Electrochemical Cells and Its Applications MCQ Quiz in मल्याळम - Objective Question with Answer for Electrochemical Cells and Its Applications - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

Electrode Charges in a Galvanic Cell

• A galvanic cell (or voltaic cell) is an electrochemical cell that generates electrical energy from a spontaneous redox reaction.
• Oxidation occurs at the anode and reduction occurs at the cathode.
• At the anode, electrons are released by the species being oxidized and flow through the external circuit to the cathode.
• This accumulation of electrons makes the anode negatively charged.
• Conversely, the cathode gains electrons and becomes positively charged.

EXPLANATION:

• In a galvanic cell:
  • Anode: Site of oxidation, negative charge.
  • Cathode: Site of reduction, positive charge.

Therefore, the charge over the anode in a galvanic cell is: Negative

Hence, no reaction will occur.

Note: The standard reduction potential for \(Cu^{2+}| Cu\) is greater than that for \(Zn^{2+}| Zn\).

Hence, Zn is better reducing agent than Cu. Hence, Zn can displace Cu from copper sulphate solution.

\(R = \dfrac{1}{k} \dfrac{l}{A}\)

Initially, only half of the electrodes were submerged but after dilution, they are completely submerged. Hence, the value of the area of electrode A in the above equation is doubled.

Due to the dilution, the volume of the solution is doubled and the molar concentration is reduced to half. Due to this, the value of specific conductance k is reduced to half.

Thus k is halved while the A is doubled. Hence R remains the same, i.e., 50 \(\Omega\).

CONCEPT:

Molar Conductivity (Λ_m)

• Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte.
• It is given by the formula:

  Λ_m = Λ_m⁰ * α

  Where:

  • Λ_m is the molar conductivity.
  • Λ_m⁰ is the molar conductivity at infinite dilution.
  • α is the degree of dissociation.

   

EXPLANATION:

• Given data:
  • Degree of dissociation (α) = 11.0% = 0.11
  • Concentration (C) = 0.02 M
  • Λ_m⁰(H⁺) = 349.68 cm² mol^-1
  • Λ_m⁰(HCOO^-) = 54.68 cm² mol^-1
• Calculate the molar conductivity at infinite dilution:
  • Λ_m⁰ = Λ_m⁰(H⁺) + Λ_m⁰(HCOO^-)
  • = 349.68 cm² mol^-1 + 54.68 cm² mol^-1
  • = 404.36 cm² mol^-1
• Now, calculate the molar conductivity:
  • Λ_m = Λ_m⁰ * α
  • = 404.36 cm² mol^-1 * 0.11
  • = 44.4796 cm² mol^-1

Therefore, the molar conductivity of the 0.02 M formic acid solution is approximately 44.46 S cm² mol^-1.

CONCEPT:

Degree of Dissociation (α)

• The degree of dissociation (α) of a weak electrolyte in solution is the fraction of the total number of moles of the electrolyte that dissociates into ions.
• It can be calculated using the specific conductance (k) and the molar conductance at infinite dilution (Λ_m⁰).
• The formula used is:

  α = Λ_m / Λ_m⁰

EXPLANATION:

• Given:
  • Specific conductance (k) = 1.65 × 10^-4 S cm^-1
  • Concentration (c) = 0.02 M
  • Λ⁰_H+ = 349.1 S cm² mol^-1
  • Λ⁰_CH3COO^- = 40.9 S cm² mol^-1
• First, calculate the molar conductance (Λ_m):
  • Λ_m = (k × 1000) / c
  • Λ_m = (1.65 × 10^-4 S cm^-1 × 1000) / 0.02 M
  • Λ_m = 8.25 S cm² mol^-1
• Next, calculate the molar conductance at infinite dilution (Λ_m⁰):
  • Λ_m⁰ = Λ⁰_H+ + Λ⁰_CH3COO^-
  • Λ_m⁰ = 349.1 S cm² mol^-1 + 40.9 S cm² mol^-1
  • Λ_m⁰ = 390 S cm² mol^-1
• Finally, calculate the degree of dissociation (α):
  • α = Λ_m / Λ_m⁰
  • α = 8.25 / 390
  • α ≈ 0.021

Therefore, the degree of dissociation of acetic acid is approximately 0.021.

The given reaction involves the reduction of oxygen gas to water under acidic conditions. This type of reaction typically occurs in galvanic cells involving oxygen gas at the cathode, hydrogen ions (H⁺) in the solution, and a platinum electrode as the inert conductor. Platinum is commonly used as the electrode for redox reactions involving gases, as it does not interfere with the chemical process.

