What is the reducing agent in the reaction #2Na + 2H_2O -> 2NaOH + H_2#?

1 Answer
Jul 29, 2016

Answer:

Sodium metal.

Explanation:

You're dealing with a redox reaction in which sodium metal, #"Na"#, is being oxidized to sodium cations, #"Na"^(+)#, and hydrogen is being reduced to hydrogen gas, #"H"_2#.

#2stackrel(color(blue)(0))("Na")_ ((s)) + 2stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(-2))("O")_ ((l)) -> 2stackrel(color(blue)(+1)) ("Na") stackrel(color(blue)(-2))("O") stackrel(color(blue)(+1))("H")_ ((aq)) + stackrel(color(blue)(0))("H") _(2(g))#

As you can see, the oxidation state of sodium goes from #color(blue)(0)# on the reactants' side, to #color(blue)(+1)# on the products' side, which implies that sodium metal is being oxidized.

On the other hand, the oxidation state of hydrogen goes from #color(blue)(+1)# on the reactants' side, to #color(blue)(0)# on the products' side, which implies that hydrogen is being reduced.

Now, a reducing agent is responsible with reducing a chemical species that takes part in a redox reaction. Similarly, an oxidizing agent is responsible with oxidizing a chemical species that takes part in a redox reaction.

You can thus say that the chemical species that is being oxidized acts as a reducing agent for the chemical species that is being reduced.

http://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Oxidizing_and_Reducing_Agents

In this case, sodium metal is being oxidized, which implies that it is acting as the reducing agent.