Electrochemistry

CN⁻ is a strong field ligand → large splitting → high excitation energy.

The high atomization enthalpy ( ) and low hydration enthalpy ( ) of copper make its standard reduction potential ( ) positive.
Explanation of Value - The electrode potential ( ) depends on: - Atomization enthalpy ( ): The energy required to convert solid Cu to Cu is high. - Hydration enthalpy ( ): Cu has low hydration energy, making it less stable in aqueous solution.

Effect on Value - Due to low hydration enthalpy, the reduction of Cu to Cu is not highly favored. - Hence, Cu Cu has a positive value of V, indicating that Cu is less reactive than expected.

Step 1: Electronic Configuration and Bonding - Zn, Cd, and Hg belong to Group 12 and have a completely filled d configuration. - The lack of unpaired d-electrons reduces the extent of metallic bonding, making these metals soft with low melting and boiling points.

Step 2: Weak Interatomic Forces - In transition metals, strong metallic bonding arises due to overlapping of d-orbitals. - However, in Zn, Cd, and Hg, fully filled d-orbitals do not contribute to bonding, resulting in weak interatomic forces and low melting/boiling points.

First, calculate the standard cell potential ( ): Next, calculate using the equation: Since the number of electrons transferred : Now, calculate using the relationship: At 25°C, R = 8.314 , , T = 298 , : Solving for : Therefore: Thus, and .

(a) The reaction involves the formation of hydroxylamine (NH₂OH) from an aldehyde (O) and hydroxylamine (HO-NH₂) in the presence of an acid catalyst (H⁺):

(b) The reaction involves ozonolysis of ethene (CH₂=CH₂) followed by reductive workup to form formaldehyde (HCHO). Heating ( ) does not change the product further:

(c) The reaction involves the conversion of an alcohol (OH) to an alkyl chloride (Cl) using thionyl chloride (SOCl₂) under heating ( ):

(d) The reaction involves the conversion of an aldehyde (CHO) to a carboxylic acid (COOH) using sodium cyanide (NaCN) and hydrochloric acid (HCl):

(e) The reaction involves the methylation of chlorobenzene (Cl₂C₆H₅) to form a methylated chlorobenzene derivative (Cl₂C₆H₄CH₃):

To solve the problem, we need to explain why chlorine gas (Cl₂) is liberated at the anode during the electrolysis of aqueous NaCl, even though oxygen gas (O₂) has a more positive standard electrode potential ( ).

1. Electrolysis of Aqueous NaCl:
In the electrolysis of aqueous NaCl, two reactions take place at the electrodes: - At the anode (positive electrode), oxidation occurs (loss of electrons). - At the cathode (negative electrode), reduction occurs (gain of electrons). The ions present in aqueous NaCl are: - Na⁺ (sodium ions) - Cl⁻ (chloride ions) - H₂O (water molecules, which dissociate into H⁺ and OH⁻ ions)

2. Standard Electrode Potentials:
We are given that oxygen gas (O₂) has a more positive standard electrode potential compared to chlorine gas (Cl₂). The standard electrode potentials are as follows: - (for the oxidation of water to oxygen) - (for the oxidation of chloride ions to chlorine gas) From these values, we see that chlorine gas (Cl₂) has a higher (more positive) electrode potential than oxygen gas (O₂), which suggests that Cl₂ should be favored at the anode.

3. Effect of Concentration and Overpotentials:
The reason chlorine gas (Cl₂) is actually liberated at the anode in the electrolysis of aqueous NaCl, rather than oxygen gas (O₂), can be explained by the following factors:
- Concentration Effect: In aqueous NaCl, the concentration of chloride ions (Cl⁻) is much higher than the concentration of water molecules (H₂O). This higher concentration of Cl⁻ ions means that the chloride ion oxidation reaction is more likely to occur at the anode, as there are more Cl⁻ ions available for oxidation.
- Overpotential: Overpotential refers to the extra voltage required to drive a particular oxidation or reduction reaction at an electrode, beyond the theoretical electrode potential. The overpotential for the evolution of oxygen gas (O₂) is relatively high, which makes the oxidation of chloride ions to chlorine gas (Cl₂) easier and more favorable under normal electrolysis conditions.

4. Final Answer:
Although oxygen gas (O₂) has a more positive standard electrode potential, chlorine gas (Cl₂) is liberated at the anode in the electrolysis of aqueous NaCl because: - Chloride ions (Cl⁻) are present in higher concentration in the solution. - The overpotential for oxygen gas (O₂) is high, making the oxidation of chloride ions to chlorine gas (Cl₂) more favorable. Thus, chlorine gas is preferentially produced at the anode.

To solve the problem, we need to calculate the cell potential of the electrochemical cell formed by connecting the Sn /Sn and Cr /Cr couples in their standard states.

1. Understanding the Standard Electrode Potentials:
The standard electrode potential (E°) for the two couples is given as:

• For Sn /Sn : E° = +0.15 V
• For Cr /Cr: E° = -0.74 V

The cell potential (E°cell) is calculated by taking the difference between the electrode potentials of the two half-reactions. The couple with the more positive potential will undergo reduction, while the other will undergo oxidation.

2. Identifying the Anode and Cathode:
In this case: - Sn will be reduced to Sn because it has the more positive electrode potential (+0.15 V), so it will act as the cathode. - Cr will be oxidized to Cr because it has the more negative electrode potential (-0.74 V), so it will act as the anode.

