Q: Q5. Calculate ΔrG° and log Kc for the following reaction at 298 K.\n2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)\nGiven: E°(Cr³⁺/Cr) = –0.74 V; E°(Cd²⁺/Cd) = –0.40 V

The standard Gibbs free energy change (ΔrG°) for the reaction is -196.98 kJ/mol, and the logarithm of the equilibrium constant (log Kc) is 34.55.

Q: Q17. Calculate the emf of the following cell at 298 K:\nMg(s) | Mg²⁺(0.13 M) || Cu²⁺(1.0 × 10⁻³ M) | Cu(s)\nGiven E°(Mg²⁺/Mg) = –2.37 V, E°(Cu²⁺/Cu) = +0.34 V

The emf of the given cell at 298 K is +2.68 V.

Question 1

Q1. Arrange the following metals in the order in which they displace each other from the solution of their salts. 
Al, Cu, Fe, Mg and Zn.

Accepted Answer

The ability of a metal to displace another metal from the solution of its salt depends on its reactivity, which is inversely related to its standard electrode potential (E°). A metal with a lower (more negative) standard electrode potential is more reactive and can displace metals with higher (less negative) standard electrode potentials from their salt solutions. This is because the more reactive metal has a greater tendency to lose electrons (get oxidized) and act as a reducing agent.

Let's list the standard electrode potentials (reduction potentials) for the given metals:
*   Mg2+/Mg: -2.37 V
*   Al3+/Al: -1.66 V
*   Zn2+/Zn: -0.76 V
*   Fe2+/Fe: -0.44 V
*   Cu2+/Cu: +0.34 V

Arranging these metals in increasing order of their standard reduction potentials (and thus decreasing order of reactivity or ability to displace others):
Mg > Al > Zn > Fe > Cu

This order indicates that:
*   Magnesium (Mg) can displace Al, Zn, Fe, and Cu from their salt solutions.
*   Aluminium (Al) can displace Zn, Fe, and Cu from their salt solutions.
*   Zinc (Zn) can displace Fe and Cu from their salt solutions.
*   Iron (Fe) can displace Cu from its salt solution.
*   Copper (Cu) cannot displace any of the other metals (Fe, Zn, Al, Mg) from their salt solutions.

Question 2

Q2. Given the standard electrode potentials:
K+/K = -2.93 V, Ag+/Ag = 0.80 V, Mg2+/Mg = -2.37 V, Cr3+/Cr = -0.74 V.
Arrange these metals in their increasing order of reducing power.

Accepted Answer

The reducing power of a metal is its tendency to lose electrons and get oxidized. A stronger reducing agent has a greater tendency to get oxidized, which corresponds to a more negative (or lower) standard reduction potential (E°).

Let's list the given standard electrode potentials (reduction potentials):
*   K+/K = -2.93 V
*   Mg2+/Mg = -2.37 V
*   Cr3+/Cr = -0.74 V
*   Ag+/Ag = +0.80 V

To arrange them in increasing order of reducing power, we need to arrange them from the least negative (or most positive) E° to the most negative E°.

*   **Ag+/Ag (+0.80 V)**: This is the most positive potential, indicating Ag+ is easily reduced, and Ag metal is a very weak reducing agent (least tendency to lose electrons).
*   **Cr3+/Cr (-0.74 V)**: This is less negative than Mg and K, so Cr is a stronger reducing agent than Ag but weaker than Mg and K.
*   **Mg2+/Mg (-2.37 V)**: This is a significantly negative potential, indicating Mg is a strong reducing agent.
*   **K+/K (-2.93 V)**: This is the most negative potential among the given, indicating K is the strongest reducing agent (greatest tendency to lose electrons).

Therefore, the increasing order of reducing power is:
**Ag < Cr < Mg < K**

Question 3

Q3. Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers

Accepted Answer

### Galvanic Cell Depiction and Explanation

The given spontaneous reaction is: Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

From this overall reaction, we can identify the oxidation and reduction half-reactions:
*   **Oxidation (Anode)**: Zn(s) → Zn2+(aq) + 2e-
*   **Reduction (Cathode)**: 2Ag+(aq) + 2e- → 2Ag(s)

This indicates that Zinc (Zn) acts as the anode (where oxidation occurs), and Silver (Ag) acts as the cathode (where reduction occurs).

#### Detailed Description of the Galvanic Cell:

1.  **Anode Compartment**: A zinc rod (electrode) is dipped into an aqueous solution containing Zn2+ ions (e.g., ZnSO4 solution). 2.  **Cathode Compartment**: A silver rod (electrode) is dipped into an aqueous solution containing Ag+ ions (e.g., AgNO3 solution). 3.  **External Circuit**: The zinc and silver electrodes are connected externally by a metallic wire, usually through a voltmeter or ammeter to measure the potential difference or current. Electrons flow from the anode (Zn) to the cathode (Ag) through this external wire. 4.  **Salt Bridge**: The two electrolyte solutions are connected by an inverted U-tube called a salt bridge. The salt bridge contains an inert electrolyte (e.g., KCl, KNO3) in a gel form. Its function is to maintain electrical neutrality in both half-cells by allowing the migration of ions, thus completing the circuit and preventing charge build-up.

#### (i) Which of the electrode is negatively charged?

*   The **zinc electrode (anode)** is negatively charged. At the anode, oxidation occurs, meaning zinc metal loses electrons (Zn → Zn2+ + 2e-). These released electrons accumulate on the zinc rod, making it the negative terminal of the cell. They then flow through the external circuit to the cathode.

#### (ii) The carriers of the current in the cell.

*   **In the external circuit (metallic wire)**: The carriers of current are **electrons**. They flow from the anode (Zn) to the cathode (Ag).
*   **In the electrolyte solutions (half-cells)**: The carriers of current are **ions**. Cations (Zn2+, Ag+) move towards the cathode, and anions (SO42-, NO3-) move towards the anode to maintain charge neutrality.
*   **In the salt bridge**: The carriers of current are **ions** (e.g., K+, Cl- from KCl). Anions from the salt bridge migrate towards the anode compartment, and cations migrate towards the cathode compartment to neutralize the excess charges built up due to the electrode reactions.

