[netpumber]

Hello out there.

I'm new member on that forum an i am undergraduate student. I'm studding now and i want some help with some redox reactions.

The exercise asks to find which of the following reactions are redox reactions.

1) BaCl_2(aq) + Na₂SO_4(aq)  BaSO_4(s) + 2NaCl_(aq)
2) 2KI_(aq) + Cl_2(g)  2KCl_(aq) + I_2(s)
3) I_2(aq) + Na₂S₂O_3(aq)  2NaI_(aq) + Na₂S₄O_6(aq)

This is how i approached that exercise.

For the first one :

BaCl_2(aq) + Na₂SO_4(aq)  BaSO_4(s) + 2NaCl_(aq)

step 1: write down the ion form where is aqueous

Ba + Cl₂ + Na + SO₄  BaSO_4(s) + 2Na + 2Cl

step 2 : Add oxidation numbers

Ba²⁺ + Cl₂^2- + Na²⁺ + SO₄^2-  BaSO_4(s)⁰ + Na²⁺ + Cl^2-

step 3 : Remove Na²⁺ and Cl^2-

Ba²⁺ +  SO₄^2-  BaSO_4(s)⁰

And conclude that there is no electron transfer so that reaction is not a redox reaction.

For the second one :

2KI_(aq) + Cl_2(g)  2KCl_(aq) + I_2(s)

Step 1: write down the ion form where is aqueous

K + I + Cl₂  K + Cl + I₂

step 2 : Add oxidation numbers

K^1- + I¹⁺ + Cl₂⁰  K^1- + Cl¹⁺ + I₂⁰

step 3 : Remove K^1-

2I¹⁺ + Cl₂⁰  2Cl¹⁺ + I₂⁰

And conclude that Iodine was reduced while the Cl was oxidized.


Is the way i approached the question right ?
I tried to work with that method in the third one too but stacked.

Any hint ?

[mjc123]

Q1: OK, except that BaCl₂ is Ba²⁺ + 2Cl^-, not Ba²⁺ + Cl₂^2-. The latter ion does not exist.

Q2: Why do you give K a negative charge and I a positive one?

Q3: For a start it's not balanced; you should have 2Na₂S₂O₃ on the LHS.
To simplify things, note that you have "S₂O₃" on one side and "S₄O₆" on the other, i.e. two units combine together. Let's call "S₂O₃" X, and write
I₂ + 2Na₂X  2NaI + Na₂X₂
Try to assign an overall oxidation number to X, and see whether you have a redox reaction. Later you can try assigning oxidation numbers to the individual atoms.

[sjb]

Hello out there.

I'm new member on that forum an i am undergraduate student. I'm studding now and i want some help with some redox reactions.

The exercise asks to find which of the following reactions are redox reactions.

1) BaCl_2(aq) + Na₂SO_4(aq)  BaSO_4(s) + 2NaCl_(aq)
2) 2KI_(aq) + Cl_2(g)  2KCl_(aq) + I_2(s)
3) I_2(aq) + Na₂S₂O_3(aq)  2NaI_(aq) + Na₂S₄O_6(aq)

This is how i approached that exercise.

For the first one :

BaCl_2(aq) + Na₂SO_4(aq)  BaSO_4(s) + 2NaCl_(aq)

step 1: write down the ion form where is aqueous

Ba + Cl₂ + Na + SO₄  BaSO_4(s) + 2Na + 2Cl

step 2 : Add oxidation numbers

Ba²⁺ + Cl₂^2- + Na²⁺ + SO₄^2-  BaSO_4(s)⁰ + Na²⁺ + Cl^2-

step 3 : Remove Na²⁺ and Cl^2-

Ba²⁺ +  SO₄^2-  BaSO_4(s)⁰

And conclude that there is no electron transfer so that reaction is not a redox reaction.

Seems sensible

For the second one :

2KI_(aq) + Cl_2(g)  2KCl_(aq) + I_2(s)

Step 1: write down the ion form where is aqueous

K + I + Cl₂  K + Cl + I₂

step 2 : Add oxidation numbers

K^1- + I¹⁺ + Cl₂⁰  K^1- + Cl¹⁺ + I₂⁰

Check your charges here...

[netpumber]

Ok here is again the Q2

2KI_(aq) + Cl_2(g)  2KCl_(aq) + I_2(s)

Step 1: write down the ion form where is aqueous

K + I + Cl₂  K + Cl + I₂

step 2 : Add oxidation numbers

K¹⁺ + I^1- + Cl₂⁰  K¹⁺ + Cl^1- + I₂⁰

step 3 : Remove the K¹⁺

2I^1- + Cl₂⁰  2Cl^1- + I₂⁰

And so the Iodine oxidized and Chlorine reduced.


As for the third one... it took me 1 hour to solve it (i believe that it's correct)

Step 1 : equalization

I_2(aq) + 2Na₂S₂O_3(aq)  2NaI_(aq) + Na₂S₄O_6(aq)

Step 2 : write down oxidation numbers

I_2(aq)⁰ + 2Na₂¹⁺S₂^x1O_3(aq)^-2  2Na¹⁺I^1-_(aq) + Na₂¹⁺S₄^x2O^-2_6(aq)

Step 3 : Calculate x1 and x2

2*(1+) + 2*x1 + 3*(-2)  x1 = 2
2*(1+) + 4*x2 + 6*(-2)  x2 = 2.5

So reaction becomes :

I_2(aq)⁰ + 2Na₂¹⁺S₂²O_3(aq)^-2  2Na¹⁺I^1-_(aq) + Na₂¹⁺S₄^2.5O^-2_6(aq)

And as a conclusion Iodine reduces and S oxidized.

Am i right ? Is there any other method to calculate them ? What you use to use ?

[mjc123]

Yes, that looks right.

[netpumber]

Hmm also here occurred a question how is it possible a half (0.5) electron to be transferred but maybe that has a more deeply answer like QM.

[mjc123]

If you look at the structures of S₂O₃^2- and S₄O₆^2-, you will see that the S atoms are not all equivalent, and will not necessarily all have the same oxidation number. 2.5 is the average oxidation number of the S atoms in S₂O₆^2-. There is no transfer of half-electrons; each S₂O₃^2- transfers 1 electron to iodine.

[netpumber]

I see i see...

Thank you very much.

[netpumber]

Hello again.

So now i want to balance that reaction in acidic solution.

Cu₂S + SO₄^2-  Cu²⁺ + SO₂ + H₂O

First step is to write down the oxidization numbers.

1. Cu₂S :
2. SO₄^2- : S has 6+ and O has 2-
3. Cu²⁺ : Cu has 2+
4. SO₂ : S has 4+ and O has 2-
5. H₂O : H has 1+ and O has 2-

What about the first one ? I don't know a rule about the oxidization number neither for Cu nor for S. How should i think of that ?

[Borek]

This is a sulfide, salt of a hydrogen sulfide (AKA hydrosulfuric aicd), H₂S. Sulfur has the same oxidation number in both compounds.

[netpumber]

So it's 2-. Thank you very much.