Oxidation-Reduction and Electrochemistry Practice Questions — JAMB UTME — Page 4

Consider the standard reduction potentials ( $E^\circ$ ) at $25^\circ\text{C}$ for the following half-reactions:

\text{Ce}^{4+}(aq) + e^- \rightarrow \text{Ce}^{3+}(aq) \quad E^\circ = +1.61\text{ V}

\text{Br}_2(l) + 2e^- \rightarrow 2\text{Br}^-(aq) \quad E^\circ = +1.07\text{ V}

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

\text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq) \quad E^\circ = +0.54\text{ V}

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

Which of the following chemical species can spontaneously oxidize $\text{I}^-(aq)$ to $\text{I}_2(s)$ under standard conditions, but is UNABLE to oxidize $\text{Br}^-(aq)$ to $\text{Br}_2(l)$ ?

\text{Fe}^{3+}(aq)

\text{Ce}^{4+}(aq)

\text{Sn}^{4+}(aq)

\text{Sn}^{2+}(aq)

Show answer & explanation

Answer: $\text{Fe}^{3+}(aq)$

Answer

The species

\text{Fe}^{3+}(aq)

is the correct choice because its standard reduction potential (+0.77 V) lies strictly between the reduction potentials of

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

(+0.54 V) and

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

(+1.07 V).

To oxidize

\text{I}^-

(

E^\circ_{\text{red}} = +0.54\text{ V}

), an oxidizing agent must have a standard reduction potential greater than

+0.54\text{ V}

. To fail to oxidize

\text{Br}^-

(

E^\circ_{\text{red}} = +1.07\text{ V}

), its reduction potential must be less than

+1.07\text{ V}

. The species

\text{Fe}^{3+}(aq)

has

E^\circ = +0.77\text{ V}

, which satisfies

+0.54\text{ V} < +0.77\text{ V} < +1.07\text{ V}

. Thus,

E^\circ_{\text{cell}}(\text{Fe}^{3+}/\text{I}^-) = +0.77 - 0.54 = +0.23\text{ V} > 0

(spontaneous) and

E^\circ_{\text{cell}}(\text{Fe}^{3+}/\text{Br}^-) = +0.77 - 1.07 = -0.30\text{ V} < 0

(non-spontaneous).

Step-by-Step Solution

1

Determine the required condition for spontaneous oxidation of

\text{I}^-(aq)

to

\text{I}_2(s)

.

The oxidation half-reaction is

2\text{I}^-(aq) \rightarrow \text{I}_2(s) + 2e^-

with

E^\circ_{\text{ox}} = -0.54\text{ V}

. For

E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{oxidant}) + E^\circ_{\text{ox}} > 0

, the oxidant must have

E^\circ_{\text{red}} > +0.54\text{ V}

.

A redox reaction is spontaneous if the cell potential

E^\circ_{\text{cell}}

is positive.

2

Determine the required condition for non-spontaneous oxidation of

\text{Br}^-(aq)

to

\text{Br}_2(l)

.

The oxidation half-reaction is

2\text{Br}^-(aq) \rightarrow \text{Br}_2(l) + 2e^-

with

E^\circ_{\text{ox}} = -1.07\text{ V}

. For

E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{oxidant}) + E^\circ_{\text{ox}} < 0

, the oxidant must have

E^\circ_{\text{red}} < +1.07\text{ V}

.

An unfeasible (non-spontaneous) reaction corresponds to a negative overall standard cell potential.

3

Combine the potential boundary constraints and evaluate the options.

The standard reduction potential of the ideal oxidant must satisfy

+0.54\text{ V} < E^\circ_{\text{red}} < +1.07\text{ V}

. Among the choices,

\text{Fe}^{3+}(aq)

has

E^\circ_{\text{red}} = +0.77\text{ V}

, which falls squarely within this range.

Comparing standard electrode potentials directly establishes which species can act as selective oxidizing agents.

Key Concept

Predicting reaction spontaneity and selective oxidation using standard electrode potentials (

E^\circ

).

Estimated Time:2m 0s

Question 62Question →

The standard reduction potentials for tin and silver half-cells are given below:

\text{Sn}^{2+}(aq) + 2e^- \rightarrow \text{Sn}(s) \quad E^\circ = -0.14\text{ V}

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

What is the standard cell potential (

E^\circ_{\text{cell}}

) for the overall redox reaction:

\text{Sn}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Sn}^{2+}(aq) + 2\text{Ag}(s)

+0.94\text{ V}

+0.66\text{ V}

-0.94\text{ V}

+1.74\text{ V}

Show answer & explanation

Answer: $+0.94\text{ V}$

Answer

+0.94\text{ V}

In the given cell reaction, silver ions undergo reduction at the cathode (

E^\circ = +0.80\text{ V}

) while tin metal undergoes oxidation at the anode (

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

). The standard cell potential is calculated using

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.80\text{ V} - (-0.14\text{ V}) = +0.94\text{ V}

. Because

E^\circ

values are intensive properties, the stoichiometric multiplier for the silver half-reaction does not scale its potential value.

Step-by-Step Solution

1

Identify the reduction and oxidation half-reactions from the given reaction equation.

Tin metal (

\text{Sn}

) is oxidized to

\text{Sn}^{2+}

at the anode, while silver ions (

\text{Ag}^+

) are reduced to silver metal at the cathode.

Oxidation occurs at the anode (loss of electrons) and reduction occurs at the cathode (gain of electrons).

2

Apply the formula for standard cell potential

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

.

