Oxidation of sulfuric acid to Marshall's acid and its half cell reaction

Question

$\ce{H2O2}$ 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.

Answer 1

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.

Answer 2

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.

Answer 3

Since oxygen ends up in a mixture of oxidation states and sulfur alone is norlt oxidized, your best bet is to render the whole reactant bisulfate ion as the reducing agent, losing the electron the whole delocalozed molecular orbitals rather than from any one atom. To wit:

$\ce{2 HSO4^- -> S2O8^{2-} + 2 H^+ + 2e^-}$

$\ce{2H^+ +2e^- -> H2(g)}$

So $n$ is two moles of electrons per mole of peroxydisulfate ion, or one mole of electrons per mole of bisulfate ion. These are the same number, of course, because you need two moles of bisulfate to bapance one mole of peroxydisulfate.