Electrochemistry

The cell reactions during charging of lead storage battery are
At anode

At cathode

So, at the anode is reduced to .

During charging, PbO2 is formed at the anode from PbSO4 (oxidation of Pb ion) and Pb is formed at the cathode from PbSO4 (reduction of Pb ion).

Therefore Options B and D are the correct answers.

Discover More From Chapter: Electrochemistry

1. Identify the half-cell reactions and the number of electrons transferred
The given cell reaction is:

H_2(g) + 2AgCl_(s) ⇌ 2Ag_(s) + 2HCl_(aq)

We can break this into two half-cell reactions:

Oxidation (Anode): H_2(g) → 2H⁺_(aq) + 2e^-
Reduction (Cathode): 2AgCl_(s) + 2e^- → 2Ag_(s) + 2Cl^-_(aq)

From these half-cell reactions, it's clear that the number of electrons transferred (n) is 2.

2. Recall the relationship between and the equilibrium constant (K)
The relationship between the standard cell potential ( ) and the equilibrium constant (K) is given by:



where:
- is the standard cell potential
- is the ideal gas constant (8.314 J/(mol·K))
- is the temperature in Kelvin
- is the number of moles of electrons transferred in the cell reaction
- is Faraday's constant (96485 C/mol)
- is the equilibrium constant

At 298 K, this equation can be simplified to:

3. Substitute the given values and solve for K
We are given:

- V
-
-

Substitute these values into the simplified equation:



Solve for :



Solve for K:

Final Answer:
(A) 2.8 x 10⁷

1. Recall the relationship between resistance, conductivity, and cell constant
The relationship between resistance (R), conductivity (κ), and cell constant (G^*) is given by:



where:
- is the resistance in ohms (Ω)
- (kappa) is the conductivity in Siemens per centimeter (S cm^-1)
- is the cell constant in cm^-1

2. Rearrange the formula to solve for the cell constant
We need to find the cell constant, so we rearrange the formula:

3. Substitute the given values
We are given:

- Ω
- S cm^-1

Substitute these values into the formula:

cm^-1

4. Calculate the cell constant
cm^-1
cm^-1

Final Answer:
(A) 0.219

We are given the following electrodes:
P = Zn²⁺ (0.0001 M) / Zn
Q = Zn²⁺ (0.1 M) / Zn
R = Zn²⁺ (0.01 M) / Zn
S = Zn²⁺ (0.001 M) / Zn

The standard electrode potential for Zn²⁺/Zn is .

Using the Nernst equation:


We calculate the electrode potentials for each electrode:

For P (Zn²⁺ = 0.0001 M):


For Q (Zn²⁺ = 0.1 M):


For R (Zn²⁺ = 0.01 M):


For S (Zn²⁺ = 0.001 M):


Therefore, the increasing order of electrode potentials is:

The correct answer is (C) : Q > R > S > P.

Correct Answer: Purine or pyrimidine base, a pentose sugar and ortho phosphoric acid

Elaborate Explanation:
Nucleic acids like DNA and RNA are long chains made of repeating units called nucleotides. When these nucleic acids are hydrolyzed (broken down using water), the final products are:
• 2 purine bases – Adenine (A) and Guanine (G)
• 2 pyrimidine bases – Cytosine (C) and Thymine (T) in DNA or Uracil (U) in RNA
• Pentose sugar – Deoxyribose (in DNA) or Ribose (in RNA)
• Ortho phosphoric acid – Also known as phosphoric acid (H₃PO₄)

During partial or intermediate hydrolysis, we get nucleotides, which are the building blocks of nucleic acids. Each nucleotide is made of:
• A nitrogenous base (purine or pyrimidine)
• A pentose sugar
• An ortho phosphoric acid group

Therefore, the intermediate product – a **nucleotide** – contains all three parts:
Purine or pyrimidine base + Pentose sugar + Ortho phosphoric acid

This is what makes a nucleotide different from a nucleoside, which has only the base and sugar (without phosphate).

Molar Conductance (Λ_m) is given by:
Λ_m = κ × 1000 / C

where,
κ = specific conductance = ohm^-1 cm^-1
C = concentration = 0.1 mol/L

Substituting the values:
Λ_m = ohm^-1 cm² mol^-1

Correct Answer: 630 ohm^-1 cm² mol^-1

Fuel cells are electrochemical devices that convert the chemical energy of a fuel, such as hydrogen, into electrical energy through an oxidation-reduction (redox) reaction.
Catalysts are used to facilitate the redox reactions occurring at the electrodes of the fuel cell. Platinum and palladium are commonly used as catalysts in fuel cells due to their high catalytic activity, stability, and resistance to corrosion.
Therefore, the correct answer is (B) Platinum - Palladium.

Calculating pH from Conductivity

Step 1: Calculate the Conductivity ( )

The conductivity ( ) can be calculated using Ohm's law: where is the resistance and is the cell constant.

Substituting the given values:

Step 2: Calculate the Molar Conductivity ( )

The molar conductivity ( ) is related to the conductivity ( ) and the concentration ( ): Given the concentration :

Step 3: Calculate the Degree of Ionization ( )

For a weak acid, the molar conductivity is related to the degree of ionization ( ): Solving for :

Step 4: Calculate the pH

The concentration of H⁺ ions ( ) can be calculated using the degree of ionization: The pH is then:

Conclusion

The closest option to our calculated pH value is (D) 3.