnO_4

.

Assign standard oxidation numbers:

K = +1

and

O = -2

. For the neutral molecule

KMnO_4

,

1(+1) + \text{Mn} + 4(-2) = 0 \implies +1 + \text{Mn} - 8 = 0 \implies \text{Mn} = +7

.

The sum of oxidation states in a neutral chemical compound must equal zero.

Key Concept

Determining oxidation states of central transition metal atoms in oxoanions and neutral salts.

Question 46Question →

In the preparation of potassium ferrate, $K_2FeO_4$ , elemental iron ( $Fe$ ) is oxidized in an alkaline medium. What is the change in the oxidation number of iron per atom during this reaction, and what is the systematic IUPAC name of $K_2FeO_4$ ?

Increase of 6; potassium tetraoxoferrate(VI)

Increase of 4; potassium tetraoxoferrate(VI)

Increase of 4; potassium tetraoxoferrate(IV)

Increase of 8; potassium tetraoxoferrate(VIII)

Show answer & explanation

Answer: Increase of 6; potassium tetraoxoferrate(VI)

Answer

The oxidation number of iron increases by 6, and the systematic IUPAC name of

K_2FeO_4

is potassium tetraoxoferrate(VI).

Free uncombined iron (

Fe

) has an oxidation number of

0

. In

K_2FeO_4

, potassium is

+1

and oxygen is

-2

. Solving

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

yields

Fe = +6

. The change in oxidation number is

+6 - 0 = +6

. The polyatomic anion

FeO_4^{2-}

contains four oxo ligands attached to iron in the

+6

oxidation state, making its systematic IUPAC name potassium tetraoxoferrate(VI).

Step-by-Step Solution

1

Determine the oxidation state of uncombined elemental iron (

Fe

).

Oxidation state of

Fe = 0

.

By IUPAC rules, elements in their free uncombined elemental state always have an oxidation number of zero.

2

Calculate the oxidation state of iron in

K_2FeO_4

.

Oxidation state of

Fe = +6

.

Potassium has an oxidation state of

+1

and oxygen has

-2

. Setting up the charge balance equation:

2(+1) + Fe + 4(-2) = 0 \implies +2 + Fe - 8 = 0 \implies Fe = +6

.

3

Calculate the change in oxidation number.

Change

= +6 - 0 = +6

(an increase of 6).

The change in oxidation state is the final state minus the initial state.

4

Formulate the systematic IUPAC name for

K_2FeO_4

.

Potassium tetraoxoferrate(VI).

The salt consists of potassium cations and the polyatomic anion

FeO_4^{2-}

. With four oxygen atoms surrounding iron in oxidation state

+6

, the IUPAC name is potassium tetraoxoferrate(VI).

Key Concept

Determination of oxidation state changes from elemental forms and IUPAC Stock nomenclature of oxoanions

Question 47Question →

What is the oxidation state of chlorine in uncombined chlorine gas, $Cl_2$ ?

0

-1

+1

-2

Show answer & explanation

Answer: $0$

Answer

The oxidation state of chlorine in uncombined chlorine gas (

Cl_2

) is

0

.

According to standard rules for assigning oxidation states, any element in its free or uncombined elemental state has an oxidation number of zero. Because

Cl_2

consists of identical chlorine atoms sharing bonding electrons equally, its oxidation state is

0

.

Step-by-Step Solution

1

Identify the chemical state of the given species.

Chlorine is present as free, uncombined diatomic chlorine gas (

Cl_2

).

Redox rules distinguish uncombined elemental states from elements bound in chemical compounds.

2

Apply the fundamental oxidation number rule for uncombined elements.

The oxidation number assigned to any atom in an uncombined element is

0

.

In a homonuclear diatomic molecule like

Cl_2

, electrons are shared equally between identical atoms, resulting in zero net charge allocation.

Key Concept

Oxidation state of uncombined elements

Question 48Question →

Ammonia gas (

\text{NH}_3

) reacts with oxygen gas (

\text{O}_2

) over a platinum-rhodium catalyst to form nitrogen(II) oxide (

\text{NO}

) and water vapor (

\text{H}_2\text{O}

) according to the equation:

4\text{NH}_{3(g)} + 5\text{O}_{2(g)} \rightarrow 4\text{NO}_{(g)} + 6\text{H}_2\text{O}_{(g)}

Which of the following statements correctly interprets the oxidation process occurring in this reaction from both classical and modern redox perspectives?

