Electrochemistry

Chemical Interference in a Daniel Cell

The cathode half-cell of a standard Daniel cell contains aqueous Cu²⁺ ions waiting to be reduced.

• The Reaction: Adding Na₂S introduces Sulphide ions (S²⁻), which aggressively bind to Copper to form Copper(II) sulphide (CuS), a highly insoluble black precipitate.
• The Consequence: This massively depletes the [Cu²⁺] concentration available for reduction. Because the active ions are removed chemically rather than being converted to electrical energy, the functional lifespan (consumption time) of the cell decreases drastically.

Calculating Cell EMF (E_cell)

To find the EMF, we must properly align the standard potentials with the actual reaction:

1. Anode (Oxidation): Cl⁻ is oxidized to Cl₂. The provided potential (+1.36 V) is for reduction. Thus, the oxidation potential is -1.36 V.
2. Cathode (Reduction): MnO₄⁻ (Mn⁷⁺) is reduced to Mn²⁺. The provided potential (-1.52 V) is an oxidation potential (from 2+ to 7+). Thus, its reduction potential is +1.52 V.
3. EMF Calculation: E_cell = E_ox(anode) + E_red(cathode) = -1.36 V + 1.52 V = +0.16 V.

Combining Standard Cell Potentials

Using the general formula: E_cell = E_ox(Anode) + E_red(Cathode)

1. Analyze the Given Cells:
   For Cell 1: E_ox(X) + E_red(Ag) = 0.80 V.
   For Cell 2: E_ox(Y) + E_red(Ag) = 1.56 V.
2. Find the Difference: Subtracting Cell 1 from Cell 2 cancels out the Silver (Ag) component:
   E_ox(Y) - E_ox(X) = 1.56 - 0.80 = 0.76 V.
3. Construct New Cell (X, Y): Since Y has a higher oxidation potential, Y will act as the anode, and X will be forced to act as the cathode. EMF = E_ox(Y) + E_red(X) = 0.76 V.

Determining the Type of Metal Protection

The activity of a metal is inversely related to its reduction potential. A lower (more negative) reduction potential means a more active metal.

• Activity Series: A (-0.44) > B (+0.34) > C (+0.80). Metal C is the least active.
• Protection Type: When a metal is coated with a less active metal, the coating acts as a physical barrier. If scratched, the underlying (more active) metal becomes the anode and rusts faster. This setup (coating is less active) is strictly defined as Cathodic Protection.

Electrolysis of Bauxite (Hall-Héroult Process)

During the industrial electrolysis of molten Bauxite (Al₂O₃ dissolved in molten cryolite), there is no water present, meaning no hydrogen gas is produced.

• At the Cathode (Negative): Aluminum ions (Al³⁺) gain electrons (reduction) and pool at the bottom as molten Aluminum metal.
• At the Anode (Positive): Oxide ions (O²⁻) lose electrons (oxidation) and form Oxygen gas (O₂). (Note: This oxygen often reacts with the carbon anodes to form CO₂, requiring regular replacement).

Electrolytic Purification of Metals

In electrolytic purification, the impure metal block is made the anode. Based on the activity series (A > B > C), here is what happens:

• Highly Active A: Because A is more active than B, it easily oxidizes and dissolves into the solution as ions.
• Target Metal B: B also oxidizes, travels through the solution, and safely reduces onto the pure cathode.
• Less Active C: Because C is less active than B, the applied voltage isn't high enough to oxidize it. It remains unreacted solid metal and simply falls down to the bottom below the anode as "anode mud".

Comparing Electrode Activities

In any galvanic cell, electrons spontaneously flow from the Anode (the more active metal) to the Cathode (the less active metal).

• Cell 1 Analysis: The green arrow shows electrons flowing from electrode A to B. Therefore, Activity A > B.
• Cell 2 Analysis: The green arrow shows electrons flowing from electrode C to B. Therefore, Activity C > B.

In both setups, metal B is bullied into acting as the cathode because it is less active than its counterpart.

Distinguishing Acids and Esters

The general formula C_nH_2nO₂ indicates the compound is either a carboxylic acid or an ester.

• Compound A (Reacts cold): Carboxylic acids react readily with cold NaOH (simple neutralization). Thus, A must be an acid, such as Butyric acid.
• Compound B (No cold reaction): Esters do not react with cold NaOH; they require heating (saponification) to break the ester bond. Thus, B must be an ester, such as Ethyl ethanoate.

