For the reaction :
I– + ClO3– + H2SO4 → Cl– + HSO4– + I2
The correct statement(s) in the balanced equation is/are
Balanced reaction will be
6I– + ClO3– + 6H2SO4 → Cl– + 6HSO4– + 3I2 + 3H2O
This question involves balancing a redox reaction and analyzing the changes in oxidation states. Let's break it down step by step.
First, we need to find which elements are undergoing oxidation and reduction. We do this by assigning oxidation numbers.
Given reaction:
Assign oxidation numbers:
Changes:
So, iodide (I⁻) is oxidized, and chlorine in ClO₃⁻ is reduced. Sulfur is not reduced.
We balance the half-reactions for oxidation and reduction.
Oxidation half-reaction (I⁻ to I₂):
Reduction half-reaction (ClO₃⁻ to Cl⁻):
This occurs in acidic medium (H₂SO₄ provides H⁺). The balanced half-reaction is:
To make electrons equal, multiply oxidation half by 3:
Now add both half-reactions:
H₂SO₄ provides H⁺ and SO₄²⁻. In the reaction, HSO₄⁻ is produced, which is the conjugate base of H₂SO₄ in acidic conditions. To incorporate H₂SO₄, note that each H₂SO₄ provides 2H⁺ and SO₄²⁻, but since we have HSO₄⁻, it means not all protons are used; some sulfate remains as HSO₄⁻.
From the half-reaction, we need 6H⁺. Each H₂SO₄ can provide 1H⁺ to form HSO₄⁻ (since H₂SO₄ → H⁺ + HSO₄⁻). So, we need 6 H₂SO₄ to provide 6H⁺, and this will produce 6 HSO₄⁻.
Thus, the balanced equation is:
Now, let's verify atom balance:
Charge: Left: 6(-1) + (-1) + 6(0) = -7; Right: (-1) + 6(-1) + 0 + 0 = -7. Balanced.
Now, check each option against the balanced equation:
So, correct statements are 1, 3, and 4.
Related Topics: Redox reactions, balancing in acidic medium, oxidation number concept.
Key Formulae: