Reactivity 3.2. Electron transfer reactions

Consider the following standard electrode potential values: 

Fe³⁺(aq) + 2e⁻ ⇌ Fe²⁺(aq); E° = +0.77 V 

Ni²⁺(aq) + 2e⁻ ⇌ Ni(s); E° = −0.26 V 

Fe²⁺(aq) + 2e⁻ ⇌ Fe(s); E° = −0.45 V 

Ca²⁺(aq) + 2e⁻ ⇌ Ca(s); E° = −2.87 V

Answer: B. Fe(s) + 2Fe³⁺(s) → 3Fe²⁺(aq) 

A spontaneous reaction occurs when

`""E_"cell"^∘=E_"reduction(cathode)"^∘-E_"reduction(anode)"^∘ > 0`

The more positive E^∘ is the stronger the oxidizing agent. The more negative E^∘ is, the  stronger the reducing agent. 

A. Incorrect:  

Fe²⁺ + Ni(s) → Fe(s) + Ni²⁺

   • Ni(s) → Ni²⁺ + 2e⁻ (oxidation, anode, E^∘ = −0.26) 

   • Fe²⁺ + 2e⁻ → Fe(s) (reduction, cathode, E^∘ = −0.45) 

`""E_"cell"^o`= −0.45 − (−0.26) = −0.19 V

→ Not spontaneous 

B. Correct:  

Fe(s) + 2Fe³⁺ → 3Fe²⁺

   • Fe(s) → Fe²⁺ + 2e⁻ (oxidation, E^∘ = −0.45) 

   • Fe³⁺ + e⁻ → Fe²⁺ (reduction,E^∘ = +0.77) 

`""E_"cell"^o` = 0.77 − (−0.45) = +1.22 V

→ Spontaneous 

C. Incorrect:  

3Fe²⁺ → Fe(s) +2Fe³⁺

Reverse of option B → non-spontaneous (E^∘ = −1.22 V) 

D. Incorrect: 

Ca²⁺ + Ni(s) → Ca(s) +Ni²⁺

   • Ni(s) → Ni²⁺ + 2e⁻ (E^∘ = −0.26) 

   • Ca²⁺ + 2e⁻ → Ca(s) (E^∘ = −2.87) 

`""E_"cell"^o` = −2.87 − (−0.26) = −2.61 V

→ Not spontaneous

3-hydroxybutanoic acid is a metabolite which the body can use to provide energy when it is low on glucose. 

a. Name the functional groups present in 3-hydroxybutanoic acid. 

b. Draw the two stereoisomers of 3-hydroxybutanoic acid. 

c. Draw the organic product formed when 3-hydroxybutanoic acid is left to react with an excess of a reducing agent and the mixture is quenched with acid.

d. i. When 3-hydroxybutanoic acid is reacted with potassium manganate(VII) (an oxidizing agent) an unstable compound with a degree of unsaturation of 2 is formed.  Suggest the structure of this unstable compound. 

ii. The unstable compound produced decomposes to form carbon dioxide and one other  product with a degree of unsaturation of 1. Identify the product formed in this  decomposition reaction. 

Structure of 3-hydroxybutanoic acid 

CH₃CH(OH)CH₂COOH 

a. Functional groups present 

   • Hydroxyl (–OH) group (on C-3) 

   • Carboxylic acid (–COOH) group 

b. Two stereoisomers

There is one chiral carbon (the carbon bearing the –OH group), so 3-hydroxybutanoic  acid exists as two enantiomers (optical isomers): 

   • (R)-3-hydroxybutanoic acid 

   • (S)-3-hydroxybutanoic acid 

c. Product from reaction with an excess reducing agent 

Reduction converts both functional groups (–COOH → –CH₂OH and –OH remains →  –CH₂OH): 

→ Product: 1,3-butanediol (CH₃CH(OH)CH₂CH₂OH) 

d. i. Reaction with KMnO₄ (oxidation) 

    • Oxidation of the secondary alcohol (–CH(OH)–) gives a carbonyl (C=O) group.

  • The resulting compound has two double bonds (one C=O and one C=C if  unstable) — degree of unsaturation = 2. 

So the unstable compound is acetoacrylic acid: 

CH₂ = C(COOH)–CO–OH 

or more simply represented as: 

CH₂ = C(COOH)–COOH 

d. ii. Decomposition 

This unstable compound decarboxylates (loses CO₂), forming a compound with one  double bond (unsaturation = 1): 

CH₂=CHCOOH

→ Product formed: Acrylic acid (propenoic acid, CH₂=CHCOOH) 

Which statement is correct for a voltaic but not for an electrolytic cell?

A. An electrolyte is required. 

B. The anode is where oxidation occurs. 

C. Ions move in the electrolyte. 

D. Electrons flow from the negative. electrode to the positive electrode.

Answer: D. Electrons flow from the negative electrode to the positive electrode.

A. Incorrect: Both voltaic and electrolytic cells require electrolytes for ion conduction.

B. Incorrect: This is true for both types of cells — oxidation always happens at the anode (by definition).

C. Incorrect: Ion movement happens in both cells to maintain charge balance.

D. Correct: (For a voltaic cell only). 

   • In a voltaic cell, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode). 

   • In an electrolytic cell, the current is forced by an external power supply and the  anode is positive — electrons flow into it instead. 

Which compound can be oxidized when heated with an acidified solution of potassium dichromate(VI)? 

A. CH₃C(O)CH₂CH₃ 

B. CH₃CH₂CH(OH)CH₃ 

C. (CH₃)₃COH 

D. CH₃(CH₂)₂COOH 

Answer: B. CH₃CH₂CH(OH)CH₃ 

Acidified potassium dichromate(VI), K₂Cr₂O₇/H⁺ → is a strong oxidizing agent,  oxidizing: 

   • Primary alcohols → aldehydes → carboxylic acids 

   • Secondary alcohols → ketones 

   • Tertiary alcohols → no reaction 

   • Ketones and carboxylic acids → generally not oxidized under normal conditions.

