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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

Limiting molar conductivity refers to the maximum molar conductivity that a solution of an electrolyte can achieve as the concentration approaches infinity (in the limit of infinite dilution). In other words, it's the molar conductivity of an electrolyte solution when it is so dilute that further... read more

Limiting molar conductivity refers to the maximum molar conductivity that a solution of an electrolyte can achieve as the concentration approaches infinity (in the limit of infinite dilution). In other words, it's the molar conductivity of an electrolyte solution when it is so dilute that further dilution doesn't significantly affect its conductivity.

At infinite dilution, the ions in the solution are effectively isolated from each other, minimizing the influence of ion-ion interactions. Thus, the limiting molar conductivity provides a measure of the inherent conductivity of the ions themselves, independent of the specific conditions of the solution.

Mathematically, the limiting molar conductivity (Λ0Λ0) of an electrolyte can be determined using the Kohlrausch's Law of independent migration of ions:

Λ0=lim⁡c→0κcΛ0=limc→0cκ

where:

• Λ0Λ0 is the limiting molar conductivity,
• κκ is the conductivity of the solution, and
• cc is the concentration of the electrolyte (which approaches zero as it tends towards infinite dilution).

Limiting molar conductivity values are often used to compare the conductivities of different electrolytes and to understand their behavior at low concentrations.

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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

The relation between conductivity (κκ) and molar conductivity (ΛmΛm) of a solution held in a cell is given by the equation: κ=Λm×Cκ=Λm×C where: κκ is the conductivity of the solution (measured in siemens per meter, S/m), ΛmΛm... read more

The relation between conductivity (κκ) and molar conductivity (ΛmΛm) of a solution held in a cell is given by the equation:

κ=Λm×Cκ=Λm×C

where:

• κκ is the conductivity of the solution (measured in siemens per meter, S/m),
• ΛmΛm is the molar conductivity of the solution (measured in siemens per meter squared per mole, S·m²/mol),
• CC is the concentration of the solution (measured in moles per cubic meter, mol/m³).

This equation demonstrates that conductivity (κκ) is directly proportional to the molar conductivity (ΛmΛm) and the concentration of the solution (CC). Therefore, an increase in molar conductivity or concentration will result in an increase in conductivity.

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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

The cell constant, resistance of the solution in the cell, and conductivity of the solution are related through the equation: Conductivity(κ)=1Cell Constant(Λ)×1Resistance of solution in cell(R)Conductivity(κ)=Cell Constant(Λ)1×Resistance of solution in cell(R)1 Where: Conductivity... read more

The cell constant, resistance of the solution in the cell, and conductivity of the solution are related through the equation:

Conductivity(κ)=1Cell Constant(Λ)×1Resistance of solution in cell(R)Conductivity(κ)=Cell Constant(Λ)1×Resistance of solution in cell(R)1

Where:

• Conductivity (κκ) is measured in siemens per meter (S/m).
• Cell Constant (ΛΛ) is measured in reciprocal meters (m−1m−1).
• Resistance of the solution in the cell (R) is measured in ohms (ΩΩ).

This equation indicates that conductivity is inversely proportional to the product of the cell constant and the resistance of the solution in the cell. Essentially, as conductivity increases, resistance decreases, and vice versa. The cell constant is a property of the conductivity cell used in the measurement, and it relates the geometry of the cell to the conductivity of the solution being measured.

The conductivity of a solution (κκ) is related to its molar conductivity (ΛmΛm) through the equation:

κ=Λm×Cκ=Λm×C

Where:

• Molar conductivity (ΛmΛm) is measured in siemens per meter per mole (S⋅m2⋅mol−1S⋅m2⋅mol−1).
• CC is the concentration of the solution in moles per liter (mol/L).

This equation shows that the conductivity of a solution is directly proportional to its molar conductivity and concentration. Molar conductivity is a measure of the conductivity of a solution containing one mole of solute per liter, and it accounts for the contribution of each ion to the solution's conductivity.

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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

The reducing power of a metal is related to its ability to lose electrons and undergo oxidation. Metals with lower standard electrode potentials (E°) have a greater tendency to lose electrons and therefore are stronger reducing agents. Here are the metals arranged in increasing order of their... read more

The reducing power of a metal is related to its ability to lose electrons and undergo oxidation. Metals with lower standard electrode potentials (E°) have a greater tendency to lose electrons and therefore are stronger reducing agents.

Here are the metals arranged in increasing order of their reducing power based on their standard electrode potentials:

1. K (Potassium) with E° = -2.93 V
2. Mg (Magnesium) with E° = -2.37 V
3. Cr (Chromium) with E° = -0.74 V
4. Fe (Iron) with E° = -0.44 V
5. Cu (Copper) with E° = 0.34 V
6. Ag (Silver) with E° = 0.80 V

So, the order of reducing power is: K < Mg < Cr < Fe < Cu < Ag.

