Redox Reactions

To determine which compound is paramagnetic, we need to examine the oxidation state of manganese and the electron configuration of the manganese ion in each compound. - KMnO (Potassium permanganate): In KMnO , the manganese ion exists in the oxidation state (Mn ). The electron configuration of Mn is , meaning that all d-orbitals are empty. Since there are no unpaired electrons, KMnO is diamagnetic.
- K MnO (Potassium manganate): In K MnO , the manganese ion exists in the oxidation state (Mn ). The electron configuration of Mn is , meaning there is one unpaired electron in the d-orbital. Since there is at least one unpaired electron, K MnO is paramagnetic.


Step 1: Examine the oxidation state of manganese in each compound.


Step 2: Check the number of unpaired electrons in the electron configuration of the manganese ion.

To convert propanoic acid to ethane, we perform a reduction reaction using a strong reducing agent like LiAlH.

The steps are: 1. Reduction of Propanoic acid: Propanoic acid is reduced by lithium aluminum hydride (LiAlH) to ethanol. 2. Reduction of Ethanol to Ethane: Ethanol is then reduced further by a reducing agent like zinc in the presence of hydrochloric acid (Zn/HCl) to ethane. Thus, Propanoic acid is first converted to ethanol and then ethanol to ethane.

To solve the problem, we need to determine what product is formed when acidified KMnO₄ oxidizes sulphite.

1. Understanding the Reaction:
In an acidic medium, potassium permanganate (KMnO₄) is a strong oxidizing agent. It oxidizes sulphite ions (SO₃^2-) to sulphate ions (SO₄^2-).

2. Oxidation of Sulphite:
The sulphite ion (SO₃^2-) gets oxidized by the KMnO₄ in the presence of an acidic medium. The manganese in KMnO₄ undergoes reduction from +7 oxidation state to +2 oxidation state, forming Mn²⁺.

3. Balanced Chemical Equation:
The balanced redox reaction in acidic solution is:

2 KMnO₄ + 5 SO₃^2- + 4 H₂O → 2 Mn²⁺ + 5 SO₄^2- + 4 OH^-

Final Answer:
Acidified KMnO₄ oxidizes sulphite (SO₃^2-) to sulphate (SO₄^2-).

Solution:

The reaction between acidified potassium permanganate (KMnO₄) and sulphite (SO₃^2-) occurs as an oxidation-reduction (redox) process. In this reaction, KMnO₄ acts as an oxidizing agent, converting the sulphite ion to another form.

The oxidation state of sulphur in SO₃^2- is +4. The goal of the oxidation process is to increase the oxidation state of sulphur. Acidified KMnO₄ effectively oxidizes sulphite ions to peroxodisulfate ions (S₂O₈^2-). This involves the formation of a compound where two sulphur atoms are bonded via an oxygen-oxygen bond, and each sulphur atom is in the +6 oxidation state.

The balanced redox reaction is:

2KMnO₄ + 5SO₃^2- + 6H⁺ → 2Mn²⁺ + 5SO₄^2- + 3H₂O

Correct answer:

S₂O₈^2-

Faraday’s First Law of Electrolysis

Faraday’s first law of electrolysis states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the amount of electricity passed through the electrolyte.

Mathematical expression:

Where:

• = mass of the substance deposited (in grams)
• = molar mass of the substance (g/mol)
• = current (in amperes)
• = time (in seconds)
• = Faraday’s constant = 96,485 C/mol

Application: Reduction of to

In the reaction, the oxidation state of Mn changes from +7 to +2. Therefore, the number of electrons required for reduction:

So, the charge required to reduce 1 mole of to is:

Thus, 482,425 C of electricity is required to reduce 1 mole of to .

Zinc as a Strong Reducing Agent in +2 Oxidation State

The element that acts as a strong reducing agent in its +2 oxidation state is Zinc (Zn).

Zinc has a relatively low standard electrode potential:

This negative value indicates that metallic zinc readily loses electrons to form ions:

The more negative the value, the stronger the reducing power, as it reflects a higher tendency to donate electrons.

Thus, in the +2 oxidation state, zinc acts as a strong reducing agent and is capable of reducing other chemical species by donating electrons.

Oxidation-Reduction Reactions

(a) Reaction of Permanganate with Iodide Ions

In an acidic solution, permanganate ions ( ) act as an oxidizing agent and oxidize iodide ions ( ) to iodine ( ). The balanced equation for this reaction is as follows:

In this reaction, permanganate ions are reduced to , and iodide ions are oxidized to iodine.

(b) Reaction of Dichromate with Iron(II) Ions

In an acidic solution, dichromate ions ( ) are reduced to chromium(III) ions ( ), while iron(II) ions ( ) are oxidized to iron(III) ions ( ). The balanced equation for this reaction is:

In this reaction, the dichromate ions are reduced to , and iron(II) is oxidized to iron(III).