SO₄²⁻ +NH₃ → SO₃²⁻ + H₂O +N₂
The balanced of the above redox reaction is as below
3SO₄²⁻ + 2NH₃ → 3SO₃²⁻ + 3 H₂O + N₂
Explanation
According to the law of mass conservation the number of atoms in the reactant side must be equal to number of atoms in product side.
Inserting coefficient 3 in front of SO₄² , 2 in front of NH₃, 3 in front of SO₃²⁻ and 3 in front of H₂O balance the equation above. This is because the number of atoms are equal in both side.
for example there are 2 atoms of N in both side of the reaction.
Equivalent reaction equation :
3SO₄²⁻ + 2NH₃ ---> 3SO₃²⁻ + N₂ + 3H₂O
In equalizing the redox reaction we can use the oxidation number method or half the reaction of the ions
Oxidation number method :
1. Determine the reducing agent and oxidizing agent
2. Equate the element that experiences a redox reaction
3. Determine the amount of reduction / increase in oxidation number (number of atoms x change in oxidation number)
4. Equate the number of changes in the oxidation number by giving a coefficient
5. equal charge (H + for acidic situations and OH- for alkaline conditions)
6. Balance the H atom with the addition of H₂O
From the reaction :
SO₄²⁻ + NH₃ ---> SO₃²⁻ + H₂O + N₂
1. SO₄²⁻ ---> SO₃²⁻ = reduction (for S : +6 to +4)
NH₃ ---> N₂ = oksidation (for N : -3 to 0)
2. add coefficient to equate atom N
SO₄²⁻ + 2NH₃ ---> SO₃²⁻ + N₂
3. SO₄²⁻ + 2NH₃ ---> SO₃²⁻ + N₂
|_+6___2 x 3__+4_|
|_-6____6 x 1___0_|
4. giving a coefficient
3SO₄²⁻ + 2NH₃ ---> 3SO₃²⁻ + N₂
5. charge already the same (left = -6, right = -6)
6. Balance the H atom with the addition of H₂O
3SO₄²⁻ + 2NH₃ ---> 3SO₃²⁻ + N₂ + 3H₂O ----> Balance
an oxidation-reduction reaction
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Keywords : Balance, redox, coefficients,Oxidation number, half reaction