A. Incorrect: Ketones are not further oxidized by acidified dichromate under mild heating. 

B. Correct: Secondary alcohols can be oxidized to ketones (butan-2-one).

C. Incorrect: Tertiary alcohols resist oxidation under these conditions (no hydrogen on the C–OH carbon). 

D. Incorrect: Already in its fully oxidized form, cannot be further oxidized by dichromate(VI). 

Which of these functional groups will react with a reducing agent?

I. Alkoxy 

II. Carboxyl 

III. Carbonyl 

A. I and II only. 

B. I and III only. 

C. II and III only. 

D. I, II and III. 

Answer: C. II and III only. 

I. Alkoxy (–OR) 

This group is part of ethers (R–O–R) — they are quite stable and not easily reduced by normal reducing agents (e.g., LiAlH₄, NaBH₄). 

⇒ Does not react with reducing agents. 

II. Carboxyl (–COOH)

Carboxylic acids can be reduced (e.g., by LiAlH₄) to primary alcohols.

⇒ Reacts with reducing agents. 

III. Carbonyl (C=O) 

Carbonyl compounds (aldehydes, ketones) readily undergo reduction:

   • Aldehydes → primary alcohols 

   • Ketones → secondary alcohols 

⇒ Reacts with reducing agents. 

All non-cyclic structural isomers of alcohols with molecular formula C₄H₁₀O are reacted with hot acidified KMnO₄. How many will decolourize the purple KMnO₄?

A. 2 

B. 3 

C. 4 

D. 5 

Answer: B. 3 

Step 1: List all non-cyclic structural isomers of alcohols with molecular formula C₄H₁₀O Possible alcohol isomers: 

1. Butan-1-ol — CH₃CH₂CH₂CH₂OH → primary alcohol 

2. Butan-2-ol — CH₃CH(OH)CH₂CH₃ → secondary alcohol 

3. 2-methylpropan-1-ol (isobutanol) — (CH₃)₂CHCH₂OH → primary alcohol

4. 2-methylpropan-2-ol (tert-butanol) — (CH₃)₃COH → tertiary alcohol → Total = 4 isomers 

Step 2: Reactivity with hot acidified KMnO₄ (oxidation) 

   • Primary alcohols → oxidized to carboxylic acids 

   • Secondary alcohols → oxidized to ketones 

   • Tertiary alcohols → no reaction (resists oxidation) 

Step 3: Which decolourise purple KMnO₄? 

What is the oxidation half-equation in the redox reaction? 

2S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + 2I⁻(aq) 

A. I₂(aq) + 2e⁻ → 2I⁻(aq) 

B. 2I⁻(aq) →. I₂(aq) + 2e⁻

C. 2S₂O₃²⁻(aq) → S₄O₆²⁻(aq) + 2e⁻ 

D. S₄O₆²⁻(aq) + 2e⁻ → 2S₂O₃²⁻(aq)

Answer: C. 2S₂O₃²⁻(aq) → S₄O₆²⁻(aq) + 2e⁻

Step 1: Identify oxidation and reduction 

   • I₂ → 2I⁻ : Reduction (gain of electrons). 

   • S₂O₃²⁻ → S₄O₆²⁻: Oxidation (loss of electrons). 

Step 2: Write half-equations 

Reduction (for iodine): 

I₂ + 2e⁻ → 2I⁻

Oxidation (for thiosulfate): 

2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻

Which compounds are susceptible to oxidation with potassium manganate(VII)?

I. CH₃CH₂CH₂CH₂OH 

II. (CH₃)₃CCH₂OH 

III. CH₃CH₂CH(OH)CH₃ 

A. I and II only. 

B. I and III only. 

C. II and III only. 

D. I, II and III.

Answer: D. I, II and III. 

Potassium manganate(VII), KMnO₄, is a strong oxidizing agent that oxidizes primary and secondary alcohols, but not tertiary alcohols. 

Compound I: CH₃CH₂CH₂CH₂OH – This is a primary alcohol (1-butanol). → Can be oxidized to butanal, and then to butanoic acid. 

Compound II: (CH₃)₃CCH₂OH – The hydroxyl group is on a primary carbon (attached  to one other carbon), so it is also a primary alcohol (neopentyl alcohol). → Can be oxidized, although it oxidizes more slowly due to steric hindrance. Compound III: CH₃CH₂CH(OH)CH₃ – The hydroxyl group is attached to a secondary carbon (attached to two other carbons). 

→ Can be oxidized to a ketone (2-butanone).

Consider the following standard electrode potential values: 

Cu⁺(aq) + e⁻ ⇌ Cu(s) E° = +0.52 V 

MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l) E° = +1.51 V

What is the cell potential for this reaction? 

MnO₄⁻(aq) + 8H⁺(aq) + 5Cu(s) → Mn²⁺(aq) + 4H₂O(l) + 5Cu⁺(aq)

A. +2.03 

B. +0.99 

C. −0.99 

D. −2.03 

Answer: B. +0.99 

   • Cathode (reduction): 

MnO₄⁻ + 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O  `""E_"red"^°` = +1.51 V

   • Anode (oxidation): 

Cu(s) ⇌ Cu⁺ + e⁻

The corresponding reduction potential for Cu+/Cu is +0.52 V. 

Cell potential:

`""E_"cell"^∘=E_"cathode"^∘-E_"anode(asreduction")^∘`= 1.51 - 0.52 = +0.99`

During a titrimetric analysis, 25.00 cm³ of an aqueous solution containing ethanol required 37.50 cm³ of 1.50 mol dm⁻³ of acidified potassium dichromate(VI) solution for complete reaction. The density of ethanol is 0.790 g cm⁻³. 

