Question

Balance the following redox reaction by inserting the appropriate coefficients. SO4^2- + NH3 = SO3^2- + H2O + N2

Answer 1

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.

Answer 2

Equivalent reaction equation :

3SO₄²⁻ + 2NH₃ ---> 3SO₃²⁻ + N₂ + 3H₂O

Further explanation

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

Learn more

an oxidation-reduction reaction

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Keywords : Balance, redox, coefficients,Oxidation number, half reaction