Anode vs. Cathode Designation

• Since electrons flow from X to Y, X is the anode (oxidation) and Y is the cathode (reduction).
• In a spontaneous galvanic cell, the anode (X) has a higher oxidation potential and a lower reduction potential than the cathode (Y).

Step 1: Determine Anode and Cathode

Both equations represent standard oxidation reactions relative to the standard hydrogen electrode (SHE). Thus:

• E°_ox(X) = +0.76 V
• E°_ox(Y) = +1.66 V

Since Y has a higher oxidation potential, Y acts as the anode and X acts as the cathode.

Step 2: Calculate Cell EMF

Electrode Reactions

• At the Anode (+) (Copper): Copper metal is oxidized to copper ions, dissolving into solution:
  Cu(s) → Cu2+(aq) + 2e-
• At the Cathode (-) (Iron): Cu²⁺ ions from the solution are reduced to copper metal, plating the iron electrode:
  Cu2+(aq) + 2e- → Cu(s)

Because the rate of oxidation at the anode equals the rate of reduction at the cathode, the concentration of copper ions in solution remains constant, and overall charge neutrality is maintained.

Faraday's Law Calculations

Since the cells are connected in series, the same charge Q passes through all of them. Let's look at the mole of electrons needed for 1 mole of each gas:

1. Oxygen: 2O2- → O2 + 4e- (Requires 4 Faradays per mole) → Moles = Q / 4
2. Nitrogen: 2N3- → N2 + 6e- (Requires 6 Faradays per mole) → Moles = Q / 6
3. Chlorine: 2Cl- → Cl2 + 2e- (Requires 2 Faradays per mole) → Moles = Q / 2

Let's find the ratio of emitted gas volumes (proportional to moles):

Mercury Cell Chemistry

In a Mercury Cell, the reduction half-reaction at the cathode produces liquid mercury:

The mercury cell is compact and steady, making it ideal for portable devices like hearing aids (ear phones).

Lithium-Ion Discharge

During discharge (spontaneous power delivery):

• Anode (Negative): LiC₆ is oxidized:
  LiC6 → C6 + Li+ + e-
• Cathode (Positive): Li⁺ ions are inserted into cobalt oxide:
  CoO2 + Li+ + e- → LiCoO2

Hence, both lithium ions (Li⁺) and electrons move toward the positive electrode (cathode) during discharge.

Electrochemical Activity Series

• Since nickel is more active than copper (Ni > Cu), nickel is oxidised and reduces copper ions. Thus, a nickel container would react and dissolve.
• Because copper is less active than nickel, it cannot displace Ni²⁺ from solution. Thus, nickel salts can be safely stored in copper containers.

Reagent Verification

• Y²⁺ concentration increases → Y is being oxidized (Y → Y²⁺ + 2e⁻) → Y is the anode.
• Since Y is the anode, X is the cathode → X²⁺ ions are being reduced: X²⁺ + 2e⁻ → X
• X²⁺ is the oxidizing agent (it gets reduced). ✓
• Electrons flow from Y (anode) to X (cathode), not X to Y → option (c) is wrong.
• Y has higher oxidation potential (it's oxidized more easily), not X → option (a) is wrong.

• Cd + Zn²⁺ → Non-spontaneous: Cd cannot reduce Zn²⁺ → Zn²⁺ is a weaker oxidizing agent than Cd²⁺. So: Cd²⁺ > Zn²⁺.
• Cd + Cu²⁺ → Spontaneous: Cd can reduce Cu²⁺ → Cu²⁺ is a stronger oxidizing agent than Cd²⁺. So: Cu²⁺ > Cd²⁺.
• Combining: Cu²⁺ > Cd²⁺ > Zn²⁺ ✓

• A sacrificial electrode must be more reactive (higher oxidation potential) than iron → it should be the anode (electrons flow FROM the sacrificial metal TO iron).
• A → Fe: Electrons flow from A to Fe → A is the anode (more reactive than Fe). Cell potential = 1.4 V ✓
• C → Fe: C is also the anode. Cell potential = 0.5 V ✓ (but weaker protection)
• Fe → B and Fe → D: Fe is the anode here → B and D are less reactive than Fe → cannot be sacrificial. ✗
• Between A (1.4V) and C (0.5V), A has higher reactivity difference → A is preferred for best cathodic protection.