During the reaction, ethanol, CH₃CH₂OH, is oxidized to ethanoic acid, CH₃COOH, while dichromate(VI) ions, Cr₂O₇²⁻, react as shown in the following half-equation:

Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l) 

a. Explain, in terms of change in oxidation state, why dichromate(VI) ions undergo  reduction. 

b. Write the half-equation for the oxidation of ethanol, CH₃CH₂OH, to ethanoic acid,  CH₃COOH. 

c. Use the half-equations to construct an ionic equation for the above redox reaction.

d. Using the titration results and relevant data, calculate the volume of ethanol in 25.00 cm³ of the alcoholic solution. 

e. Determine the concentration as a percentage by volume of the alcoholic solution.

a. Why dichromate(VI) is reduced (oxidation-state view) 

In Cr₂O₇²⁻, chromium is +6, and it becomes +3 in Cr³⁺; the oxidation state decreases (it gains electrons), so dichromate(VI) is reduced.

b. Oxidation half-equation for ethanol → ethanoic acid (acidic solution)

CH₃CH₂OH + H₂O → CH₃COOH + 4H⁺ + 4e⁻ 

c. Overall ionic equation 

Match electrons: 2×(dichromate half-equation) gives 12 e⁻; 3×(ethanol half-equation)  gives 12 e⁻. 

Add and simplify: 

2Cr₂O₇²⁻ + 16H⁺ + 3CH₃CH₂OH → 4Cr³⁺+ 3CH₃COOH + 11H₂O

d. Volume of ethanol in the 25.00 cm³ sample 

Moles of dichromate used: 

`""n_(Cr_2O_7^2-)` = 1.50 moldm⁻³ × 0.03750dm³ = 0.05625 mol 

Electron balance shows Cr₂O₇²⁻ : ethanol = 2 : 3. 

So moles of ethanol: 

n_ethanol = `frac{3}{2}`× 0.05625 = 0.084375 mol 

Mass of ethanol: 

m = n × M = 0.084375 × 46.07 ≈ 3.887 g 

Volume of ethanol (density = 0.790 g cm⁻³): 

V = `frac{m}{rho}=frac{3.887}{0.790}`≈ 4.92 cm³

e. Concentration as % (v/v) 

%`frac{v}{v}=frac{4.92}{25.00}*100`≈ 19.7% (v/v)

For which of the reactions below will the Gibbs energy change ΔG° be the most  negative? 

A. Cu(s) + 2Ag⁺(aq) → 2Ag(s) + Cu²⁺(aq) E° = +0.46 V 

B. Co(s) + Cu²⁺(aq) → Cu(s) + Co²⁺(aq) E° = +0.62 V 

C. Fe²⁺(aq) + Cu²⁺(aq) → Fe³⁺(aq) + Cu⁺(aq) E° = −0.61 V 

D. H₂(g) + Cr²⁺(aq) → Cr(s) + 2H⁺(aq) E° = −0.74 V 

Answer: B. Co(s) + Cu²⁺(aq) → Cu(s) + Co²⁺(aq) E° = +0.62 V

Reason: 

ΔG^o = −nF`""E_"cell"^o`

The most negative ΔG^o occurs for the largest positive value of nE^o.

Compute n (electrons transferred) and E^o: 

A. Incorrect:  

n = 2 (Cu⁰ → +2; 2×Ag⁺ 1e⁻ each). 

nE^o = 2(0.46) = 0.92 

B. Correct:  

n = 2 (Co⁰ → +2; Cu²⁺ → Cu). 

nE^o = 2(0.62) = 1.24 (largest) → most negative ΔG^o. 

C. Incorrect: 

n = 1, E^o = −0.61 → nE^o = −0.61 (overall ΔG^o > 0, not spontaneous).

D. Incorrect: 

n = 2, E^o = −0.74 → nE^o = −1.48 (also ΔG^o > 0). 

Using the standard electrode potentials given, calculate the standard cell potential when an I₂, I⁻ half-cell (E° = +0.54 V) is connected to a Cl₂/Cl⁻ half-cell (E° = +1.36 V).

A. +3.80 V 

B. +1.90 V 

C. +1.64 V 

D. +0.82 V 

Answer: D. +0.82 V 

Step 1. Identify which half-cell is oxidized and which is reduced 

The more positive E^o value means the species is a stronger oxidizing agent (more likely  to be reduced). 

• Cl₂/Cl⁻ has higher E^o= +1.36 V→ Cl₂ is reduced. 

• I₂/ I⁻ will be oxidized (the reverse of its half-equation). 

Step 2. Write half-equations 

• Anode (oxidation): 

2I₂ → I⁻ + 2e⁻

`""E_"oxidation"^o`= -0.54V

• Cathode (reduction):

Cl₂ + 2e⁻ → 2Cl⁻

`""E_"reduction"^o = +1.36`

Step 3. Calculate cell

`""E_"cell"^o=E_"cathode"^o-E_"anode (reduction)"^o`

`""E_"cell"^o`= 1.36 - 0.54 = +0.82 V

→ Cell potential = +0.82 V, spontaneous reaction. 

a. i. Draw a diagram for the voltaic cell formed by connecting the following standard  half-cells: 

Ni(s) | Ni²⁺(aq) || Mn²⁺(aq) | Mn(s) 

ii. Describe the key features of the hydrogen half-cell. 

b. i. Write an equation for the reaction in each half-cell, identifying the species which is  oxidized and the oxidizing agent.

ii. State which electrode is the anode and state the direction of electron flow in the  external circuit. 

iii. For the overall cell, calculate its voltage and state the sign of ΔG. 

a. i. Draw the cell diagram 

The cell is: 

Ni(s) | Ni²⁺(aq) || Mn²⁺(aq) | Mn(s) 

Diagram description: 

• Left-hand side (anode) → Ni(s) electrode dipping into Ni²⁺(aq)

• Right-hand side (cathode) → Mn(s) electrode dipping into Mn²⁺(aq)

• Salt bridge connects both solutions 

• Voltmeter connected externally between the two electrodes

a. ii. Describe the key features of the hydrogen half-cell 

The standard hydrogen electrode (SHE) has: 

1. A platinum electrode coated with platinum black (inert, conducts electrons).

2. A solution containing 1.0 mol dm⁻³ H⁺ ions (usually from HCl).

3. Hydrogen gas at 1 atm bubbled over the electrode. 

4. Temperature at 298 K (25 °C). 

5. Assigned a standard potential of 0.00 V, used as the reference electrode.

b. i. Write half-equations and identify species 

Given standard electrode potentials (approximate typical values):

Ni²⁺ + 2e⁻ → Ni(s), E^o = −0.25 V

Mn²⁺ + 2e⁻ → Mn(s), E^o = −1.18 V

The more positive E^o value means reduction occurs there (cathode). Hence: 

   • Ni²⁺/Ni is reduced (cathode reaction). 

