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Offline netpumber

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Help with some redox reactions
« on: September 08, 2015, 09:57:10 AM »
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) BaCl2(aq) + Na2SO4(aq)  :rarrow: BaSO4(s) + 2NaCl(aq)
2) 2KI(aq) + Cl2(g)  :rarrow: 2KCl(aq) + I2(s)
3) I2(aq) + Na2S2O3(aq)  :rarrow: 2NaI(aq) + Na2S4O6(aq)

This is how i approached that exercise.

For the first one :

BaCl2(aq) + Na2SO4(aq)  :rarrow: BaSO4(s) + 2NaCl(aq)

step 1: write down the ion form where is aqueous

Ba + Cl2 + Na + SO4  :rarrow: BaSO4(s) + 2Na + 2Cl

step 2 : Add oxidation numbers

Ba2+ + Cl22- + Na2+ + SO42-  :rarrow: BaSO4(s)0 + Na2+ + Cl2-

step 3 : Remove Na2+ and Cl2-

Ba2+ +  SO42-  :rarrow: BaSO4(s)0

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

For the second one :

2KI(aq) + Cl2(g)  :rarrow: 2KCl(aq) + I2(s)

Step 1: write down the ion form where is aqueous

K + I + Cl2  :rarrow: K + Cl + I2

step 2 : Add oxidation numbers

K1- + I1+ + Cl20  :rarrow: K1- + Cl1+ + I20

step 3 : Remove K1-

2I1+ + Cl20  :rarrow: 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 ?


Offline mjc123

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Re: Help with some redox reactions
« Reply #1 on: September 08, 2015, 10:47:46 AM »
Q1: OK, except that BaCl2 is Ba2+ + 2Cl-, not Ba2+ + Cl22-. 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 + 2Na2:rarrow: 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.

Offline sjb

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Re: Help with some redox reactions
« Reply #2 on: September 08, 2015, 10:53:26 AM »
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) BaCl2(aq) + Na2SO4(aq)  :rarrow: BaSO4(s) + 2NaCl(aq)
2) 2KI(aq) + Cl2(g)  :rarrow: 2KCl(aq) + I2(s)
3) I2(aq) + Na2S2O3(aq)  :rarrow: 2NaI(aq) + Na2S4O6(aq)

This is how i approached that exercise.

For the first one :

BaCl2(aq) + Na2SO4(aq)  :rarrow: BaSO4(s) + 2NaCl(aq)

step 1: write down the ion form where is aqueous

Ba + Cl2 + Na + SO4  :rarrow: BaSO4(s) + 2Na + 2Cl

step 2 : Add oxidation numbers

Ba2+ + Cl22- + Na2+ + SO42-  :rarrow: BaSO4(s)0 + Na2+ + Cl2-

step 3 : Remove Na2+ and Cl2-

Ba2+ +  SO42-  :rarrow: BaSO4(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) + Cl2(g)  :rarrow: 2KCl(aq) + I2(s)

Step 1: write down the ion form where is aqueous

K + I + Cl2  :rarrow: K + Cl + I2

step 2 : Add oxidation numbers

K1- + I1+ + Cl20  :rarrow: K1- + Cl1+ + I20

Check your charges here...

Offline netpumber

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Re: Help with some redox reactions
« Reply #3 on: September 08, 2015, 03:23:28 PM »
Ok here is again the Q2

2KI(aq) + Cl2(g)  :rarrow: 2KCl(aq) + I2(s)

Step 1: write down the ion form where is aqueous

K + I + Cl2  :rarrow: K + Cl + I2

step 2 : Add oxidation numbers

K1+ + I1- + Cl20  :rarrow: K1+ + Cl1- + I20

step 3 : Remove the K1+

2I1- + Cl20  :rarrow: 2Cl1- + 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

I2(aq) + 2Na2S2O3(aq)  :rarrow: 2NaI(aq) + Na2S4O6(aq)

Step 2 : write down oxidation numbers

I2(aq)0 + 2Na21+S2x1O3(aq)-2  :rarrow: 2Na1+I1-(aq) + Na21+S4x2O-26(aq)

Step 3 : Calculate x1 and x2

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

So reaction becomes :

I2(aq)0 + 2Na21+S22O3(aq)-2  :rarrow: 2Na1+I1-(aq) + Na21+S42.5O-26(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 ?

Offline mjc123

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Re: Help with some redox reactions
« Reply #4 on: September 09, 2015, 05:06:45 AM »
Yes, that looks right.

Offline netpumber

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Re: Help with some redox reactions
« Reply #5 on: September 09, 2015, 05:14:15 AM »
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.

Offline mjc123

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Re: Help with some redox reactions
« Reply #6 on: September 09, 2015, 05:20:18 AM »
If you look at the structures of S2O32- and S4O62-, 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 S2O62-. There is no transfer of half-electrons; each S2O32- transfers 1 electron to iodine.

Offline netpumber

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Re: Help with some redox reactions
« Reply #7 on: September 09, 2015, 05:26:56 AM »
I see i see...

Thank you very much.

Offline netpumber

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Re: Help with some redox reactions
« Reply #8 on: September 10, 2015, 01:22:54 PM »
Hello again.

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

Cu2S + SO42-  :rarrow: Cu2+ + SO2 + H2O

First step is to write down the oxidization numbers.

1. Cu2S :
2. SO42- : 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 ?

Offline Borek

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Re: Help with some redox reactions
« Reply #9 on: September 10, 2015, 01:57:31 PM »
This is a sulfide, salt of a hydrogen sulfide (AKA hydrosulfuric aicd), H2S. Sulfur has the same oxidation number in both compounds.
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Offline netpumber

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Re: Help with some redox reactions
« Reply #10 on: September 10, 2015, 02:35:54 PM »
So it's 2-. Thank you very much.

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