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the quantity of antimony in an ore can be determined by an oxidation-reduction titration with an oxidizing agent. The ore is dissolved in hot, concentrated acid and passed over a reducing agent so that all the antimony is in the form Sb^3+(aq). The Sb^3+ (aq) is completely oxidized by an aqueous solution BrO(3)^- + Sb^3+ ---> Br^- + Sb^5+. Complete and balance the equation for this reaction in acidic solution.

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Answer:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

Explanation:

At a redox equation, one substance is being oxidized (losing electrons), and the other is being reduced (gaining electrons). In the given reaction:

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

When it's at an acidic solution, it must be ions H⁺ on the reactant, which will form water with the oxygen, so the complete reaction is:

BrO₃⁻(aq) + Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq) + H₂O(l)

As we can see, the antimony is being oxidized (go from +3 to +5), and the Bromo is being reduced. The oxidation number of brome in the reactant, knowing that the oxidation number of O is -2, is:

x + 3*(-2) = -1

x = +5

So, it's going from +5 to -1, and the half-reactions are:

BrO₃⁻(aq) + 6e⁻ → Br⁻(aq)

Sb³⁺(aq) → Sb⁵⁺(aq) + 2e⁻

The number of electrons must be the same, so the second equation must be multiplied by 3:

3Sb³⁺(aq) → 3Sb⁵⁺(aq) + 6e⁻

Thus, the equation will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + H₂O(l)

Now, we verify the amount of the elements, which must be equal on both sides. So, we multiply H₂O by 3, and H⁺ by 6, and the balanced reaction will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

The balanced chemical equation:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

Redox reaction:

At a redox equation, one substance is being oxidized (losing electrons), and the other is being reduced (gaining electrons).

In the given reaction:

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

When it's at an acidic solution, it must be ions H⁺ on the reactant, which will form water with the oxygen, so the complete reaction is:

BrO₃⁻(aq) + Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq) + H₂O(l)

As we can see, the antimony is being oxidized (go from +3 to +5), and the Bromine is being reduced. The oxidation number of bromine in the reactant, knowing that the oxidation number of O is -2, is:

x + 3*(-2) = -1

x = +5

So, it's going from +5 to -1, and the half-reactions are:

BrO₃⁻(aq) + 6e⁻ → Br⁻(aq)

Sb³⁺(aq) → Sb⁵⁺(aq) + 2e⁻

The number of electrons must be the same, so the second equation must be multiplied by 3:

3Sb³⁺(aq) → 3Sb⁵⁺(aq) + 6e⁻

Thus, the equation will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + H₂O(l)

The balanced reaction will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

Find more information about Balanced chemical equation here:

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