Redox Reactions

Correct Answer: NH₄NO₃

Explanation:
In ammonium nitrate (NH₄NO₃), nitrogen exists in two different compounds within the same molecule:

• In the ammonium ion (NH₄⁺), nitrogen has an oxidation state of -3.
• In the nitrate ion (NO₃^-), nitrogen has an oxidation state of +5.

So, nitrogen exhibits two different oxidation states (-3 and +5) in this single compound.

The other options contain nitrogen in only one oxidation state.

The given chemical reactions are:

Reaction 1:

In this reaction, silver ion ( ) is reduced to , and oxygen ( ) is involved as the oxidizing agent. Since silver ions are reduced, **cyanide ions (CN⁻)** are oxidized, making them the reducing agent.

Reaction 2:

Here, zinc metal ( ) is oxidized to , while the silver ions ( ) are reduced to metallic silver. Since zinc is losing electrons, it is the reducing agent in this reaction.

Conclusion:

In both reactions, the **reducing agent is CN⁻ (cyanide ion)**, as it is responsible for reducing the metal ions ( and ).

Correct Answer: Option (A): CN⁻

The correct answer is Option 1: H₃PO₂.

To reduce AgNO₃ to metallic silver, the oxoacid of phosphorus must have a reducing property. Let's analyze the given options:

• H₃PO₂ (Hypophosphorous acid): This acid is a strong reducing agent.
• H₄P₂O₇ (Pyrophosphoric acid): This acid is not a strong reducing agent.
• H₄P₂O₆ (Hypophosphoric acid): This acid is not a strong reducing agent.
• H₃PO₄ (Phosphoric acid): This acid is not a strong reducing agent.

H₃PO₂ (Hypophosphorous acid) has the strongest reducing ability because it contains the highest number of P-H bonds. These P-H bonds can easily donate electrons, making it a good reducing agent. This characteristic allows it to reduce AgNO₃ to metallic silver.

Therefore, the oxoacid of phosphorus that can reduce AgNO₃ to metallic silver is H₃PO₂.

The correct answer is Option 1: H₃PO₂.

We are given the following chemical reaction:

aMnO₄⁻ + bSO₃²⁻ + H₂O → xMnO₂ + ySO₄²⁻ + zOH⁻

Let's balance the equation step by step:

Step 1: Balance manganese (Mn)

The manganese on the left-hand side is in the form of MnO₄⁻ (with oxidation state +7). On the right-hand side, it is in the form of MnO₂ (oxidation state +4). Thus, to balance manganese atoms, we need:

a = 2, x = 2

Step 2: Balance sulfur (S)

The sulfur on the left-hand side is in the form of SO₃²⁻, and on the right-hand side, it is in the form of SO₄²⁻. To balance sulfur, we need:

b = 5, y = 5

Step 3: Balance oxygen (O)

To balance oxygen, we need to ensure that the number of oxygen atoms on both sides is the same. The left-hand side has 4 from MnO₄⁻ and 3 from SO₃²⁻, giving a total of 7 oxygen atoms, and the right-hand side has 4 from MnO₂ and 4 from SO₄²⁻. We adjust the oxygen atoms by balancing with water (H₂O) molecules.

We determine that:

a = 8, y = 6

Conclusion:

The values of a and y are: a = 8, y = 6

The question pertains to the redox reaction between sulfur dioxide ( ) and acidified permanganate ions ( . To solve this, we need to understand the oxidation and reduction processes occurring in the reaction:

1. The reaction involves two half-reactions: one for the oxidation of and another for the reduction of
2. Oxidation: is oxidized to . In this process, sulfur's oxidation state increases from +4 in to +6 in .
3. Reduction: (in the presence of acid) is reduced to . Manganese's oxidation state decreases from +7 in to +2 in .
4. Balance the half-reactions considering the number of electrons transferred and combine them to get the overall reaction.

Given these explanation points, the correct option is: SO₂ is oxidised to SO₄²⁻, MnO₄⁻ is reduced to Mn²⁺.

The other options are incorrect because:

• SO₂ is reduced to S, MnO₄⁻ is oxidised to MnO₄: This does not represent a legitimate redox reaction. SO₂ is generally oxidized, not reduced to elemental sulfur.
• SO₂ is oxidised to SO₃²⁻, MnO₄⁻ is reduced to MnO₂: SO₂ oxidation results in SO₄²⁻ rather than SO₃²⁻, and MnO₄⁻ reduction in acid medium yields Mn²⁺, not MnO₂.
• SO₂ is reduced to H₂S, MnO₄⁻ is oxidised to MnO₄: This configuration is incorrect for reactions with permanganate in acidic media, as this medium oxidizes rather than reduces sulfur compounds like SO₂.

Hence, the step-by-step analysis confirms why the correct answer is the stated option.

To determine how many Coulombs are required to oxidize 0.1 mole of H₂O to oxygen, we follow these steps:

1. First, consider the balanced chemical reaction for the oxidation of water to produce oxygen gas:

1. This equation shows that 2 moles of water produce 1 mole of oxygen gas, releasing 4 moles of electrons. Therefore, for the oxidation of 2 moles of water, 4 moles of electrons are needed.
2. We need to determine the number of electrons required to oxidize 0.1 mole of water. Since 4 moles of electrons are required for 2 moles of H₂O, for 0.1 mole of H₂O, the moles of electrons required are:

1. Next, we'll convert the moles of electrons to Coulombs using Faraday's constant, which is approximately 96,500 C/mol. Thus, the charge in Coulombs is:

1. Therefore, the number of Coulombs required to oxidize 0.1 mole of H₂O to oxygen is .

This corresponds to the option 1.93 × 10⁴ C, which is the correct answer.

To solve the given problem of finding out the mass of calcium deposited, we need to use Faraday's laws of electrolysis. The relevant concepts involved here are:

1. The amount of substance deposited during electrolysis is directly proportional to the charge passed through the electrolyte.

Let's proceed with the solution:

1. Determine the total charge (Q) passed through the electrolyte: The charge passed can be calculated using the formula: , where is the current and is the time in seconds.
   • Current,
   • Time,
   • Convert the time into seconds:
     • Total time
2. Calculate the charge:
3. Use Faraday's second law to find the mass of calcium deposited: The mass deposited is calculated using the formula: where:
   • is the mass deposited,
   • is the charge (19299 C),
   • is the molar mass of calcium (40.08 g/mol),
   • is the number of electrons exchanged (2 for calcium, as will gain two electrons to form Ca),
   • is Faraday's constant (approximately 96500 C/mol).
4. Calculate the mass of calcium deposited:

Thus, the mass of calcium deposited is 4.0 g, confirming that the correct answer is the option "4.0 g".

- (a) Combination reaction: is a combination reaction as two reactants combine to form one product.
- (b) Decomposition reaction: is a decomposition reaction, where one reactant breaks down into two products.
- (c) Displacement reaction: is a displacement reaction where one element displaces another from a compound.
- (d) Disproportionation reaction: is a disproportionation reaction where the same species undergoes both oxidation and reduction.
Thus, the correct answer is 2) a-iii, b-i, c-iv, d-ii.