#### (iii) Individual reaction at each electrode.

*   **At the Anode (Negative Electrode - Oxidation)**:
    Zn(s) → Zn2+(aq) + 2e-
    (Zinc metal is oxidized to zinc ions, releasing electrons.)

*   **At the Cathode (Positive Electrode - Reduction)**:
    2Ag+(aq) + 2e- → 2Ag(s)
    (Silver ions are reduced to silver metal by accepting electrons.)

Question 4

Q4. Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)
(ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(a

Accepted Answer

The standard cell potentials (E°cell), standard Gibbs free energy (ΔᵣG°), and equilibrium constant (K) for the given reactions are calculated as follows:

**Standard Electrode Potentials Used:**
*   E°(Cr³⁺/Cr) = -0.74 V
*   E°(Cd²⁺/Cd) = -0.40 V
*   E°(Fe³⁺/Fe²⁺) = +0.77 V
*   E°(Ag⁺/Ag) = +0.80 V

**Part (i): 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)**

**1. Standard Cell Potential (E°cell):**
*   Oxidation half-reaction: Cr(s) → Cr³⁺(aq) + 3e⁻ (Anode)
    E°oxidation = -E°(Cr³⁺/Cr) = -(-0.74 V) = +0.74 V
*   Reduction half-reaction: Cd²⁺(aq) + 2e⁻ → Cd(s) (Cathode)
    E°reduction = E°(Cd²⁺/Cd) = -0.40 V

To balance electrons, multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3. The number of electrons transferred (n) = 6.

E°cell = E°cathode - E°anode = E°(Cd²⁺/Cd) - E°(Cr³⁺/Cr)
E°cell = (-0.40 V) - (-0.74 V) = +0.34 V

**2. Standard Gibbs Free Energy (ΔᵣG°):**
ΔᵣG° = -nFE°cell
Where, n = 6 (electrons transferred)
F = 96485 C/mol (Faraday's constant)
E°cell = +0.34 V

ΔᵣG° = -(6 mol) × (96485 C/mol) × (0.34 J/C)
ΔᵣG° = -196829.4 J
ΔᵣG° = -196.83 kJ

**3. Equilibrium Constant (K):**
E°cell = (0.0592/n) log K (at 298 K)
0.34 = (0.0592/6) log K
log K = (0.34 × 6) / 0.0592 = 2.04 / 0.0592 = 34.459
K = 10³⁴·⁴⁵⁹
K ≈ 2.88 × 10³⁴

**Part (ii): Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)**

**1. Standard Cell Potential (E°cell):**
*   Oxidation half-reaction: Fe²⁺(aq) → Fe³⁺(aq) + e⁻ (Anode)
    E°oxidation = -E°(Fe³⁺/Fe²⁺) = -(+0.77 V) = -0.77 V
*   Reduction half-reaction: Ag⁺(aq) + e⁻ → Ag(s) (Cathode)
    E°reduction = E°(Ag⁺/Ag) = +0.80 V

Number of electrons transferred (n) = 1.

E°cell = E°cathode - E°anode = E°(Ag⁺/Ag) - E°(Fe³⁺/Fe²⁺)
E°cell = (+0.80 V) - (+0.77 V) = +0.03 V

**2. Standard Gibbs Free Energy (ΔᵣG°):**
ΔᵣG° = -nFE°cell
Where, n = 1 (electron transferred)
F = 96485 C/mol
E°cell = +0.03 V

ΔᵣG° = -(1 mol) × (96485 C/mol) × (0.03 J/C)
ΔᵣG° = -2894.55 J
ΔᵣG° = -2.89 kJ

**3. Equilibrium Constant (K):**
E°cell = (0.0592/n) log K (at 298 K)
0.03 = (0.0592/1) log K
log K = 0.03 / 0.0592 = 0.5067
K = 10⁰·⁵⁰⁶⁷
K ≈ 3.21

Question 5

Q5. Calculate ΔrG° and log Kc for the following reaction at 298 K.
2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)
Given: E°(Cr³⁺/Cr) = –0.74 V; E°(Cd²⁺/Cd) = –0.40 V

Accepted Answer

The standard Gibbs free energy change (ΔrG°) for the reaction is -196.98 kJ/mol, and the logarithm of the equilibrium constant (log Kc) is 34.55.

Question 6

Q6. Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe³⁺(aq) and I⁻(aq)
(ii) Ag⁺(aq) and Cu(s)
(iii) Fe³⁺(