Based on the electrochemical system provided in the figure, here is the comprehensive step-by-step chemical breakdown and analysis of the electrolysis of molten silver bromide (AgBr).

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Detailed Step-by-Step Analysis

1. Identifying the Poles of the Cell

• Looking closely at the battery symbol at the top:
  • Electrode B is connected to the longer vertical plate of the battery power supply, representing the positive terminal (+). In an electrolytic cell, the positive pole is the Anode.
  • Electrode A is connected to the shorter vertical plate of the battery power supply, representing the negative terminal (-). In an electrolytic cell, the negative pole is the Cathode.

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2. Half-Cell Reaction at Electrode B (Anode / Positive Pole)

• The Process: Oxidation occurs at the anode. Negatively charged bromide ions (Br^-) from the melt migrate towards the positive Electrode B.
• The Observation: Each bromide ion loses electrons to form reddish-brown bromine gas vapors (Br₂) escaping from the electrode surface.
• The Reaction Box:


2Br^-(l) → Br₂(g) ↑ + 2e^-



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3. Half-Cell Reaction at Electrode A (Cathode / Negative Pole)

• The Process: Reduction occurs at the cathode. Positively charged silver ions (Ag⁺) from the melt migrate towards the negative Electrode A.
• The Observation: Silver ions gain electrons and are reduced, depositing as solid silver metal (Ag) covering the surface of Electrode A.
• The Reaction Box:


Ag⁺(l) + e^- → Ag(s) ↓



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Summary of Expected Exam Questions

This standard curriculum setup typically asks you to match the correct observation/mass change with each electrode. You can directly conclude:

Electrode	Polarity & Role	Chemical Reaction Type	Visible Observation / Consequence
Electrode B	Positive (+) Anode	Oxidation	Reddish-brown bromine gas vapors evolve; Mass stays constant.
Electrode A	Negative (-) Cathode	Reduction	Silver metal is deposited; Mass of the electrode increases.

• In an electrolytic cell: the electrode connected to the negative terminal of the battery is the cathode (reduction), and the one connected to the positive terminal is the anode (oxidation).
• Electrode B = Cathode: Ag⁺ + e⁻ → Ag (silver is deposited). ✓
• Electrode A = Anode: 2Br⁻ → Br₂ + 2e⁻ (oxidation of bromide). ✓

1. Identifying Electrode Polarities

• Looking closely at the battery symbol at the top of the image:
  • The longer vertical line represents the positive terminal (+), which is connected to Electrode X on the right. Therefore, Electrode X is the Anode.
  • The shorter vertical line represents the negative terminal (-), which is connected to Electrode Y on the left. Therefore, Electrode Y is the Cathode.

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2. Electrochemistry Reactions in Aqueous CuCl₂

• At Electrode X (Anode / Positive Pole): Negatively charged chloride ions (Cl⁻) from the solution migrate towards the positive Electrode X. They lose electrons (undergo oxidation) to form chlorine gas bubbles.
  
  
  2Cl⁻(aq) → Cl₂(g) ↑ + 2e⁻
  
  Observation: Pungent chlorine gas (Cl₂) bubbles evolve directly at electrode X. This makes statement (d) the absolute correct choice.
• At Electrode Y (Cathode / Negative Pole): Positively charged copper ions (Cu²⁺) migrate towards the negative Electrode Y. They gain electrons (undergo reduction) and deposit as metallic copper.
  
  
  Cu²⁺(aq) + 2e⁻ → Cu(s) ↓

Faraday's Law: m = (M × I × t) / (n × F)

Where: M = 23, I = 10 A, t = 10 × 3600 = 36000 s, n = 1 (Na⁺ + e⁻ → Na), F = 96500 C/mol

m = (23 × 10 × 36000) / (1 × 96500) = 8,280,000/96500 = 85.8 g ✓

• W: E^°_red = +1.42 V (Cathode relative to Zn)
• Z: E^°_red = -2.375 V (Much more active than Zn)
• Y: E^°_red = +1.2 V (Cathode relative to Zn)
• X: E^°_red = +0.34 V (Cathode relative to Zn)

• Cell 3 (Electrode E): Ni²⁺ → Equivalent weight = (58.7)/(2) = 29.35 g.
• Cell 2 (Electrode C): Ag⁺ → Equivalent weight = (108)/(1) = 108 g.
• Cell 1 (Electrode A): Cu²⁺ → Equivalent weight = (63.5)/(2) = 31.75 g.

