Balance the following redox reaction by inserting the appropriate coefficients. so4^2- + nh3 = so3^2- + h2o + n2

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_|

                  |_-66 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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.