Accepted Answer

A reaction is feasible (spontaneous) if its standard cell potential (E°cell) is positive. We will calculate E°cell for each pair using standard reduction potentials from NCERT Table 3.1 (or commonly accepted values). Standard Reduction Potentials (approximate values): E°(Fe³⁺/Fe²⁺) = +0.77 V E°(I₂/I⁻) = +0.54 V E°(Ag⁺/Ag) = +0.80 V E°(Cu²⁺/Cu) = +0.34 V E°(Br₂/Br⁻) = +1.09 V (i) **Fe³⁺(aq) and I⁻(aq)** Proposed reaction: Fe³⁺(aq) + I⁻(aq) → Fe²⁺(aq) + ½I₂(s) Here, Fe³⁺ is reduced to Fe²⁺, and I⁻ is oxidized to I₂. Reduction: Fe³⁺(aq) + e⁻ → Fe²⁺(aq) ; E°red = +0.77 V Oxidation: 2I⁻(aq) → I₂(s) + 2e⁻ ; E°ox = -E°(I₂/I⁻) = -0.54 V E°cell = E°red (cathode) + E°ox (anode) = +0.77 V + (-0.54 V) = +0.23 V Since E°cell > 0, the reaction is **feasible**. (ii) **Ag⁺(aq) and Cu(s)** Proposed reaction: 2Ag⁺(aq) + Cu(s) → 2Ag(s) + Cu²⁺(aq) Here, Ag⁺ is reduced to Ag, and Cu is oxidized to Cu²⁺. Reduction: Ag⁺(aq) + e⁻ → Ag(s) ; E°red = +0.80 V Oxidation: Cu(s) → Cu²⁺(aq) + 2e⁻ ; E°ox = -E°(Cu²⁺/Cu) = -0.34 V E°cell = E°red (cathode) + E°ox (anode) = +0.80 V + (-0.34 V) = +0.46 V Since E°cell > 0, the reaction is **feasible**. (iii) **Fe³⁺(aq) and Br⁻(aq)** Proposed reaction: Fe³⁺(aq) + Br⁻(aq) → Fe²⁺(aq) + ½Br₂(aq) Here, Fe³⁺ is reduced to Fe²⁺, and Br⁻ is oxidized to Br₂. Reduction: Fe³⁺(aq) + e⁻ → Fe²⁺(aq) ; E°red = +0.77 V Oxidation: 2Br⁻(aq) → Br₂(aq) + 2e⁻ ; E°ox = -E°(Br₂/Br⁻) = -1.09 V E°cell = E°red (cathode) + E°ox (anode) = +0.77 V + (-1.09 V) = -0.32 V Since E°cell < 0, the reaction is **not feasible**. (iv) **Ag(s) and Fe³⁺(aq)** Proposed reaction: Ag(s) + Fe³⁺(aq) → Ag⁺(aq) + Fe²⁺(aq) Here, Ag is oxidized to Ag⁺, and Fe³⁺ is reduced to Fe²⁺. Reduction: Fe³⁺(aq) + e⁻ → Fe²⁺(aq) ; E°red = +0.77 V Oxidation: Ag(s) → Ag⁺(aq) + e⁻ ; E°ox = -E°(Ag⁺/Ag) = -0.80 V E°cell = E°red (cathode) + E°ox (anode) = +0.77 V + (-0.80 V) = -0.03 V Since E°cell < 0, the reaction is **not feasible**. (v) **Br₂(aq) and Fe²⁺(aq)** Proposed reaction: Br₂(aq) + 2Fe²⁺(aq) → 2Br⁻(aq) + 2Fe³⁺(aq) Here, Br₂ is reduced to Br⁻, and Fe²⁺ is oxidized to Fe³⁺. Reduction: Br₂(aq) + 2e⁻ → 2Br⁻(aq) ; E°red = +1.09 V Oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻ ; E°ox = -E°(Fe³⁺/Fe²⁺) = -0.77 V E°cell = E°red (cathode) + E°ox (anode) = +1.09 V + (-0.77 V) = +0.32 V Since E°cell > 0, the reaction is **feasible**.

Question 7

Q7. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes

Accepted Answer

The products of electrolysis depend on the nature of the electrolyte, the electrodes used (inert or active), and the standard electrode potentials of the species present, including water. At the cathode, reduction occurs, and the species with a higher (less negative) reduction potential will be preferentially reduced. At the anode, oxidation occurs, and the species with a lower (less positive) reduction potential (or higher oxidation potential) will be preferentially oxidized. Overpotential effects can also influence the products, especially for gas evolution.

### (i) An aqueous solution of AgNO₃ with silver electrodes.

**Ions present in solution:** Ag⁺, NO₃⁻, H⁺, OH⁻ (from water dissociation)
**Electrodes:** Silver (Ag) electrodes (active electrodes)

**At Cathode (Reduction):**
Possible reduction reactions:
1.  Ag⁺(aq) + e⁻ → Ag(s) ; E° = +0.80 V
2.  2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E° = -0.83 V (at pH 7) or 2H⁺(aq) + 2e⁻ → H₂(g) ; E° = 0.00 V

Since E°(Ag⁺/Ag) (+0.80 V) is much higher than E°(H₂O/H₂) or E°(H⁺/H₂), Ag⁺ ions will be preferentially reduced.
**Product at Cathode: Silver (Ag) metal is deposited.**

**At Anode (Oxidation):**
Possible oxidation reactions:
1.  Oxidation of Ag electrode: Ag(s) → Ag⁺(aq) + e⁻ ; E°ox = -E°(Ag⁺/Ag) = -0.80 V
2.  Oxidation of NO₃⁻: NO₃⁻ is difficult to oxidize, usually requires very high potentials.
3.  Oxidation of H₂O: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; E°ox = -E°(O₂/H₂O) = -1.23 V

Since the silver electrode is active, it will be preferentially oxidized over water or nitrate ions because its oxidation potential is less negative (or reduction potential is less positive) compared to water, and NO₃⁻ is very difficult to oxidize.
**Product at Anode: Silver (Ag) metal from the electrode dissolves to form Ag⁺ ions.**

**Overall Reaction:** Ag(s) (anode) + Ag⁺(aq) (cathode) → Ag⁺(aq) (anode) + Ag(s) (cathode)
This process is essentially electrorefining or electroplating, where silver from the anode dissolves and deposits onto the cathode.

### (ii) An aqueous solution of AgNO₃ with platinum electrodes.

**Ions present in solution:** Ag⁺, NO₃⁻, H⁺, OH⁻
**Electrodes:** Platinum (Pt) electrodes (inert electrodes)

**At Cathode (Reduction):**
Possible reduction reactions:
1.  Ag⁺(aq) + e⁻ → Ag(s) ; E° = +0.80 V
2.  2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E° = -0.83 V (at pH 7)

Ag⁺ has a much higher reduction potential than water. So, Ag⁺ ions will be preferentially reduced.
**Product at Cathode: Silver (Ag) metal is deposited.**

**At Anode (Oxidation):**
Possible oxidation reactions:
1.  Oxidation of NO₃⁻: Difficult to oxidize.
2.  Oxidation of H₂O: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; E°ox = -1.23 V

Between NO₃⁻ and H₂O, water has a lower oxidation potential (easier to oxidize) than NO₃⁻. Therefore, water will be oxidized.
**Product at Anode: Oxygen (O₂) gas is evolved.**

**Overall Reaction:** 4AgNO₃(aq) + 2H₂O(l) → 4Ag(s) + O₂(g) + 4HNO₃(aq)

### (iii) A dilute solution of H₂SO₄ with platinum electrodes.

**Ions present in solution:** H⁺, SO₄²⁻, OH⁻ (from water dissociation)
**Electrodes:** Platinum (Pt) electrodes (inert electrodes)

**At Cathode (Reduction):**
Possible reduction reactions:
1.  2H⁺(aq) + 2e⁻ → H₂(g) ; E° = 0.00 V
2.  2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E° = -0.83 V (at pH 7)

