# What is the n factor of h2o2 undergoing disproportionation reaction liberating oxygen [closed]

There is a lot of confusion about the n-factor of hydrogen peroxide in its disproportionation into water and oxygen.

$$\ce{2H2O2 -> 2H2O + O2}$$

Some sources(1,2) say n-factor of $$\ce{H2O2}$$ is 2 and other say n-factor of $$\ce{H2O2}$$ is 1 for this reaction.

Actually my main problem is "the relationship between normality and volume strength of $$\ce{H2O2}$$". According to my respected teacher:
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$ and
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$
implying that n-factor of $$\ce{H2O2}$$ is 2 but I am calculating n-factor of $$\ce{H2O2}$$ as 1 because the definition of n-factor is "the no of electron gained or lost per mole".

As we can see 2 electrons are participating in this reaction so electron exchanged per mole is 1 which is equal to n-factor. Please clarify.

Sources

• biggest source my teacher and some people on quora say 2 accoding to me n factor should be 1 beacause n factor means loss or gain of electron per mole so in this reaction 2 electrons are perticipating in the reaction so 2 electron/2 mole = 1 – gurdeep singh Feb 26 at 11:08
• What is the definition of that "n" factor? What do you do with it? – Karl Feb 26 at 11:17
• there is a problem with Indian education system people dont want to confront anyone even if they are teaching wrong concepts there are teachers on youtube you can go just type volume strength of h202 on the search bar you will find teacher telling n factor of h 2o2 in disprotionation reaction is 2 I – gurdeep singh Feb 26 at 12:03
• "n factors" are really old form of chemistry, nowadays stoichiometry + molarity is enough. – orthocresol Feb 26 at 16:21
• Yea. Tell your teacher chem.SE says hi and he's welcome to skip the 20th and come directly into the 21st century. – Karl Feb 26 at 23:18

I want to elaborate the answer of Ashish Ahuja because he is also perplexed by the n-factors. First of all for general readers n-factors are of historical interest only. Due to my interest in historical analytical methods, the following points would clarify this situation:

A) Equivalent weights do not require any knowledge of molarity, moles, electrons because it pre-dates all of these.

B) Equivalent weight is totally dependent on the reaction, so half reaction for an equivalent weight is meaningless without a context of how it is being used.

C) In the comments, I showed that the equivalent weight of $$\ce{H2O2}$$ can be calculated from the very original definition of equivalent weights without invoking the useless n-factors.

Equivalent weight of X (oxidizing agent) is the weight of X that will furnish 1 gram-equivalent of oxygen = 8g. So we can write

$$\ce{H2O2-> H2O +0.5O2}$$

Which means that 8 grams of oxygen as [O] would be provided by 34.01 g of $$\ce{H2O2}$$. There was no need to involve moles

This means that the molarity and normality of $$\ce{H2O2}$$ for a disproportionation reaction must be the same.

D) The volume strength is another archaic unit, which was very practical. The OP wrote

the relationship between normality and volume strength of $$\ce{H2O2}$$". According to my respected teacher:
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$
$$\pu{Molaritymality = \frac{Volume~strength}{11.2}}$$

Please gently tell your respected teacher that is both incorrect AND correct! The reason is that volume strength was determined by decomposition of hydrogen peroxide into water and oxygen (exactly by this disproportionation reaction). So if you are quoting normality with respect to this disproportionation reaction, then normality and molarity are the same.

Now if you are using the hydrogen peroxide in a redox titration reaction, it "consumes" two hydrogen ions. Again equivalent weight is the amount that furnishes/consumes 1 gram equivalent of oxygen or 1 gram equivalent hydrogen.

$$\ce{H2O2 + 2H+ + 2e → 2H2O}$$

Here 34.01 g of $$\ce{H2O2}$$ consumes two gram equivalents of hydrogen as an oxidizing agent. So the equivalent weight of this compound as an oxidizer in acidic medium is 34.01/2.

The moral of the story is that normality is solely dependent on what reaction is being used! The examiner or the teacher must provide a context. It is meaningless otherwise!