Ammonia is oxidized because nitrogen loses hydrogen classically, which corresponds to an increase in the oxidation state of nitrogen from

-3

to

+2

.

Oxygen gas is reduced because elemental

\text{O}_2

starts with an oxidation state of

-2

and gains hydrogen to form

\text{H}_2\text{O}

.

Ammonia acts as an oxidizing agent because nitrogen loses electrons, resulting in a decrease in its oxidation state.

Oxygen gas acts as a reducing agent because its oxidation number increases from

-2

to

0

as it forms water.

Show answer & explanation

Answer: Ammonia is oxidized because nitrogen loses hydrogen classically, which corresponds to an increase in the oxidation state of nitrogen from $-3$ to $+2$ .

Answer

Ammonia is oxidized because nitrogen loses hydrogen classically, which corresponds to an increase in the oxidation state of nitrogen from

-3

to

+2

.

In classical terminology, oxidation involves the removal of hydrogen or addition of oxygen. Ammonia (

\text{NH}_3

) loses hydrogen to form nitrogen(II) oxide (

\text{NO}

), fulfilling classical oxidation. In modern electronic terms, the oxidation number of nitrogen increases from

-3

in

\text{NH}_3

to

+2

in

\text{NO}

, representing a loss of 5 electrons per nitrogen atom, which directly aligns with the modern definition of oxidation.

Step-by-Step Solution

1

Analyze classical redox definitions for ammonia

In

\text{NH}_3

, nitrogen is combined with hydrogen. In forming

\text{NO}

, hydrogen is removed and oxygen is added. Classically, loss of hydrogen and gain of oxygen both define oxidation.

Classical redox concepts define oxidation as the gain of oxygen or loss of hydrogen.

2

Determine oxidation states for nitrogen in reactants and products

In

\text{NH}_3

, hydrogen is

+1

, so

N + 3(+1) = 0 \Rightarrow N = -3

. In

\text{NO}

, oxygen is

-2

, so

N + (-2) = 0 \Rightarrow N = +2

.

Sum of oxidation numbers in a neutral molecule equals zero.

3

Correlate modern redox definition with classical observation

The change in oxidation state of nitrogen from

-3

to

+2

is an increase of

5

, which corresponds to loss of electrons (oxidation).

Modern redox concepts define oxidation as an increase in oxidation state due to loss of electrons.

Key Concept

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

Estimated Time:2m 0s

Question 49Question →

During the disproportionation reaction of white phosphorus (

P_4

) in an alkaline aqueous medium, phosphorus reacts according to the equation:

P_4 + 3OH^- + 3H_2O \rightarrow PH_3 + 3H_2PO_2^-

What are the oxidation numbers of phosphorus in

P_4

,

PH_3

, and

H_2PO_2^-

respectively, and what is the systematic IUPAC name of the

H_2PO_2^-

ion?

0

,

-3

, and

+1

; Dihydrogendioxophosphate(I) ion

+4

,

-3

, and

+1

; Dihydrogendioxophosphate(III) ion

0

,

+3

, and

+3

; Dihydrogentrioxophosphate(III) ion

-4

,

+3

, and

+5

; Dioxophosphate(V) ion

Show answer & explanation

Answer: $0$ , $-3$ , and $+1$ ; Dihydrogendioxophosphate(I) ion

Answer

The oxidation numbers of phosphorus in

P_4

,

PH_3

, and

H_2PO_2^-

are

0

,

-3

, and

+1

respectively, and the systematic IUPAC name of

H_2PO_2^-

is the Dihydrogendioxophosphate(I) ion.

The option stating '

0

,

-3

, and

+1

; Dihydrogendioxophosphate(I) ion' is correct because uncombined phosphorus (

P_4

) has an oxidation number of

0

. In phosphine (

PH_3

), phosphorus has an oxidation state of

-3

because hydrogen is

+1

. In

H_2PO_2^-

, solving

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

gives

x = +1

. Following IUPAC rules for oxoanions with hydrogen ligands, two hydrogens ('Dihydrogen'), two oxygens ('dioxo'), and phosphorus(I) in an anion ('phosphate(I)') combine to form the name Dihydrogendioxophosphate(I) ion.