In a dilute acidic solution, the concentration of H⁺ is high. H⁺ ions have a higher reduction potential than water (or rather, water reduction is less favorable in acidic conditions compared to H⁺ reduction). So, H⁺ ions will be preferentially reduced.
**Product at Cathode: Hydrogen (H₂) gas is evolved.**

**At Anode (Oxidation):**
Possible oxidation reactions:
1.  Oxidation of SO₄²⁻: SO₄²⁻ is very difficult to oxidize (E° for S₂O₈²⁻/SO₄²⁻ is +2.01 V, implying very high oxidation potential for SO₄²⁻).
2.  Oxidation of H₂O: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; E°ox = -1.23 V

Water has a much lower oxidation potential than sulfate ions. Therefore, water will be oxidized.
**Product at Anode: Oxygen (O₂) gas is evolved.**

**Overall Reaction:** 2H₂O(l) → 2H₂(g) + O₂(g)
(H₂SO₄ acts as an electrolyte, facilitating conduction but not consumed in the overall reaction).

### (iv) An aqueous solution of CuCl₂ with platinum electrodes.

**Ions present in solution:** Cu²⁺, Cl⁻, H⁺, OH⁻
**Electrodes:** Platinum (Pt) electrodes (inert electrodes)

**At Cathode (Reduction):**
Possible reduction reactions:
1.  Cu²⁺(aq) + 2e⁻ → Cu(s) ; E° = +0.34 V
2.  2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E° = -0.83 V (at pH 7)
3.  2H⁺(aq) + 2e⁻ → H₂(g) ; E° = 0.00 V

Cu²⁺ has a higher reduction potential (+0.34 V) than H⁺ (0.00 V) and water (-0.83 V). So, Cu²⁺ ions will be preferentially reduced.
**Product at Cathode: Copper (Cu) metal is deposited.**

**At Anode (Oxidation):**
Possible oxidation reactions:
1.  Oxidation of Cl⁻: 2Cl⁻(aq) → Cl₂(g) + 2e⁻ ; E°ox = -E°(Cl₂/Cl⁻) = -1.36 V
2.  Oxidation of H₂O: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; E°ox = -E°(O₂/H₂O) = -1.23 V

Comparing the oxidation potentials, water (-1.23 V) appears to be easier to oxidize than Cl⁻ (-1.36 V). However, due to the **overpotential** for oxygen evolution on platinum electrodes, the actual potential required to oxidize water is higher than the theoretical 1.23 V. This often makes the oxidation of Cl⁻ ions more favorable, especially if the concentration of Cl⁻ is high.

In concentrated solutions of halides, the halide ion is preferentially oxidized. In dilute solutions, water is oxidized. Assuming a typical aqueous solution of CuCl₂, the concentration of Cl⁻ is usually sufficient for its oxidation.
**Product at Anode: Chlorine (Cl₂) gas is evolved.**

**Overall Reaction:** CuCl₂(aq) + H₂O(l) → Cu(s) + Cl₂(g) + H₂O (remaining)
(Note: The question implies an aqueous solution, so water is present. The overall reaction would be 2CuCl₂(aq) + 2H₂O(l) → 2Cu(s) + 2Cl₂(g) + O₂(g) if water was oxidized, but with overpotential, Cl₂ is preferred.)

Final products:
(i) Cathode: Ag, Anode: Ag⁺
(ii) Cathode: Ag, Anode: O₂
(iii) Cathode: H₂, Anode: O₂
(iv) Cathode: Cu, Anode: Cl₂

Question 8

Q8. Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)
(ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(a

Accepted Answer

The standard cell potentials (E°cell), standard Gibbs free energy (ΔᵣG°), and equilibrium constants (K) for the given reactions are calculated below using standard electrode potentials:

**Standard Electrode Potentials (E° values) used:**
*   E°(Cr³⁺/Cr) = -0.74 V
*   E°(Cd²⁺/Cd) = -0.40 V
*   E°(Fe³⁺/Fe²⁺) = +0.77 V
*   E°(Ag⁺/Ag) = +0.80 V

**(i) Reaction: 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)**

**1. Half-reactions:**
*   Anode (Oxidation): Cr(s) → Cr³⁺(aq) + 3e⁻
*   Cathode (Reduction): Cd²⁺(aq) + 2e⁻ → Cd(s)

To balance electrons, multiply the anode reaction by 2 and the cathode reaction by 3:
*   2Cr(s) → 2Cr³⁺(aq) + 6e⁻
*   3Cd²⁺(aq) + 6e⁻ → 3Cd(s)

Overall reaction: 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)
Number of electrons transferred, n = 6.

**2. Standard Cell Potential (E°cell):**
E°cell = E°cathode - E°anode
E°cell = E°(Cd²⁺/Cd) - E°(Cr³⁺/Cr)
E°cell = (-0.40 V) - (-0.74 V)
E°cell = -0.40 V + 0.74 V
E°cell = +0.34 V

**3. Standard Gibbs Free Energy (ΔᵣG°):**
ΔᵣG° = -nFE°cell
Where F = 96485 C/mol (Faraday's constant)
ΔᵣG° = -(6 mol e⁻) × (96485 C/mol e⁻) × (0.34 J/C)
ΔᵣG° = -196829.4 J
ΔᵣG° = -196.83 kJ

**4. Equilibrium Constant (K):**
ΔᵣG° = -RTlnK
Alternatively, E°cell = (0.0592/n)logK (at 298 K)
logK = (n × E°cell) / 0.0592
logK = (6 × 0.34 V) / 0.0592 V
logK = 2.04 / 0.0592
logK = 34.459
K = 10^(34.459)
K = 2.87 × 10³⁴

---

**(ii) Reaction: Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)**

**1. Half-reactions:**
*   Anode (Oxidation): Fe²⁺(aq) → Fe³⁺(aq) + e⁻
*   Cathode (Reduction): Ag⁺(aq) + e⁻ → Ag(s)

Overall reaction: Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)
Number of electrons transferred, n = 1.