• your answer is much clear thankyou – gurdeep singh Feb 28 at 7:57
• we can also get oxygen by the oxidation and disproptionation reaction of h2o2 .....but he always stressed disproptionation as the reaction for finding volume strength but your point cleary got through – gurdeep singh Feb 28 at 8:02

We can split the reaction

$$\ce{H2O2 -> H2O + O2}$$

into the respective reduction and oxidation half-reactions.

$$\ce{H2O2 -> O2 + 2H+ + 2e-}$$

$$\ce{H2O2 + 2H+ + 2e- -> 2H2O}$$

Since the n-factor of $$\ce{H2O2}$$ for both these half-reactions is 2, the n-factor is:

$$\frac{1}{n_f} = \frac{1}{2} + \frac{1}{2} = 1$$

$$n_f = 1$$

Based on the comments under your post I believe you have gotten this far, but some sources claim that the n-factor for this reaction is 2. Well, how do you verify what is correct? In such a case, I'd simply write the balanced chemical reaction and verify whether the law of chemical equivalence holds or not.

$$\ce{2H2O2 -> 2H2O + O2}$$

The n-factor for $$\ce{O2}$$ should be $$2$$ as can be seen in the first half-reaction, and for $$\ce{H2O}$$ the n-factor should be $$1$$, since we have $$2$$ moles of $$\ce{e-}$$ in the second half-reaction and $$2$$ moles of $$\ce{H2O}$$, and n-factor is defined per-mole of the substance.

From the law of chemical equivalence, the number of equivalents of $$\ce{H2O2}$$, $$\ce{H2O}$$ and $$\ce{O2}$$ should all be equal. If the claim I make above that the n-factor of $$\ce{H2O2}$$ is $$1$$ is true, this should hold.

Plugging in the moles from the balanced chemical equation and using the n-factors described above,

Equivalents of $$\ce{H2O2} = 2 \times 1$$, $$\ce{H2O} = 2 \times 1$$, $$\ce{O2} = 1 \times 2$$ (moles $$\times$$ n-factor)

Thus the n-factor is indeed $$1$$. The n-factor of $$\ce{H2O2}$$ for the half-reactions is $$2$$.

(Note that all the n-factors above are with respect to the reaction $$\ce{H2O2 -> H2O + O2}$$ and they (most probably) will change for a different reaction)

Update (the above part answered the question before the edit):

From [1]:

The "volume strength" of hydrogen peroxide is defined as the number of times its own volume of oxygen a sample of hydrogen peroxide solution will evolve if decomposed naturally. $$\ce{H2O2 -> H2O + O2}$$

As you've mentioned in your question,

$$\pu{Normality = \frac{Volume~strength}{5.6}}$$ and
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$

This does not match with the reasoning that the n-factor of $$\ce{H2O2}$$ for the above reaction is $$1$$, which would imply that the normality and molarity should be equal.

From [2]:

Now, equivalent weight of $$\ce{H2O2} = \frac{34}{2} = 17$$

A (N) $$\ce{H2O2}$$ solution contains 17 gms of $$\ce{H2O2}$$ per litre.

So in fact when we are relating the normality of an $$\ce{H2O2}$$ solution to it's volume strength, we are not referring to the normality of $$\ce{H2O2}$$ in the decomposition reaction. You would be right in saying that the normality of $$\ce{H2O2}$$ in the decomposition reaction is equal to its molarity.

One way of justifying the n-factor of $$2$$ of an $$\ce{H2O2}$$ solution (not undergoing any reaction) is due to the $$-2$$ charge on the $$\ce{O2^{2-}}$$ peroxide ion.

I can't really explain why this has been done, since it would make just as much sense to consider the n-factor of $$\ce{H2O2}$$ in the decomposition reaction, it is just a historical fact. Since n-factor is not a constant for a particular compound, it should be really important to specify context about normality when we relate it to volume strength, which is rarely done when the relation is taught which is what leads to this confusion. Both the n-factor's are correct in different contexts, so I would just recommend sticking to the n-factor of $$2$$ when talking about volume strength since that is what you have been taught.

[1]: Chemical Calculations by King A archive.org

[2]: An Introduction To Chemistry Vol. 1 Ed. 3rd by Das, Ranajit archive.org

• ashish ahujha thankyou for your answer i want to ask one more thing i am preparing for jee for engineering course in the paper a question about volume strength is asked in which they take the n factor 2 to get the formula (Normality = Volume strength/5.6) and (Molarity = volume strength /11.2) if you can give me some clarity about this because if the n factor is 1 for this reaction molarity and normality should be equal – gurdeep singh Feb 26 at 13:57
• @gurdeepsingh That's actually an interesting question. I've learnt the same thing, and I never realized that till now. I'm afraid I don't have the answer right now, I personally feel that you should ask a separate question for that since that deviates from the original question here and then other users will also be able to answer. I'll let you know if I find the answer. – Ashish Ahuja Feb 26 at 14:02
• chemistry.stackexchange.com/q/125269/92638 – gurdeep singh Feb 26 at 14:04
• I am against these n-factor calculations, because not only these are confusing, but nobody uses them anymore. Actually, I have not seen n-factor being discussed in any modern chemistry book by Western authors. – M. Farooq Feb 26 at 15:37
• Ashish, you answer is numerically correct. The approach of n-factor is bad. It is not your fault, this is the way it is taught. – M. Farooq Feb 26 at 15:46