Question

$\bf{\fbox\red{REDOX}}$

Calculate equivalent mass of:-

1) FeC2O4 → Fe^3+ + CO2

&

2)FeS2 → Fe3+ + SO3

⤥(Provide short trick/Method if possible )

Thank You !!♠​

Answer 1

FeS2+O2⟶Fe2O3+SO2

Balanced reaction is

4FeS2+11O2⟶2Fe2O3+8SO2

1 mole of FeS2=11/4 moles of O2

Each O− atom takes 2e− for reduction

∴1O2 molecule takes 4e−

∴ Number of e− gained by O2=4/11×4=11

∴ Number of e− lost by FeS2=11

∴ Equivalent weight of FeS2 in this reaction= 11/M

Answer 2

Answer:

1) M/3 [Assuming mass of FeC2O4 as M]

2) N/9 [Assuming mass of FeS as N]

Explanation:

Hi dear,

These are redox reactions where we need to find out the equivalent weight of the reactant with respect to itself. We can do it with the help of n - factor which means the number of electrons replaced in a reaction (only in case of a redox reaction) [As n-factor has different meanings for different reactions. For eg, If we consider a acidic reaction, n-factor is the number of H⁺ ions replaced there.]

Method of doing these questions :

• Split the reactant
• Write the reactions
• Find the total number of electrons coming out (name it as n - factor)
• Finally, find the equivalent weight by the formula $\boxed {\Large {\textsf {Equivalent weight = \frac{\textsf{Molecular weight}}{\textsf{n - factor}} }}}$

1) FeC₂O₄ → Fe⁺³ + CO₂

Let us split the reactant as it is ionic in nature.

FeC₂O₄ can be written as Fe⁺² and C₂O₄⁻²

So, by the reaction we understand that,

Fe⁺² turns into Fe⁺³ and C₂O₄⁻² turns into CO₂

So, writing these reactions, we get,

$Fe^{+2} \longrightarrow Fe^{+3} + e^- ----[1]\\ C_2O_4^{-2} \longrightarrow 2CO_2 + 2e^- ----[2]$

Adding equations [1] and [2], we get

$Fe^{+2} + C_2O_4^{-2} \longrightarrow Fe^{+3} + e^- + 2CO_2 + 2e^- \\\text{which finally turns} \\FeC_2O_4 \longrightarrow Fe^{+3} + 2CO_2 + 3e^-$

So, in this reaction we finally get 3e- out which means n-factor = 3

We know that,

$\Large {\textsf {Equivalent weight = \frac{\textsf{Molecular weight}}{\textsf{n - factor}} }}$

Assuming the molecular weight of FeC₂O₄ be M, then equivalent weight will be M/3

Equivalent weight of FeC₂O₄ is M/3

[In these type of questions, in finding equivalent weights, don't waste your time in finding the mass of the given reactant. Find out the mass only if required. If not required, just substitute as M and solve the problem]

-------------------------------------

2) FeS → Fe⁺³ + SO₃

[Note : There is a mistake in the question. It should be 'FeS' not FeS₂]

Let us split the reactant as it is ionic in nature

FeS can be written as Fe⁺² and S⁻²

So, by the reaction we understand that,

Fe⁺² turns into Fe⁺³ and S⁻² turns into S⁺⁶ (In SO₃, S has +6 oxidation state)

So, writing these reactions, we get,

$Fe^{+2} \longrightarrow Fe^{+3} + e^- ----[3]\\ S^{-2} \longrightarrow S^{+6} + 8e^- ----[4]$

Adding equations [3] and [4], we get

$Fe^{+2} + S^{-2} \longrightarrow Fe^{+3} + e^- + S^{+6} + 8e^-$

which finally turns

$Fe^{+2} + S^{-2} \longrightarrow Fe^{+3} + S^{+6} + 9e^-$

So, in this reaction we finally get 9e- out which means n-factor = 9

We know that,

$\Large {\textsf {Equivalent weight = \frac{\textsf{Molecular weight}}{\textsf{n - factor}} }}$

Assuming the molecular weight of FeS be N, then equivalent weight will be N/9

Equivalent weight of FeS is N/9

Hope it helps you!!