E^\circ_{\text{cell}} = (+0.80\text{ V}) - (-0.14\text{ V}) = +0.80\text{ V} + 0.14\text{ V} = +0.94\text{ V}

.

Standard electrode potentials are intensive properties and are not multiplied by stoichiometric coefficients.

Key Concept

Calculation of Standard Cell Potential and Reaction Spontaneity

Question 63Question →

Given the standard reduction potentials ( $E^\circ$ ) below, arrange the metals in order of increasing reducing strength (from weakest reducing agent to strongest reducing agent):

\begin{aligned} \text{Ag}^+_{(\text{aq})} + \text{e}^- &\rightarrow \text{Ag}_{(\text{s})} \quad E^\circ = +0.80\text{ V} \\ \text{Cu}^{2+}_{(\text{aq})} + 2\text{e}^- &\rightarrow \text{Cu}_{(\text{s})} \quad E^\circ = +0.34\text{ V} \\ \text{Fe}^{2+}_{(\text{aq})} + 2\text{e}^- &\rightarrow \text{Fe}_{(\text{s})} \quad E^\circ = -0.44\text{ V} \\ \text{Zn}^{2+}_{(\text{aq})} + 2\text{e}^- &\rightarrow \text{Zn}_{(\text{s})} \quad E^\circ = -0.76\text{ V} \end{aligned}

Which sequence represents the correct order of increasing reducing strength?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of increasing reducing strength is Silver (

\text{Ag}

), Copper (

\text{Cu}

), Iron (

\text{Fe}

), and Zinc (

\text{Zn}

).

Reducing power increases as standard reduction potential (

E^\circ

) becomes more negative, because metals with negative reduction potentials readily lose electrons. Silver has the most positive reduction potential (

+0.80\text{ V}

), followed by copper (

+0.34\text{ V}

), iron (

-0.44\text{ V}

), and zinc (

-0.76\text{ V}

). Therefore, the order from weakest to strongest reducing agent is Silver, Copper, Iron, and Zinc.

Step-by-Step Solution

1

Understand the relationship between standard reduction potential (

E^\circ

) and reducing strength.

A species with a more negative standard reduction potential has a greater tendency to undergo oxidation (lose electrons) and is therefore a stronger reducing agent. A species with a more positive

E^\circ

is a weaker reducing agent.

Reducing strength is inversely proportional to standard reduction potential.

2

Compare the given

E^\circ

values.

E^\circ(\text{Ag}^+/\text{Ag}) = +0.80\text{ V} > E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V} > E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\text{ V} > E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}

Listing potentials from most positive to most negative orders the elements from weakest to strongest reducing agent.

3

Arrange the metals in increasing order of reducing power.

Silver (

\text{Ag}

) < Copper (

\text{Cu}

) < Iron (

\text{Fe}

) < Zinc (

\text{Zn}

)

Silver has the highest

E^\circ

and is the weakest reducing agent, while zinc has the lowest

E^\circ

and is the strongest reducing agent.

Key Concept

Reducing Strength and Standard Electrode Potentials

Question 64Question →

Complete the statement describing the redox reaction and observation when hydrogen sulfide gas acts as a reducing agent with iron(III) chloride solution.

Fill in the blanks below

When hydrogen sulfide gas (

H_2S

) is bubbled through a reddish-brown aqueous solution of iron(III) chloride (

FeCl_3

), the solution changes color to due to the reduction of

Fe^{3+}

to

Fe^{2+}

ions, while the hydrogen sulfide is oxidized to produce a yellow precipitate of elemental .

Show answer & explanation

Answer

The solution turns green (or pale green) due to the formation of

Fe^{2+}

ions, and a yellow precipitate of sulfur (or sulphur) is deposited.

In the chemical reaction

2Fe^{3+}_{(aq)} + H_2S_{(g)} \rightarrow 2Fe^{2+}_{(aq)} + S_{(s)} + 2H^{+}_{(aq)}

, iron(III) chloride acts as an oxidizing agent and is reduced to green iron(II) ions. Simultaneously, hydrogen sulfide acts as a reducing agent and is oxidized to elemental sulfur, which forms a yellow precipitate.

Step-by-Step Solution

1

Analyze the oxidation state changes in the reaction between

Fe^{3+}

ions and

H_2S

.

Iron is reduced from

+3

in

Fe^{3+}

to

+2

in

Fe^{2+}

. Sulfur is oxidized from

-2

in

H_2S

to

0

in elemental sulfur (

S

).

Iron(III) ions act as an oxidizing agent and undergo reduction, while hydrogen sulfide acts as a reducing agent and undergoes oxidation.

2

Relate the chemical species formed to their characteristic physical observations.

Aqueous

Fe^{3+}

ions (reddish-brown/yellow-brown) are converted into aqueous

Fe^{2+}

ions (green). The oxidation of hydrogen sulfide yields elemental sulfur, which appears as a pale yellow solid precipitate.

Observing distinct color changes and solid deposition is the primary laboratory method for identifying redox species.

Key Concept

Redox testing of hydrogen sulfide as a reducing agent with iron(III) salts

Estimated Time:1m 0s

Question 65Question →

The standard reduction potentials for two half-cell reactions at

25^\circ\text{C}

are given below:

\text{Mn}^{2+}(aq) + 2e^- \rightarrow \text{Mn}(s) \quad E^\circ = -1.18\text{ V}

\text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \quad E^\circ = -0.13\text{ V}

Calculate the standard cell potential ( $E^\circ_{\text{cell}}$ ), in volts, for the spontaneous electrochemical reaction between these two half-cells.