The correct answer is a) Pb²⁺ ions are converted to Pb⁴⁺ ions at the anode (positive plate).

🔋 Chemical Explanation

• Cell Type Shift: During charging, the battery switches from a galvanic cell to an electrolytic cell.
• Anode Identity: The positive plate behaves as the anode where oxidation occurs during the recharging phase.
• Oxidation State Change: Lead sulfate (PbSO₄, where lead is Pb²⁺) is oxidized into lead dioxide (PbO₂, where lead is Pb⁴⁺).
• Reaction Equation: PbSO₄(s) + 2H_2O(l) → PbO₂(s) + SO₄^2-(aq) + 4H⁺(aq) + 2e^-.

💡 Why Other Options Fail

• Option b: The conversion of Pb²⁺ to metallic Pb is a reduction process, which occurs exclusively at the cathode (negative plate) during charging.
• Option c: Since charging is forced by an external power source, electrons are pumped from the positive electrode to the negative electrode via the external circuit.
• Option d: Within the internal solution of the cell, electrical charge is carried by moving ions (H⁺ and SO₄^2-), not free electrons.

Step 1: Determine EMF.
The reaction shows Cu²⁺ is reduced (Cathode) and Cr is oxidized (Anode).
Reduction potential of Cu = +0.340 V. Oxidation potential of Cr = +0.740 V.
EMF = E°(oxidation) + E°(reduction) = 0.740 V + 0.340 V = +1.08 V.

Step 2: Determine Cell Type.
Because the EMF is positive, the reaction is spontaneous. A spontaneous electrochemical cell is a Galvanic cell.

Connecting vessel (1) [Zn/Zn²⁺] and (4) [Cu/Cu²⁺] creates a classic Daniell cell. Zinc is more active than Copper, so Zinc acts as the anode (oxidation) and Copper acts as the cathode (reduction). Electrons flow through the wire from Anode (Zn, vessel 1) to Cathode (Cu, vessel 4).

Correction Note: In the user-provided prompt 'Answer Key', the answer is marked as (C) for 19. Let's recalculate based on standard chemistry.

First cell: X is Anode. E_cell = E_ox(X) + E_red(H) => 0.23 = E_ox(X) + 0. So, Oxidation Potential of X = +0.23 V.

Second cell: Y is Cathode. E_cell = E_ox(H) + E_red(Y) => 0.80 = 0 + E_red(Y). So, Reduction Potential of Y = +0.80 V. (Oxidation Potential of Y = -0.80 V).

Comparing Oxidation Potentials: X (+0.23 V) > Y (-0.80 V).
Because X has a higher oxidation potential, X is the Anode and Y is the Cathode.

EMF = E_ox(Anode) + E_red(Cathode) = 0.23 V + 0.80 V = +1.03 V.

(Thus, statement (a) is chemically correct based on the provided text, despite what a raw key might say. X oxidizes, Y reduces.)

Understanding Protection Types

In standard curriculum terminology:

• Anodic Protection (Sacrificial Anode): Coating a metal with a more active metal. The coating has a higher oxidation potential.
• Cathodic Protection (Cathodic Coating): Coating a metal with a less active metal. The coating has a lower oxidation potential.

Ordering the Elements by Activity

Higher oxidation potential = More active.

W (+1.67) > X (+0.126) > Y (-0.401) > Z (-1.420)

Evaluating the Options

Reasoning: In an electrolytic cell, the Anode is the positive electrode where Oxidation (loss of electrons) takes place.

Molten potassium bromide (KBr) contains K⁺ and Br⁻ ions.

• At the Cathode (Reduction): K⁺ + e⁻ → K(l)
• At the Anode (Oxidation): 2Br⁻ → Br₂(g) + 2e⁻

In a Hydrogen-Oxygen fuel cell (alkaline type), the electrolyte is typically hot aqueous KOH. Let's look at the half-reactions:

Anode (Oxidation): 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
Here, OH⁻ ions are consumed.

Cathode (Reduction): O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Here, OH⁻ ions are formed.

Therefore, the hydroxide ions migrate from the cathode to the anode, being produced at the cathode (reduction site) and used up at the anode (oxidation site).

From the balanced half-reaction, producing 1 mole of MnO₄⁻ requires 5 moles of electrons (5 Faradays).