   • Mn(s) is oxidized (anode reaction). 

Half-equations: 

At anode (oxidation): 

Mn(s) → Mn²⁺(aq) + 2e⁻

At cathode (reduction): 

Ni²⁺ (aq) + 2e⁻ → Ni(s) 

Overall equation: 

Mn(s) + Ni²⁺(aq) →Mn²⁺(aq)  + Ni(s) 

   • Oxidized species: Mn 

   • Oxidizing agent: Ni²⁺ 

b. ii. Which electrode is the anode and electron flow direction

   • Anode: Mn electrode (oxidation occurs here). 

   • Cathode: Ni electrode (reduction occurs here). 

   • Electron flow: from Mn → Ni in the external circuit. 

b. iii. Calculate cell voltage and sign of ΔG. 

`""E_"cell"^o = E_"cathode"^o - E_"anode"^o`

`""E_"cell"^o`= (-0.25) - (-1.18) = +0.93V

`""E_"cell"^o` positive → reaction is spontaneous. 

• Therefore, ΔG° is negative.

ΔG° = -nF`""E_"cell"^o` < 0

In copper-plating, orchids are coated with a thin layer of graphite paste before placing  them in a bath of aqueous copper(II) sulfate and electroplating with copper as the anode.

a. Suggest a reason why orchids are first coated with graphite. 

b. Deduce the half equations at the cathode and anode. 

c. To ensure high standards of electroplated orchids, the copper coating must be at least  0.5 mm thick. Given that the total surface area of a typical orchid is 10 cm² and the  operating current is 20 A, calculate the time required to electroplate an orchid. [Density  of copper = 8.96 g cm⁻³] 

a. Graphite makes the orchid surface conductive and inert, so it can act as the cathode without reacting with the electrolyte. 

b. Half-equations (Cu anode | CuSO₄(aq) | graphite-coated orchid cathode)

   • Cathode (on the orchid): 

Cu²⁺(aq) + 2e⁻ → Cu(s) 

   • Anode (copper): 

Cu(s) → Cu2+(aq) + 2e−

c. Thickness required = 0.5 mm = 0.05 cm 

Area = 10 cm² 

Volume of Cu deposited: 

V = A × thickness = 10 × 0.05 = 0.50 cm³ 

Mass of Cu: 

m = pV = 8.96 × 0.50 = 4.48 g 

Moles of Cu: 

n =`frac{m}{M}`= `frac{4.48}{63.55}`≈ 0.0705 mol 

Charge needed (z = 2 for Cu²⁺): 

Q = nzF = 0.0705 × 2 × 96485 ≈ 1.37 × 10⁴ C

Time at I = 20 A:

`t = frac{Q}{I}=frac{1.37*10^4}{20}`≈ 6.8 × 10² s ≈ 11.4 min 

Which components are used to make the standard hydrogen electrode?

A. H₂(g), H⁺(aq), Pt(s). 

B. H₂(g), H⁺(aq), Ni(s). 

C. H₂(g), HO⁻(aq), Pt(s). 

D. H₂(g), HO⁻(aq), Ni(s). 

Answer: A. H₂(g), H⁺(aq), Pt(s). 

The Standard Hydrogen Electrode (SHE) is composed of: 

• Hydrogen gas (H₂) at 1 atm pressure, 

• 1.0 mol dm⁻³ H⁺ ions (usually from a strong acid like HCl)

• An inert platinum (Pt) electrode coated with platinum black to catalyze the redox reaction. 

Half-equation: 

2H⁺(aq) + 2e⁻ ↔ 2H₂(g)

z mol of copper is deposited from CuSO₄ (aq) by a current I in time t. What is the amount of silver, in mol, deposited by electrolysis from AgNO₃ (aq) by a current `frac{I}{2}` in time 2t?

A. `frac{z}{4}`

B. `frac{z}{2}`

C. z 

D. 2z

Answer: D. 2z 

From Faraday’s Law: 

`n = frac{It}{z_eF}`

where 

n = moles of metal deposited, 

I = current, 

t = time, 

z_e = number of electrons per ion (valency), 

F = Faraday constant. 

Step 1: For copper 

Cu²⁺ + 2e^- → Cu(s)

So, z_e = 2.

`""n_"Cu"=frac{It}{2F}=z`

Step 3: For silver 

Ag⁺ + e⁻ → Ag(s)

So, z_e = 1. 

Now the conditions: Current =`frac{I}{2}`, time = 2t

`""n_"Ag"=frac{(frac{I}{2})*(2t)}{1*F}=frac{It}{F}`

Step 4: Express n_Ag in terms of z

From the copper case: 

z = `frac{It}{2F}`⇒ `frac{It}{F}=2z`

Thus,

n_Ag = 2z

a. Electrolysis can be used to obtain chlorine from molten sodium chloride. Write an equation for the reaction occurring at each electrode and describe the two different ways  in which electricity is conducted when the cell is in operation. 

b. In one experiment involving the electrolysis of molten sodium chloride, 0.1 mol of chlorine was formed. Deduce, giving a reason, the amount of sodium formed at the same  time. 

c. In another experiment involving the electrolysis of molten sodium chloride, the time of the electrolysis was halved and the current increased from 1 amp to 5 amp, compared  to the experiment in (b). Deduce the amount of chlorine formed, showing your working.

d. If dilute aqueous sodium chloride is electrolyzed, a different product is obtained at each electrode. Identify the product formed at each electrode and write an equation  showing its formation. 

a. Electrolysis of molten sodium chloride

   • At cathode (reduction):

Na⁺ + e⁻ → Na(l)

   • At anode (oxidation):

2Cl⁻ → Cl₂(g) + 2e⁻

Overall reaction:

2NaCl(l) → 2Na(l) + Cl₂(g)

How electricity is conducted:

• In molten NaCl, conduction is by movement of ions (Na⁺ and Cl⁻).

