10.3 Flashcards
are halides reducing agents? what does that mean? and what reduction trend in present in group 7?
Halide IONS can act as reducing agents. The ions can lose an outer electron so they can become halogen molecules. The reducing power of group 7 halogens decrease as you go down the group.
Describe the reactions of solid sodium chloride with concentrated sulfuric acid in detail.
Concentrated sulfuric acid is added drop by drop to sodium chloride crystals:
H2SO4 (l) + NaCl (s) → HCl (g) + NaHSO4 (s)
(the very complicated-looking element is sodium hydrogen sulfate)
The HCl gas produced is seen as steamy fumes
This reaction is not a redox reaction as chlorine is too weak to reduce sulfur.
Describe the reaction of solid sodium bromine with concentrated sulfuric acid.
The reaction that takes place initially is:
H2SO4 (l) + NaBr (s) → HBr (g) + NaHSO4 (s)
The HBr gas produced is seen as steamy fumes. Solid sodium hydrogensulfate is also observed.
Then, bromide ions reduce sulfuric acid to sulfur dioxide gas:
2HBr (g) + H2SO4 (l) → Br2 (g) + SO2 (g) + 2H2O (l)
Brown fumes of bromine are observed. Colourless sulfur dioxide is formed.
This is a redox reaction because sulfur is reduced and bromine is oxidized.
Describe the reaction of solid sodium iodide with concentrated sulfuric acid.
The reaction that takes place initially is the same as that of sodium chloride and sodium bromide:
H2SO4 (l) + NaI (s) → HI (g) + NaHSO4 (s)
The HI gas produced is seen as steamy fumes. Solid sodium hydrogensulfate is also observed
Iodide ions are much stronger reducing agents than chloride and bromide ions
They are able to reduce the sulfur in H2SO4 (oxidation state = +6) to sulfur dioxide (oxidation state = +4) then to sulfur (oxidation state = 0) and finally to hydrogen sulfide (oxidation state = -2)
The equation for the formation of hydrogen sulfide is:
8HI (g) + H2SO4 (l) → 4I2 (s) + H2S (g) + 4H2O (l)
Iodine is seen as a black solid. Hydrogen sulfide has a strong smell of bad eggs. Sulfur is seen as a yellow solid as the sulfur passes through the oxidation state 0