To produce 0.2 moles of MnO₄⁻, the moles of electrons required = 0.2 × 5 = 1 mole of electrons (1 Faraday).

We know that 1 Faraday is equal to 96500 Coulombs (C). Therefore, the required electricity is 96500 C.

The formula C₃H₈O represents saturated alcohols and ethers. Let's analyze the clues:

• Compound X: Oxidizes to an acid (reacts with Na₂CO₃). Therefore, X is a primary alcohol. Structure: CH₃-CH₂-CH₂OH (Propan-1-ol). It has 1 methyl group.
• Compound Y: Oxidizes to a compound that isn't an acid (it's a ketone). Therefore, Y is a secondary alcohol. Structure: CH₃-CH(OH)-CH₃ (Propan-2-ol). It has 2 methyl groups.
• Compound Z: Does not oxidize. Therefore, Z is an ether. Structure: CH₃-O-CH₂-CH₃ (Ethyl methyl ether). It has 2 methyl groups.

This perfectly aligns with Option D: 1, 2, 2.

Electrolysis Rules

• Anode (+): Oxidation occurs; anions discharge

• Cathode (−): Reduction occurs; cations discharge

• Pt electrodes are inert (don't participate)

For CuCl₂ Solution

Zinc Half-Cell Equilibrium

Zn(s) ⇌ Zn²⁺(aq) + 2e⁻

Zinc has high oxidation tendency:

• Zn atoms leave the rod as Zn²⁺ ions → solution contains positive Zn²⁺ ions

• Electrons remain on the rod → rod becomes negatively charged

Step 1: Determine the State of Each Battery

The density of the electrolyte (H₂SO₄) indicates the battery's charge state:

Battery Y (charged) acts as the energy source and charges battery X (discharged).

Step 2: Battery X — Electrolytic Cell (Charging Process)

During charging, reactions are reversed from discharge:

At − electrode (cathode in electrolytic cell):

Pb²⁺ + 2e⁻ → Pb (reduction) ✓

At + electrode (anode in electrolytic cell):

Pb²⁺ → Pb⁴⁺ + 2e⁻ (oxidation, forms PbO₂)

Step 3: Battery Y — Galvanic Cell (Discharging Process)

At + electrode (cathode in galvanic cell):

Pb⁴⁺ + 2e⁻ → Pb²⁺ (reduction) ✓

At − electrode (anode in galvanic cell):

Pb → Pb²⁺ + 2e⁻ (oxidation)

Conclusion

Option (d) correctly describes:

• X: Pb²⁺ → Pb at − electrode (reduction during charging) ✓

• Y: Pb⁴⁺ → Pb²⁺ at + electrode (reduction during discharge) ✓

Faraday's Law: mass ∝ M/n (same charge)

Ratio = 65/32.5 = 2 : 1 ✓ (matches condition)

Both deposit metal at the cathode (X = Au, Y = Zn).

Identifying Electrodes

Iron has higher oxidation potential (+0.409 V) → Fe is anode (oxidized)

Tin: Sn²⁺ + 2e⁻ → Sn (reduction) → Sn is cathode

EMF Calculation

EMF Calculation

EMF = E°(oxidation, anode) + E°(reduction, cathode)

EMF = 0.23 + 0.80 = +1.03 V

Type of Cell

Positive EMF → reaction is spontaneous → Galvanic cell ✓

Step 1: Mass of Gold Deposited

Au³⁺ + 3e⁻ → Au

Moles Au = (moles e⁻) / 3 = 0.5 / 3 = 0.1667 mol

Mass = 0.1667 × 197 = 32.83 g ≈ 32.8 g ✓

Step 2: Concentration Change

In electroplating, the anode is the same metal (Au) as the coating.

At anode: Au → Au³⁺ + 3e⁻ (replenishes ions)

At cathode: Au³⁺ + 3e⁻ → Au (deposits on spoon)

Ions consumed = ions produced → concentration remains constant ✓

Bifunctional Nature of Glycolic Acid

Glycolic acid has both -OH (alcohol) and -COOH (acid) groups.

Experiment (A): With Methanol

Methanol is an alcohol → reacts with the -COOH group of glycolic acid (esterification)

HO-CH₂-COOH + CH₃OH → HO-CH₂-COOCH₃ + H₂O ✓

Experiment (B): With Acetic Acid

Acetic acid is acid → reacts with the -OH group of glycolic acid (esterification)

HO-CH₂-COOH + CH₃COOH → CH₃COO-CH₂-COOH + H₂O ✓

Anode (Cu): Cu → Cu²⁺ + 2e⁻ (dissolves, mass decreases, releases Cu²⁺ into solution).

