This question involves balancing a redox reaction and analyzing the changes in oxidation states. Let's break it down step by step.

Step 1: Identify the Redox Pair

First, we need to find which elements are undergoing oxidation and reduction. We do this by assigning oxidation numbers.

Given reaction: $I^{-}+{\mathrm{ClO}}_3^{-}+H_2\mathrm{SO}4\to {\mathrm{Cl}}^{-}+{\mathrm{HSO}}^4-+I_2$

Assign oxidation numbers:

• In I⁻: Iodine has oxidation number -1.
• In ClO₃⁻: Oxygen is -2, so chlorine is +5 (since -2*3 + Cl = -1).
• In H₂SO₄: Hydrogen is +1, oxygen is -2, so sulfur is +6 (since +1*2 + S + (-2)*4 = 0).
• In Cl⁻: Chlorine is -1.
• In HSO₄⁻: Hydrogen is +1, oxygen is -2, so sulfur is +6 (since +1 + S + (-2)*4 = -1).
• In I₂: Iodine is 0.

Changes:

• Iodine: from -1 (in I⁻) to 0 (in I₂) → oxidation (loss of electrons).
• Chlorine: from +5 (in ClO₃⁻) to -1 (in Cl⁻) → reduction (gain of electrons).
• Sulfur: remains +6 in both H₂SO₄ and HSO₄⁻ → no change, not reduced or oxidized.

So, iodide (I⁻) is oxidized, and chlorine in ClO₃⁻ is reduced. Sulfur is not reduced.

Step 2: Balance the Redox Reaction

We balance the half-reactions for oxidation and reduction.

Oxidation half-reaction (I⁻ to I₂):

$2I^{-}\to I_2+2e^{-}$

Reduction half-reaction (ClO₃⁻ to Cl⁻):

This occurs in acidic medium (H₂SO₄ provides H⁺). The balanced half-reaction is:

${\mathrm{ClO}}_3^{-}+6H^{+}+6e^{-}\to {\mathrm{Cl}}^{-}+3H_{2}O$

To make electrons equal, multiply oxidation half by 3:

$6I^{-}\to 3I_2+6e^{-}$

Now add both half-reactions:

$6I^{-}+{\mathrm{ClO}}_3^{-}+6H^{+}\to {\mathrm{Cl}}^{-}+3H_{2}O+3I_2$

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:

$6I^{-}+{\mathrm{ClO}}_3^{-}+6H_2\mathrm{SO}4\to {\mathrm{Cl}}^{-}+6{\mathrm{HSO}}^4-+3H_{2}O+3I_2$

Now, let's verify atom balance:

• I: 6 on left, 6 on right (in 3I₂).
• Cl: 1 on left, 1 on right.
• O: 3 (from ClO₃⁻) + 24 (from 6H₂SO₄) = 27; on right: 24 (from 6HSO₄⁻) + 3 (from 3H₂O) = 27.
• H: 12 (from 6H₂SO₄) on left; on right: 6 (from 6HSO₄⁻) + 6 (from 3H₂O) = 12.
• S: 6 on left, 6 on right.

Charge: Left: 6(-1) + (-1) + 6(0) = -7; Right: (-1) + 6(-1) + 0 + 0 = -7. Balanced.

Step 3: Analyze the Statements

Now, check each option against the balanced equation:

1. Iodide is oxidized. True, as I⁻ loses electrons to form I₂.
2. Sulphur is reduced. False, sulfur oxidation state remains +6.
3. Stoichiometric coefficient of HSO₄⁻ is 6. True, as seen in the balanced equation.
4. H₂O is one of the products. True, with coefficient 3.

So, correct statements are 1, 3, and 4.

Related Topics and Formulae

Related Topics: Redox reactions, balancing in acidic medium, oxidation number concept.

Key Formulae:

• Oxidation number calculation: Sum of oxidation numbers equals charge on species.
• Half-reaction method: Balance atoms, then charges by adding electrons.
• In acidic medium: Add H⁺ and H₂O to balance oxygen and hydrogen.