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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

The corrosion of iron in the atmosphere primarily involves the following electrochemical reactions: Formation of Iron Oxide (Rusting): When iron is exposed to atmospheric oxygen and moisture, it undergoes oxidation to form iron oxide, commonly known as rust. The overall reaction is: 4Fe+3O2+6H2O→4Fe(OH)34Fe+3O2+6H2O→4Fe(OH)3 This... read more

The corrosion of iron in the atmosphere primarily involves the following electrochemical reactions:

1. Formation of Iron Oxide (Rusting): When iron is exposed to atmospheric oxygen and moisture, it undergoes oxidation to form iron oxide, commonly known as rust. The overall reaction is:

4Fe+3O2+6H2O→4Fe(OH)34Fe+3O2+6H2O→4Fe(OH)3

This reaction occurs in the presence of water and oxygen. Iron atoms lose electrons to oxygen molecules, forming iron ions (Fe²⁺) and hydroxide ions (OH⁻). These ions then react with water molecules to form rust (Fe(OH)₃).

2. Electron Transfer (Oxidation Reaction): The oxidation of iron can be represented as:

Fe→Fe2++2e−Fe→Fe2++2e

Iron atoms lose electrons to form ferrous ions (Fe²⁺). This process is known as oxidation. These electrons released during oxidation are involved in the reduction half-reaction.

3. Reduction Reaction (Oxygen Reduction): Atmospheric oxygen reacts with water and electrons to form hydroxide ions. This reaction is represented as:

O2+2H2O+4e−→4OH−O2+2H2O+4e→4OH

Oxygen molecules gain electrons and react with water molecules to form hydroxide ions (OH⁻). This process is called reduction.

4. Overall Reaction: Combining the oxidation and reduction half-reactions, we get the overall reaction for the corrosion of iron in the atmosphere:

4Fe+3O2+6H2O→4Fe(OH)34Fe+3O2+6H2O→4Fe(OH)3

This reaction shows the formation of rust (iron hydroxide) from iron, oxygen, and water.

In summary, the corrosion of iron in the atmosphere is an electrochemical process involving the oxidation of iron and reduction of oxygen. Moisture and oxygen from the atmosphere facilitate the formation of rust, leading to the degradation of iron structures over time.

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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

The cell constant (denoted as KK) of an electrolytic cell is defined as the ratio of the distance between the electrodes (ll) to the cross-sectional area of the electrodes (AA). It is mathematically represented as: K=lAK=Al The resistance of the solution in the cell (RR) can be determined using Ohm's... read more

The cell constant (denoted as KK) of an electrolytic cell is defined as the ratio of the distance between the electrodes (ll) to the cross-sectional area of the electrodes (AA). It is mathematically represented as:

K=lAK=Al

The resistance of the solution in the cell (RR) can be determined using Ohm's law, which states that resistance (RR) is equal to the voltage (VV) applied across the cell divided by the current (II) passing through it:

R=VIR=IV

The conductivity of the solution (κκ) is the reciprocal of resistance (RR), represented as:

κ=1Rκ=R1

Thus, the relation among cell constant, resistance of the solution in the cell, and conductivity of the solution can be expressed as:

κ=1K×VIκ=K1×IV

Or more simply:

κ=VK×Iκ=K×IV

Now, the molar conductivity of a solution (ΛmΛm) is related to its conductivity (κκ) and the concentration of the solution (CC) by the equation:

Λm=κCΛm=Cκ

This equation shows that molar conductivity is directly proportional to the conductivity of the solution. As the conductivity of the solution increases, its molar conductivity also increases, provided the concentration remains constant.

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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

To calculate the conductivity (κ) of the solution, we can use the formula: κ=Λm×Cκ=Λm×C where: κκ is the conductivity of the solution, ΛmΛm is the molar conductivity, and CC is the concentration of the solution in mol/L. Given: Λm=138.9 S... read more

To calculate the conductivity (κ) of the solution, we can use the formula:

κ=Λm×Cκ=Λm×C

where:

• κκ is the conductivity of the solution,
• ΛmΛm is the molar conductivity, and
• CC is the concentration of the solution in mol/L.

Given:

• Λm=138.9 S cm2mol−1Λm=138.9S cm2mol−1
• C=1.5 MC=1.5M

Substitute the given values into the formula:

κ=138.9 S cm2mol−1×1.5 mol/Lκ=138.9S cm2mol−1×1.5mol/L

κ=208.35 S/cmκ=208.35S/cm

So, the conductivity of the solution is 208.35 S/cm208.35S/cm.