• In the external circuit, conduction is by movement of electrons through the wires.

b. Given: 0.1 mol Cl₂ is formed.

From the anode reaction:

2Cl⁻ → Cl₂ + 2e⁻

1 mol Cl₂ corresponds to 2 mol e⁻.

→ 0.1 mol Cl₂ corresponds to 0.2 mol e⁻.

At the cathode, each Na⁺ gains 1 e⁻t o form Na.

→ 0.2 mol e⁻ forms 0.2 mol Na.

`=>` Amount of Na formed = 0.2 mol (For every 1 mol Cl₂ formed, 2 mol Na are produced.)

c. If time is halved and current is increased from 1 A to 5 A, the charge (Q = It) changes by:

Q₂ = I₂t₂ = 5 × `frac{t}{2}`= 2.5I₁t

So the charge (and moles of electrons) increase by a factor of 2.5.

Hence, chlorine formed:

0.1 mol × 2.5 = 0.25 mol Cl₂

d. Electrolysis of dilute aqueous NaCl (Brine)

Because water is present, H⁺ and OH⁻ ions compete with Na⁺ and Cl⁻.

At cathode (reduction):

Hydrogen is produced (since water is reduced more easily than Na⁺):

2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)

At anode (oxidation):

Chloride is oxidized (not OH⁻ under these conditions):

2Cl⁻ → Cl₂(g) + 2e⁻

Products:

   • Cathode: Hydrogen gas (H₂)

   • Anode: Chlorine gas (Cl₂)

   • Solution left behind: Sodium hydroxide (NaOH)

a. The standard electrode potentials for three electrode systems are given below.

Ti³⁺(aq) + e⁻ → Ti²⁺(aq) E° = −0.37 V 

Fe³⁺(aq) + e⁻ → Fe²⁺(aq) E° = +0.77 V 

Ce⁴⁺(aq) + e⁻ → Ce³⁺(aq) E° = +1.45 V 

i. Using the data above, deduce which species is the best reducing agent, giving a reason in terms of electrons for your answer. 

ii. Write an equation, including state symbols, for the overall reaction with the greatest cell potential. 

iii. State and explain the sign of ΔG° for the reaction in (a)(ii).

b. State the name of a solution that would produce only hydrogen and oxygen when electrolyzed using platinum electrodes.

a. i. Best reducing agent: Ti²⁺(aq). 

Reason: It donates electrons most readily (is most easily oxidized) because its couple  has the most negative reduction potential: 

Ti³⁺ + e⁻ → Ti²⁺, E° = –0.37 V. 

ii. Pair the strongest oxidizing agent with the strongest reducing agent:

   • Cathode (reduction, largest E°): Ce⁴⁺ + e⁻ → Ce³⁺ (E° = +1.45 V).

   • Anode (oxidation): Ti²⁺ → Ti³⁺ + e⁻ (reverse of –0.37 V) 

   • Overall: 

Ce⁴⁺(aq) + Ti²⁺(aq) → Ce³⁺(aq) + Ti³⁺(aq) 

E°cell = 1.45 – (–0.37) = +1.82 V 

iii. ΔG° = –nFE° 

Here n = 1 and E°cell = +1.82 V → ΔG° < 0. 

So the reaction is spontaneous and releases free energy.

b. A solution that gives only H₂ and O₂ with Pt electrodes: dilute sulfuric acid (H₂SO₄  (aq)).

a. Molten sodium chloride can be electrolyzed using graphite electrodes.

i. Draw the essential components of this electrolytic cell and identify the products that  form at each electrode. 

ii. State the half-equations for the oxidation and reduction processes and deduce the  overall cell reaction, including state symbols. 

b. Explain why solid sodium chloride does not conduct electricity. 

a. i. Essential cell + products 

  • Diagram (describe): a container/crucible of molten NaCl(l) with two graphite electrodes dipped in. Connect the positive terminal of a DC supply to the anode, and the negative terminal to the cathode. 

   • Ions move: Na⁺ → cathode, Cl⁻ → anode. 

   • Product at positive electrode (anode): Cl₂(g). 

• Product at negative electrode (cathode): Na(l) (sodium metal; molten at operating temperature). 

a. ii. Half-equations and overall 

   • Cathode (reduction): Na⁺(l) + e⁻ → Na(l) 

   • Anode (oxidation): 2Cl⁻(l) → Cl₂(g) + 2e⁻ 

   • Overall reaction: 

2NaCl(l) → 2Na(l) + Cl₂(g) 

b. Why solid NaCl doesn’t conduct

In the solid, ions are locked in a giant ionic lattice and cannot move, so there are no  mobile charge carriers. (When molten or dissolved, the ions are free to move and it  conducts.) 

Chromium(III) oxide, Cr₂O₃, can undergo different reactions to give other chromium containing species as shown in the diagram below:

Which statement correctly describe these reactions? 

A. The formation of Cr₂O₇²⁻ from CrO₄²⁻ is a redox reaction. 

B. Aluminium is acting as an oxidizing agent. 

C. CrO₃, CrO₄²⁻, Cr₂O₇²⁻ contains chromium in its highest oxidation state.

D. Cr₂O₃ reacts with CrO₃ in a disproportionation reaction to give CrO₂. 

Answer: C. CrO₃, CrO₄²⁻, Cr₂O₇²⁻ contains chromium in its highest oxidation state.

A. Incorrect: 

   • In acid: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. 

   • Cr stays +6 in both chromate and dichromate → acid–base/condensation  equilibrium, not redox. 

B. Incorrect:

   • Thermite-type reduction: Cr₂O₃ + 2Al → 2Cr + Al₂O₃. 

   • Al is oxidized to Al³⁺, so it is the reducing agent, not oxidizing.

C. Correct: 

In all three, chromium is +6, which is Cr’s maximum common oxidation state.

D. Incorrect: 

Cr₂O₃ (Cr³⁺) + CrO₃ (Cr⁶⁺) → 3CrO₂ (Cr⁴⁺). 