Show answer & explanation

Answer: 1.05

Answer

The standard cell potential for the spontaneous reaction is

+1.05\text{ V}

.

For a spontaneous redox reaction in a galvanic cell, the half-cell with the higher standard reduction potential acts as the cathode, and the one with the lower standard reduction potential acts as the anode. Lead(II) ions (

\text{Pb}^{2+}

,

E^\circ = -0.13\text{ V}

) have a higher reduction potential than manganese(II) ions (

\text{Mn}^{2+}

,

E^\circ = -1.18\text{ V}

). Substituting these values into

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

yields

E^\circ_{\text{cell}} = -0.13\text{ V} - (-1.18\text{ V}) = +1.05\text{ V}

.

Step-by-Step Solution

1

Identify cathode and anode roles for a spontaneous galvanic cell.

Cathode (reduction):

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

with

E^\circ = -0.13\text{ V}

. Anode (oxidation):

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

with

E^\circ = -1.18\text{ V}

.

For a spontaneous process (

E^\circ_{\text{cell}} > 0

), the species with the more positive reduction potential acts as the oxidizing agent and undergoes reduction at the cathode.

2

Use the cell potential formula

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

.

E^\circ_{\text{cell}} = -0.13\text{ V} - (-1.18\text{ V})

Subtracting the standard reduction potential of the anode from that of the cathode gives the overall electromotive force of the cell.

3

Perform the subtraction to find the final numerical answer.

E^\circ_{\text{cell}} = +1.05\text{ V}

-0.13 + 1.18 = 1.05

, confirming a positive standard cell potential for the spontaneous reaction.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Question 66Question →

The standard reduction potentials for four half-cell reactions at $25^\circ\text{C}$ are given below:

1. $\text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd}(s) \quad E^\circ = -0.40\text{ V}$
2. $\text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \quad E^\circ = -0.13\text{ V}$
3. $\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E^\circ = +0.77\text{ V}$
4. $\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ = +0.80\text{ V}$

Which of the following reaction processes is non-spontaneous under standard conditions?

Oxidation of lead metal by aqueous cadmium ions

Oxidation of cadmium metal by aqueous lead ions

Oxidation of iron(II) ions by aqueous silver ions

Reduction of iron(III) ions by cadmium metal

Show answer & explanation

Answer: Oxidation of lead metal by aqueous cadmium ions

Answer

Oxidation of lead metal by aqueous cadmium ions is non-spontaneous because its standard cell potential is negative (

E^\circ_{\text{cell}} = -0.27\text{ V}

).

The reaction process involving the oxidation of lead metal by aqueous cadmium ions requires cadmium ions to act as the oxidizing agent (reduced at the cathode) and lead metal to act as the reducing agent (oxidized at the anode). Calculating the standard cell potential gives

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.40\text{ V} - (-0.13\text{ V}) = -0.27\text{ V}

. Because

E^\circ_{\text{cell}}

is negative, the reaction is non-spontaneous under standard conditions.

Step-by-Step Solution

1

Identify the reduction and oxidation half-reactions for the proposed reaction.

For the oxidation of lead metal by cadmium ions: Cathode (reduction):

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

, Anode (oxidation):

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

.

Reaction spontaneity depends on the overall standard cell potential, calculated as

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

.

2

Substitute the standard reduction potentials into the cell potential formula.

E^\circ_{\text{cell}} = E^\circ(\text{Cd}^{2+}/\text{Cd}) - E^\circ(\text{Pb}^{2+}/\text{Pb}) = -0.40\text{ V} - (-0.13\text{ V}) = -0.27\text{ V}

.

The cathode potential is the reduction potential of the species being reduced, and the anode potential is the reduction potential of the species being oxidized.

3

Evaluate reaction spontaneity based on the sign of

E^\circ_{\text{cell}}

.

Since

E^\circ_{\text{cell}} = -0.27\text{ V} < 0

, this reaction cannot occur spontaneously under standard conditions.

A chemical reaction is spontaneous under standard conditions if and only if

E^\circ_{\text{cell}} > 0

.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Question 67Question →

Complete the statement below regarding the characteristic laboratory test observation and chemical role of acidified potassium tetraoxomanganate(VII) when reacted with iron(II) tetraoxosulfate(VI).

Fill in the blanks below

When acidified potassium tetraoxomanganate(VII) solution is added to a solution containing iron(II) tetraoxosulfate(VI), the purple color of the permanganate solution turns , because the permanganate ion acts as an agent.

Show answer & explanation

Answer

The purple solution turns colorless because potassium tetraoxomanganate(VII) acts as an oxidizing agent.

Acidified potassium tetraoxomanganate(VII) contains the intensely purple

MnO_4^-

ion. Upon reacting with a reducing agent like

Fe^{2+}

, the

MnO_4^-

ion accepts electrons and is reduced to the colorless

Mn^{2+}

ion. Because it removes electrons from iron(II) ions and oxidizes them to iron(III), potassium tetraoxomanganate(VII) functions as an oxidizing agent.

Step-by-Step Solution

1

Determine the color change of acidified potassium tetraoxomanganate(VII) when reacted with a reducing agent such as iron(II) ions.

The manganese in

MnO_4^-

(purple) is reduced to

Mn^{2+}

, which is colorless in dilute aqueous solution.

Permanganate ions undergo reduction from an oxidation state of +7 to +2 in acidic media.

2

Identify the chemical role of potassium tetraoxomanganate(VII) in this redox reaction.

Potassium tetraoxomanganate(VII) accepts electrons from iron(II) ions, converting

Fe^{2+}

to

Fe^{3+}

, making it an oxidizing agent.