Cathode: Cu²⁺ + 2e⁻ → Cu (deposits, mass increases).

Rate of dissolution = rate of deposition → [Cu²⁺] remains constant. This is the principle of electrorefining.

• NaCl — forms PbCl₂↓ → not suitable.
• Na₂CO₃ — forms PbCO₃↓ and FeCO₃↓ → not suitable.
• Ethyl alcohol — non-electrolyte → not suitable.
• NaNO₃ — all nitrate salts are soluble; does not react with Fe²⁺ or Pb²⁺ → suitable.

Same current → same moles of electrons through each cell. Mass deposited ∝ (M/n):

Highest M/n = 108 (Ag) → greatest cathode mass increase in cell ③.

X is oxidised (anode); Cu²⁺ is reduced (cathode), E°(Cu²⁺/Cu) = +0.34 V.

E_cell = E°_cathode − E°_anode

+1.8 = +0.34 − E°(X) → E°(X) = 0.34 − 1.8 = −1.46 V

During discharge:

• Anode (Pb): Pb + SO₄²⁻ → PbSO₄ + 2e⁻ → anode mass increases.
• Cathode (PbO₂): PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O
• H₂SO₄ is consumed and water is produced → solution density decreases.

A sacrificial anode must be more active (more negative reduction potential) than the protected metal.

B = −1.70 V. Elements more negative: D (−2.52 V) and C (−2.87 V) → both can serve as sacrificial anodes.

A (−0.45 V) is less active than B → cannot protect it.

n(NaCl) = 5.85/58.5 = 0.1 mol; requires 0.1 mol e⁻ = 0.1 F for complete decomposition.

Cathode: Na⁺ + e⁻ → Na → m(Na) = 0.1 × 23 = 2.3 g

Anode: 2Cl⁻ → Cl₂ + 2e⁻ → n(Cl₂) = 0.05 mol → m(Cl₂) = 0.05 × 71 = 3.55 g

0.1 F is exactly enough → complete decomposition.

1-chloropropane + NaOH(aq)/Δ → propan-1-ol (primary alcohol), NOT isopropyl alcohol. Isopropyl alcohol comes from 2-chloropropane. So option (a) is incorrect.

The correct option is d).

The correct option is c).

The anode is the site of oxidation. The reaction 4OH⁻ → O₂ + 2H₂O + 4e⁻ is an oxidation (loss of electrons) and is non-spontaneous in an electrolytic cell. ✓

• (a), (d) ✗ Reductions occurring at the cathode.
• (b) ✗ Copper plating is reduction at the cathode.

The correct option is d).

In cell X (galvanic): electrons flow B → A, so B is the anode (more active) and the spontaneous reaction is: B + A²⁺ → B²⁺ + A.

In cell Y: the given reaction A + B²⁺ → A²⁺ + B is the exact reverse → it is non-spontaneous ⇒ an electrolytic cell with a negative emf.

• A is oxidised → A is the anode (positive) in an electrolytic cell.
• B²⁺ is reduced at B → B is the cathode, which in an electrolytic cell is the negative electrode. ✓

(c) is wrong: in an electrolytic cell the anode is the positive electrode, not negative.

The correct option is b).

• Pipe = Z (−2.71 V): Most active → sacrificial anode for both tank and tap.
• Tap = Y (+0.30 V): Most noble → provides cathodic protection to the tank.
• Tank = X (−0.40 V): Protected cathodically by both. ✓

The correct option is c).

At the impure anode, Cu, Fe, and Ag all dissolve → large mass decrease. At the pure cathode, only Cu²⁺ deposits → smaller mass gain (Ag falls as anode sludge, Fe²⁺ stays in solution). Therefore decrease in anode mass > increase in cathode mass. ✓

The correct option is d).

Both the fuel cell and the mercury cell use an alkaline (KOH) electrolyte whose concentration does not change during operation. ✓

• (a) ✗ Fuel cells use externally supplied energy, not stored chemical energy.
• (b) ✗ Reduction occurs at the positive electrode (cathode).

The correct option is b).

The correct option is c).