Step-by-Step Solution

1

Determine the oxidation state of phosphorus in

P_4

.

Oxidation state =

0

By definition, any element in its free, uncombined elemental state has an oxidation number of zero.

2

Determine the oxidation state of phosphorus in

PH_3

.

Oxidation state =

-3

Hydrogen bonded to non-metals has an oxidation state of

+1

. Setting up the equation:

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

.

3

Determine the oxidation state of phosphorus in

H_2PO_2^-

.

Oxidation state =

+1

Hydrogen is

+1

and oxygen is

-2

. Setting up the ion charge equation:

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

.

4

Derive the systematic IUPAC name for

H_2PO_2^-

.

Dihydrogendioxophosphate(I) ion

The polyatomic ion contains 2 hydrogen atoms ('Dihydrogen'), 2 oxygen ligands ('dioxo'), the central phosphorus atom in an anion suffix ('phosphate'), and Roman numeral '(I)' indicating its

+1

oxidation state.

Key Concept

Calculation of oxidation states in polyatomic ions and application of inorganic IUPAC nomenclature rules.

Estimated Time:2m 0s

Question 50Question →

Consider the redox reaction taking place in an acidic medium:

\text{MnO}_4^-(\text{aq}) + \text{H}_2\text{O}_2(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + \text{O}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

When this chemical equation is balanced using the smallest whole-number coefficients, what is the stoichiometric coefficient of

\text{H}^+(\text{aq})

?

16

6

8

3

Show answer & explanation

Answer: 6

Answer

6

The balanced net redox equation is

2\text{MnO}_4^-(\text{aq}) + 5\text{H}_2\text{O}_2(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{O}_2(\text{g}) + 8\text{H}_2\text{O}(\text{l})

. Thus, the stoichiometric coefficient of

\text{H}^+(\text{aq})

is 6.

Step-by-Step Solution

1

Write and balance the reduction half-reaction

\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Manganese is reduced from oxidation state +7 in

\text{MnO}_4^-

to +2 in

\text{Mn}^{2+}

, requiring 5 electrons. Charge and mass are balanced using

\text{H}^+

and

\text{H}_2\text{O}

.

2

Write and balance the oxidation half-reaction

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

Oxygen in

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

is oxidized from -1 to 0 in

\text{O}_2

, releasing 2 electrons per molecule.

3

Equalize electron loss and gain

Multiply reduction half-reaction by 2 and oxidation half-reaction by 5:

2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

5\text{H}_2\text{O}_2 \rightarrow 5\text{O}_2 + 10\text{H}^+ + 10\text{e}^-

The least common multiple of 5 and 2 electrons transferred is 10.

4

Combine the half-reactions and simplify redundant species

2\text{MnO}_4^- + 5\text{H}_2\text{O}_2 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}

Subtracting

10\text{H}^+

and

10\text{e}^-

from both sides yields a net coefficient of 6 for

\text{H}^+

on the reactant side.

Key Concept

Ion-Electron Method for Balancing Redox Equations in Acidic Medium

Question 51Question →

In the reduction half-reaction

\text{Fe}^{3+}(\text{aq}) + n e^- \rightarrow \text{Fe}(\text{s})

what is the value of

n

required to balance the electrical charge?

1

2

3

6

Show answer & explanation

Answer: $3$

Answer

The value of

n

is

3

.

To reduce one

\text{Fe}^{3+}

ion with a

+3

charge to a neutral iron atom with a

0

charge, exactly

3

electrons (each carrying a charge of

-1

) must be added to the reactant side so that the net charge on both sides equals

0

.

Step-by-Step Solution

1

Determine the charge on the reactant side and product side

The left side has one

\text{Fe}^{3+}

ion with a charge of

+3

. The right side has neutral

\text{Fe}(\text{s})

with a charge of

0

.

Charge conservation must be satisfied in a balanced half-reaction.

2

Calculate the number of negative electrons (

e^-

) required to balance net charge

+3 + n(-1) = 0 \implies n = 3

.

Each electron carries a single negative charge (

-1

).