A chemical species that causes another substance to be oxidized while undergoing reduction itself is defined as an oxidizing agent.

Key Concept

Laboratory test for reducing agents using acidified potassium tetraoxomanganate(VII)

Question 68Question →

Match each chemical system in Column A with its correct outcome and rationale based on the electrochemical series in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potassium with zinc sulfate matches spontaneous displacement due to a more negative reduction potential; Copper with hydrochloric acid matches no reaction due to a positive reduction potential relative to hydrogen; Fluorine with sodium chloride matches spontaneous halide oxidation due to a higher reduction potential; Silver with gold(III) nitrate matches spontaneous metal displacement because silver acts as a stronger reducing agent than gold.

Each chemical system correctly pairs with its electrochemical behavior: Potassium displaces Zinc because it possesses a more negative reduction potential; Copper does not react with dilute acid because its reduction potential is positive relative to hydrogen; Fluorine displaces chloride ions because it has a higher reduction potential and thus greater oxidizing power; Silver displaces gold ions because silver has a lower reduction potential than gold, making it the stronger reducing agent.

Step-by-Step Solution

1

Analyze position of Potassium and Zinc in the electrochemical series.

Potassium (

E^\circ = -2.93\text{ V}

) has a more negative reduction potential than Zinc (

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

), making it a stronger reducing agent.

Metals with more negative standard reduction potentials spontaneously displace ions of metals below them in the series.

2

Evaluate the reactivity of Copper in non-oxidizing acid.

Copper (

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

) lies below Hydrogen (

E^\circ = 0.00\text{ V}

) in the electrochemical series.

Metals with positive reduction potentials cannot spontaneously reduce hydrogen ions to evolve

\text{H}_2

gas.

3

Compare oxidizing strengths of Fluorine and Chlorine.

Fluorine (

E^\circ = +2.87\text{ V}

) has a higher reduction potential than Chlorine (

E^\circ = +1.36\text{ V}

).

A halogen with a higher reduction potential acts as a stronger oxidizing agent and displaces halide ions with lower reduction potentials.

4

Determine feasibility of displacement between Silver and Gold ions.

Silver (

E^\circ = +0.80\text{ V}

) is more easily oxidized than Gold (

E^\circ = +1.50\text{ V}

).

The metal with the smaller reduction potential acts as the reducing agent, resulting in a positive standard cell potential (

E^\circ_{\text{cell}} > 0

).

Key Concept

Electrochemical Series and Reaction Spontaneity

Estimated Time:1m 30s

Question 69Question →

A steady electric current is passed through an aqueous solution of zinc tetraoxosulfate(VI) for $4825\text{ s}$ . If $3.25\text{ g}$ of zinc is deposited at the cathode, what is the magnitude of the electric current, in Amperes, used?

$[\text{Zn} = 65,\text{ 1 F} = 96500\text{ C mol}^{-1}]$

Show answer & explanation

Answer: 2

Answer

The magnitude of the electric current required is

2.0\text{ A}

.

Depositing

3.25\text{ g}

of

\text{Zn}

(atomic mass

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

) requires

0.05\text{ mol}

of zinc metal. Since each

\text{Zn}^{2+}

ion requires

2

electrons to be reduced,

0.10\text{ mol}

of electrons (

9650\text{ C}

) must pass through the electrolyte. Dividing this charge by time (

4825\text{ s}

) gives

2.0\text{ A}

.

Step-by-Step Solution

1

Calculate the moles of zinc deposited at the cathode.

Moles of

\text{Zn} = \frac{3.25\text{ g}}{65\text{ g mol}^{-1}} = 0.05\text{ mol}

.

Dividing the mass of metal deposited by its relative atomic mass gives the number of moles deposited.

2

Determine the quantity of electricity in Coulombs needed for the deposition.

Reduction half-reaction:

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

. Moles of

e^- = 2 \times 0.05\text{ mol} = 0.10\text{ mol}

. Quantity of electricity

Q = 0.10\text{ mol} \times 96500\text{ C mol}^{-1} = 9650\text{ C}

.

Faraday's second law relates the mole ratio of electrons to metal ion charge.

3

Calculate the steady electric current in Amperes.

I = \frac{Q}{t} = \frac{9650\text{ C}}{4825\text{ s}} = 2.0\text{ A}

.

Electric current is defined as the rate of charge flow over time (

I = \frac{Q}{t}

).

Key Concept

Faraday's Laws of Electrolysis and Quantitative Calculations

Question 70Question →

An aqueous solution of chromium(III) tetraoxosulfate(VI) is electrolyzed using inert platinum electrodes. A steady current of $5.00\text{ A}$ is passed through the electrolyte for $96.5\text{ minutes}$ . If the cathodic current efficiency for chromium deposition is $75.0\%$ , calculate the mass, in grams, of chromium metal deposited at the cathode. [Molar mass of $\text{Cr} = 52.0\text{ g/mol}$ , $1\text{ F} = 96500\text{ C/mol}$ ]

Show answer & explanation

Answer: 3.9

Answer

3.90\text{ g}

To find the mass of chromium deposited, calculate total charge (

Q = I \times t = 5.00 \times 5790 = 28950\text{ C}

), adjust for

75.0\%

current efficiency (

Q_{\text{eff}} = 21712.5\text{ C}

), convert to Faradays (

0.225\text{ F}

), divide by the valency of 3 for

\text{Cr}^{3+}

to find moles of chromium (

0.075\text{ mol}

), and multiply by molar mass (

52.0\text{ g/mol}

) to yield

3.90\text{ g}

.

