7
$\begingroup$

I think it's Hydrogen, because,

$\ce{PbO + H2 -> Pb + H2O}$

Here $\ce{PbO}$ is reduced to $\ce{Pb}$ and $\ce{H2}$ acts as a reducing agent.

But, in $\ce{2K + H2 -> 2KH}$, hydrogen gains a electron and undergoes reduction Thus, $\ce{K}$ is oxidized to $\ce{KH}$ and hydrogen works as oxidizing agent.

My teacher says it's wrong but doesn't give any explanation.

Can someone explain why my reasoning is wrong.

$\endgroup$
  • 1
    $\begingroup$ Considering you made a typo here, and the first reaction is $\ce{PbO + H2 -> Pb + H2O},$ it looks OK to me. Probably you got a picky teacher and they also expected you to add conditions and states of aggregation, but this is something you need to sort out with them. $\endgroup$ – andselisk Dec 19 '19 at 6:43
  • 1
    $\begingroup$ Like Einstein "said", "Everything is relative..." $\endgroup$ – MaxW Dec 19 '19 at 6:49
  • 1
    $\begingroup$ @andselisk The question on the test asked for a one word answer. As far as I know everyone else wrote Sulphur Dioxide as the answer. I think I was absent the day she explained redox reactions, she probably mentioned it in class. $\endgroup$ – Padmanava Dec 19 '19 at 8:00
  • 1
    $\begingroup$ By the way, "Sulphur Dioxide" is a two-word answer! :-) $\endgroup$ – Mathew Mahindaratne Dec 19 '19 at 8:16
  • 3
    $\begingroup$ Some teachers are like a tram: they go on rails and won't swerve either way, no matter what. Indeed, SO2 is an oxidizing and a reducing agent. So is H2, and so is HI, and so are many other gases (arguably, more than those that aren't). $\endgroup$ – Ivan Neretin Dec 19 '19 at 8:26
11
$\begingroup$

I'm guessing your teacher is looking for sulfur dioxide as the answer, but I don't see how or why you're supposed to be able to arrive to this answer logically. Either you'd need to read about it specifically, or maybe you're supposed to stare at a table of standard reduction potentials and notice that $\ce{SO2}$ appears as both a reagent and as a product with comparable voltages:

$$ \begin{array}{} \ce{SO2(aq) + 4 H+(aq) + 4 e- &<=>& S(s) + 2 H2O(l)} & E^\circ = \pu{+0.50 V}\\ \ce{SO2(aq) + 2 H2O(l) &<=>& SO4^2-(aq) + 4 H+(aq) + 2 e-} & E^\circ = \pu{-0.17 V} \end{array} $$

(The bottom equation is the inverse of what you would find in a reduction potential table)

Extrapolating from these aqueous potentials, it appears $\ce{SO2}$ is a modest reductant and a weak oxidiser, but both reactions can be accessed in reasonable conditions.

I think you're technically correct (as Ivan mentions), but what makes hydrogen gas a "less correct" answer is that it only rarely acts an oxidiser (in the presence of very strong reducing agents, such as alkali metals), whereas it can act as a reductant much more commonly. Compare these equations with the ones above:

$$ \begin{array}{} \ce{H2(g) + 2 e- &<=>& 2 H^-(aq, extrapolated)} & E^\circ = \pu{-2.3 V}\\ \ce{H2(g) &<=>& 2 H+(aq) + 2 e-} & E^\circ = \pu{+0.00 V} \end{array} $$

(Reference for the hydride potential)

Note the much larger gap in potential between the two reactions. In this sense, the reducing nature of $\ce{H2}$ vastly overwhelms its oxidising nature, so most people will think of it as a reducing gas.

| improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ My first thought was also hydrogen, and I was once a chemistry teacher myself. If the question was "Name a gas that can both reduce and oxidize reactants," then H2 is a correct answer. $\endgroup$ – CarlF Dec 19 '19 at 15:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.