**2. Standard Cell Potential (E°cell):**
E°cell = E°cathode - E°anode
E°cell = E°(Ag⁺/Ag) - E°(Fe³⁺/Fe²⁺)
E°cell = (0.80 V) - (0.77 V)
E°cell = +0.03 V

**3. Standard Gibbs Free Energy (ΔᵣG°):**
ΔᵣG° = -nFE°cell
ΔᵣG° = -(1 mol e⁻) × (96485 C/mol e⁻) × (0.03 J/C)
ΔᵣG° = -2894.55 J
ΔᵣG° = -2.89 kJ

**4. Equilibrium Constant (K):**
E°cell = (0.0592/n)logK (at 298 K)
logK = (n × E°cell) / 0.0592
logK = (1 × 0.03 V) / 0.0592 V
logK = 0.03 / 0.0592
logK = 0.5067
K = 10^(0.5067)
K = 3.21

**Summary of Results:**
**(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)**
*   E°cell = +0.34 V
*   ΔᵣG° = -196.83 kJ
*   K = 2.87 × 10³⁴

**(ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)**
*   E°cell = +0.03 V
*   ΔᵣG° = -2.89 kJ
*   K = 3.21

Question 9

Q9. The standard electrode potential for Daniel cell is 1.1 V. Calculate the standard Gibbs energy for the reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).

Accepted Answer

The standard Gibbs energy for the reaction is -212.3 kJ/mol.

Question 10

Q10. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Accepted Answer

The potential of the hydrogen electrode in contact with a solution of pH 10 is -0.591 V.

Question 11

Q11. Calculate the emf of the cell in which the following reaction takes place: Ni(s) + 2Ag⁺(0.002 M) → Ni²⁺(0.16 M) + 2Ag(s) Given that E°_cell = 1.05 V

Accepted Answer

The emf of the cell (E_cell) is 0.914 V.

Question 12

Q12. Predict if the reaction between the following is feasible:
(i) Fe³⁺(aq) and I⁻(aq)
(ii) Ag⁺(aq) and Cu(s)
(iii) Fe³⁺(aq) and Br⁻(aq)
(iv) Ag(s) and Fe³⁺(aq)
(v) Br₂(aq) and Fe

Accepted Answer

A reaction is feasible (spontaneous) if the standard cell potential (E°_cell) for the reaction is positive (E°_cell > 0). E°_cell is calculated as E°_cathode - E°_anode, where the cathode is the reduction half-cell and the anode is the oxidation half-cell. Alternatively, E°_cell = E°_reduction + E°_oxidation. **Given standard reduction potentials:** E°_(I₂/I⁻) = +0.54 V E°_(Cu²⁺/Cu) = +0.34 V E°_(Ag⁺/Ag) = +0.80 V E°_(Fe³⁺/Fe²⁺) = +0.77 V E°_(Br₂/Br⁻) = +1.09 V Let's analyze each case: **(i) Fe³⁺(aq) and I⁻(aq)** * **Possible Reaction:** Fe³⁺ will be reduced, and I⁻ will be oxidized. * Reduction: Fe³⁺(aq) + e⁻ → Fe²⁺(aq) ; E°_red = +0.77 V * Oxidation: 2I⁻(aq) → I₂(s) + 2e⁻ ; E°_ox = -E°_(I₂/I⁻) = -0.54 V * **Overall Reaction:** 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s) * **E°_cell Calculation:** E°_cell = E°_red (Fe³⁺/Fe²⁺) + E°_ox (I⁻/I₂) E°_cell = +0.77 V + (-0.54 V) = +0.23 V * **Conclusion:** Since E°_cell is positive (+0.23 V > 0), the reaction is **feasible**. **(ii) Ag⁺(aq) and Cu(s)** * **Possible Reaction:** Ag⁺ will be reduced, and Cu will be oxidized. * Reduction: Ag⁺(aq) + e⁻ → Ag(s) ; E°_red = +0.80 V * Oxidation: Cu(s) → Cu²⁺(aq) + 2e⁻ ; E°_ox = -E°_(Cu²⁺/Cu) = -0.34 V * **Overall Reaction:** 2Ag⁺(aq) + Cu(s) → 2Ag(s) + Cu²⁺(aq) * **E°_cell Calculation:** E°_cell = E°_red (Ag⁺/Ag) + E°_ox (Cu/Cu²⁺) E°_cell = +0.80 V + (-0.34 V) = +0.46 V * **Conclusion:** Since E°_cell is positive (+0.46 V > 0), the reaction is **feasible**. **(iii) Fe³⁺(aq) and Br⁻(aq)** * **Possible Reaction:** Fe³⁺ will be reduced, and Br⁻ will be oxidized. * Reduction: Fe³⁺(aq) + e⁻ → Fe²⁺(aq) ; E°_red = +0.77 V * Oxidation: 2Br⁻(aq) → Br₂(aq) + 2e⁻ ; E°_ox = -E°_(Br₂/Br⁻) = -1.09 V * **Overall Reaction:** 2Fe³⁺(aq) + 2Br⁻(aq) → 2Fe²⁺(aq) + Br₂(aq) * **E°_cell Calculation:** E°_cell = E°_red (Fe³⁺/Fe²⁺) + E°_ox (Br⁻/Br₂) E°_cell = +0.77 V + (-1.09 V) = -0.32 V * **Conclusion:** Since E°_cell is negative (-0.32 V < 0), the reaction is **not feasible**. **(iv) Ag(s) and Fe³⁺(aq)** * **Possible Reaction:** Fe³⁺ will be reduced, and Ag will be oxidized. * Reduction: Fe³⁺(aq) + e⁻ → Fe²⁺(aq) ; E°_red = +0.77 V * Oxidation: Ag(s) → Ag⁺(aq) + e⁻ ; E°_ox = -E°_(Ag⁺/Ag) = -0.80 V * **Overall Reaction:** Ag(s) + Fe³⁺(aq) → Ag⁺(aq) + Fe²⁺(aq) * **E°_cell Calculation:** E°_cell = E°_red (Fe³⁺/Fe²⁺) + E°_ox (Ag/Ag⁺) E°_cell = +0.77 V + (-0.80 V) = -0.03 V * **Conclusion:** Since E°_cell is negative (-0.03 V < 0), the reaction is **not feasible**. **(v) Br₂(aq) and Fe²⁺(aq)** * **Possible Reaction:** Br₂ will be reduced, and Fe²⁺ will be oxidized. * Reduction: Br₂(aq) + 2e⁻ → 2Br⁻(aq) ; E°_red = +1.09 V * Oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻ ; E°_ox = -E°_(Fe³⁺/Fe²⁺) = -0.77 V * **Overall Reaction:** Br₂(aq) + 2Fe²⁺(aq) → 2Br⁻(aq) + 2Fe³⁺(aq) * **E°_cell Calculation:** E°_cell = E°_red (Br₂/Br⁻) + E°_ox (Fe²⁺/Fe³⁺) E°_cell = +1.09 V + (-0.77 V) = +0.32 V * **Conclusion:** Since E°_cell is positive (+0.32 V > 0), the reaction is **feasible**.

