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I was reading the first answer to Equivalent weight in case of disproportionation reaction there it is said

Take a disproportionation reaction:- $\ce{nA -> xB + $(n - x)$C}$ where A is being oxidised to B and rest reduced to C. Also let $n_1$ and $n_2$ be the n factors of A to B and C respectively We have 2 equations: one by conservation of electrons, $$\begin{align*} n_1 x &= n_2(n - x) \\[7pt] \implies x &= \frac{n_2\cdot n}{n_1 + n_2} \end{align*}$$ and the other from the definition of $n$-factor: $$\begin{align*} n_f &= \frac{\text{moles of electrons transferred}}{\text{moles of reactant}} \\ &= \frac{n_1 x}{n} \\ &= \frac{n_1 n_2}{n_1 + n_2} \end{align*}$$

In the second last line $n_f = \frac{n_1x}{n}$ shouldn't it be $\frac{n_1x+n_2(n-x)}{n}$ which is equal to $\frac{2n_1n_2}{n_1 + n_2}$ by expanding?
The reason for my thinking is because equivalent is defined as the weight of a substance that reacts with an arbitrary amount of another substance (eg:- 1g H). Hydrogen can both supply and take electrons therefore both the oxidation and reduction half reactions of A can take place with H and so I'm reasoning that n factor = adding up the electrons for both oxidation and reduction half reactions

Saif
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    Why would you use this concept here at all? One downside is apparent: it causes confusion. What are the upsides? When this reacts with that, you may use equivalents to find out how much "this" you need per certain amount of "that". Not the case here, though. – Ivan Neretin Aug 16 '23 at 15:27
  • @IvanNeretin Can you explain why it cannot be useful? I'm learning this in school, and my teacher told me that we will eventually use these concepts to make solving titration and some mole concept problems easier. I've also been bugged by this for a week so i want to understand it. – Saif Aug 16 '23 at 15:30
  • I saw some comments on other posts by how it is rarely used in titration but I dont understand why – Saif Aug 16 '23 at 15:30
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    It is fine to use equivalents for titration problems. But your imaginary example has nothing to do with titration. It is about *one* compound reacting with itself. And that's where equivalents are useless. – Ivan Neretin Aug 16 '23 at 15:33
  • Yes you're right. I was thinking what if a double titration was done with the one decomposing substance where the titration can help identify the amounts of the products of the decomposition, but then too you won't need the n factor of the one decomposing substance, you will only need the n factor of the products – Saif Aug 28 '23 at 03:59
  • Because it's the same as if you started out titrating the products. – Saif Aug 28 '23 at 04:21

1 Answers1

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If one insists on the formal usage of the n-factor and equivalent weight on disproportionation reactions (for which it is not intended), one must accept the fact the compound has simultaneously two n-factors (which can be incidentally equal):

  • one as the oxidant
  • one as the reductant.

Then the procedure is the same as for ordinary redox reactions where oxidants and reductants are different molecular entities.

But as Ivan and others have suggested, using n-factors for similar cases is rather useless and just complicates things. Be aware that the N-factor, equivalent, gram-equivalent and equivalent weight are obsolete concepts that should be taught only as what they are and how they were used.

The disproportionation redox reaction

$$\ce{n A -> x B + $(n - x)$C}$$

or alternatively written with small natural numbers $i,j$ as

$$\ce{$(i + j)$ A -> i B + j C}$$

is better to rewrite for easier analysis to 2 redox half-reactions:

\begin{align} \ce{ i A &-> i B + n e-} \\ \ce{ j A + n e- &-> j C} \end{align}

Then for n-factors of A as oxidant resp. reductant:

\begin{align} n_\mathrm{f,A,ox} = \frac{n}{j} \tag{oxidant}\\ n_\mathrm{f,A,red} = \frac{n}{i} \tag{reductant} \end{align}

It could be better understood if we reverse the reaction:

$$\ce{i B + j C -> $(i + j)$ A }$$

Poutnik
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