Question #044ba

Sodium metal, #"Na"#, will react with chlorine gas, #"Cl"_2#, to form sodium chloride, #"NaCl"#, according to the following balanced chemical equation

#2"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s))#

To check that this is indeed a redox reaction, assign oxidation numbers to the atoms that take part in the reaction

#2stackrel(color(blue)(0))("Na")_ ((s)) + stackrel(color(blue)(0))("Cl")_ (2(g)) -> 2stackrel(color(blue)(+1))("Na")stackrel(color(blue)(-1))("Cl")_ ((s))#

Notice that sodium metal goes from an oxidation state of #color(blue)(0)# on the reactants' side to an oxidation state of #color(blue)(+1)# on the products' side. This means that sodium metal is being oxidized, since its oxidation number is increasing

Similarly, chlorine goes from an oxidation state of #color(blue)(0)# on the reactants' side to an oxidation state of #color(blue)(-1)# on the products' side. This means that chlorine is being reduced, since its oxidation number is decreasing.

So, you know that sodium is being oxidized, which can only mean that chlorine is doing the oxidizing, i.e. chlorine acts as an oxidizing agent.

On the other hand, chlorine is being reduced, which can only mean that sodium is doing the reducing, i.e. sodium acts as a reducing agent.

So, to sum this up, sodium acts as a reducing agent because it is giving its valence electron to a chlorine atom, thus reducing chlorine to chloride anions, #"Cl"^(-)#.

Chlorine acts as an oxidizing agent because it is taking away the valence electron of a sodium atom, thus oxidating sodium metal to sodium cations, #"Na"^(+)#.