Question 13

Q13. Calculate the standard Gibbs energy change and the equilibrium constant for the reaction:
Cd²⁺(aq) + Zn(s) → Cd(s) + Zn²⁺(aq)
Given: E°Cd²⁺/Cd = -0.40 V, E°Zn²⁺/Zn = -0.76 V

Accepted Answer

The standard Gibbs energy change (ΔG°) for the reaction is -69.48 kJ/mol, and the equilibrium constant (K_c) for the reaction is approximately 1.02 x 10¹². This indicates that the reaction is spontaneous and highly favors product formation at equilibrium.

Question 14

Q14. Calculate the standard Gibbs energy change for the reaction:
Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s)
Given: E°(Mg²⁺/Mg) = –2.37 V, E°(Cu²⁺/Cu) = +0.34 V

Accepted Answer

The standard Gibbs energy change (ΔG°) for the given reaction is -599.91 kJ/mol.

Question 15

Q15. Read the following data:
E°(Cr₂O₇²⁻/Cr³⁺) = 1.33 V; E°(Cl₂/Cl⁻) = 1.36 V; E°(MnO₄⁻/Mn²⁺) = 1.51 V; E°(Fe³⁺/Fe²⁺) = 0.77 V
Arrange these in increasing order of strength as oxidisi

Accepted Answer

The strength of an oxidizing agent is directly proportional to its standard reduction potential (E°). A higher positive E° value indicates a greater tendency for reduction, and thus a stronger oxidizing agent.

Given standard reduction potentials:
1. E°(Fe³⁺/Fe²⁺) = 0.77 V
2. E°(Cr₂O₇²⁻/Cr³⁺) = 1.33 V
3. E°(Cl₂/Cl⁻) = 1.36 V
4. E°(MnO₄⁻/Mn²⁺) = 1.51 V

Arranging these in increasing order of their E° values (and thus increasing order of strength as oxidizing agents):
Fe³⁺ < Cr₂O₇²⁻ < Cl₂ < MnO₄⁻

Therefore, the increasing order of strength as oxidizing agents is:
Fe³⁺ < Cr₂O₇²⁻ < Cl₂ < MnO₄⁻

Question 16

Q16. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes

Accepted Answer

The products of electrolysis depend on the nature of the electrolyte, the electrodes used, and the standard electrode potentials of the species involved. For aqueous solutions, water can also be oxidized or reduced.

### (i) An aqueous solution of AgNO₃ with silver electrodes.

**Species present:** Ag⁺, NO₃⁻, H₂O, Ag (from electrodes)

**At Cathode (Reduction):**
Possible reactions:
1. Ag⁺(aq) + e⁻ → Ag(s) ; E° = +0.80 V
2. 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E° = -0.83 V (at pH 7, or -0.41 V at 1M H⁺)

Since E°(Ag⁺/Ag) (+0.80 V) is much higher (more positive) than E° for water reduction, Ag⁺ ions will be preferentially reduced.

**At Anode (Oxidation):**
Possible reactions:
1. 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; E° = -1.23 V (at pH 7, or -0.82 V at 1M H⁺)
2. 2NO₃⁻(aq) → (no simple oxidation product, requires very high potential)
3. Ag(s) → Ag⁺(aq) + e⁻ ; E° = -0.80 V (oxidation of electrode material)

Since the electrodes are active (silver), the silver electrode itself will be oxidized because its oxidation potential (-0.80 V) is less negative (more easily oxidized) than water (-1.23 V). Nitrate ions are difficult to oxidize.

**Products:**
*   **At Cathode:** Silver (Ag) metal is deposited.
*   **At Anode:** Silver (Ag) metal from the electrode dissolves to form Ag⁺ ions.

### (ii) An aqueous solution of AgNO₃ with platinum electrodes.

**Species present:** Ag⁺, NO₃⁻, H₂O, Pt (inert electrodes)

**At Cathode (Reduction):**
Possible reactions:
1. Ag⁺(aq) + e⁻ → Ag(s) ; E° = +0.80 V
2. 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E° = -0.83 V (at pH 7)

Ag⁺ ions will be preferentially reduced due to their higher standard reduction potential.

**At Anode (Oxidation):**
Possible reactions:
1. 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; E° = -1.23 V (at pH 7)
2. 2NO₃⁻(aq) → (difficult to oxidize)

Between water and nitrate ions, water has a lower (less negative) oxidation potential and is preferentially oxidized. (Note: The actual potential for O₂ evolution is often higher due to overpotential, but it's still preferred over nitrate oxidation).

**Products:**
*   **At Cathode:** Silver (Ag) metal is deposited.
*   **At Anode:** Oxygen (O₂) gas is evolved.

### (iii) A dilute solution of H₂SO₄ with platinum electrodes.

**Species present:** H⁺, SO₄²⁻, H₂O, Pt (inert electrodes)

**At Cathode (Reduction):**
Possible reactions:
1. 2H⁺(aq) + 2e⁻ → H₂(g) ; E° = 0.00 V (in acidic solution)
2. 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E° = -0.83 V (at pH 7)

Since the solution is acidic (H₂SO₄), H⁺ ions are present in significant concentration. H⁺ ions will be preferentially reduced.

**At Anode (Oxidation):**
Possible reactions:
1. 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; E° = -1.23 V (at pH 7, or -0.82 V at 1M H⁺)
2. 2SO₄²⁻(aq) → S₂O₈²⁻(aq) + 2e⁻ ; E° = -2.01 V (for S₂O₈²⁻/SO₄²⁻)

Water has a much lower (less negative) oxidation potential than sulfate ions, so water will be preferentially oxidized. Even with overpotential for oxygen evolution, it is still preferred over sulfate oxidation.