Key Concept

Charge conservation in half-reactions

Question 52Question →

In hot, concentrated alkaline solutions, chlorine gas undergoes a disproportionation redox reaction according to the unbalanced equation:

\text{Cl}_2(\text{g}) + \text{OH}^-(\text{aq}) \rightarrow \text{ClO}_3^-(\text{aq}) + \text{Cl}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})

When this equation is balanced using the smallest set of whole-number coefficients, what is the stoichiometric coefficient of hydroxide ions (

\text{OH}^-

) and the total number of moles of electrons transferred in the balanced equation?

6 hydroxide ions and 5 moles of electrons

6 hydroxide ions and 6 moles of electrons

12 hydroxide ions and 10 moles of electrons

3 hydroxide ions and 5 moles of electrons

Show answer & explanation

Answer: 6 hydroxide ions and 5 moles of electrons

Answer

The balanced equation requires 6 hydroxide ions and involves the transfer of 5 moles of electrons.

In the balanced redox equation

3\text{Cl}_2(\text{g}) + 6\text{OH}^-(\text{aq}) \rightarrow \text{ClO}_3^-(\text{aq}) + 5\text{Cl}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l})

, the stoichiometric coefficient of hydroxide ions is 6, and 5 moles of electrons are transferred per mole of reaction as written.

Step-by-Step Solution

1

Assign oxidation numbers to determine the oxidation and reduction half-reactions.

Elemental chlorine

\text{Cl}_2

has an oxidation number of

0

. In

\text{ClO}_3^-

, chlorine has an oxidation state of

+5

(oxidation). In

\text{Cl}^-

, chlorine has an oxidation state of

-1

(reduction).

Disproportionation involves the simultaneous oxidation and reduction of the same element.

2

Write and balance the oxidation half-reaction in basic medium.

\frac{1}{2}\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 3\text{H}_2\text{O} + 5e^-

One chlorine atom increases in oxidation state from

0

to

+5

, releasing

5e^-

. Six

\text{OH}^-

ions balance the charge and oxygen/hydrogen mass.

3

Write and balance the reduction half-reaction.

\frac{1}{2}\text{Cl}_2 + e^- \rightarrow \text{Cl}^-

One chlorine atom decreases in oxidation state from

0

to

-1

, accepting

1e^-

.

4

Equalize electron transfer between half-reactions and combine.

Multiply the reduction half-reaction by 5:

\frac{5}{2}\text{Cl}_2 + 5e^- \rightarrow 5\text{Cl}^-

. Combine with the oxidation half-reaction:

3\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 5\text{Cl}^- + 3\text{H}_2\text{O}

. Total electrons transferred

n_e = 5

.

The number of electrons lost in oxidation must equal the number gained in reduction.

Key Concept

Balancing Disproportionation Redox Reactions in Basic Medium

Question 53Question →

An aqueous solution of a sodium salt $X$ completely decolorizes acidified potassium tetraoxomanganate(VII) solution. Subsequent addition of barium chloride solution to the resulting reaction mixture yields a white precipitate that is insoluble in dilute hydrochloric acid. What role does salt $X$ play in the redox reaction, and what is the chemical formula of the sulfur-containing anion in $X$ ?

Salt

X

acts as a reducing agent, and the anion is

SO_3^{2-}

.

Salt

X

acts as an oxidizing agent, and the anion is

SO_3^{2-}

.

Salt

X

acts as a reducing agent, and the anion is

SO_4^{2-}

.

Salt

X

acts as an oxidizing agent, and the anion is

SO_4^{2-}

.

Show answer & explanation

Answer: Salt $X$ acts as a reducing agent, and the anion is $SO_3^{2-}$ .

Answer

Salt

X

acts as a reducing agent, and the anion is

SO_3^{2-}

.

In the reaction with acidified potassium tetraoxomanganate(VII), the manganate(VII) ion (

MnO_4^-

) is reduced from oxidation state +7 to +2, causing the purple solution to become colorless. The species that causes this reduction must be a reducing agent. Trioxosulfate(IV) ions (

SO_3^{2-}

) contain sulfur in the +4 oxidation state and are oxidized by

KMnO_4

to tetraoxosulfate(VI) ions (

SO_4^{2-}

). The formed

SO_4^{2-}

ions then react with

Ba^{2+}

from barium chloride to form a white insoluble precipitate of

BaSO_4

. Thus, salt

X

is a reducing agent containing

SO_3^{2-}

.

