Electrochemistry

Step 1: Faraday’s law. One equivalent of Ag (108 g) requires one Faraday.
Step 2: Charge required. Q = 1F = 96500 C

Step 1: Reduction potential order. Higher reduction potential oxidises lower ones.
Step 2: Reaction feasibility. Y oxidises X but cannot oxidise Z.

Step 1: Principle of iodometric titration. Oxidizing agents liberate iodine from iodide.
Step 2: Fe³+ behaviour. Fe³⁺ does not liberate iodine quantitatively.

Step 1: Calculate equivalent conductance.
The equivalent conductance of a salt at infinite dilution is the sum of the conductances of its ions: Thus, the correct answer is (3).

Step 1: Use the relation for electrolysis.
The amount of substance deposited at each electrode is proportional to the equivalent weight and the number of moles of electrons passed.
Step 2: Conclusion.
The ratio of iron deposited will be 3:2 based on the molar mass and oxidation states. Final Answer:

Step 1: Use the formula for electrolysis.
The number of moles of metal deposited is . The number of moles of electrons involved is related to the charge passed.
Step 2: Calculate the oxidation state.
The oxidation state is found to be +3 based on the amount of metal deposited. Final Answer:

Step 1: Useful electrical energy obtained from fuel cell:
Step 2: Total energy released:
Step 3: Efficiency:

Step 1: Faraday’s first law of electrolysis: One Faraday deposits one gram equivalent weight.

Step 1: Cell (A): Pt electrodes in CuSO₄ Anode reaction produces H⁺ ions ⟹ pH decreases.
Step 2: Cell (B): Cu electrodes in CuSO₄ Copper dissolves at anode and deposits at cathode ⟹ no net change in H⁺ ⟹ pH constant.
Step 3: Cell (C): Pt electrodes in KCl Cathode produces OH⁻ ions ⟹ pH increases.

Step 1: Relation between equilibrium constant and standard emf:

Step 2:

Step 3:

Thus statements I and IV are correct.

Step 1: Charge passed: Q=It=1×(16×60+5)=965C
Step 2: Equivalent weight of Cu²+: (63.5)/(2)=31.75
Step 3: Normality: N=(Q)/(F× V)=(965)/(96500×1)=0.01N
Step 4: Since CuCl₂ has valency factor 2: Strength=0.02N

Step 1: Charge passed: Q = It = 1.25×10⁻3×(15×60)=1.125C
Step 2: Moles of electrons: n = (Q)/(F)=(1.125)/(96500)=1.165×10⁻5
Step 3: For Ag⁺ + e⁻ → Ag, moles of Ag⁺ = moles of electrons.
Step 4: [Ag⁺] = frac1.165×10⁻50.1 =1.165×10⁻4≈2.32×10⁻4

By Faraday’s first law of electrolysis, one Faraday of charge deposits one gram equivalent of a substance.

Cell (A): Pt electrodes in CuSO₄. Cathode: Cu²++2e⁻ → Cu Anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ ⟹ H⁺ increases, pH decreases. Cell (B): Cu electrodes in CuSO₄. Cathode: Cu²++2e⁻ → Cu Anode: Cu → Cu²++2e⁻ ⟹ No change in H⁺, pH constant. Cell (C): Pt electrodes in KCl. Cathode: 2H₂O+2e⁻ → H₂+2OH⁻ Anode: 2Cl⁻ → Cl₂+2e⁻ ⟹ OH⁻ increases, pH increases.

Step 1: An oxidising agent causes oxidation of other species by accepting electrons. Step 2: Greater tendency to accept electrons means greater tendency to get reduced. Step 3: This corresponds to a higher (more positive) standard reduction potential. ⟹ Stronger oxidising agent Longleftrightarrow Higher Eᶜircᵣₑdᵤctᵢₒₙ

Step 1: The relation between standard electrode potential and equilibrium constant is: Δ G^∘ = -nF E^∘ and Δ G^∘ = -RT ln K Step 2: Equating: RT ln K = nF E^∘ ⟹ ln K = (nF E^∘)/(RT) Step 3: Converting to base-10 logarithm: log K = (nF E^∘)/(2.303RT) Step 4: Since log₁0 e = 0.4342, log K = (0.4342 nF E^∘)/(RT) Hence, statements I, II and IV are correct.

To determine the electromotive force (EMF) of the cell represented by the notation:

we use the standard electrode potentials of the two half-cells. The overall cell EMF is given by the formula:

Here:

Substituting the given values:

Therefore, the EMF of the cell is .

The given half reactions and their reduction potentials are: In the cell reaction: Oxidation potential of Zn is the negative of its reduction potential: Standard cell potential is: Hence, the standard cell potential is .

Step 1: Understanding the Concept:
A reducing agent is a species that loses electrons (undergoes oxidation).
The tendency of a species to be reduced is measured by its standard reduction potential ( ).
The species with the lowest (most negative) value has the greatest tendency to undergo oxidation and therefore acts as the strongest reducing agent.

Step 2: Key Formula or Approach:
1. List all given standard reduction potentials.
2. Identify the lowest value.
3. The reduced form (the species on the right side of the reduction half-reaction) corresponding to the lowest is the strongest reducing agent.

Step 3: Detailed Explanation:
Given potentials:
1.
2.
3.
4.
The lowest value is for the couple .
This means metal has the highest tendency to oxidize to among the given species.
Hence, is the strongest reducing agent.

Step 4: Final Answer:
Based on the minimum reduction potential, is the strongest reducing agent.