Question

Find the n factor of fes2 in half reaction fes2 ----->fe2o3 +so2

Answer 1

Answer:

In FeS2,both Fe and S are getting oxidised. Fe from +2 to +3 therefore n factor of Fe=1 and S from -1 to +4 therefore n factor of S=2*5=10(because there are 2atoms of S) therefore n factor of FeS2 is 10+1=11

Answer 2

Answer:

The n- factor of the FeS₂ in the given reaction is equal to eleven.

Explanation:

n- factor determines the number of electrons gained or lost by any species during the reaction.

We use n- factor to determine the equivalent weight of the compound.

Equivalent weight = molecular weight/n-factor

The given reaction is :

FeS₂  +  O₂  $\longrightarrow$  Fe₂O₃   +   SO₂

The given reaction of FeS₂ and  O₂  is a redox reaction. In which O₂ acts as an oxidising agent and FeS₂  is a reducing agent.

The oxidation of Fe²⁺  from +2  to +3 as:

Fe²⁺   $\longrightarrow$    Fe³⁺  + e⁻

The oxidation of S from -1 to +4 as :  

S⁻   $\longrightarrow$    S⁺⁴ +  5e⁻

The reduction of O₂ from 0  to -2 as:  

O₂  + 4e⁻  $\longrightarrow$   O²⁻  

Therefore the total number of electrons FeS₂ is losing is one from Fe abd 10 from two sulphur.

The n-factor of FeS₂ = 10 +1 =11

Therefore, the n-factor of FeS₂ for this reaction is 11.