Step-by-Step Solution

1

Analyze the color change of acidified potassium tetraoxomanganate(VII).

The change from purple (

MnO_4^-

) to colorless (

Mn^{2+}

) indicates that

MnO_4^-

is reduced.

Manganese is reduced from oxidation state +7 to +2 by gaining electrons from a reducing agent.

2

Determine the role of salt

X

.

Salt

X

provides electrons to reduce

MnO_4^-

, meaning salt

X

is oxidized and acts as a reducing agent.

A substance that undergoes oxidation and causes another species to be reduced is a reducing agent.

3

Identify the sulfur-containing anion from the precipitation test.

The oxidation product of salt

X

reacts with

Ba^{2+}

to form insoluble

BaSO_4

. Therefore, the oxidized species formed in solution is

SO_4^{2-}

.

Trioxosulfate(IV) ion (

SO_3^{2-}

) has sulfur in oxidation state +4 and is oxidized by acidified

KMnO_4

to tetraoxosulfate(VI) ion (

SO_4^{2-}

), which gives a white precipitate of

BaSO_4

insoluble in HCl.

Key Concept

Oxidizing and Reducing Agents and Qualitative Redox Tests

Question 54Question →

Match each redox reaction mixture (left) with its corresponding characteristic laboratory observation (right).

Click a left item, then click its matching right item

Show answer & explanation

Answer

Acidified

KMnO_4

with

SO_2

changes from purple to colorless (

Mn^{2+}

); Acidified

K_2Cr_2O_7

with

FeSO_4

changes from orange to green (

Cr^{3+}

); Chlorine gas with

KI

forms brown iodine (

I_2

); Hydrogen sulfide gas with

FeCl_3

reduces

Fe^{3+}

to

Fe^{2+}

(pale green) with a yellow sulfur precipitate.

Each test pair matches an oxidizing or reducing reagent with its definitive qualitative laboratory test observation: permanganate decolorizes upon reduction, dichromate turns green upon reduction, iodide oxidizes to brown iodine, and iron(III) reduces to pale green iron(II) alongside yellow sulfur precipitation.

Step-by-Step Solution

1

Determine the redox behavior of

SO_2

with acidified

KMnO_4

SO_2

is oxidized while reducing purple

MnO_4^-

to colorless

Mn^{2+}

.

Potassium permanganate test for reducing agents involves reduction of manganese from oxidation state +7 to +2.

2

Determine the redox behavior of

FeSO_4

with acidified

K_2Cr_2O_7

Fe^{2+}

ions oxidize to

Fe^{3+}

while

Cr_2O_7^{2-}

(orange) is reduced to

Cr^{3+}

(green).

Dichromate(VI) is a standard oxidizing agent whose reduced form contains green chromium(III) ions.

3

Determine the reaction of

Cl_2

with

KI

Cl_2

oxidizes colorless

I^-

ions to elemental iodine (

I_2

), turning the solution brown.

Chlorine is a stronger oxidizing agent than iodine and displaces iodide from solution.

4

Determine the reaction of

H_2S

with

FeCl_3

H_2S

reduces yellow-brown

Fe^{3+}

to pale green

Fe^{2+}

while forming a insoluble yellow deposit of sulfur.

Hydrogen sulfide acts as a reducing agent and deposits elemental sulfur upon oxidation.

Key Concept

Laboratory tests and characteristic color changes for oxidizing and reducing agents.

Question 55Question →

Match each redox reaction scenario involving an oxidizing or reducing agent on the left with its corresponding characteristic laboratory observation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match each redox reagent with its specific electron-transfer observation:

SO_2

with acidified

KMnO_4

produces a purple to colorless change;

Cl_2

with aqueous

KI

turns colorless solution brown; conc.

HNO_3

with

FeSO_4

converts pale green solution to brown; and

H_2S

with

FeCl_3

converts yellow/brown solution to pale green with yellow sulfur deposit.

