Moles of Reduction Product in the Oxidation of Aluminium by Nitric Acid

Question

If, in 2 M $\ce{HONO2}$, 2 mols $\ce{Al (s)}$ are oxidized to $\ce{Al(OH2)6^3+ (aq)}$, mols of reduction product hydrogen is
(a) two;
(b) three;
(c) four;
(d) five;
(e) six.

Regarding the above question, how would one solve it?

Would it be valid to simply say that I know that 6 electrons are being transferred because we have this skeleton equation (balanced for charge):

$$\ce{Al -> Al(OH2)6^3+ + 6 e-}$$

And given that 6 electrons are being transferred and that hydrogen is a reduction product, we must obtain hydrogen from some source in this system. The source would probably be hydronium ion, since hydronium ion is less stable than water. Plus abstracting protons from water to form hydrogen gas wouldn't make much sense in strongly acidic solution (we'd be forming hydroxide ions as well)! And knowing the following reaction:

$$\ce{2 H+ + 2 e- -> H2}$$

Can we just say since we know that 6 electrons are being moved, and that hydrogen gas is a product, we know that the immediately above reaction is occurring, and according to its stoichiometry, 2 moles of electrons yields 2 moles of hydrogen gas?

Answer 1

You simply build the half reactions and add them up so you can see how many moles of hydrogen gas are produced by a mole of Aluminium being oxidised. (Note: Your first equation is unbalanced.) \begin{aligned}\ce{ (1)&& Al + 6H2O &-> Al(OH2)_{6}^{3+} + 3e^{-}&|\cdot2\\ + (2)&& 2H3O+ + 2e^{-} &-> H2 ^ + 2H2O &|\cdot3\\\hline (3)&& 2Al + 12H2O + 6H3O+ &-> 2Al(OH2)_{6}^{3+} + 3H2 ^ + 6H2O\\ (4)\equiv(3)&& 2Al + 6H2O + 6H3O+ &-> 2Al(OH2)_{6}^{3+} + 3H2 ^ \\ (5)\equiv(3)&& Al + 3H2O + 3H3O+ &-> Al(OH2)_{6}^{3+} + \color{\green}{3/2}H2 ^ \\ }\end{aligned}

There are $\color{\green}{\text{one and a half}}$ mole Hydrogen gas formed for each mole of Aluminium.
Hence, three mole of Hydrogen gas are formed by two mole of Aluminium.

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Edit upon Dissenters follow up question "Why exactly would hydronium ion be the source?"
It says in the question that the solution is $2\mathrm{M}~\ce{HONO2}$ (nitric acid, although this is a strange way of writing it). Therefore there will be an excess of $\ce{H3O+}$. And cations travel to the cathode. Water could be a proton source, but not in that scenario.

It is necessary to acidify the solution, because otherwise you would passify the aluminium surface with $\ce{Al(OH)3}$

Answer 2

2 moles of electrons yields 2 moles of hydrogen gas

Doesn't your second equation say that 2 moles of electrons yields 1 mole of $\ce{H2}$? Following your reasoning, wouldn't 2 moles of $\ce{Al}$ yield 12 moles of electrons, which in turn would yield 6 moles of $\ce{H2}$? In any case, the balanced equation is $$\ce{12H2O + 6HNO3 + 2Al -> 2Al(H2O)_6^{+3} + 3H2 + 6NO3^-}$$ so 2 moles of $\ce{Al}$ would produce 3 moles of $\ce{H2}$ (answer B).