Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: netpumber on September 08, 2015, 09:57:10 AM
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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 ?
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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 + 2Na2X :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.
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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...
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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 ?
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Yes, that looks right.
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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.
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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.
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I see i see...
Thank you very much.
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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 ?
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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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So it's 2-. Thank you very much.