Each pair correctly links the specific chemical species undergoing oxidation or reduction to its empirical qualitative test result. Sulfur(IV) oxide decolorizes acidified potassium tetraoxomanganate(VII); chlorine oxidizes iodide ions to brown iodine; concentrated trioxonitrate(V) acid converts green iron(II) to brown iron(III); and hydrogen sulfide reduces brown iron(III) to green iron(II) with yellow sulfur precipitation.

Step-by-Step Solution

1

Analyze the redox roles of the reagents in each left item.

SO_2

and

H_2S

act as reducing agents;

Cl_2

and conc.

HNO_3

act as oxidizing agents.

Identifying whether a species donates or accepts electrons determines the expected chemical transformation of the target solution.

2

Determine the oxidation state change and color change for

SO_2

+ acidified

KMnO_4

.

MnO_4^-

(oxidation state +7, purple) is reduced to

Mn^{2+}

(oxidation state +2, colorless).

Manganate(VII) reduction is the standard test for reducing agents like

SO_2

.

3

Determine the oxidation state change and color change for

Cl_2

+ aqueous

KI

.

I^-

(oxidation state -1, colorless) is oxidized to

I_2

(oxidation state 0, brown).

Halogen displacement shows chlorine's higher electronegativity and oxidizing strength compared to iodine.

4

Determine the oxidation state change for conc.

HNO_3

+

FeSO_4

and

H_2S

+

FeCl_3

.

Conc.

HNO_3

oxidizes pale green

Fe^{2+}

to brown

Fe^{3+}

.

H_2S

reduces yellow/brown

Fe^{3+}

to pale green

Fe^{2+}

with precipitate of sulfur.

Iron transitions between +2 (pale green) and +3 (yellow/brown) depending on whether an oxidant or reductant is introduced.

Key Concept

Laboratory identification of oxidizing and reducing agents via characteristic color changes and oxidation state transitions

Estimated Time:2m 0s

Question 56Question →

Consider the unbalanced redox reaction taking place in an acidic medium:

\text{a MnO}_4^-(\text{aq}) + \text{b SO}_3^{2-}(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d Mn}^{2+}(\text{aq}) + \text{e SO}_4^{2-}(\text{aq}) + \text{f H}_2\text{O}(\text{l})

When this chemical equation is balanced using the smallest set of whole-number coefficients, what is the value of the stoichiometric coefficient

c

for

\text{H}^+(\text{aq})

?

Show answer & explanation

Answer: 6

Answer

The value of the stoichiometric coefficient c for H+(aq) is 6.

Balancing the reduction half-reaction (

2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

) and oxidation half-reaction (

5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-

) gives a combined total of

16\text{H}^+

on the reactant side and

10\text{H}^+

on the product side. Subtracting

10\text{H}^+

from both sides leaves a net coefficient of 6 for

\text{H}^+(\text{aq})

on the reactant side.

Step-by-Step Solution

1

Write the balanced reduction half-reaction for permanganate ion in acidic medium.

\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5e^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})

Manganese goes from oxidation state +7 to +2, requiring 5 electrons, 8 H+ ions to balance oxygen atoms, forming 4 H2O molecules.

2

Write the balanced oxidation half-reaction for sulfite ion to sulfate ion.

\text{SO}_3^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{SO}_4^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) + 2e^-

Sulfur goes from oxidation state +4 to +6, releasing 2 electrons and 2 H+ ions while consuming 1 H2O molecule.

3

Equalize the number of transferred electrons by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 5.

2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-

The least common multiple of 5 and 2 transferred electrons is 10.

4

Combine the half-reactions and subtract common species (

10e^-

,

10\text{H}^+

, and

5\text{H}_2\text{O}

) from both sides.

2\text{MnO}_4^-(\text{aq}) + 5\text{SO}_3^{2-}(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{SO}_4^{2-}(\text{aq}) + 3\text{H}_2\text{O}(\text{l})

Subtracting

10\text{H}^+

from

16\text{H}^+

leaves

6\text{H}^+

on the reactant side, giving

c = 6

.

Key Concept

Balancing Redox Equations using the Ion-Electron Method in Acidic Medium

Question 57Question →

Match each redox reagent mixture or reaction system with its corresponding characteristic diagnostic observation.