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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

To calculate the electrode potential of the zinc rod dipped in a 0.1 M solution of ZnSO₄, we can use the Nernst equation: E=E0−0.0592nlog⁡QE=E0−n0.0592logQ Where: EE = Electrode potential E0E0 = Standard electrode potential nn = Number of moles of electrons transferred QQ = Reaction quotient The... read more

To calculate the electrode potential of the zinc rod dipped in a 0.1 M solution of ZnSO₄, we can use the Nernst equation:

E=E0−0.0592nlog⁡QE=E0n0.0592logQ

Where:

• EE = Electrode potential
• E0E0 = Standard electrode potential
• nn = Number of moles of electrons transferred
• QQ = Reaction quotient

The half-reaction for the zinc electrode can be written as:

Zn2++2e−→ZnZn2++2eZn

Given that the solution is 95% dissociated, the concentration of Zn2+Zn2+ ions will be 0.1 M * 0.95 = 0.095 M.

The reaction quotient, QQ, is given by:

Q=[Zn][Zn2+]=1[Zn2+]Q=[Zn2+][Zn]=[Zn2+]1

Since the stoichiometric coefficient of Zn2+Zn2+ in the half-reaction is 1, and the stoichiometric coefficient of ZnZn is also 1, the concentration of Zn2+Zn2+ ions is equal to the concentration of ZnZn atoms.

Q=10.095=10.526Q=0.0951=10.526

The standard electrode potential (E0E0) for the zinc electrode is given as -0.76 V.

The number of moles of electrons transferred (nn) is 2 (as per the balanced half-reaction).

Now, plug in these values into the Nernst equation:

E=−0.76−0.05922log⁡(10.526)E=−0.76−20.0592log(10.526)

Now, calculate the electrode potential EE:

E=−0.76−0.05922×1.022E=−0.76−20.0592×1.022

E=−0.76−0.0302E=−0.76−0.0302

E=−0.7902E=−0.7902

So, the electrode potential is approximately -0.7902 V.

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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

In a lead-acid storage battery, such as the lead-acid car battery, the reactions at the cathode and anode during discharge (when the battery is in use) and charging are as follows: During Discharge: At the cathode (positive electrode): PbO2(s) + 4H+(aq) + SO4^2-(aq) + 2e^- → PbSO4(s) + 2H2O(l) At... read more

In a lead-acid storage battery, such as the lead-acid car battery, the reactions at the cathode and anode during discharge (when the battery is in use) and charging are as follows:

During Discharge:

At the cathode (positive electrode): PbO2(s) + 4H+(aq) + SO4^2-(aq) + 2e^- → PbSO4(s) + 2H2O(l)

At the anode (negative electrode): Pb(s) + SO4^2-(aq) → PbSO4(s) + 2e^-

In simpler terms, during discharge, lead dioxide (PbO2) at the cathode combines with sulfuric acid (H2SO4) and protons (H+) from the acid to form lead sulfate (PbSO4) and water. At the anode, lead (Pb) reacts with sulfate ions (SO4^2-) to form lead sulfate and releases electrons.

During Charging:

At the cathode (positive electrode): PbSO4(s) + 2H2O(l) → PbO2(s) + 4H+(aq) + SO4^2-(aq) + 2e^-

At the anode (negative electrode): PbSO4(s) + 2e^- → Pb(s) + SO4^2-(aq)

During charging, the reactions are essentially reversed. Lead sulfate on the cathode is converted back to lead dioxide, sulfuric acid, and water, while lead sulfate on the anode is converted back to lead and sulfate ions.

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Answered on 07 Apr Learn Unit III: Electrochemistry

Nazia Khanum

To calculate the molar conductivity (ΛmΛm) of a solution, you can use the formula: Λm=κcΛm=cκ Where: κκ is the conductivity of the solution in Siemens per centimeter (S cm−1Scm−1). cc is the concentration of the solution in moles... read more

To calculate the molar conductivity (ΛmΛm) of a solution, you can use the formula:

Λm=κcΛm=cκ

Where:

• κκ is the conductivity of the solution in Siemens per centimeter (S cm−1Scm−1).
• cc is the concentration of the solution in moles per liter (MM).

Given:

• Conductivity (κκ) = 0.025 S cm^-1
• Concentration (cc) = 0.20 M

Substituting the given values into the formula:

Λm=0.025 S cm−10.20 M=0.0250.20 S cm2 mol−1Λm=0.20M0.025Scm−1=0.200.025Scm2mol−1
Λm=0.125 S cm2 mol−1Λm=0.125Scm2mol−1

Therefore, the molar conductivity of the 0.20 M solution of KCl at 298 K is 0.125 S cm2 mol−10.125Scm2mol−1.

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