Answer:
[tex]\large \boxed{\text{3. Na}}[/tex]
Explanation:
We can use oxidation numbers to decide which substance is reduced.
[tex]\rm 2\stackrel{\hbox{0}}{\hbox{Na}} + 2\stackrel{\hbox{+1}}{\hbox{ H}_{2}}\stackrel{\hbox{-2}}{\hbox{O}}\longrightarrow \rm 2\stackrel{\hbox{+1}}{\hbox{Na}^{+}} + 2\stackrel{\hbox{-2}}{\hbox{O}}\stackrel{\hbox{+1}}{\hbox{H}^{-}} + \stackrel{\hbox{0}}{\hbox{H}_{2}}[/tex]
The oxidation number of Na changes from 0 in Na to +1 in Na⁺.
The oxidation number of H changes from +1 in H₂O to 0 in H₂.
[tex]\text{An increase in oxidation number is oxidation, so $\large \boxed{\textbf{Na}}$ is the substance oxidized.}[/tex]
1 and 4 are wrong because H₂ and Na⁺ are products.
2. is wrong because there is no H⁺ to be oxidized or reduced.