Z = Toluene (formed from reduction of X and dry distillation of Y). X = para-cresol (para methyl phenol), Y = sodium p-methylbenzoate. Z + H₂ → W = methylcyclohexane. ✓

🎯 The Core Chemical Rule: In electrochemistry, a displacement reaction is non-spontaneous if the added elemental metal has a lower oxidation potential (less active) than the metal ion already dissolved in the solution. The more active metal easily undergoes oxidation, serving as a **strong reducing agent**.

✅ Ranking the Elements by Activity (Oxidation Potentials):
• From Reaction 1 (Non-spontaneous): The solid metal Z cannot displace Y²⁺ ions. This implies that Z is less active than Y. Therefore: Y > Z.
• From Reaction 2 (Non-spontaneous): The solid metal Y cannot displace X²⁺ ions. This means that Y is less active than X. Therefore: X > Y.
• Final Activity Series: Combining both observations gives a descending activity trend of: X > Y > Z.

📊 Evaluating the Target Reaction:
• The target equation describes elemental metal X reacting with Z²⁺ ions. Since **X is more active than Z** (positioned higher in the electromotive series), it easily displaces Z, meaning the reaction occurs **spontaneously**.
• During this change, metal X loses electrons and undergoes oxidation (X → X²⁺ + 2e^-). By supplying electrons to reduce Z²⁺, the neutral element **X acts as a reducing agent**.

🎯 The Core Chemical Rule: During the **recharging** process, a lithium-ion battery behaves as an electrolytic cell. The chemical reactions that occurred spontaneously during discharge are reversed by an external power source. Consequently, lithium ions (Li⁺) migrate from the positive electrode (acting as the anode during charging) to the negative electrode (acting as the cathode during charging).

✅ Analyzing Electrode Behavior During Charging:
• At the Anode (Positive Electrode): The structural material consists of lithium cobalt oxide (LiCoO₂). When recharging, lithium ions leave the lattice layer and exit into the liquid medium. This loss causes a noticeable **decrease in the mass of this anode**.
• Oxidation Mechanism: To preserve overall charge balance as positive Li⁺ ions leave, the transition metal framework experiences oxidation, where cobalt(III) ions (Co³⁺) are oxidized to cobalt(IV) ions (Co⁴⁺) as shown in the charging half-reaction:
    LiCoO₂ → Li_1-xCoO2 + xLi⁺ + xe^-
• At the Cathode (Negative Electrode): The graphite matrix (C₆) intercalates the arriving lithium ions alongside free circuit electrons to form lithium graphite (LiC₆), increasing its mass. The total concentration of lithium ions inside the electrolyte remains overall constant because the rate of ionic entry matches the rate of insertion.

📊 Conclusion: Option (d) perfectly describes the structural physics of the charging state, showing that the active anode mass decreases while the cobalt core is oxidized from +3 to +4.

🎯 The Core Chemical Rule: Faraday's laws of electrolysis state that the quantity of electricity passed determines the number of moles of electrons transferred. By comparing the moles of electrons to the moles of products formed at each electrode, we can deduce the accurate stoichiometric ratio and valency of the ions involved.

✅ Step-by-Step Quantitative Analysis:
• 1. Calculate the Moles of Electrons Transferred: Using Faraday's constant (96500 C/mol e^-):
    Moles of e^- = (Quantity of electricity)/(Faraday's Constant) = (9650 C)/(96500 C/mol) = 0.1 mol of e^-
• 2. Determine the Cathode Valence Ratio: The problem states that 0.1 mol of copper is deposited.
    Ratio = (Moles of e^-)/(Moles of Cu) = (0.1 mol)/(0.1 mol) = 1 : 1
    This indicates 1 mol of electrons reduces 1 mol of copper ions, confirming the presence of the monovalent copper ion (Cu⁺). This narrows our options to rows (a) or (c).
• 3. Determine the Anode Gas Ratio: The problem states that 0.05 mol of gas is released.
    Ratio = (Moles of e^-)/(Moles of gas) = (0.1 mol)/(0.05 mol) = 2 : 1
    This means 2 mol of electrons are lost per 1 mol of gas produced. Looking at the options:
    - For chlorine gas (Cl₂): 2Cl^- → Cl₂ + 2e^- (Ratio e^- : gas = 2 : 1). This perfectly matches our calculation!
    - For oxygen gas (O₂): 2O^2- → O₂ + 4e^- (Ratio e^- : gas = 4 : 1).