Click a left item, then click its matching right item

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Answer

Acidified potassium heptaoxodichromate(VI) with sulfur(IV) oxide matches the orange to green color change. Freshly prepared iron(II) tetraoxosulfate(VI) with concentrated trioxonitrate(V) acid matches the pale green to reddish-brown change. Aqueous potassium iodide with chlorine gas and starch matches the blue-black complex formation. Acidified potassium tetraoxomanganate(VII) with hydrogen peroxide matches decolorization from purple to colorless with oxygen gas effervescence.

Each test reagent matches its specific chemical observation based on electron transfer and oxidation state changes:

K_2Cr_2O_7

turns green when reduced by

SO_2

;

Fe^{2+}

turns brown when oxidized by conc.

HNO_3

;

KI

yields free

I_2

which gives a blue-black color with starch when oxidized by

Cl_2

; and acidified

KMnO_4

is decolorized with

O_2

evolution when reduced by

H_2O_2

.

Step-by-Step Solution

1

Analyze the reduction of acidified

K_2Cr_2O_7

by

SO_2

Chromium decreases in oxidation number from

+6

in

Cr_2O_7^{2-}

to

+3

in

Cr^{3+}

, causing a distinct color shift from orange to green.

SO_2

acts as a reductant and

K_2Cr_2O_7

as an oxidant.

2

Analyze the oxidation of

FeSO_4

by concentrated

HNO_3

Iron increases in oxidation state from

+2

(

Fe^{2+}

, pale green) to

+3

(

Fe^{3+}

, brown/yellowish-brown).

Concentrated

HNO_3

is a strong oxidizing agent.

3

Analyze halogen displacement of

KI

by

Cl_2

Chlorine oxidizes

I^-

to

I_2

. Liberated

I_2

forms a blue-black starch-iodine inclusion complex.

Chlorine has a higher standard reduction potential than iodine.

4

Analyze the redox reaction between acidified

KMnO_4

and

H_2O_2

Manganese is reduced from

+7

(

MnO_4^-

, purple) to

+2

(

Mn^{2+}

, colorless), while peroxide oxygen is oxidized from

-1

to

0

(

O_2

gas bubbles).

In the presence of a stronger oxidant (

KMnO_4

),

H_2O_2

behaves as a reducing agent.

Key Concept

Oxidizing and Reducing Agents and Diagnostic Chemical Tests

Estimated Time:1m 30s

Question 58Question →

Fill in the blanks to complete the chemical observation and oxidation state change during the test for sulfur(IV) oxide gas. What are the correct terms to fill in the blanks?

Fill in the blanks below

When

SO_2

gas is passed through an acidified solution of potassium heptaoxodichromate(VI),

K_2Cr_2O_7

, the solution turns from orange to because the dichromate ion is reduced, changing the oxidation state of chromium from

+6

to .

Show answer & explanation

Answer

The solution turns green because chromium is reduced from an oxidation state of +6 to +3.

Sulfur(IV) oxide (

SO_2

) is a strong reducing agent. When passed into an acidified solution of potassium heptaoxodichromate(VI), it reduces the orange dichromate ion (

Cr_2O_7^{2-}

, where

Cr

is in the

+6

oxidation state) to the green chromium(III) ion (

Cr^{3+}

, where

Cr

is in the

+3

oxidation state). Thus, the color turns green and the final oxidation state is

+3

.

Step-by-Step Solution

1

Identify the role of

SO_2

and

K_2Cr_2O_7

in the redox reaction.

SO_2

acts as a reducing agent and is oxidized to

SO_4^{2-}

, while

K_2Cr_2O_7

acts as an oxidizing agent.

Reducing agents cause the reduction of other species while being oxidized themselves.

2

Determine the color change of the acidified

K_2Cr_2O_7

solution.

The orange

Cr_2O_7^{2-}

ion is reduced to the green

Cr^{3+}

ion.

The formation of hydrated chromium(III) ions in solution imparts a distinct green color.

3

Determine the initial and final oxidation states of chromium.

In

Cr_2O_7^{2-}

, chromium has an oxidation state of

+6

. Upon reduction to

Cr^{3+}

, its oxidation state becomes

+3

.

The half-reaction is

Cr_2O_7^{2-} + 14H^+ + 6e^- \nrightarrow 2Cr^{3+} + 7H_2O

.

