Unit 2 - Redox question
The question I have is
Solid sodium iodide undergoes a redox reaction with concentrated sulfuric acid.
Give the formula for each of the following in this reaction.
Formula of the solid reduction product -
Formula of the oxidation product -
That is question 3e on http://filestore.aqa.org.uk/subjects/AQA-CHEM2-W-QP-JUN11.PDF
I don't understand how to get the answer
Sulfuric acid is H2SO4. Sodium Iodide is NaI.
How am I supposed to react these two together? They could make anything?
What i wrote down was HI for the first one
and NaHSO4 for the second
Any help will be extremely appreciated!
Thank you
Solid sodium iodide undergoes a redox reaction with concentrated sulfuric acid.
Give the formula for each of the following in this reaction.
Formula of the solid reduction product -
Formula of the oxidation product -
That is question 3e on http://filestore.aqa.org.uk/subjects/AQA-CHEM2-W-QP-JUN11.PDF
I don't understand how to get the answer
Sulfuric acid is H2SO4. Sodium Iodide is NaI.
How am I supposed to react these two together? They could make anything?
What i wrote down was HI for the first one
and NaHSO4 for the second
Any help will be extremely appreciated!
Thank you
Original post by user1-4
The question I have is
Solid sodium iodide undergoes a redox reaction with concentrated sulfuric acid.
Give the formula for each of the following in this reaction.
Formula of the solid reduction product -
Formula of the oxidation product -
That is question 3e on http://filestore.aqa.org.uk/subjects/AQA-CHEM2-W-QP-JUN11.PDF
I don't understand how to get the answer
Sulfuric acid is H2SO4. Sodium Iodide is NaI.
How am I supposed to react these two together? They could make anything?
What i wrote down was HI for the first one
and NaHSO4 for the second
Any help will be extremely appreciated!
Thank you
Solid sodium iodide undergoes a redox reaction with concentrated sulfuric acid.
Give the formula for each of the following in this reaction.
Formula of the solid reduction product -
Formula of the oxidation product -
That is question 3e on http://filestore.aqa.org.uk/subjects/AQA-CHEM2-W-QP-JUN11.PDF
I don't understand how to get the answer
Sulfuric acid is H2SO4. Sodium Iodide is NaI.
How am I supposed to react these two together? They could make anything?
What i wrote down was HI for the first one
and NaHSO4 for the second
Any help will be extremely appreciated!
Thank you
Neither of these are reduced or oxidised.
Original post by user1-4
The question I have is
Solid sodium iodide undergoes a redox reaction with concentrated sulfuric acid.
Give the formula for each of the following in this reaction.
Formula of the solid reduction product -
Formula of the oxidation product -
That is question 3e on http://filestore.aqa.org.uk/subjects/AQA-CHEM2-W-QP-JUN11.PDF
I don't understand how to get the answer
Sulfuric acid is H2SO4. Sodium Iodide is NaI.
How am I supposed to react these two together? They could make anything?
What i wrote down was HI for the first one
and NaHSO4 for the second
Any help will be extremely appreciated!
Thank you
Solid sodium iodide undergoes a redox reaction with concentrated sulfuric acid.
Give the formula for each of the following in this reaction.
Formula of the solid reduction product -
Formula of the oxidation product -
That is question 3e on http://filestore.aqa.org.uk/subjects/AQA-CHEM2-W-QP-JUN11.PDF
I don't understand how to get the answer
Sulfuric acid is H2SO4. Sodium Iodide is NaI.
How am I supposed to react these two together? They could make anything?
What i wrote down was HI for the first one
and NaHSO4 for the second
Any help will be extremely appreciated!
Thank you
The reaction is: NaI + H2SO4 ---> I2 + Na2SO4 + H2S + H2O
(it's not balanced)
(edited 11 years ago)
Original post by V95
The reaction is: NaI + H2SO4 ---> I2 + Na2SO4 + H2S + H2O
(it's not balanced)
(it's not balanced)
Solid sulphur is the product in this reaction. H2S is a gas.
As for the oxidation product, think about which species has lost electrons, or which species' oxidation state has gone up
Another way to think about it is that its been oxidised to form the product, therefore its a reducing agent as a reactant. Using your knowledge you should have gained from (big hint here) group 7 metals, and how they react with concentrated sulphurice acid. (Only 2 halide ions are reducing agents in this instant as they reduce H2SO4)
(edited 11 years ago)
I'm sorry, but i still don't understand how you work out the products in the equation
NaI + H2SO4 ---> I2 + Na2SO4 + H2S + H2O
Or if it is meant to be S instead of H2S
Once I have the equation I think I can work out what is the reducing agent and what is the oxidising agent, however I can't get to the equation in the first place.
How would I attain this equation
NaI + H2SO4 ---> I2 + Na2SO4 + H2S + H2O
Or if it is meant to be S instead of H2S
Once I have the equation I think I can work out what is the reducing agent and what is the oxidising agent, however I can't get to the equation in the first place.
How would I attain this equation
Original post by user1-4
I'm sorry, but i still don't understand how you work out the products in the equation
NaI + H2SO4 ---> I2 + Na2SO4 + H2S + H2O
Or if it is meant to be S instead of H2S
Once I have the equation I think I can work out what is the reducing agent and what is the oxidising agent, however I can't get to the equation in the first place.
How would I attain this equation
NaI + H2SO4 ---> I2 + Na2SO4 + H2S + H2O
Or if it is meant to be S instead of H2S
Once I have the equation I think I can work out what is the reducing agent and what is the oxidising agent, however I can't get to the equation in the first place.
How would I attain this equation
In the question they told you a solid was formed. I- ions reduce sulfur(VI) into 3 possible products; SO2, S and H2S. Out of the possible products, which is solid? Therefore that's what the sulfur is reduced to.
2Cl- -> Cl2 + 2e-
H2SO4 + 6H+ + 6e- -> S + 4H2O
6Cl- + H2SO4 + 6H+ -> 3Cl2 + S +4H2O
That's your redox reaction producing solid sulfur
(edited 11 years ago)
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