**Products:**
*   **At Cathode:** Hydrogen (H₂) gas is evolved.
*   **At Anode:** Oxygen (O₂) gas is evolved.

### (iv) An aqueous solution of CuCl₂ with platinum electrodes.

**Species present:** Cu²⁺, Cl⁻, H₂O, Pt (inert electrodes)

**At Cathode (Reduction):**
Possible reactions:
1. Cu²⁺(aq) + 2e⁻ → Cu(s) ; E° = +0.34 V
2. 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) ; E° = -0.83 V (at pH 7)

Cu²⁺ ions will be preferentially reduced due to their higher standard reduction potential.

**At Anode (Oxidation):**
Possible reactions:
1. 2Cl⁻(aq) → Cl₂(g) + 2e⁻ ; E° = -1.36 V
2. 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ ; E° = -1.23 V (at pH 7)

Comparing the standard oxidation potentials, water (-1.23 V) appears to be more easily oxidized than chloride ions (-1.36 V). However, due to the overpotential for oxygen evolution on platinum electrodes, the oxidation of Cl⁻ ions to Cl₂ gas often occurs preferentially, especially if Cl⁻ concentration is high. The actual potential required for Cl₂ evolution is around 1.36 V, while for O₂ it can be 1.4-1.7 V due to overpotential. Therefore, Cl⁻ is oxidized.

**Products:**
*   **At Cathode:** Copper (Cu) metal is deposited.
*   **At Anode:** Chlorine (Cl₂) gas is evolved.

Question 17

Q17. Calculate the emf of the following cell at 298 K:
Mg(s) | Mg²⁺(0.13 M) || Cu²⁺(1.0 × 10⁻³ M) | Cu(s)
Given E°(Mg²⁺/Mg) = –2.37 V, E°(Cu²⁺/Cu) = +0.34 V

Accepted Answer

The emf of the given cell at 298 K is +2.68 V.

Question 18

Q18. Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe³⁺(aq) and I⁻(aq)
(ii) Ag⁺(aq) and Cu(s)
(iii) Fe³⁺(

Accepted Answer

A reaction is feasible (spontaneous) if the standard cell potential (E°cell) for the overall reaction is positive (E°cell > 0). We calculate E°cell for each pair by identifying the species that gets oxidized and reduced, and then using the formula E°cell = E°cathode - E°anode.

Question 19

Q19. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes

Accepted Answer

The products of electrolysis depend on the nature of the electrolyte, the electrodes used, and the standard electrode potentials of all species present, including water. Overpotential effects also play a role, especially for gas evolution.

Question 20

Q20. Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.

Accepted Answer

The ability of a metal to displace another metal from its salt solution depends on its reactivity, which is inversely related to its standard reduction potential. A metal with a lower (more negative) standard reduction potential is a stronger reducing agent and can displace metals with higher (less negative or positive) standard reduction potentials from their salt solutions. In other words, a more reactive metal can displace a less reactive metal. Let's list the standard reduction potentials (E° values) for these metals: Mg²⁺/Mg: -2.37 V Al³⁺/Al: -1.66 V Zn²⁺/Zn: -0.76 V Fe²⁺/Fe: -0.44 V Cu²⁺/Cu: +0.34 V Arranging these metals in increasing order of their standard reduction potentials (i.e., decreasing order of reactivity/ability to displace others): Mg < Al < Zn < Fe < Cu Therefore, the order in which they displace each other from the solution of their salts (most reactive to least reactive, or strongest reducing agent to weakest reducing agent) is: **Mg > Al > Zn > Fe > Cu** This means: - Mg can displace Al, Zn, Fe, Cu. - Al can displace Zn, Fe, Cu (but not Mg). - Zn can displace Fe, Cu (but not Mg, Al). - Fe can displace Cu (but not Mg, Al, Zn). - Cu cannot displace any of the others.

Question 21

Q21. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10⁻³ S

Accepted Answer

The cell constant (G*) of a conductivity cell is calculated using the relationship between conductivity (κ), resistance (R), and cell constant. The formula is:
Cell Constant (G*) = Conductivity (κ) × Resistance (R)

Given:
Resistance (R) = 1500 Ω
Conductivity (κ) = 0.146 × 10⁻³ S cm⁻¹

Substituting the given values into the formula:
G* = (0.146 × 10⁻³ S cm⁻¹) × (1500 Ω)

Since S (Siemens) is the reciprocal of Ω (Ohm), i.e., S = Ω⁻¹, the units S and Ω will cancel each other out, leaving the unit for cell constant as cm⁻¹.

G* = 0.146 × 10⁻³ × 1500 cm⁻¹
G* = 0.146 × 1.5 cm⁻¹
G* = 0.219 cm⁻¹

Therefore, the cell constant is 0.219 cm⁻¹.

Question 22

Q22. Calculate the potential of the following cell at 298 K:
Mg(s) | Mg²⁺(0.13 M) || Cu²⁺(1.0 × 10⁻³ M) | Cu(s)
[Given E°(Mg²⁺/Mg) = –2.37 V, E°(Cu²⁺/Cu) = +0.34 V]

Accepted Answer

The potential of the given cell at 298 K is +2.68 V.

Question 23

Q23. Write the reactions occurring in the lead storage battery (a) during charging (b) during discharging.

Accepted Answer

The lead storage battery is a secondary cell, meaning it can be recharged. The reactions occurring depend on whether it is discharging (acting as a galvanic cell) or charging (acting as an electrolytic cell).

Question 24

Q24. Write the cell reactions for the fuel cell (H₂ – O₂ fuel cell).

Accepted Answer

The H₂ – O₂ fuel cell is a type of fuel cell that converts the chemical energy from a fuel (hydrogen) and an oxidant (oxygen) into electrical energy through a redox reaction. The reactions typically occur in an alkaline or acidic medium.

Question 25

Q25. Conductivity of 0.00241 M acetic acid is 7.896 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity. If Λ°m for acetic acid is 390.5 S cm² mol⁻¹, what is its dissociation constant?

