When the following redox equation is balanced with the smallest whole number coefficients, what is the coefficient of Sb3 + (aq) ? BrO3 - (aq) + Sb3 + (aq) → Br - (aq) + Sb5 + (aq) (acidic solution)

The balanced reaction is:

6H⁺ + BrO₃⁻ + 3Sb³⁺ → 3Sb⁵⁺ + Br⁻ + 3H₂O

so the coefficient for Sb³⁺ is 3

Explanation:

BrO₃⁻ (aq) + Sb³⁺ (aq) → Br⁻ (aq) + Sb⁵⁺ (aq)

First of all, let's determine the half reactions

BrO₃⁻ acts with + 5; Br⁻ with - 1.

Oxidation state has decreased, this is reduction

Sb³⁺ → Sb⁵⁺

This oxidation state has increased from + 3 to + 5, then this is the oxidation

In acidic medium, we complete with water, the same amount of oxygen at the opposite side in the reaction. And, in the other side we complete with protons, the same amount as hydrogen we have.

6H⁺ + BrO₃⁻ → Br⁻ + 3H₂O

As BrO₃⁻ has change from + 5 o - 1, it had to gain 6 electrons, so:

6H⁺ + BrO₃⁻ + 6e⁻ → Br⁻ + 3H₂O (reduction)

Sb³⁺ → Sb⁵⁺ + 2e⁻ (oxidation)

The Sb had to release 2 electrons, to change from + 3 to + 5

As electron are disbalanced, we have to multiply the two half reactions:

(6H⁺ + BrO₃⁻ + 6e⁻ → Br⁻ + 3H₂O). 2

(Sb³⁺ → Sb⁵⁺ + 2e⁻). 6

12H⁺ + 2BrO₃⁻ + 12e⁻ + 6Sb³⁺ → 6Sb⁵⁺ + 12e⁻ + 2Br⁻ + 6H₂O

Electrons are cancelled, so the balance equation is:

12H⁺ + 2BrO₃⁻ + 6Sb³⁺ → 6Sb⁵⁺ + 2Br⁻ + 6H₂O

We can divide by 2, to have the minimum stoichiometry

6H⁺ + BrO₃⁻ + 3Sb³⁺ → 3Sb⁵⁺ + Br⁻ + 3H₂O