This is comproportionation (different oxidation states combine to an intermediate), not  disproportionation. 

Consider the following reactions which all occur in solution at room temperature:

Fe (s) + Cu²⁺ (aq) → Fe²⁺ (aq) + Cu (s) 

Mg (s) + Zn²⁺ (aq) → Mg²⁺ (aq) + Zn (s) 

Zn (s) + Fe²⁺ (aq) → Zn²⁺ (aq) + Fe (s) 

Which is the correct combination of the strongest oxidizing agent and the strongest  reducing agent?

Answer: B. 

Given reactions: 

   • Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s) 

→ Fe is oxidized (reducing agent), Cu²⁺ is reduced (oxidizing agent).

   • Mg(s) + Zn²⁺(aq) → Mg²⁺(aq) + Zn(s) 

→ Mg is oxidized (reducing agent), Zn²⁺ is reduced (oxidizing agent).

   • Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s) 

→ Zn is oxidized (reducing agent), Fe²⁺ is reduced (oxidizing agent).

Step 1: Determine relative reactivity 

From the above: 

   • Mg can reduce Zn²⁺ → Mg is stronger reducing agent than Zn.

   • Zn can reduce Fe²⁺ → Zn is stronger reducing agent than Fe. 

   • Fe can reduce Cu²⁺ → Fe is stronger reducing agent than Cu. 

So the reducing power order (strongest → weakest): Mg > Zn > Fe > Cu The oxidizing power order (reverse, for the ions): Cu²⁺ > Fe²⁺ > Zn²⁺ > Mg²⁺

Step 2: Identify strongest agents 

   • Strongest oxidizing agent: the one that’s most easily reduced → Cu²⁺(aq)

   • Strongest reducing agent: the one that’s most easily oxidized → Mg(s)

Ethanedioic acid (oxalic acid),(COOH)₂, reacts with acidified potassium permanganate  solution, KMnO₄, according to the following equation: 

5(COOH)₂(aq) + 2MnO₄⁻(aq) + 6H⁺(aq) → 10CO₂(g) + 2Mn²⁺(aq) + 8H₂O(l)

The reaction is a redox reaction. 

a. Define oxidation in terms of electron transfer. 

b. Calculate the change in oxidation numbers of carbon and manganese.

c. Identify the oxidizing and reducing agents. 

a. Oxidation (in terms of electron transfer): loss of electrons. 

b. Oxidation-number changes 

   • Carbon in ethanedioic (oxalic) acid, (COOH)₂: each carbon is +3 (carboxyl  carbon). 

In the product CO₂, carbon is +4. 

→ Change per C: +3 → +4 (+1, i.e., oxidation). 

With 10 carbons total (from 5(COOH)₂), total electrons lost = 10 e⁻.

   • Manganese in MnO₄⁻: +7 → in Mn²⁺: +2. 

→ Change per Mn: +7 → +2 (−5, i.e., reduction). 

With 2 Mn, total electrons gained = 10 e⁻ (balances the carbon loss).

c. Agents 

   • Oxidizing agent: MnO₄⁻ (acidified KMnO₄); it is reduced to Mn²⁺.

   • Reducing agent: (COOH)₂ (ethanedioic/oxalic acid); it is oxidized to CO₂. 

Chemical energy can be converted to electrical energy in the voltaic cell below.

a. i. State the electron arrangement of a magnesium atom. 

ii. State the half-equation which describes the change at the Mg electrode and deduce  which metal is the positive electrode (cathode) of the cell. 

b. Deduce the equation for the overall reaction occurring in the cell.

a. i. Electron arrangement of magnesium atom 

Magnesium has atomic number 12, so its electron arrangement is: 2, 8, 2 or in subshell  notation: 1s² 2s² 2p⁶ 3s² 

a. ii. Half-equation at the Mg electrode and identification of electrodes At the magnesium electrode, magnesium loses electrons (oxidation): Mg(s) → Mg²⁺(aq)  + 2e⁻ 

   • Since oxidation occurs at magnesium, the Mg electrode is the anode (negative  electrode). 

   • The Fe electrode therefore is the cathode (positive electrode) because reduction  occurs there. 

At the iron electrode (cathode): Fe²⁺(aq) + 2e⁻ → Fe(s) 

b. Overall cell reaction 

Combine the two half-equations: 

Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)

Which is the species oxidized and the oxidizing agent in the reaction?

MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + Cl₂(g) + 2H₂O(l) 

Answer: D. 

Step 1: Determine oxidation states 

Step 2: Identify species oxidized and oxidizing agent 

   • Species oxidized: Chloride ion (Cl⁻) because it goes from -1 → 0.

   • Species reduced: Manganese in MnO₂ (from +4 → +2). 

   • Oxidizing agent: The substance that causes oxidation (is reduced itself).

→ MnO₂ is reduced, so MnO₂ is the oxidizing agent. 

Where does oxidation occur in a voltaic cell? 

A. Positive electrode and anode. 

B. Negative electrode and anode. 

C. Positive electrode and cathode. 

D. Negative electrode and cathode. 

Answer: B. Negative electrode and anode. 

For voltaic (galvanic) cells: 

   • Oxidation occurs at the anode. 

   • Reduction occurs at the cathode. 

   • In a voltaic cell, the anode is negative and the cathode is positive (because electrons flow spontaneously from anode → cathode). 

Which describes the flow of electrons in a voltaic cell? 

A. From the cathode (positive electrode) to the anode (negative electrode) through the external circuit. 

B. From the anode (negative electrode) to the cathode (positive electrode) through the external circuit. 

C. From the oxidizing agent to the reducing agent through the salt bridge.

D. From the reducing agent to the oxidizing agent through the salt bridge. 

Answer: B. From the anode (negative electrode) to the cathode (positive electrode) through the external circuit. 

