Oxidation Number to balance a redox reaction

[agnibho]

Homework Statement

Balance Cl₂+NaOH -> NaCl + NaClO₃ + H₂O

 

[chemisttree]

Start by determining the oxidation numbers of all species.

 

[agnibho]

Cl⁰₂ + Na⁺¹O^-2H⁺¹ -> Na⁺¹Cl + Na⁺¹Cl^-1O₃^-2 + H₂⁺¹O^-2

To equalize the Oxidation Numbers of Cl and Na, 3 atoms of Cl and 6 atoms of Na are required. Thus,
3Cl₂ + 6NaOH -> 5NaCl + NaClO₃ + xH₂O
By inspection x = 3
∴ 3Cl₂ + 6NaOH -> 5NaCl + NaClO₃ + 3H₂O

I think this is correct, but if I have done something wrong please tell me so, as I'm new to this topic.
Thanks in advance.

 

[Borek]

No idea what you mean by "equalize the ON", specially in the context of sodium - ON for sodium doesn't change, and Na⁺ is only a spectator.

You have Cl(0) on the left and Cl(+1) and Cl(-1) on the right. This is disproportionation - teh same element becomes both oxidized and reduced at the same time. Can you tell how many atoms of Cl get oxidized if 1 atom gets reduced?

 

[DeeNos]

In order to balance a redox reaction, we must first determine the oxidation numbers of each element involved. In this reaction, chlorine (Cl) has an oxidation number of 0 in its elemental form. Once it combines with sodium hydroxide (NaOH), it becomes -1 in NaCl and +5 in NaClO3. On the other hand, sodium (Na) has an oxidation number of +1 in both NaOH and NaCl, and hydrogen (H) has an oxidation number of +1 in H2O.

To balance the reaction, we need to make sure that the total charge on both sides of the equation is equal. In this case, we can see that the total charge on the reactant side is 0 (2xCl2 + NaOH = 0) while the total charge on the product side is +1 (NaCl + NaClO3 + H2O = +1). To balance this, we can add 3 more NaOH on the reactant side, which will give us a total charge of -3 (2xCl2 + 3NaOH = -3). To balance the charge on the product side, we can add 3 more H2O, which will give us a total charge of 0 (+3NaCl + NaClO3 + 3H2O = 0).

Now, we can see that the equation is balanced in terms of charge, but we also need to make sure that the number of atoms on both sides is equal. To do this, we can add a coefficient of 2 in front of NaClO3 on the product side, which will give us a total of 2 NaClO3. This will also give us a total of 6 Cl atoms on the product side, which is balanced with the 6 Cl atoms on the reactant side.

Therefore, the balanced equation is 2Cl2 + 6NaOH -> 3NaCl + 2NaClO3 + 3H2O. This balanced equation satisfies both the charge balance and the atom balance, and represents the correct stoichiometry of the reaction.