Reaction of Halides with Sulfuric Acid Flashcards
why are halogens oxidising agents
- using OILRIG, oxidation is the loss of electrons
- halogens need 1 more electron to complete their shells, so they attract electrons
- this means that the element reacting to them loses an electron, thereby being oxidised
- as the halogens (usually) cause what they react with to be oxidised, they are oxidising agents
what is the trend in oxidising power down the halogens and why
- their oxidising power decreases down the group
- for the same reason their reactivity decreases down the group
- they have larger atoms and greater shielding effects
are halogen ions oxidising or reducing agents
they are reducing agents
why
- they have a full shell of electrons
- but as they are charged they are ready to give away the extra electron they have
- meaning the reacting element gains an electron
what is the trend in the halogen ions reducing power down the group
it increases down the group
why does their reducing power increase down the group
- because their reactivity decreases down the group
- aka their ability to attract electrons
- then their ability to let electrons go increases down the group
- as the halogen ions are releasing electrons, their effectiveness increases down the group
what is the general half equation for this happening
2X- = X2 + 2e-
what would be the equations for the partial ionisation of sulfuric acid
- H2SO4 = H+ + HSO4-
- HSO4- = H+ + SO4^2-
what needs to be noted about the nature of this partial ionisation
- its a reversible reaction
- so therefore has an equilibrium position (which lies to the right)
when concentrated, what can sulfuric acid act as other than an acid
an oxidising agent
when concentrated sulfuric acid is reduced, what are the three possible products
- sulfur dioxide
- sulfur
- hydrogen sulfide
how would sulfuric acid be reduced to from sulfur dioxide
- taking away the SO2 from H2SO4, the other product would have to be water
- to complete the water you need 2H+ ions to form 2H2O (as taking SO2 leaves you with 2 Os)
- but to balance it out you need 2e-
- giving H2SO4 + 2H+ + 2e = 2H2O + SO2
where would the 2H+ ions come from
the partial ionisation of other H2SO4 molecules
how would you figure out that this is reduction
- by working out the oxidation numbers of sulfur
- in H2SO4 it is +6
- but in SO2 it is +4
- the decrease in oxi number means it has gained 2 electrons
- meaning it has been reduced
when sulfuric acid is added to sodium chloride, bromide and iodide, what is a similar observation that can be made from all of their reactions
misty fumes
what observations other than misty fumes would be made from sodium bromide
- brown fumes
- colourless gas with choking smell
what observations are unique to sodium iodide
- purple fumes or black solid
- yellow solid
- colourless gas with rotten egg smell
what does sulfuric acid act as with sodium chloride and why
- just an acid
- because chloride ions have low reducing power
what does sulfuric acid act as with sodium bromide and why
- an acid and oxidising agent
- because bromine ions have decent reducing power
what does sulfuric acid act as with sodium iodide and why
- an acid and oxidising agent
- because iodide ions have the greatest reducing power
what is the equation for the reaction between sodium chloride and concentrated sulfuric acid
NaCl + H2SO4 = NaHSO4 + HCl
what is the equation for the formation of misty fumes of hydrogen bromide, using sodium bromide and sulfuric acid
NaBr + H2SO4 = NaHSO4 + HBr
what is the equation for the oxidation of the misty fums of hydrogen bromide using sulfuric acid
2HBr + H2SO4 = 2H2O + SO2 + Br2
what is the equation for the formation of misty fumes of hydrogen iodide using sodium iodide and sulfuric acid
NaI + H2SO4 = NaHSO4 + HI
what is the equation for the oxidation of the misty fumes of hydrogen on iodide using sulfuric acid, other than the method for bromine
6HI + H2SO4 = 4H2O + S + 3I2
