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!!!
