How do you balance #NaClO_3 -> NaCl + O_2#?

1 Answer
Feb 16, 2016

Answer:

#2NaClO_3(s)->2NaCl(s)+3O_2(g)#
or
#NaClO_3(aq)+Na_2SO_3(aq)->NaCl(aq)+O_2(g)+Na_2SO_4(aq)#

Explanation:

To balance the reaction, if it is happening at the solid state (decomposition):

#2NaClO_3(s)->2NaCl(s)+3O_2(g)#

However, to balance the reaction, if it is happening in aqueous medium:

#NaClO_3(aq)->NaCl(aq)+O_2(g)#

We should recognize that this is a reduction half equation since the oxidation number of oxygen is increasing from #(-2)# in #NaClO_3# to #(0)# in #O_2#.

The oxidation numbers for #Na# and #Cl# are not changing and they are #(+1)# and #(-1)# respectively.

To balance this reaction then:
#NaClO_3(aq)+2H^(+)(aq)+2e^(-)->NaCl(aq)+O_2(g)+H_2O(l)#

Therefore, in order for this reaction to occur, you will need a reducing agent which will reduce the oxidizing agent #NaClO_3# such as for example #Na_2SO_3#.

The oxidation half equation will then be:

#SO_3^(2-)(aq)+H_2O(l)->SO_4^(2-)(aq)+2H^(+)(aq)+2e^(-)#

The redox reaction will thus be:

Oxidation: #SO_3^(2-)(aq)+cancel(color(green)(H_2O(l)))->SO_4^(2-)(aq) +cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))#

Reduction: #NaClO_3(aq)+cancel(color(blue)(2H^(+)(aq)))+cancel(color(red)(2e^(-)))->NaCl(aq)+O_2(g)+cancel(color(green)(H_2O(l)))#

RedOx:
#NaClO_3(aq)+SO_3^(2-)(aq)->NaCl(aq)+O_2(g)+SO_4^(2-)(aq)#
or
#NaClO_3(aq)+Na_2SO_3(aq)->NaCl(aq)+O_2(g)+Na_2SO_4(aq)#

Here is a video that explains how to balance a redox reaction in acidic medium:
Balancing Redox Reactions | Acidic Medium.

And here is another one on how to balance a redox reaction in basic medium:
Balancing Redox Reactions | Basic Medium.