A. Incorrect: Wrong direction. Electrons move toward the cathode, not away from it.

B. Correct: This is exactly how electrons flow in a voltaic cell. 

C. Incorrect: The salt bridge carries ions, not electrons. 

D. Incorrect: The salt bridge allows ion movement, not electrons. 

Propan-2-ol can be used as a fuel in the fuel cell. At the anode propan-2-ol is oxidized  to carbon dioxide. The electrons pass around the external circuit to the cathode. The protons formed from the oxidation move through the electrolyte to the cathode, where  they react with oxygen to produce water.

a. Formulate half-equations for the reactions at the anode and cathode respectively.

b. Formulate the equation for the overall reaction. 

c. The fuel cell has a cell potential (under standard conditions) of 1.56 V. By using  suitable data from the data booklet suggest a value for the E° of the CO₂/CH₃CHOHCH₃ electrode reaction. 

d. Suggest a possible advantage of using the propan-2-ol fuel cell compared to a  hydrogen fuel cell. 

a. Half-equations

   • At the anode (oxidation): 

Propan-2-ol (CH₃CHOHCH₃) is oxidized to carbon dioxide. For simplicity, you can write the overall oxidation as: 

CH₃CHOHCH₃ + 9H₂O → 3CO₂ + 24H⁺ + 24e⁻

This shows oxidation (loss of electrons). 

   • At the cathode (reduction): 

Oxygen is reduced to water: 

6O₂ + 24H⁺ + 24e⁻ → 12H₂O 

b. Overall cell reaction 

Add both half-equations, canceling electrons and balancing water and protons:

CH₃CHOHCH₃ + 3O₂ → 3CO₂ + 4H₂O 

c. Cell potential 

`""E_"cell"^°`= 1.56 V

and we know: 

`E_(O_2"/"H_2O)^o`= +1.23V

Use the formula: 

E_cell = E_cathode - E_anode

So: 

1.56 = 1.23 - `""E_"anode"^o`

`""E_"anode"^o`= -0.33V

→ `""E_(CO_2"/"CH_3CHOHCH_3)^o`= -0.33 V

d. Advantage of using propan-2-ol fuel cell over hydrogen fuel cell

   • Propan-2-ol is a liquid fuel, easier and safer to store and transport than hydrogen gas.

   • It has a higher energy density, giving longer operation for the same volume of fuel. 

   • It can be produced from renewable biomass sources.

Consider the following reaction: 

3H₂Se + 8HFeO₄⁻ + 6H₂O → 8Fe(OH)₃ + 3SeO₄²⁻ + 2OH⁻

Which statement is correct? 

A. HFeO₄⁻ is the oxidizing agent because it undergoes oxidation. 

B. HFeO₄⁻ is the oxidizing agent because it undergoes reduction. 

C. H₂Se is the oxidizing agent because it undergoes reduction. 

D. H₂Se is the oxidizing agent because it undergoes oxidation. 

Answer: B. HFeO₄⁻ is the oxidizing agent because it undergoes reduction.

Step 1: Identify agents 

   • Oxidizing agent = gets reduced 

→ HFeO₄⁻ (Fe⁶⁺ → Fe³⁺) 

   • Reducing agent = gets oxidized 

→ H₂Se (Se²⁻ → Se⁶⁺) 

Step 2: Evaluate each option 

A. Incorrect:  

HFeO₄⁻ is the oxidizing agent, but it undergoes reduction, not oxidation. An oxidizing  agent causes oxidation in another species by being reduced itself. 

B. Correct:  

Fe in HFeO₄⁻ goes from +6 to +3, meaning it gains electrons (reduction). Therefore,  HFeO₄⁻ acts as the oxidizing agent. 

C. Incorrect: 

H₂Se is oxidized, not reduced. Its Se atom goes from −2 → +6, losing electrons, so it  cannot be the oxidizing agent. It’s actually the reducing agent. 

D. Incorrect: 

While H₂Se does undergo oxidation, that makes it the reducing agent, not the oxidizing  agent. The oxidizing agent must gain electrons, not lose them. 

Which statements are correct for a voltaic cell? 

I. A spontaneous redox chemical reaction produces electrical energy.

II. Oxidation occurs at the cathode (negative electrode). 

III. Electrons flow from anode (negative electrode) to cathode (positive electrode).

A. I and II only. 

B. I and III only. 

C. II and III only. 

D. I, II and III. 

Answer: B. I and III only. 

   • Statement I: A spontaneous redox chemical reaction produces electrical energy.

   → Correct.  

That’s the definition of a voltaic (galvanic) cell – a spontaneous redox reaction converts  chemical energy into electrical energy. 

   • Statement II: Oxidation occurs at the cathode (negative electrode). → Incorrect.  In a voltaic cell, oxidation occurs at the anode, which is the negative electrode. The  cathode (positive electrode) is where reduction happens. 

   • Statement III: Electrons flow from anode (negative electrode) to cathode  (positive electrode). → Correct. 

Electrons are released during oxidation at the anode and travel through the external  circuit to the cathode. 

A voltaic cell is constructed from zinc and copper half-cells. Zinc is more reactive than  copper.

Which statement is correct when this cell produces electricity? 

A. Electrons flow from the copper half-cell to the zinc half-cell. 

B. The concentration of Cu²⁺ (aq) increases. 

C. Electrons flow through the salt bridge. 

D. Negative ions flow through the salt bridge from the copper half-cell to the zinc half cell. 

Answer: D. Negative ions flow through the salt bridge from the copper half-cell to the zinc half-cell. 

A zinc–copper voltaic cell 

   • Zinc is more reactive → oxidized → anode (negative electrode)

   • Copper is less reactive → reduced → cathode (positive electrode) Half equations: 

Anode: Zn(s) → Zn²⁺(aq) + 2e⁻ 

Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s) 

Electrons move through the wire from Zn to Cu. 

Now check each option 

A. Incorrect: Electrons flow from zinc → copper, not the reverse.

B. Incorrect: Cu²⁺ ions are used up (reduced to Cu(s)), so concentration decreases.

C. Incorrect: Ions, not electrons, move through the salt bridge. 

D. Correct: Anions (negative ions, e.g., NO₃⁻ or Cl⁻) move toward the anode (Zn half cell) to balance the positive charge of Zn²⁺ being formed. 

