What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l)H2O(l) and H+(aq)H+(aq) in the appropriate blanks. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e. g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

The balanced equation is then given as

BrO₃⁻ + 3Sn²⁺ + 6H⁺ → Br⁻ + 3Sn⁴⁺ + 3H₂O

The coefficients written in a bracket separated by commas is then given as

(1, 3, 6, 1, 3, 3)

Explanation:

The equation for the question is

BrO₃⁻ (aq) + Sn²⁺ + __ → Br⁻ + Sn⁴⁺ + __

The H⁺ and H₂O presented in the question shows that this redox reaction takes place in acidic medium.

We first identify which reactant specie is being oxidized or reduced.

Oxidation is the increase in oxidation number of reactant species due to the loss of electrons while Reduction is the reduction in oxidation number of reactant species due to the gain of electrons.

From the reaction given, it is evident that the Tin ion is oxidized as its oxidation number increases from +2 to +4.

And the Br in BrO₃⁻ undergoes reduction to have its oxidation number change from +5 to -1.

Note that +5 was calculated for as thus.

If the oxidation number of Br in BrO₃⁻ is unknown and called x, the oxidation number of the O in BrO₃⁻ is -2.

x + (-2)(3) = -1

x = -1 + 6 = +5

So, back to the question, since we know which reactant species are oxidized and reduced, we can then write the reduction and oxidation half reactions. Balanced reduction and oxidation half reactions at that showing the number of electrons lost or gained.

Reduction half reaction

BrO₃⁻ → Br⁻

This isn't possible, so we add the spectator compounds/elements/ions provided in the form of H⁺ and H₂O

BrO₃⁻ + H⁺ → Br⁻ + H₂O

We then balance this reaction stoichiometrically,

BrO₃⁻ + 6H⁺ → Br⁻ + 3H₂O

Now, we can check the oxidation numbers on both sides to know the number of electrons gained.

-1 + 6 → -1 + 0

+5 → -1

Hence, it is evident that 6 electrons are gained in this reduction half reaction.

So, we rewrite the reduction half reaction finally to be

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

We then repeat this for the oxidation half reaction.

Sn²⁺ → Sn⁴⁺

This is possible, So, no need to add the spectator compounds/elements/ions provided in the form of H⁺ and H₂O

The equation is also balance stoichiometrically, So, we just proceed to balance the charges/oxidation numbers.

+2 → +4

This shows that 2 electrons are lost, So, we rewrite our oxidation half reaction as

Sn²⁺ - 2e⁻ → Sn⁴⁺

It is more appropriate to write it in the form

Sn²⁺ → Sn⁴⁺ + 2e⁻

So, we can then write the two balanced half reactions (stoichiometrically and charge balance) on top of each other.

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

Sn²⁺ → Sn⁴⁺ + 2e⁻

In order to add, we need the number of electrons on both of these to be the same, so We multiply the reduction half reaction by 1 and multiple the oxidation half reaction by 3

[BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O] × 1

[Sn²⁺ → Sn⁴⁺ + 2e⁻ ] × 3

We then get

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

3Sn²⁺ → 3Sn⁴⁺ + 6e⁻

We can then add the two half reactions now

BrO₃⁻ + 6H⁺ + 6e⁻ + 3Sn²⁺ → Br⁻ + 3H₂O + 3Sn⁴⁺ + 6e⁻

Taking out the 6 electrons that appear on both sides, we have

BrO₃⁻ + 6H⁺ + 3Sn²⁺ → Br⁻ + 3H₂O + 3Sn⁴⁺

Written more properly as

BrO₃⁻ + 3Sn²⁺ + 6H⁺ → Br⁻ + 3Sn⁴⁺ + 3H₂O

Hope this Helps!!!

your answer would be “dry”