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
c.
12OH^- + Cl2 ==> 2ClO3^- + 10e + 6H2O
(Note: This is twice what you will need when you balance the entire equation so you just divide all of the coefficients by 2. That will leave just Cl (and not Cl2 but when you add it to the other half cell--likewise divided by 2--the Cl from the oxidized half cell and the Cl from the reduced half cell you get the Cl2 you're familiar with.)
Here is a link that shows how to do this as well as most any other item regarding redox equations.
http://www.chemteam.info/Redox/Redox.html
e. Reduction is the gain of electrons.Which half cell gained electrons?
f. The oxidizing agent is the substance being reduced.
d. What type reaction? It's a redox equation; however, when the same substance is both oxidized and reduced it is still a redox reaction but it is given a special name of disproportionation.
By the way, when I have an equation like this, I just separate them into to the two half cells, balanced them with Cl2 ==> 2Cl^- and
Cl2 ==> ClO3^-, then add the two half cells and divide all of the coefficients by 2.
There is no point in me redoing those parts you understand (and I assume you know how to do parts of it). What is it you're having trouble with?
c, d, e and f are where I am struggling.
Thank you, thank you thank you
a. To calculate the oxidation number for chlorine in each of the species, we need to follow some rules:
- The oxidation number of chlorine in its elemental form (Cl2) is 0.
- In a compound, the sum of the oxidation numbers is zero.
- Oxygen is typically assigned an oxidation number of -2 in compounds, unless it is bonded to fluorine or in a peroxide (O2^2-).
Using these rules, we can determine the oxidation numbers:
In Cl2(g): Each chlorine atom has an oxidation number of 0.
In Cl-: Each chlorine atom has an oxidation number of -1 since it has gained one electron.
In ClO3-: The oxidation number of oxygen is -2; therefore, 3 oxygen atoms contribute -6. Since the overall charge of the ion is -1, the oxidation number of chlorine must be +5 to balance it out.
b. The formation of Cl- from Cl2 can be represented by the half equation:
Cl2(g) + 2 e- → 2 Cl-(aq)
This equation shows that two electrons are gained by each molecule of chlorine gas to form two chloride ions.
c. The formation of ClO3- from Cl2 can be represented by the half equation:
3 Cl2(g) + 6 OH-(aq) → 5 Cl-(aq) + ClO3-(aq) + 3 H2O(l)
In this equation, chloride ions are formed along with chlorate ions, ClO3-, when chlorine reacts with hydroxide ions.
d. The reaction described is a redox (reduction-oxidation) reaction. Redox reactions involve the transfer of electrons between reactants. In this case, chlorine atoms in Cl2 are being both reduced (gaining electrons to form chloride ions) and oxidized (losing electrons to form chlorate ions).
e. The species being reduced in this reaction is Cl2. Each chlorine atom in Cl2 gains two electrons to form two chloride ions (Cl-).
f. The oxidizing agent is the species that causes the oxidation of another compound. In this reaction, the oxidizing agent is the hydroxide ion (OH-) because it causes the oxidation of chlorine, resulting in the formation of chlorate ions (ClO3-). The hydroxide ions lose electrons, which are gained by the chlorine atoms.