Question

Given the reaction: 2na + 2h2o → 2na+ + 2oh− + h2

which substance is oxidized?




1.





h2


2.





h+


3.





na


4.





na+

Answer 1

`\large \boxed{\text{3. Na}}`

Step-by-step explanation:

We can use oxidation numbers to decide which substance is reduced.

`\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)}`

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.

`\text{An increase in oxidation number is oxidation, so $\large \boxed{\textbf{Na}}$ is the substance oxidized.}`

1 and 4 are wrong because H₂ and Na⁺ are products.

2. is wrong because there is no H⁺ to be oxidized or reduced.