Oxidation number in acidic medium

In summary, the student calculates the oxidation number of chromium in a acidic medium by assuming the number of the less obvious atom and then solving for the oxidization state of the Cr atom.
  • #1
Bandarigoda
36
0
Hi everyone,
In my school I was given a problem. It's that Cr2O7-2 + H2S in a acidic medium. And my teacher told that Chromium becomes Cr+3 .

I want to know is there such a way to calculate the oxidation number in acidic medium. Or is it required the experience?
 
Chemistry news on Phys.org
  • #2
First, experience. When it fails, there are always Pourbaix diagrams.
 
  • #3
Bandarigoda said:
Hi everyone,
In my school I was given a problem. It's that Cr2O7-2 + H2S in a acidic medium. And my teacher told that Chromium becomes Cr+3 .?

I've recently been trying to teach myself this stuff. Somebody let me know if this is way off base but this is a method that seems to work for me.

A neutral molecule has an overall oxidization number of zero, and a charged radical or ion has an overall oxidization number equal to it's charge.

So Cr2O7(2-) has an overall oxidization number of -2. The way I've been dealing with ones like this is to just assume the number of the simple atom (oxygen) and thereby deduce the number of the less obvious one (transition metal in this case).

Let the oxidization number of O be -2 and that of Cr be "n". Then,

[tex] 2n + 7\times(-2) = -2[/tex]

Solving for "n" the oxidization state of the Cr atom in Cr2O7(2-) must be +6.

If this becomes Cr(3+) on the right hand side of the balanced equation then the oxidization number of Cr is decreasing from 6 to 3, therefore Cr is being reduced in this process.

Disclaimer. I'm still on my Chemistry "L" plates :). So anyone please feel free to critic this answer if I'm wrong.
 
  • #4
chemnoob said:
So anyone please feel free to critic this answer if I'm wrong.

You are for two reasons - one, it doesn't address the OP problem (which is how to predict ON of a product of the reaction), two, it is wrong as oxidation number is a property of an atom, not of the molecule. Cr2O72- has a _charge_ of -2, not oxidation number. Cr in Cr2O72- has oxidation number of +6 (which you calculated correctly).

Please note oxidation numbers are only an accounting device, handy when balancing redox reactions, but there is no measurable property of atoms that can be connected with ON. Which is why I prefer half reactions approach to balancing redox.
 
  • #5
Borek said:
it is wrong as oxidation number is a property of an atom, not of the molecule. Cr2O72- has a _charge_ of -2, not oxidation number.
Ok. So is it still correct to say that the sum of the oxidization numbers (of the individual atoms) in a molecule must equal the charge, -2 in the above example?

it doesn't address the OP problem (which is how to predict ON of a product of the reaction),
To be honest I don't even know what the reaction products would be for that reaction, apart from the [itex]Cr^{\,3+}[/itex] as that was given.

Does the OP need to complete a balanced equation something like as follows?

[tex]Cr_2O_7^{\,\,2-} + \, ? H_2S \, + \, ?? H_3O^{+} \rightarrow 2 Cr^{\,3+} + \, ?[/tex]
 
  • #6
I see. So it's up to my experience. I'll have to find more equations and solve them. Thanks for help guys.

Yes the overall charge is equal to -2
Cr is +6
So, 12 + (-14)=-2
 
  • #7
@chemnoob balancing is not a problem for me as long as I know the oxidation number.
 
  • #8
chemnoob said:
Ok. So is it still correct to say that the sum of the oxidization numbers (of the individual atoms) in a molecule must equal the charge, -2 in the above example?

Yes, that part is perfectly OK.

Trick is, sometimes assigning individual ONs is quite difficult. Try S2O32-.

To be honest I don't even know what the reaction products would be for that reaction, apart from the [itex]Cr^{\,3+}[/itex] as that was given.

Does the OP need to complete a balanced equation something like as follows?

[tex]Cr_2O_7^{\,\,2-} + \, ? H_2S \, + \, ?? H_3O^{+} \rightarrow 2 Cr^{\,3+} + \, ?[/tex]

Yes. In such cases hydrogen sulfide gets oxidized to elemental sulfur (when the oxidizer is mild) or to SO42- (when the oxidizer is strong). The important part is

Cr2O72- + H2S -> Cr3+ + SO42-

as all other molecules required for balancing (H2O, H+ and OH-) are always present in water solutions and they can be used freely whenever needed to balance charge, hydrogen and oxygen.
 
  • Like
Likes 1 person

1. What is meant by "oxidation number" in acidic medium?

In acidic medium, the oxidation number refers to the hypothetical charge that an atom would have if all of its bonds were completely ionic. It is a useful tool for keeping track of electron transfers and identifying the oxidizing and reducing agents in a chemical reaction.

2. How is the oxidation number of an element determined in acidic medium?

The oxidation number of an element in acidic medium is determined by following a set of rules. These rules include assigning a +1 charge to hydrogen atoms and a -2 charge to oxygen atoms, and adjusting the charges based on the overall charge of the compound and the known oxidation numbers of other elements.

3. Can the oxidation number of an element change in acidic medium?

Yes, the oxidation number of an element can change in acidic medium. This is because acidic conditions can facilitate the transfer of electrons between atoms, resulting in changes in oxidation numbers.

4. How does the oxidation number in acidic medium differ from that in basic medium?

The main difference between oxidation numbers in acidic and basic medium is the presence of hydrogen ions. In basic medium, hydrogen ions are often replaced by hydroxide ions, which can affect the overall charge and therefore the oxidation numbers of elements.

5. Why is it important to consider oxidation numbers in acidic medium?

Considering oxidation numbers in acidic medium is important because it helps us understand the electron transfer processes taking place in a chemical reaction. This information is crucial in predicting the products of a reaction and understanding its overall mechanism.

Similar threads

Replies
3
Views
2K
Replies
4
Views
1K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
9
Views
2K
Replies
16
Views
2K
Replies
4
Views
1K
Replies
6
Views
4K
Back
Top