Accepted Answer

The molar conductivity (Λm) of the acetic acid solution is 32.76 S cm² mol⁻¹. The dissociation constant (Ka) for acetic acid is 1.85 × 10⁻⁵ mol L⁻¹.

### Detailed Calculation:

**Given:**
Concentration of acetic acid (C) = 0.00241 M
Conductivity (κ) = 7.896 × 10⁻⁵ S cm⁻¹
Limiting molar conductivity (Λ°m) = 390.5 S cm² mol⁻¹

**Part 1: Calculation of Molar Conductivity (Λm)**
The formula for molar conductivity is:
Λm = (κ × 1000) / C
where κ is in S cm⁻¹ and C is in mol L⁻¹.

Substituting the given values:
Λm = (7.896 × 10⁻⁵ S cm⁻¹ × 1000 cm³ L⁻¹) / 0.00241 mol L⁻¹
Λm = 0.07896 / 0.00241 S cm² mol⁻¹
Λm = 32.76 S cm² mol⁻¹

**Part 2: Calculation of Dissociation Constant (Ka)**
First, we need to calculate the degree of dissociation (α) for acetic acid, which is a weak electrolyte.
The formula for the degree of dissociation is:
α = Λm / Λ°m

Substituting the values:
α = 32.76 S cm² mol⁻¹ / 390.5 S cm² mol⁻¹
α = 0.0839

Now, we can calculate the dissociation constant (Ka) for the equilibrium:
CH₃COOH ⇌ CH₃COO⁻ + H⁺
Ka = [CH₃COO⁻][H⁺] / [CH₃COOH]
Ka = (Cα)(Cα) / C(1-α)
Ka = Cα² / (1 - α)

Substituting the values of C and α:
Ka = (0.00241 mol L⁻¹ × (0.0839)²) / (1 - 0.0839)
Ka = (0.00241 × 0.007039) / 0.9161
Ka = 0.00001696 / 0.9161
Ka = 1.85 × 10⁻⁵ mol L⁻¹

Question 26

Q26. How much charge is required for the following reductions:
(i) 1 mol of Al³⁺ to Al.
(ii) 1 mol of Cu²⁺ to Cu.
(iii) 1 mol of MnO₄⁻ to Mn²⁺.

Accepted Answer

The charge required for the given reductions is:
(i) 289500 C or 3 F for 1 mol of Al³⁺ to Al.
(ii) 193000 C or 2 F for 1 mol of Cu²⁺ to Cu.
(iii) 482500 C or 5 F for 1 mol of MnO₄⁻ to Mn²⁺.

### Detailed Calculation:

We use the relationship that the charge on one mole of electrons is equal to one Faraday (F), which is approximately 96500 Coulombs (C).
Charge (Q) = n × F, where 'n' is the number of moles of electrons transferred.

**(i) 1 mol of Al³⁺ to Al**
The reduction half-reaction is: Al³⁺(aq) + 3e⁻ → Al(s)
From the stoichiometry, 1 mole of Al³⁺ ions requires 3 moles of electrons for reduction.
Therefore, n = 3.
Charge required = 3 × F = 3 × 96500 C = 289500 C.

**(ii) 1 mol of Cu²⁺ to Cu**
The reduction half-reaction is: Cu²⁺(aq) + 2e⁻ → Cu(s)
From the stoichiometry, 1 mole of Cu²⁺ ions requires 2 moles of electrons for reduction.
Therefore, n = 2.
Charge required = 2 × F = 2 × 96500 C = 193000 C.

**(iii) 1 mol of MnO₄⁻ to Mn²⁺**
First, we need to determine the number of electrons involved in this reduction. We can do this by balancing the half-reaction in an acidic medium.
Step 1: Determine the oxidation state of Mn.
In MnO₄⁻, let the oxidation state of Mn be x. Then, x + 4(-2) = -1, which gives x = +7.
In Mn²⁺, the oxidation state of Mn is +2.
Step 2: Calculate the change in oxidation state.
Change = Initial state - Final state = (+7) - (+2) = 5.
This means 5 electrons are gained per Mn atom.
The balanced half-reaction is: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
From the stoichiometry, 1 mole of MnO₄⁻ ions requires 5 moles of electrons for reduction.
Therefore, n = 5.
Charge required = 5 × F = 5 × 96500 C = 482500 C.

Question 27

Q27. How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl₂.
(ii) 40.0 g of Al from molten Al₂O₃.

Accepted Answer

The electricity required in terms of Faraday is:
(i) 1 F to produce 20.0 g of Ca.
(ii) 4.44 F to produce 40.0 g of Al.

### Detailed Calculation:

**(i) 20.0 g of Ca from molten CaCl₂**
The electrode reaction for the production of Ca is the reduction of Ca²⁺ ions:
Ca²⁺ + 2e⁻ → Ca(s)
This equation shows that 1 mole of Ca (molar mass ≈ 40 g/mol) is produced by 2 moles of electrons.
Charge required for 1 mole (40 g) of Ca = 2 Faraday (2 F).

We need to produce 20.0 g of Ca.
First, calculate the moles of Ca in 20.0 g:
Moles of Ca = Mass / Molar Mass = 20.0 g / 40 g mol⁻¹ = 0.5 mol.

Now, calculate the required electricity:
Since 1 mol of Ca requires 2 F of electricity,
0.5 mol of Ca will require = 0.5 mol × 2 F/mol = 1 F.

**(ii) 40.0 g of Al from molten Al₂O₃**
The electrode reaction for the production of Al is the reduction of Al³⁺ ions:
Al³⁺ + 3e⁻ → Al(s)
This equation shows that 1 mole of Al (molar mass ≈ 27 g/mol) is produced by 3 moles of electrons.
Charge required for 1 mole (27 g) of Al = 3 Faraday (3 F).

We need to produce 40.0 g of Al.
First, calculate the moles of Al in 40.0 g:
Moles of Al = Mass / Molar Mass = 40.0 g / 27 g mol⁻¹ ≈ 1.481 mol.

Now, calculate the required electricity:
Since 1 mol of Al requires 3 F of electricity,
1.481 mol of Al will require = 1.481 mol × 3 F/mol ≈ 4.44 F.

Chapter 3: Electrochemistry