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.

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    $\begingroup$ 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
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    $\begingroup$ Like Einstein "said", "Everything is relative..." $\endgroup$ – MaxW Dec 19 '19 at 6:49
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    $\begingroup$ @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
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    $\begingroup$ By the way, "Sulphur Dioxide" is a two-word answer! :-) $\endgroup$ – Mathew Mahindaratne Dec 19 '19 at 8:16
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    $\begingroup$ 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

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.

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    $\begingroup$ 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

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