Key Concept

Laboratory identification test for reducing agents using acidified potassium heptaoxodichromate(VI)

Question 59Question →

Match each redox testing reagent or indicator on the left with its characteristic diagnostic observation on the right when reacting with an oxidizing or reducing agent.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Acidified potassium tetraoxomanganate(VII) turns from purple to colorless with reducing agents; starch-iodide paper turns blue-black with oxidizing agents; acidified potassium heptaoxodichromate(VI) turns from orange to green with reducing agents; and freshly prepared iron(II) sulfate changes from pale green to reddish-brown with oxidizing agents.

Each testing reagent displays a distinct diagnostic color change depending on whether it reacts with an oxidizing or reducing agent. Acidified

\text{KMnO}_4

changes from purple to colorless in the presence of a reducing agent due to reduction of

\text{MnO}_4^-

to

\text{Mn}^{2+}

. Starch-iodide paper turns blue-black in the presence of an oxidizing agent as iodide is oxidized to iodine. Acidified

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

changes from orange to green in the presence of a reducing agent as

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

is reduced to

\text{Cr}^{3+}

. Freshly prepared

\text{FeSO}_4

changes from pale green to reddish-brown when an oxidizing agent oxidizes

\text{Fe}^{2+}

to

\text{Fe}^{3+}

.

Step-by-Step Solution

1

Identify the role of acidified

\text{KMnO}_4

solution in redox testing.

Acidified

\text{KMnO}_4

contains manganese in the

+7

oxidation state (purple). When it oxidizes a reducing agent, manganese is reduced to

\text{Mn}^{2+}

(colorless).

This is the classic quantitative and qualitative test for reducing agents.

2

Determine the response of moist starch-iodide paper to oxidizing gases.

Oxidizing agents liberate free iodine (

I_2

) from iodide ions (

I^-

). Free iodine reacts with starch to yield a distinctive blue-black color.

This tests specifically for oxidizing agents such as chlorine or ozone.

3

Identify the color change associated with acidified

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

.

Dichromate ions (orange, chromium oxidation state

+6

) are reduced to

\text{Cr}^{3+}

ions (green, oxidation state

+3

) by reducing agents.

The reduction of dichromate(VI) to chromium(III) causes the orange-to-green color change.

4

Determine the oxidation behavior of iron(II) sulfate solution.

Iron(II) ions (pale green) act as a reducing agent and are oxidized to iron(III) ions (reddish-brown/yellow) by oxidizing agents.

The oxidation of

\text{Fe}^{2+}

to

\text{Fe}^{3+}

shifts the solution color from green to brown/yellow.

Key Concept

Laboratory Diagnostic Tests for Oxidizing and Reducing Agents

Question 60Question →

Consider the half-reaction representing the oxidation of thiosulfate ions to tetrathionate ions:

2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + n e^-

What is the number of electrons,

n

, required to balance the charge in this half-reaction?

Show answer & explanation

Answer: 2

Answer

The number of electrons required to balance the charge in the half-reaction is 2.

To balance a half-reaction, both atom counts and net electric charges must be equal on both sides of the equation. The reactant side contains 2 thiosulfate ions (

2\text{S}_2\text{O}_3^{2-}

), giving a net charge of

2 \times (-2) = -4

. The product side contains 1 tetrathionate ion (

\text{S}_4\text{O}_6^{2-}

), giving a net charge of

-2

. Adding 2 electrons (

2 e^-

) to the product side lowers its total charge to

-4

, equalizing the charge on both sides.

Step-by-Step Solution

1

Calculate the total charge of the reactant species.

Reactant charge = 2 * (-2) = -4.

There are 2 thiosulfate ions, each with an ionic charge of -2.

2

Calculate the net charge of the ionic product species.

Product charge (excluding electrons) = -2.

There is 1 tetrathionate ion with an ionic charge of -2.

3

Equate the overall charges on both sides to solve for the number of electrons n.

-4 = -2 - n, giving n = 2.

Adding 2 electrons (each carrying a -1 charge) to the product side brings the total product charge to -4, matching the reactant side.

Key Concept

Balancing electric charge in oxidation half-reactions

Oxidation-Reduction and Electrochemistry

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