• Galvanic Cell Components: In galvanic cells, oxidation takes place at the anode and reduction occurs at the cathode. The oxygen reduction reaction (ORR) occurs at the cathode, while the anode typically involves a different redox process, depending on the setup of the cell.

• Platinum Electrode: In this reaction, platinum is used as the inert electrode, and H⁺ ions in the acidic medium are reduced to form water, making the oxygen gas reduce at the cathode.

Explanation: 

• In the reaction \(\frac{1}{2} O_2(g) + 2H^+(aq) + 2e^- \rightarrow H_2O(l)\), oxygen is being reduced at the cathode in an acidic medium. Therefore, the correct setup involves an oxygen gas supply, a platinum electrode (since it is inert and used for gas-phase reactions), and an acidic electrolyte containing H⁺ ions.

• Among the options, the correct one is \(Pt[O_2(g)]|H_2SO_4(sol^n)|H_2O(l)\) as it has the required oxygen gas, platinum electrode, and acidic medium (sulfuric acid), where oxygen is reduced to water.

Conclusion:

The correct answer is: (1) Pt[O₂(g)]|H₂SO₄(solⁿ)|H₂O(l), as it involves the reduction of oxygen in an acidic medium with platinum as the electrode.

CONCEPT:

Molar Conductivity at Infinite Dilution

• Molar conductivity is a measure of the ability of an electrolyte solution to conduct electricity, based on the number of moles of solute.
• At infinite dilution, the ions are far apart, and there is minimal ion-ion interaction, meaning molar conductivity approaches a maximum, constant value.
• In an infinitely dilute solution, adding more water does not affect the molar conductivity because the solution is already infinitely diluted.

EXPLANATION:

• In the given problem, the volume of the solution is doubled by adding more water to an infinitely dilute solution.
• Since the solution is already infinitely dilute, further addition of water will not change the molar conductivity. It may remain the same or be difficult to measure accurately due to the extreme dilution.
• Therefore, molar conductivity is unaffected in this scenario.

Conclusion:-

The correct answer is Option 2: "remain same or cannot be measured accurately."

The Correct Answer is Electrode Potential.

Key Points 

• The potential difference that develops between the electrode and electrolyte is known as the Electrode potential.
• This potential arises due to the tendency of metal ions to move into the solution or for ions in the solution to deposit on the metal electrode.
• It is a measure of the ability of an electrode to donate or accept electrons from the electrolyte.
• Electrode potential is crucial in determining the direction of electron flow in an electrochemical cell.

Additional Information 

The correct answer is Liquid junction potential

Concept:-

• Liquid Junction Potential: A liquid junction potential is a potential difference that arises at the boundary between two solutions of different concentrations.
• Mobility of Ions: The mobility of an ion is its ability to move through a solution.

Explanation:-

When two different concentrations of hydrochloric acids are used as electrolytes in a system, it contributes as an additional source of potential difference at the interface. This is an example of liquid junction potential.

• A liquid junction potential is a potential difference that arises at the boundary between two solutions of different concentrations.
• The potential difference is caused by the difference in the mobilities of the ions in the two solutions.
• In the case of two different concentrations of hydrochloric acid, the hydrogen ions (H+) will move more quickly from the more concentrated solution to the less concentrated solution.
• This movement of ions creates a potential difference at the boundary between the two solutions.

Conclusion:-

When two different concentrations of hydrochloric acids are used as electrolytes in a system, it contributes as an additional source of potential difference at the interface. This is an example of liquid junction potential.

Correct answer: 2)

Concept:

• Normal or standard hydrogen electrode (SHE) is a primary standard reference electrode. The SHE is the universal electrode.
• Standard Hydrogen Electrode is used as a reference electrode when calculating the standard electrode potential of a half cell.
• Reference electrode: The electrode which acts as both anode and cathode is known as reference electrode.

Explanation:

For hydrogen electrode, 

\(2H_{aq}^{+} + 2e^{-}\rightarrow H_{2}(g)(1 atm)\)

\(E_{Cell}^{o}= 0\)

\(E_{Cell}= E_{Cell}^{o}-\frac{0.059}{2}log \frac{pH_{2}}{[H^{+}]^{2}}\)

As pH=3

[H⁺]= 10^-3 M

\(E_{Cell}= -\frac{0.059}{2}log \frac{1}{[10^{-3}]^{2}}\)

\(E_{Cell}= -\frac{0.059}{2}log (10)^{6}\)

\(E_{Cell}= \)-0.177 V

Conclusion:

Thus, the potential of the electrode at this PH is -0.177 V.