Step-by-Step Solution

1

Convert the electrolysis time into seconds

t = 96.5 \times 60 = 5790\text{ s}

Standard SI unit of time (seconds) is required for charge calculation (

Q = I \times t

).

2

Calculate the total charge transferred

Q = 5.00\text{ A} \times 5790\text{ s} = 28950\text{ C}

Determines total quantity of electricity delivered by the current source.

3

Apply the current efficiency percentage

Q_{\text{eff}} = 28950\text{ C} \times 0.750 = 21712.5\text{ C}

Only 75% of the total current is utilized specifically for reducing chromium ions.

4

Convert effective charge into moles of electrons

n_e = \frac{21712.5\text{ C}}{96500\text{ C/mol}} = 0.225\text{ mol } e^-

Faraday's constant gives the charge carried per mole of electrons.

5

Relate moles of electrons to moles of chromium deposited

n_{\text{Cr}} = \frac{0.225}{3} = 0.075\text{ mol}

Reduction of one mole of

\text{Cr}^{3+}

requires three moles of electrons (3 Faradays).

6

Calculate the mass of chromium deposited

m = 0.075\text{ mol} \times 52.0\text{ g/mol} = 3.90\text{ g}

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

Key Concept

Faraday's Laws of Electrolysis and Current Efficiency

Question 71Question →

Given the following standard reduction potentials at

25^\circ\text{C}

:

\text{Al}^{3+}_{\text{(aq)}} + 3\text{e}^- \rightarrow \text{Al}_{\text{(s)}} \quad E^\circ = -1.66\text{ V}

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

Calculate the standard electromotive force ( $E^\circ_{\text{cell}}$ ), in volts, of the galvanic cell formed by coupling these two half-cells.

Show answer & explanation

Answer: 2

Answer

The standard electromotive force (

E^\circ_{\text{cell}}

) of the galvanic cell is

2.00\text{ V}

.

The standard electromotive force of a galvanic cell is defined as

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

. Since copper has a higher standard reduction potential (

+0.34\text{ V}

) than aluminium (

-1.66\text{ V}

), reduction occurs at the copper electrode (cathode) and oxidation occurs at the aluminium electrode (anode). Evaluating the potential difference yields

E^\circ_{\text{cell}} = 0.34\text{ V} - (-1.66\text{ V}) = 2.00\text{ V}

.

Step-by-Step Solution

1

Determine which half-cell undergoes reduction (cathode) and which undergoes oxidation (anode).

Copper is the cathode (

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

) and aluminium is the anode (

E^\circ = -1.66\text{ V}

).

The half-cell with the higher standard reduction potential spontaneously undergoes reduction at the cathode.

2

Apply the formula for calculating standard electromotive force (

E^\circ_{\text{cell}}

).

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

The cell EMF is the standard potential difference between the reduction half-reaction and the oxidation half-reaction.

3

Perform the subtraction to evaluate

E^\circ_{\text{cell}}

.

E^\circ_{\text{cell}} = 0.34 - (-1.66) = 2.00\text{ V}

Subtracting a negative quantity is mathematically equivalent to adding its positive value.

Key Concept

Calculation of standard cell potential (

E^\circ_{\text{cell}}

) from standard electrode reduction potentials.

Question 72Question →

The standard reduction potentials for aluminium and nickel half-cells at

25^\circ\text{C}

are given below:

\text{Al}^{3+}(aq) + 3e^- \rightarrow \text{Al}(s) \quad E^\circ = -1.66\text{ V}

\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \quad E^\circ = -0.25\text{ V}

Calculate the standard electromotive force ( $E^\circ_{\text{cell}}$ ) in volts for the spontaneous reaction between these two half-cells.

Show answer & explanation

Answer: 1.41

Answer

The standard electromotive force (

E^\circ_{\text{cell}}

) for the spontaneous galvanic cell reaction is +1.41 V.

For a spontaneous electrochemical reaction, the standard cell electromotive force (

E^\circ_{\text{cell}}

) must be positive. The half-reaction with the more positive standard reduction potential (

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

at

-0.25\text{ V}

) proceeds as a reduction at the cathode. The half-reaction with the less positive potential (

\text{Al}^{3+}/\text{Al}

at

-1.66\text{ V}

) proceeds as an oxidation at the anode. Calculating

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.25\text{ V} - (-1.66\text{ V}) = +1.41\text{ V}

. Because

E^\circ

is an intensive property, the stoichiometric coefficients used to balance electrons (

2\text{Al} + 3\text{Ni}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Ni}

) do not affect the numerical values of the half-cell potentials.

Step-by-Step Solution

1

Determine which electrode undergoes reduction (cathode) and which undergoes oxidation (anode)

Nickel ion reduction occurs at the cathode (

E^\circ = -0.25\text{ V}

), and aluminium metal oxidation occurs at the anode (

E^\circ = -1.66\text{ V}

).

A galvanic cell operates spontaneously (

E^\circ_{\text{cell}} > 0

) when the half-cell with the more positive standard reduction potential acts as the cathode.

2

State the equation for standard cell potential

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

The cell potential measures the potential difference between the reduction half-reaction and the oxidation half-reaction under standard conditions.

3

Substitute the reduction potential values into the equation

E^\circ_{\text{cell}} = -0.25\text{ V} - (-1.66\text{ V}) = +1.41\text{ V}

Standard electrode potentials are intensive properties; hence, balancing electron stoichiometry does not scale

E^\circ

values.