3. Calculating the Cell Potential:
The cell potential (E°cell) is given by the equation: E°cell = E°(cathode) - E°(anode)
Substituting the values: E°cell = (+0.15 V) - (-0.74 V)
E°cell = 0.15 V + 0.74 V = 0.89 V

Final Answer:
The cell potential will be 0.89 V.

To solve the problem, we need to determine the charge required for the reduction of 1 mole of MnO₄⁻ to MnO₂.

1. Understanding the Reduction Process:
The reduction of MnO₄⁻ to MnO₂ involves a change in the oxidation state of manganese (Mn). We start by identifying the oxidation states:

• In MnO₄⁻, the oxidation state of Mn is +7 (since each oxygen is -2, and the total charge is -1).
• In MnO₂, the oxidation state of Mn is +4 (since each oxygen is -2, and the compound is neutral).

2. Calculating the Change in Oxidation State:
The change in the oxidation state of Mn is:


This means each Mn atom gains 3 electrons during the reduction process.

3. Determining the Total Charge:
For 1 mole of MnO₄⁻, the number of electrons transferred is 3 moles. Since 1 Faraday (F) corresponds to the charge of 1 mole of electrons, the total charge required is:

4. Final Answer:
The charge required for the reduction of 1 mole of MnO₄⁻ to MnO₂ is .

To reduce MnO to Mn , we need to consider the change in oxidation state of Mn. In MnO , Mn is in the +2 oxidation state, and in Mn , Mn is also in the +2 state. Thus, no change in oxidation state occurs. However, the reduction process involves the addition of 2 electrons per ion, and for 1 mole of MnO , 4 moles of electrons are required.
This corresponds to a charge of 4 Faradays.

Step 1: Understanding the cell potential. The cell potential is calculated by subtracting the anode potential from the cathode potential. The two given standard electrode potentials are for the Sn^{4+}/Sn^{2+} couple (+0.15 V) and the Cr^{3+}/Cr couple (-0.74 V).

Step 2: Calculation. The cell potential is given by:

Step 3: Conclusion. Thus, the cell potential is +0.89 V, corresponding to option (B). \vspace{10pt}

1. Complete and Balance the Following Chemical Equations:

(a) Equation:
2MnO₄⁻(aq) + 10I⁻(aq) + 16H⁺(aq) →

Solution:
In acidic medium, MnO₄⁻ gets reduced to Mn²⁺ and I⁻ gets oxidized to I₂.

Half-reactions:
Reduction half-reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation half-reaction: 2I⁻ → I₂ + 2e⁻

Balancing the half-reactions:
Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5 to equalize the number of electrons transferred.

Reduction: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O

Oxidation: 10I⁻ → 5I₂ + 10e⁻

Balanced Equation:
2MnO₄⁻(aq) + 10I⁻(aq) + 16H⁺(aq) → 2Mn²⁺(aq) + 5I₂(aq) + 8H₂O(l)

(b) Equation:
Cr₂O₇²⁻(aq) + 6Fe²⁺(aq) + 14H⁺(aq) →

Solution:
In acidic medium, Cr₂O₇²⁻ gets reduced to Cr³⁺ and Fe²⁺ gets oxidized to Fe³⁺.

Half-reactions:
Reduction half-reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Oxidation half-reaction: 6Fe²⁺ → 6Fe³⁺ + 6e⁻

Balancing the half-reactions:
The number of electrons is already balanced in the half-reactions.

Balanced Equation:
Cr₂O₇²⁻(aq) + 6Fe²⁺(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 6Fe³⁺(aq) + 7H₂O(l)

To solve the problem, we need to write the cell reaction and calculate the e.m.f. of the given cell at 298 K.

1. Understanding the Cell Setup:
The given cell is represented as:

2. Identifying the Half-Reactions:
The cell consists of two half-cells: - The first half-cell involves the reduction of tin (Sn) to tin ions ( ) at the anode:
- The second half-cell involves the reduction of hydrogen ions ( ) to hydrogen gas ( ) at the cathode:

3. Standard Cell Potential (E°):
The standard electrode potential for the Sn /Sn half-reaction is given as , and for the H /H half-reaction, .
Therefore, the standard cell potential is:

4. Nernst Equation:
The Nernst equation is used to calculate the cell potential under non-standard conditions: Where: - - (the number of electrons transferred) - is the reaction quotient, which is given by: Substitute the given concentrations and pressures:

5. Calculating the Cell Potential:
Now, substitute the values into the Nernst equation: Since , we get:

6. Final Answer:
The cell reaction is:

The e.m.f. of the cell at 298 K is 0.11045 V.

(i) Cell potential, also known as the electrode potential or electromotive force (EMF), is the difference in electric potential between two electrodes in a galvanic or electrochemical cell. It is a measure of the ability of a cell to drive an electrochemical reaction and is usually measured in volts (V). The greater the cell potential, the greater the ability of the cell to perform work. The standard cell potential is calculated under standard conditions, typically 25°C, 1 M concentration for solutions, and 1 atm pressure for gases.
(ii) Fuel cell
A fuel cell is an electrochemical device that converts the chemical energy of a fuel (usually hydrogen) and an oxidant (such as oxygen) into electrical energy. This conversion occurs through a redox reaction at the electrodes. The fuel cell operates continuously as long as the fuel and oxidant are supplied, making it a clean and efficient source of power. Common examples include hydrogen fuel cells used in electric vehicles and other energy applications.