Reactions of halides with sulfuric acid

Question 1

the oxidising power of HALOGENS………….down the group

Answer 1

decreases

Question 2

the reducing power of HALIDE IONS…………………..down the group

Answer 2

increases

Question 3

the half equation showing the reducing action of halide ions is

Answer 3

2X- = X2 + 2e-

Question 4

when sulfuric acid is concentrated, it contains……………….ions (number of ions)

Answer 4

very few

Question 5

the two-stage equations showing the partial ionisation of sulfuric acid is

Answer 5

H2SO4 ⇌ HSO4- + H+

HSO4- ⇌ SO4 2- + H+

Question 6

sulfuric acid, especially when concentrated, can act as a …………………….agent

Answer 6

oxidising agent, so is reduced

Question 7

when sulfuric acid acts as an oxidising agent, it is reduced, but the extent of its reduction and the products formed depend on

Answer 7

the species being oxidised

Question 8

the three possible products of reduction of sulfuric acid are:

Answer 8

• sulfur dioxide
• sulfur
• hydrogen sulfide

Question 9

write the half equation for the reduction of sulfuric acid when sulfur dioxide is formed:
show the oxidation of sulfur in the equation

Answer 9

H2SO4 + 2H+ + 2e- = 2H2O + SO2

+6 +4
difference: 2

Question 10

write the half equation for the reduction of sulfuric acid when sulfur is formed:
show the oxidation of sulfur in the equation

Answer 10

H2SO4 + 6H+ + 6e- = 4H2O + S

+6 0
difference: 6

Question 11

write the half equation for the reduction of sulfuric acid when hydrogen sulfide is formed:
show the oxidation of sulfur in the equation

Answer 11

H2SO4 + 8H+ = 8e- = 4H2O + H2S

+6 -2
difference: 8

Question 12

• the trend for the reducing power of the halide ions in a SOLID ionic compound down the group is:
• this can can simply be explained using:
• however, we can also explain the trend more convincingly by looking at the change in……………….. of the halide ionic SOLID use in the experiment

Answer 12

• trend is an increase in reducing power (meaning the ion gives up its electron more ‘easiy’)
• this can be simply explained using the fall in the amount of energy required down the group to remove an electron as the ions get bigger, so the outer electron is further away and so easier to remove
• however, the trend is more due to the fall in the CHANGE OF LATTICE ENTHALPY which means the halide ion is easier to separate from the solid ionic compound as you progress down the group. this fall from each each member to the next is greater than the fall in the energy required to remove an electron, so is therefore the greater contributing factor in explaining the trend in depth

Question 13

• the trend for the reducing power of the halide ions IN SOLUTION down the group is:
• this is because

Answer 13

• increase in reducing power
• the fluorine is very small, so are good oxdising agents and easily form their negative ions in water. this makes it difficult to reverse the process and turn fluoride ions into fluorine molecules
• however, for iodine, forming its negative ion in water is more difficult, so it is easier to form iodine molecules by reversing the process, making it a stronger reducing agent (it is easier to oxidise)