Key Concept

Calculating standard cell electromotive force (

E^\circ_{\text{cell}}

) and predicting spontaneity from standard reduction potentials

Question 73Question →

What mass of copper is deposited at the cathode when a steady current of

0.50\text{ A}

is passed through an aqueous solution of copper(II) tetraoxosulfate(VI) for

1930\text{ s}

? [

\text{Molar mass of Cu} = 64\text{ g mol}^{-1}, 1\text{ F} = 96,500\text{ C mol}^{-1}

]

0.16\text{ g}

0.32\text{ g}

0.64\text{ g}

1.28\text{ g}

Show answer & explanation

Answer: $0.32\text{ g}$

Answer

The correct mass of copper deposited is

0.32\text{ g}

.

Passing

0.50\text{ A}

for

1930\text{ s}

transfers

965\text{ C}

of electricity, which equals

0.01\text{ mol}

of electrons. Because reduction of

Cu^{2+}

requires

2

moles of electrons per mole of copper metal deposited (

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

),

0.005\text{ mol}

of copper is produced. Multiplying

0.005\text{ mol}

by the molar mass of copper (

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

) gives

0.32\text{ g}

.

Step-by-Step Solution

1

Calculate the total quantity of electricity (

Q

) transferred in Coulombs.

Q = I \times t = 0.50\text{ A} \times 1930\text{ s} = 965\text{ C}

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

2

Determine the number of moles of electrons transferred.

\text{Moles of } e^- = \frac{965\text{ C}}{96,500\text{ C mol}^{-1}} = 0.01\text{ mol}

One Faraday (

96,500\text{ C}

) corresponds to one mole of electrons.

3

Use the cathode half-reaction stoichiometry to find the moles of copper deposited.

Cu^{2+} + 2e^- \rightarrow Cu(s) \implies \text{Moles of Cu} = \frac{0.01\text{ mol } e^-}{2} = 0.005\text{ mol}

Copper(II) ions require 2 moles of electrons per mole of copper metal deposited.

4

Calculate the mass of copper deposited.

\text{Mass} = \text{moles} \times \text{molar mass} = 0.005\text{ mol} \times 64\text{ g mol}^{-1} = 0.32\text{ g}

Mass is found by multiplying the quantity in moles by the molar mass.

Key Concept

Faraday's First Law of Electrolysis and Quantitative Stoichiometry of Electrode Reactions

Question 74Question →

Match each chemical testing reagent or reaction system on the left with its corresponding diagnostic observation and redox transformation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Acidified potassium dichromate(VI) solution turns from orange to green (

Cr_2O_7^{2-} \rightarrow Cr^{3+}

); starch-potassium iodide paper turns blue-black (

I^- \rightarrow I_2

); iron(III) chloride solution turns from reddish-brown to pale green with yellow sulfur precipitate (

Fe^{3+} \rightarrow Fe^{2+}

); acidified potassium tetraoxomanganate(VII) turns from purple to colorless (

MnO_4^- \rightarrow Mn^{2+}

).

Each laboratory reagent exhibits a specific color change reflecting its underlying redox reaction: potassium dichromate(VI) changes from orange to green (

Cr_2O_7^{2-} \rightarrow Cr^{3+}

); starch-potassium iodide paper turns blue-black due to iodine liberation (

I^- \rightarrow I_2

); iron(III) chloride changes from reddish-brown to pale green with yellow sulfur precipitation (

Fe^{3+} \rightarrow Fe^{2+}

); and potassium tetraoxomanganate(VII) changes from purple to colorless (

MnO_4^- \rightarrow Mn^{2+}

).

Step-by-Step Solution

1

Determine the diagnostic color change for acidified potassium dichromate(VI).

Orange

Cr_2O_7^{2-}

ions are reduced to green

Cr^{3+}

ions when reacting with a reducing agent.

Chromium undergoes a reduction in oxidation state from

+6

to

+3

.

2

Analyze the chemistry of the starch-potassium iodide test paper.

Oxidizing agents convert

I^-

(iodide) into

I_2

(iodine), producing a characteristic blue-black complex with starch.

Iodide ions act as the reducing agent in the indicator paper and undergo oxidation.

3

Identify the reaction between iron(III) ions and hydrogen sulfide.

Reddish-brown

Fe^{3+}

is reduced to pale green

Fe^{2+}

, while sulfide (

S^{2-}

) is oxidized to free yellow sulfur (

S

).

H_2S

is a classic laboratory reducing agent for iron(III) salts.

4

Determine the diagnostic observation for acidified potassium tetraoxomanganate(VII).

Purple

MnO_4^-

ions are reduced to virtually colorless

Mn^{2+}

ions in acid medium.

Manganese undergoes reduction from

+7

to

+2

oxidation state.

Key Concept

Qualitative laboratory tests and color changes associated with common oxidizing and reducing agents.

Question 75Question →

During the electrolysis of concentrated sodium chloride solution (brine) using inert platinum electrodes, chlorine gas is liberated at the anode instead of oxygen gas. Which factor primarily accounts for the preferential discharge of chloride ions over hydroxide ions in this process?

The high concentration of chloride ions relative to hydroxide ions in the solution

The lower position of chloride ions relative to hydroxide ions in the electrochemical series

The chemical affinity between the platinum electrode and chloride ions

The greater ionic mass of chloride ions compared to hydroxide ions

Show answer & explanation

Answer: The high concentration of chloride ions relative to hydroxide ions in the solution

Answer

The high concentration of chloride ions relative to hydroxide ions in the solution

Under standard conditions, hydroxide ions are discharged preferentially to chloride ions because hydroxide ions lie lower in the electrochemical series of anions. However, in concentrated brine, the concentration of chloride ions is vastly higher than that of hydroxide ions, causing concentration to become the decisive factor leading to chlorine gas evolution.

