Question

\(aMnO^{-}_4+bS_2O_{3}^-+H_2O\rightarrow xMnO_2+ySO_{4}^-+zOH^-\)
a and y respectively are

• A. 8; 3
• B. 8; 6
• C. 3; 6
• D. 8; 8

Answer 1

The Correct Option is B

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

Answer 2

To balance the reaction, we need to ensure that both mass and charge are balanced.

1. First, balance the manganese (Mn) atoms. In MnO₄⁻ (permanganate), Mn is in the +7 oxidation state, while in MnO₂ (manganese dioxide), Mn is in the +4 oxidation state. Therefore, the reduction of MnO₄⁻ to MnO₂ involves a 3-electron gain.

2. Next, balance the sulfur atoms. In S₂O₃²⁻ (thiosulfate), sulfur has an oxidation state of +2, and in SO₄²⁻ (sulfate), sulfur has an oxidation state of +6. The oxidation of thiosulfate to sulfate involves a 4-electron loss.

3. From the stoichiometry of the reaction, we find that to balance the electrons involved in the oxidation and reduction processes, \(a = 8\) and \(y = 6\).

Thus, the correct values for \(a\) and \(y\) are 8 and 6, respectively.