Question

n-factor for the following reaction is:
${{Fe}}{{{S}}_{{2}}}\xrightarrow{{}}{{F}}{{{e}}_{{2}}}{{{O}}_{{3}}}{{ + S}}{{{O}}_{{2}}}$
A. 8
B. 9
C. 10
D. 11

Answer 1

Hint: n- factor for acids is the number of ${{{H}}^{{ + }}}$ ions that is produced in water and n-factor for the base is the number of ${{O}}{{{H}}^{{ - }}}$ ions that is produced in water, it is possible to obtain equivalent mass by dividing molecular weight by n-factor.

Complete step by step answer:
Acids are defined as the ones that produce ${{{H}}^{{ + }}}$ ions when replaced by ${{1 mole}}$ of acid in the reaction. Bases are the ones that produce ${{O}}{{{H}}^{{ - }}}$ ions when replaced by ${{1 mole}}$ of base solution. For salts n-factor is the total moles of anionic/cationic charge replaced in ${{1 mole}}$ of the salt.
Consider the given reaction
${{Fe}}{{{S}}_{{2}}}\xrightarrow{{}}{{F}}{{{e}}_{{2}}}{{{O}}_{{3}}}{{ + S}}{{{O}}_{{2}}}$
${{2F}}{{{e}}^{{{2 + }}}}\xrightarrow{{}}{{F}}{{{e}}_{{2}}}^{{{3 + }}}{{ + 2}}{{{e}}^{{ - }}}{{ }}{{{n}}_{{f}}}{{ = }}\dfrac{{{2}}}{{{2}}}{{ = 1}}$
${{{S}}_{{2}}}^{{{2 - }}}\xrightarrow{{}}{{F}}{{{e}}_{{2}}}^{{{3 + }}}{{ + 2}}{{{e}}^{{ - }}}{{ }}{{{n}}_{{f}}}{{ = 10}}$

Thus the total n-factor=10+1=11

So, the correct answer is Option D.

Additional Information:
N-factor is used specially in physical chemistry to solve the numerical. N-factor for redox reaction is the change in the oxidation and reduction number on either side of the chemical reaction. In redox reaction n-factor is equal to the number of moles of electron lost or gained per mole
In case of acid, n-factor is the basicity of acid i.e., how many ${{{H}}^{{ + }}}$ ions it gives in the water is its n-factor
Eg: ${{HCl}}\xrightarrow{{}}{{{H}}^{{ + }}}{{ + C}}{{{l}}^{{ - }}}$here the ${{{n}}_{{f}}}{{ = 1}}$
${{{H}}_{{2}}}{{S}}{{{O}}_{{4}}}\xrightarrow{{}}{{2}}{{{H}}^{{ + }}}{{ + S}}{{{O}}_{{4}}}^{{{2 - }}}$${{{n}}_{{f}}}{{ = 2}}$
In case of base, now many ${{O}}{{{H}}^{{ - }}}$ is given in water is its n-factor
${{NaOH}}\xrightarrow{{}}{{N}}{{{a}}^{{ + }}}{{ + O}}{{{H}}^{{ - }}}$
${{S}}{{{O}}_{{2}}}{{ + 2NaOH}}\xrightarrow{{}}{{{H}}_{{2}}}{{O + N}}{{{a}}_{{2}}}{{S}}{{{O}}_{{3}}}$${{{n}}_{{f}}}{{ = 2}}$

Note: If an equivalent mass is to be calculated then we need the n- factor because ${{{E}}_{{m}}}{{ = }}\dfrac{{{M}}}{{{{{n}}_{{f}}}}}$. Inorder to calculate the normality then we need to know the n- factor because ${{Normality = Molarity \times n - factor}}$. N-factor concept is important in titrations too because it involves equivalent mass.