Step-by-Step Solution

1

Identify the ions present at the anode during electrolysis of concentrated sodium chloride solution.

The anions attracted to the anode are chloride ions (

Cl^-

) and hydroxide ions (

OH^-

).

Anions migrate to the positively charged anode.

2

Compare the factors governing preferential discharge of these anions.

Based on position in the electrochemical series,

OH^-

is discharged more easily than

Cl^-

. However, because the solution is concentrated,

Cl^-

ions far outnumber

OH^-

ions.

High concentration of a particular ion can overcome its disadvantage in electrochemical series position.

3

Deduce the discharged product and the dominant factor.

Chloride ions (

Cl^-

) are discharged to form chlorine gas (

Cl_2

) due to the concentration effect.

The concentration factor overrides the standard electrochemical series order in concentrated brine.

Key Concept

Effect of concentration on preferential discharge of ions during electrolysis

Question 76Question →

Consider a standard galvanic cell constructed using cobalt and silver half-cells under standard conditions:

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

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

Which of the following statements correctly describes the operational mechanics of this cell?

Cobalt acts as the cathode and undergoes oxidation, leading to a decrease in electrode mass.

Electrons flow spontaneously from the cobalt electrode to the silver electrode through the external conductor.

The standard electromotive force (

E^\circ_{\text{cell}}

) generated by the cell is

+0.52\text{ V}

.

Nitrate ions (

\text{NO}_3^-

) from the salt bridge migrate into the silver half-cell to maintain electrical neutrality.

Show answer & explanation

Answer: Electrons flow spontaneously from the cobalt electrode to the silver electrode through the external conductor.

Answer

Electrons flow spontaneously from the cobalt electrode to the silver electrode through the external conductor.

In a galvanic cell, oxidation occurs at the electrode with the lower standard reduction potential (cobalt electrode, making it the anode). Reduction occurs at the electrode with the higher standard reduction potential (silver electrode, making it the cathode). Electrons are generated at the anode by oxidation and travel through the external circuit to the cathode.

Step-by-Step Solution

1

Identify the anode and cathode based on standard reduction potentials (

E^\circ

).

The half-cell with the more negative standard reduction potential (

E^\circ(\text{Co}^{2+}/\text{Co}) = -0.28\text{ V}

) undergoes oxidation at the anode. The half-cell with the more positive potential (

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

) undergoes reduction at the cathode.

Species with higher standard reduction potentials are more easily reduced, while those with lower reduction potentials are more easily oxidized.

2

Determine the direction of electron flow in the external circuit.

Oxidation at the cobalt anode releases electrons:

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

. These electrons flow through the wire toward the silver cathode.

Electrons always flow spontaneously from the anode (site of oxidation) to the cathode (site of reduction) in a galvanic cell.

3

Calculate the cell potential (

E^\circ_{\text{cell}}

) to verify consistency.

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.80\text{ V} - (-0.28\text{ V}) = +1.08\text{ V}

.

A positive standard cell electromotive force confirms that the cell reaction is spontaneous in the stated direction.

Key Concept

Galvanic Cell Mechanics and Standard Electrode Potentials

Question 77Question →

The standard reduction potentials for gallium and nickel half-reactions at

25^\circ\text{C}

are given below:

\text{Ga}^{3+}(aq) + 3e^- \rightarrow \text{Ga}(s) \quad E^\circ = -0.56\text{ V}

\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \quad E^\circ = -0.25\text{ V}

What is the standard electromotive force (

E^\circ_{\text{cell}}

) for the spontaneous reaction between these two electrochemical systems?

+0.31\text{ V}

-0.31\text{ V}

+0.37\text{ V}

+0.81\text{ V}

Show answer & explanation

Answer: $+0.31\text{ V}$

Answer

+0.31\text{ V}

For a redox reaction to be spontaneous under standard conditions, the cell potential (

E^\circ_{\text{cell}}

) must be positive. In the electrochemical series, species with higher reduction potentials undergo reduction (cathode). Comparing

-0.25\text{ V}

(nickel) and

-0.56\text{ V}

(gallium), nickel has the higher potential and acts as the cathode. Applying

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

yields

-0.25\text{ V} - (-0.56\text{ V}) = +0.31\text{ V}

.

Step-by-Step Solution

1

Identify cathode and anode based on standard reduction potentials for spontaneity

Cathode:

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

(

E^\circ = -0.25\text{ V}

), Anode:

\text{Ga}^{3+}/\text{Ga}

(

E^\circ = -0.56\text{ V}

)

The half-cell with the more positive (less negative) standard reduction potential undergoes reduction at the cathode in a spontaneous cell.

2

Calculate the standard cell potential using

E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}}

E^\circ_{\text{cell}} = -0.25\text{ V} - (-0.56\text{ V}) = +0.31\text{ V}

Standard electromotive force is the difference between the reduction potential of the cathode and the reduction potential of the anode.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Estimated Time:1m 30s

Question 78Question →

During the electrolysis of an aqueous solution containing equal concentrations of cations using inert carbon electrodes, ions migrate to the cathode where discharge occurs based on their position in the electrochemical series. Arrange the following cations in increasing order of their ease of preferential discharge at the cathode (from the least easily discharged to the most easily discharged):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

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

K^+

), Zinc ion (

Zn^{2+}

), Hydrogen ion (

H^+

), and Copper(II) ion (

Cu^{2+}

).

