$\ce{H2O2}$ can be prepared by successive reactions:

$\ce{2NH4HSO4 -> H2 + (NH4)2S2O8}$
$\ce{(NH4)2S2O8 + 2H2O -> 2NH4HSO4 + H2O2}$

The first reaction is an electrolytic reaction and second is steam distillation. What amount of current would have to be used in first reaction to produce enough intermediate to yield $\pu{102g}$ pure $\ce{H2O2}$ per hour. Assume efficiency 50%.

I was solving this question on Faraday's Laws Of Electrolysis when I stumbled across a conceptual flaw of mine. I realized that the sulfate anion in the first reaction is at a +6 Oxidation State and so is Marshall's acid, as far as what I know of Redox Reactions and their balancing we look at the number of electrons exchanged and thus formulate the half-cell reaction.

However, this logic of mine failed in the above question as the oxidation state of the central atom is unchanged, which makes me wonder how to calculate the valency-factor/N-factor and correspondingly the equivalent weight. I know this is a conceptual shortcoming of mine and that the aforementioned logic is very 'methodical' per se, which is why it fails. If someone could please point out where I am going wrong it'd be highly appreciated.

  • $\begingroup$ It is oxidation state of oxygen, not sulphur, that is changing. $\endgroup$
    – Poutnik
    Dec 23, 2021 at 16:45
  • $\begingroup$ Using image of text is highly discouraged. Please, type the text instead of pasting a photo now onwards. Images cannot be properly indexed by the search engines. $\endgroup$ Dec 24, 2021 at 6:06

2 Answers 2


Forget about the equivalent weights. Work with moles, and only with moles. Look how it goes.

$102$ g $\ce{H2O2}$ is $\ce{\frac{102 g}{34 g/mol} = 3.00 mol H2O2}$. The production of $1$ mol $\ce{H2O2}$ requires $1$ mol $\ce{(NH4)2S2O8}$. Now we will show that the production of $1$ mol $\ce{(NH4)2S2O8}$ requires $2$ moles electrons. Before doing this, we will first show that the ion $\ce{HSO4-}$ from $\ce{NH4HSO4}$ is at least partly decomposed into the following two ions according to the following equation : $$\ce{HSO4^- <=> H+ + SO4^{2-}}$$ One of these ions ($\ce{H+}$) is reduced in $\ce{H2}$ at the cathode, and the other one ($\ce{SO4^{2-}}$) is oxidized at the anode according to : $$\ce{2H+ + 2 e- -> H2}$$ $$\ce{2 SO4^{2-} -> S2O8^{2-} + 2 e-}$$ This shows that $2$ electrons are needed to produce $1$ mole $\ce{(NH4)2S2O8}$, and later on $1$ mole $\ce{H2O2}$.

Then Faraday law gives you the intensity I needed to produce $3.00$ mol $\ce{(NH4)2S2O8}$ in $1$ hour = $3600$ s. It is :

$$\pu{I = \frac{3.00 ~mol~·~2~·~96500~ As/mol}{3600~ s} = 160 A}$$ This value is obtained if the yield is $100$%. As the yield is $50$%, the intensity must be twice the previous value. This is $320$ A.


In the event that this question is actually more than just a theoretical exercise, the general electrolysis half-reactions, as cited by Maurice in my opinion, may not actually be in accord with this particular's complex reaction system per a review of the possible underlying mechanics.

More precisely, here is a suggested overview of reaction mechanics that support my comment:

$\ce{H2O = H+ + OH-}$

$\ce{Electrolysis of HSO4- => .H + .SO4-}$

$\ce{NH4+ = H+ + NH3}$

$\ce{.H + .H = H2}$

$\ce{.H + .SO4- = HSO4-}$

$\ce{.SO4- + .SO4- = S2O8^{2-} }$

$\ce{ 2 NH4+ + S2O8^{2-} = (NH4)2S2O8 }$

Also, a slow reaction, introducing a powerful radical:

$\ce{.SO4- + H2O = .OH + H+ + SO4^{2-} }$

$\ce{HSO4- +.OH = H2O + .SO4- }$

So, while one may claim seemingly that only that 2 electrons are required, my analysis as outlined above which is subject to kinetics, suggests some possible reversed reactions resulting in reduced efficiency.

As a result, not surprisingly, more than 2 electrons may actually be required to produce the single mole of $\ce{(NH4)2S2O8}$ in an experiment, and as such, I would recommend qualifying Maurice analysis with the words "at least", if one is writing up this experiment to account for observed results. More interesting is the provided statement "Assume efficiency 50%", which appears supportive of my take on the reaction system.

  • $\begingroup$ Hmm intensive anodic oxidation at high current density is not friendly for producing hydrogen radicals. $\endgroup$
    – Poutnik
    Dec 25, 2021 at 10:11
  • $\begingroup$ Albeit, the somehow formed sulfate radical anion, per this source Eq (8) at acp.copernicus.org/articles/18/2809/2018/acp-18-2809-2018.pdf is the cited path to $\ce{S2O8^{2-}}$ via a self reaction with the sulfate radical anion. Interesting, the slow action of H2O on $\ce{.SO4^{-}}$ per Eq (7) does introduce the $\ce{.OH}$ radical. Also, I am unclear if any formed NH3 is converted into the $\ce{.NH2}$ radical. $\endgroup$
    – AJKOER
    Dec 25, 2021 at 14:21
  • $\begingroup$ OK, I have edited my thread to correct for the formation of the sulfate radical. Per the electrolysis of HSO4- to H+ + e- + .SO4- with the implication in concentrated solutions the formation of S2O8(2-) as confirmed in this reference toppr.com/ask/en-us/question/… . $\endgroup$
    – AJKOER
    Dec 25, 2021 at 16:20

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