Identify the oxidant and the number of electrons transferred (n) in each redox reaction below.3cl2 + 2fe > 2fecl3h2o + hg^2+ + no2^1- > 2h^1+ + hg + no3^1-3so3^2- + 2mno4^1- + > 3so4^2- + 2mno2 + 2oh^1-

1. Oxidant - Cl₂, n = 6

2. Oxidant - Hg²⁺, n = 2

3. Oxidant - MnO₄⁻, n = 6

Explanation:

Oxidant or Oxidizing agent, is a species that gets reduced by gaining electrons and oxidizes the other species.

1. 2Fe + 3Cl₂ → 2FeCl₃

In this reaction, Fe atom loses 3 electrons each and gets oxidized from 0 oxidation state in Fe to +3 oxidation state in FeCl₃. Whereas, Cl atom gains 1 electron each and gets reduced from 0 oxidation state in Cl₂ to -1 oxidation state in FeCl₃. Therefore, Cl₂ is the oxidizing agent.

Now the total number of electrons lost by 2 Fe atoms = total number of electrons gained by 6 Cl atoms = 6.

Therefore, number of electrons transferred (n) = 6.

2. H₂O + Hg²⁺ + NO₂⁻ → 2H⁺ + Hg + NO₃⁻

In this reaction, N loses 2 electrons and gets oxidized from +3 oxidation state in NO₂⁻ to +5 oxidation state in NO₃⁻. Whereas, Hg gains 2 electrons and gets reduced from +2 oxidation state in Hg²⁺ to 0 oxidation state in Hg. Therefore, Hg²⁺ is the oxidizing agent.

Now the total number of electrons lost by N = total number of electrons gained by Hg = 2

Therefore, number of electrons transferred (n) = 2

3. 3SO₃²⁻ + 2MnO₄⁻ + H₂O → 3SO₄²⁻ + 2MnO₂ + 2OH⁻

In this reaction, each S atom loses 2 electrons and gets oxidized from +4 oxidation state in SO₃²⁻ to +6 oxidation state in SO₄²⁻. Whereas, each Mn atom gains 3 electrons and gets reduced from +7 oxidation state in MnO₄⁻ to +4 oxidation state in MnO₂. Therefore, MnO₄⁻ is the oxidizing agent.

Now the total number of electrons lost by 3 S atoms = total number of electrons gained by 2 Mn atoms = 6

Therefore, number of electrons transferred (n) = 6

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)