In electrolysis at inert electrodes under equal concentrations, the primary factor determining preferential discharge of cations at the cathode is their position in the electrochemical series. Cations positioned lower in the series gain electrons more easily (have higher reduction potentials). Therefore, Potassium ion (

K^+

) is the hardest to discharge, followed by Zinc ion (

Zn^{2+}

), Hydrogen ion (

H^+

), and finally Copper(II) ion (

Cu^{2+}

), which is the most easily discharged.

Step-by-Step Solution

1

Recall the electrochemical series position for cations

The order of cations from top (most reactive / hardest to discharge) to bottom (least reactive / easiest to discharge) is

K^+

,

Zn^{2+}

,

H^+

,

Cu^{2+}

.

Ions lower in the electrochemical series have higher standard reduction potentials and accept electrons more readily at the cathode.

2

Arrange the cations in increasing order of ease of discharge

Potassium ion (

K^+

) < Zinc ion (

Zn^{2+}

) < Hydrogen ion (

H^+

) < Copper(II) ion (

Cu^{2+}

).

Ease of discharge increases down the electrochemical series.

Key Concept

Position of cations in the electrochemical series determines their relative ease of discharge at the cathode.

Question 79Question →

Consider the unbalanced redox reaction occurring in acidic solution:

\text{a AsO}_3^{3-}(\text{aq}) + \text{b MnO}_4^-(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d AsO}_4^{3-}(\text{aq}) + \text{e Mn}^{2+}(\text{aq}) + \text{f H}_2\text{O}(\text{l})

When this ionic equation is balanced using the smallest possible whole-number coefficients, what is the value of the coefficient

c

for

\text{H}^+

?

Show answer & explanation

Answer: 6

Answer

The coefficient c for hydrogen ions (H⁺) in the balanced redox equation is 6.

Balancing the oxidation half-reaction shows that each arsenite ion produces 2 electrons and 2 H⁺ ions. The reduction half-reaction shows that each permanganate ion consumes 5 electrons and 8 H⁺ ions. Multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2 balances the total electron transfer at 10 electrons. Combining the equations gives 16 H⁺ on the left and 10 H⁺ on the right, which simplifies to 6 H⁺ on the reactant side.

Step-by-Step Solution

1

Balance the oxidation half-reaction (arsenite to arsenate)

AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻

Arsenic changes oxidation state from +3 to +5, releasing 2 electrons. Oxygen is balanced with H₂O and hydrogen with H⁺.

2

Balance the reduction half-reaction (permanganate to manganese(II))

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Manganese changes oxidation state from +7 to +2, consuming 5 electrons in acidic medium.

3

Equalize the electrons transferred in both half-reactions

5(AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻) and 2(MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O)

The total number of electrons gained and lost must equal 10 electrons.

4

Combine the half-reactions and cancel common terms

5 AsO₃³⁻ + 2 MnO₄⁻ + 6 H⁺ → 5 AsO₄³⁻ + 2 Mn²⁺ + 3 H₂O

Subtracting 10 H⁺ and 5 H₂O from both sides leaves 6 H⁺ on the reactant side.

Key Concept

Balancing Ion-Electron Redox Half-Reactions in Acidic Medium

Question 80Question →

An aqueous solution of copper(II) sulfate ( $CuSO_4$ ) undergoes electrolysis using copper sheets as both the anode and cathode. Which statement correctly describes the reaction occurring at the anode and identifies the primary factor governing this process?

The copper anode dissolves into solution as

Cu^{2+}

ions, governed by the nature of the electrode material.

Hydroxide ions (

OH^-

) are discharged to liberate oxygen gas, governed strictly by their higher position in the electrochemical series relative to sulfate ions (

SO_4^{2-}

).

Sulfate ions (

SO_4^{2-}

) are preferentially discharged to produce sulfur dioxide gas, governed by the high concentration of sulfate ions relative to hydroxide ions (

OH^-

).

Hydrogen ions (

H^+

) are discharged to liberate hydrogen gas, governed by the low electrode potential of water molecules.

Show answer & explanation

Answer: The copper anode dissolves into solution as $Cu^{2+}$ ions, governed by the nature of the electrode material.

Answer

The copper anode dissolves into solution as

Cu^{2+}

ions, governed by the nature of the electrode material.

In the electrolysis of copper(II) sulfate solution using active copper electrodes, oxidation occurs at the anode. Because copper is an active metal electrode, dissolving copper metal into copper(II) ions (

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

) requires less energy than discharging hydroxide ions or sulfate ions. Thus, the nature of the electrode material is the governing factor.

Step-by-Step Solution

1

Identify the ions present in aqueous

CuSO_4

solution.

Cations:

Cu^{2+}

and

H^+

. Anions:

SO_4^{2-}

and

OH^-

.

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

2

Analyze the nature of the electrode at the anode.

The anode is made of copper, which is an active (reactive) electrode, not an inert electrode like platinum or graphite.

Active electrodes participate directly in the electrochemical reaction.

3

Determine the preferred anodic oxidation reaction.

Copper atoms from the anode oxidize preferentially (

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

) instead of discharging

OH^-

or

SO_4^{2-}

anions.

The energy required to oxidize metallic copper atoms is lower than that needed to discharge anions from the solution, demonstrating that the nature of the electrode overrides electrochemical series position and ion concentration.

Key Concept

Influence of the Nature of Electrodes on Preferential Discharge

Estimated Time:2m 0s

Oxidation-Reduction and Electrochemistry

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