In the presence of an alkali, Cl2 reacts to form Cl- and ClO3- , as follows:
3 Cl2(g) + 6 OH−(aq) → 5 Cl−(aq) + ClO3−(aq) + 3 H2O(l)
a. Calculate the oxidation number for chlorine in each of the species.
b. Write a half equation for the formation of Cl- from Cl2
c. Write a half equation for the formation of ClO3- from Cl2
d. Name and explain what type of reaction this is.
e. Identify which species is being reduced
f. identify which is the oxidising agent
a. To calculate the oxidation number for chlorine in each of the species, we need to follow the rules for assigning oxidation numbers. In this reaction, chlorine is present as Cl2 (g), Cl- (aq), and ClO3- (aq).
- In Cl2 (g), since it is a diatomic molecule, each Cl atom has an oxidation number of 0.
- In Cl- (aq), since it is an ion with a charge of -1, the oxidation number for chlorine is -1.
- In ClO3- (aq), since oxygen usually has an oxidation number of -2 in compounds (unless it's in peroxides or superoxides), and since there are three oxygen atoms present, the total oxidation number contributed by the oxygen atoms is -6. The sum of oxidation numbers in ClO3- must equal its overall charge, which is -1. Therefore, the oxidation number for chlorine is +5.
b. The half equation for the formation of Cl- from Cl2 is:
Cl2(g) + 2e- → 2Cl-(aq)
c. The half equation for the formation of ClO3- from Cl2 is:
3Cl2(g) + 6H2O(l) → 5ClO3-(aq) + 6H+(aq) + 6e-
d. This reaction is a redox reaction. Specifically, it is a combination of a reduction reaction (formation of Cl-) and an oxidation reaction (formation of ClO3-).
e. In this reaction, Cl2 is being reduced because it gains electrons to form Cl-.
f. The oxidizing agent is OH- as it gets oxidized and loses electrons to form water.
a. To calculate the oxidation number for chlorine in each of the species, we need to follow the rules for assigning oxidation numbers.
In Cl2, since it is a diatomic molecule, each chlorine atom has an oxidation number of 0.
In Cl-, since it is an anion, the oxidation number of chlorine is -1.
In ClO3-, the oxidation number of oxygen is -2, and since there are three oxygen atoms, the total oxidation number contributed by oxygen is -6. So, we can calculate the oxidation number of chlorine by setting the total oxidation number equal to the charge of the ion. Therefore, chlorine must have an oxidation number of +5 in ClO3-.
b. To write a half equation for the formation of Cl- from Cl2, we need to focus on the chlorine atoms. The oxidation state of Cl2 is 0 and the oxidation state of Cl- is -1. Therefore, the half equation for the reduction of Cl2 to Cl- would be:
Cl2 + 2 e- → 2 Cl-
c. To write a half equation for the formation of ClO3- from Cl2, we need to focus on the chlorine atoms. The oxidation state of Cl2 is 0, and the oxidation state of ClO3- is +5. Therefore, the half equation for the oxidation of Cl2 to ClO3- would be:
Cl2 → ClO3- + 6 e-
d. This reaction is a redox reaction, as there is a simultaneous oxidation and reduction occurring. In this reaction, chlorine is both being oxidized (from oxidation state 0 in Cl2 to +5 in ClO3-) and reduced (from oxidation state 0 in Cl2 to -1 in Cl-).
e. The species being reduced in this reaction is Cl2, as it accepts electrons and forms Cl-.
f. The oxidizing agent is the species that gets reduced. In this reaction, Cl2 is the oxidizing agent as it accepts electrons and becomes Cl-.