📊 Conclusion: Combining the matched conditions (Cu⁺ at the cathode and Cl₂ at the anode), Row (c) represents the fully accurate electrolysis mechanism for this cell.

🎯 The Core Chemical Rule: In structural storage problems, standard activity strings are written as decreasing hierarchies. A salt cannot be safely contained if the vessel possesses a higher oxidation potential than the solute, whereas an electrochemical mismatch where the vessel is less reactive allows successful storage.

✅ Step-by-Step Activity Series Deduction:
• Analyzing Metal X: Storing the ionic salt of X inside a container made of Z is impossible due to a spontaneous redox displacement reaction. This shows that the vessel metal Z is more reactive than the solute metal X (oxidation potential: Z > X).
• Analyzing Metal Y: Storing the ionic salt of Y inside a container made of Z is fully successful because no spontaneous displacement takes place. This confirms that the vessel metal Z is less reactive than the solute metal Y (oxidation potential: Y > Z).
• Final Structural Order Matrix: Combining these potential barriers proves that Y holds the highest oxidation capacity, Z occupies the intermediate space, and X remains the least active component. Expressed as a descending reactivity string, it reads: X < Z < Y.

📊 Conclusion: The mathematical formatting tracking this precise operational reactivity descending sequence aligns perfectly with choice (b).

🎯 The Core Chemical Rule: The electrolytes chosen for a salt bridge must contain ions that do not react or form insoluble precipitates with the ions present in either the anode or cathode half-cells. Precipitating out essential ions disrupts electrical neutrality, halting the electric current.

✅ Step-by-Step Precipitation Analysis:
• 1. Components of a Daniell Cell: A standard Daniell cell utilizes a zinc anode dipped in a zinc sulfate (ZnSO₄) solution, and a copper cathode dipped in a copper(II) sulfate (CuSO₄) solution.
• 2. Incompatibility of Calcium Chloride (CaCl₂): When CaCl₂ is placed in the salt bridge, its calcium cations (Ca²⁺) migrate toward the cathode compartment to balance the negative charge. However, the cathode compartment is filled with sulfate ions (SO₄^2-).
• 3. Chemical Precipitation Reaction: The Ca²⁺ ions immediately combine with SO₄^2- ions to form a dense white precipitate of calcium sulfate (CaSO₄ ↓), which is highly insoluble:
    Ca²⁺(aq) + SO₄^2-(aq) → CaSO₄(s) ↓
• This fast precipitation reaction blocks the pores of the salt bridge and depletes the free ions needed to neutralize the accumulation of charges. As a result, the flow of electrons is blocked, and the cell stops operating after a very short time.

📊 Conclusion: Because calcium forms an insoluble precipitate with the sulfate ions of the cell solutions, option (b) represents the correct outcome.

🎯 The Core Chemical Rule: Iron (Fe) only corrodes when it acts as an **Anode** (i.e., when connected to a less active metal with a lower oxidation potential). The rate of iron corrosion increases as the oxidation potential of the connected metal decreases, creating a larger potential difference.

✅ Analyzing the Electrodes and Corrosion Rates:
• Sacrificial Cathodic Protection: In tubes (1) and (3), metals W (+1.27 V) and Y (+0.74 V) have higher oxidation potentials than Iron (+0.44 V). They act as anodes and corrode instead of Iron, meaning **Iron does not corrode at all in tubes (1) and (3)**.
• Iron Corrosion Condition: In tubes (2) and (4), metals Z (−0.4 V) and X (+0.12 V) have lower oxidation potentials than Iron (+0.44 V). Therefore, Iron acts as the anode and undergoes corrosion.
• Comparing Corrosion Speed:
    - **Tube (4) with metal X:** Potential difference = 0.44 − 0.12 = 0.32 V.
    - **Tube (2) with metal Z:** Potential difference = 0.44 − (−0.4) = 0.84 V.
• Since tube (2) has a larger potential difference, Iron corrodes faster in tube (2) than in tube (4). Thus, the rate of corrosion follows the order: (4) < (2).

📊 Conclusion: The rate of corrosion in tube (4) is less than that in tube (2), making option (a) the correct statement.

n(Cu)=0.12 mol → e⁻=0.24 mol → Q=0.24×96500=23160 C → t=23160/96.5=240 sec.

Y=propanone; reduction → 2-propanol (C₃H₈O); isomer = ethyl methyl ether.