Halide Ions with Sulfuric Acid
Halide ions are reducing agents, and so are oxidised (lose e-)
Sulfuric acid is an oxidising agent, and so the S is reduced (gains e-)
Reaction of NaF with H2SO4
• No redox reaction; it is a displacement reaction.
• Overall equation:
2NaF (s) + H2SO4 (l) ==> Na2SO4 (s) + 2HF (g)
OR
NaF (s) + H2SO4 (l) ==> NaHSO4 (s) + HF (g)
• Ionic equation:
F- (s) + H+ (l) ==> HF (g)
Reaction of NaCl with H2SO4
• No redox reaction; it is a displacement reaction.
• Overall equation:
2NaCl (s) + H2SO4 (l) ==> Na2SO4 (s) + 2HCl (g)
OR
NaCl (s) + H2SO4 (l) ==> NaHSO4 (s) + HCl (g)
• Ionic equation:
Cl- (s) + H+ (l) ==> HCl (g)
Reaction of NaBr with H2SO4
• Redox reaction in which the oxidation state of bromine changes from
-1 in Br- to 0 in Br2 (oxidation, increase in oxidation state by 1). The
oxidation state of sulfur changes from +6 in H2SO4 to +4 in SO2
(reduction, decrease in oxidation state by 2).
• Overall equation:
2NaBr (s) + 2H2SO4 (l) ==> Na2SO4 (s) + Br2 (g) + SO2 (g) + 2H2O (l)
OR
2NaB (s) + 3H2SO4 (l) ==> 2NaHSO4 (s) + Br2 (g) + SO2 (g) + 2H2O (l)
Reaction of NaBr with H2SO4
• The overall reaction can be broken down into two equations (as found
in CGP book):
NaBr (s) + H2SO4 (l) ==> NaHSO4 (s) + HBr (g)
2HBr (g) + H2SO4 (l) ==> Br2 (g) + SO2 (g) + 2H2O (l)
Reaction of NaBr with H2SO4
• Half-equations:
2Br- ==> Br2 + 2e- (oxidation, 2 e- lost)
H2SO4 + 2H+ + 2e- ==> SO2 + 2H2O (reduction, 2e- gain)
• Overall net ionic redox equation:
2Br- + H2SO4 + 2H+ ==> Br2 + SO2 + 2H2O
Reaction of NaI with H2SO4
• Redox reaction in which the oxidation state of iodine changes from -1
in I- to 0 in I2 (oxidation, increase by 1 in oxidation state). The
oxidation state of sulfur changes from +6 in H2SO4 to -2 in H2S
(reduction, decrease in oxidation state by 8)
• Overall equation:
8NaI (s) + 5H2SO4 (l) ==> 4Na2SO4 (s) + 4I2 (g/s) + H2S (g) + 4H2O (l)
OR
8NaI (s) + 9H2SO4 (l) ==> 8NaHSO4 (s) + 4I2 (g/s) + H2S (g) + 4H2O (l)
Reaction of NaI with H2SO4
• The overall reaction can be broken down into three equations (as
found in CGP book):
NaI (s) + H2SO4 (l) ==> NaHSO4 (s) + HI (g)
2HI (g) + H2SO4 (l) ==> I2 (g) + SO2 (g) + 2H2O (l)
6HI (g) + SO2 (g) ==> H2S (g) + 3I2 (s) + 2H2O (l)
Reaction of NaI with H2SO4
• Half-equations:
2I- ==> I2 + 2e- (oxidation, 2 e- loss)
H2SO4 + 2H+ + 2e- ==> SO2 + 2H2O (reduction, 2e- gain)
H2SO4 + 8H+ + 8e– ==> H2S + 4H2O (reduction, 8 e- gained)
• Overall net ionic redox equation:
8I– + H2SO4 + 8H+ ==> 4I2 + H2S + 4H2O
Reaction of NaI with H2SO4
• Iodide is such a strong reducing agent that it can reduce the SO2 to
H2S or S.
• The half equation for S is:
2I- ==> I2 + 2e- (oxidation, 2 e- loss)
H2SO4 + 6H+ + 6e- ==> S + 4H2O (reduction, 6 e- gain)
• Overall net ionic redox equation for S:
6I- + H2SO4 + 6H+ ==> 3I2 + S + 4H2O