The equations below represent reactions involved in the Winkler method for  determining the concentration of dissolved oxygen in water: 

2Mn(OH)₂(s) + O₂(aq) → 2MnO(OH)₂(s) 

MnO(OH)₂(s) + 2H₂SO₄(aq) → Mn(SO₄)₂(s) + 3H₂O(l) 

Mn(SO₄)₂(s) + 2I⁻(aq) → Mn²⁺(aq) + I₂(aq) + 2SO₄²⁻(aq) 

2S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + 2I⁻(aq) 

What is the amount, in mol, of thiosulfate ions, S₂O₃²⁻(aq), needed to react with the  iodine, I₂(aq), formed by 1.00 mol of dissolved oxygen? 

A. 2.00 

B. 3.00 

C. 4.00 

D. 6.00

Answer: C. 4.00 

Step 1: Relationship between O₂ and I₂

From (1): 

2Mn(OH)₂(s) + O₂(aq) → 2MnO(OH)₂(s) 

So 1 mol of O₂ produces 2 mol of MnO(OH)₂. 

From (3): 

Mn(SO₄)₂ + 2I⁻ → Mn²⁺ + I₂ + 2SO₄²⁻

Each 1 mol of Mn compound produces 1 mol of I₂. 

Hence, 1 mol of O₂ → 2 mol of I₂. 

Step 2: Relationship between I₂ and S₂O₃²⁻ 

From (4): 

2S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻

So 1 mol of I₂ reacts with 2 mol of S₂O₃²⁻. 

Step 3: Combine ratios 

1 mol O₂  ⇒ 2 mol I₂ ⇒ 2 × 2 = 4 mol S₂O₃²⁻

What are the products when molten sodium chloride is electrolyzed? 

Answer: C. 

Molten sodium chloride contains: 

   • Na⁺ (sodium ions) 

   • Cl⁻ (chloride ions) 

Since it’s molten, there’s no water – only these ions. 

Step 2: Reactions at each electrode 

   • At the cathode (negative electrode): Reduction occurs – Na⁺ gains electrons:

Na⁺ + e⁻ → Na(l) 

→ Product: sodium metal 

   • At the anode (positive electrode): Oxidation occurs – Cl⁻ loses electrons: 2Cl⁻ → Cl₂(g) + 2e⁻

→ Product: chlorine gas 

A reaction takes place when a rechargeable battery is used: 

Pb(s) + PbO₂(s) + 4H⁺(aq) + 2SO₄²⁻(aq) → 2PbSO₄(s) + 2H₂O(l)

Which statements are correct? 

I. H⁺ is reduced. 

II. The oxidation state of Pb metal changes from 0 to +2. 

III. PbO₂ is the oxidising agent. 

A. I and II only. 

B. I and III only. 

C. II and III only. 

D. I, II and III.

Answer: C. II and III only. 

I. H⁺ is reduced → Incorrect 

H⁺ is not reduced — it remains part of the water molecules formed, but the real reduction occurs in PbO₂ (Pb⁴⁺ → Pb²⁺) 

II. The oxidation state of Pb metal changes from 0 to +2 → Correct Pb(s) is oxidized to Pb²⁺ in PbSO₄, so oxidation state increases from 0 → +2.

III. PbO₂ is the oxidizing agent → Correct 

PbO₂ contains Pb in +4 oxidation state; it accepts electrons and is reduced to Pb²⁺ – hence, it acts as the oxidizing agent. 

What are the relative volumes of gas given off at E and F during electrolysis of the two cells in series? Assume all electrodes are inert. 

A. 1 : 1 

B. 1 : 2

C. 2 : 1 

D. 5 : 2

Answer: C. 2 : 1 

Cell 1: CuSO₄(aq) (inert electrodes) 

   • Cathode (−): Cu²⁺ + 2e⁻ → Cu(s) (copper deposited) 

   • Anode (+): 2H₂O → O₂(g) + 4H⁺ + 4e⁻ (since electrodes are inert and sulfate ions are not oxidized) 

→ Gas evolved = O₂ at electrode E (anode) 

Cell 2: Dilute H₂SO₄(aq) (inert electrodes) 

   • Cathode (−): 2H⁺ + 2e⁻ → H₂(g) 

   • Anode (+): 2H₂O → O₂(g) + 4H⁺ + 4e⁻ 

But since it’s in series with the CuSO₄ cell, the same current (same number of electrons) passes through both.

At F (cathode), H₂ gas is produced. 

→ Gas evolved = H₂ at electrode F (cathode) 

From stoichiometry: 

   • O₂ formation → 4e⁻ per O₂ 

   • H₂ formation → 2e⁻ per H₂ 

For the same current, electrons passing are equal in both cells. 

`frac{V_(H_2)}{V_(O_2)}=frac{4}{2}=2:1`

A voltaic cell is constructed from zinc and copper half-cells. Zinc is more reactive than copper. Which statement is correct when this cell produces electricity?

A. Electrons flow from the copper half-cell to the zinc half-cell. 

B. The concentration of Cu²⁺(aq) increases. 

C. Electrons flow through the salt bridge. 

D. Negative ions flow through the salt bridge from the copper half-cell to the zinc half cell. 

Answer: D. Negative ions flow through the salt bridge from the copper half-cell to the zinc half-cell. 

Cell facts (Zn–Cu voltaic cell): 

    • Zinc is more reactive → anode (−): Zn(s) → Zn²⁺(aq) + 2e⁻ (oxidation).

   • Copper is cathode (+): Cu²⁺(aq) + 2e⁻ → Cu(s) (reduction). 

   • Electrons travel in the external wire from Zn → Cu. 

   • Salt bridge carries ions: anions → anode (Zn half-cell), cations → cathode (Cu  half-cell). 

   • Concentrations: [Zn²⁺] increases, [Cu²⁺] decreases. 

A. Incorrect: Electrons flow from Zn (anode) to Cu (cathode). 

B. Incorrect: Cu²⁺ is consumed at the cathode (it’s reduced to Cu), so its concentration  decreases. 

C. Incorrect: The salt bridge conducts ions, not electrons. 

D. Correct: Anions migrate toward the anode (Zn half-cell) to balance the build-up of  Zn²⁺.