Group 7, the Halogens: Trends in Properties - Reducing Ability

Question 1

What are halides?

Answer 1

• 1- ions made by halogens

Question 2

What type of agents are halide ions in redox reactions?

Answer 2

• Reducing agent

- Oxidised itself

Question 3

How does the reducing ability of the halides change as you descend the group?

Answer 3

• Increases

Question 4

Why does the reducing ability of the halides increase as you descend the group?

Answer 4

• Ionic radius and shielding increases

- Attraction between positive nucleus and outer electrons becomes weaker

Question 5

What do the reactions of halides with sulphuric acid tell us about the halides?

Answer 5

• Tells us the relative reducing abilities of the halides

Question 6

What are the 2 stages that can occur during reaction of halides and sulphuric acid?

Answer 6

• Displacement

- Redox

Question 7

Which halides undergo displacement reactions with sulphuric acid?

Answer 7

• All halides from F to I

Question 8

Give the general word equation for the reaction between sodium halides with sulphuric acid

Answer 8

• Sodium halide + sulphuric acid → sodium hydrogen sulphate + hydrogen halide

Question 9

Give the general symbol equation for the reaction between sodium halides with sulphuric acid, including state symbols

Answer 9

• NaX (s)+ H2SO4 (l) → NaHSO4 (s) + HX (g)

Question 10

Which halide ions undergo a further redox reaction with sulphuric acid?

Answer 10

• Bromide ions

- Iodide ions

Question 11

What do bromide ions reduce sulphuric acid into?

Answer 11

• Bromide ions reduce H2SO4 to SO2 only

Question 12

Give:

• A half-equation for the oxidation of bromide ions to bromine
• A half-equation for the reduction of sulphuric acid to sulphur dioxide
• An overall equation for the reaction between bromide ions and sulphuric acid

Answer 12

• 2Br- → Br^2+ 2e-
• H2SO4 + 2H+ + 2e- → SO2 + 2H2O
• 2Br- + H2SO4 + 2H+ → 2Br- + SO2+ 2H2O

Question 13

What do iodide ions reduce sulphuric acid into?

Answer 13

• Iodide ions reduce H2SO4 to SO2, S and H2S

Question 14

Give:

• A half-equation for the oxidation of iodide ions to iodine
• A half-equation for the reduction of sulphuric acid to sulphur
• An overall equation for the reaction between iodide ions and sulphuric acid to produce sulphur

Answer 14

• (2I- → I2+ 2e-) x 3
• H2SO4 + 6H+ + 6e- → S+ 4H2O
• 6I- + H2SO4+ 6H+ → 3I2+ S + 4H2O

Question 15

Give:

• A half-equation for the oxidation of iodide ions to iodine
• A half-equation for the reduction of sulphuric acid to hydrogen sulphide
• An overall equation for the reaction between iodide ions and sulphuric acid to produce hydrogen sulphide

Answer 15

• (2I-→ I2+ 2e-) x 4
• H2SO4+ 8H+ + 8e- → H2S+ 4H2O
• 8I- + H2SO4+ 8H+ → 4I2 + H2S+ 4H2O

Question 16

Give the symbol equation, including state symbols, for:

• Displacement reaction of NaF with H2SO4
• Displacement reaction of NaCl with H2SO4

Answer 16

• NaF (s) + H2SO4 (l) → NaHSO4 (s) + HF (g)

- NaCl (s) + H2SO4 (l) → NaHSO4 (s) + HCl (g)

Question 17

What observations can be made for:

• Displacement reaction of all sodium halides with H2SO4

Answer 17

• Misty white fumes for all

Question 18

What products are formed when Br- reduces H2SO4? Do equations for all 2

Answer 18

• H2SO4 + 2Br- → SO42- + 2HBr

- 2Br- + H2SO4 + 2H+ → Br2 + SO2 + 2H2O

Question 19

What observations can be made during the displacement, oxidation and reduction reactions between NaBr and H2SO4?

Answer 19

• Displacement

• HBr
• Misty white fumes

•. Oxidation

• Br2
• Orange-brown vapour / orange-brown liquid

•. Reduction

• SO2
• Choking gas

Question 20

What observations can be made during the displacement, oxidation and reduction reactions between NaI and H2SO4?

Answer 20

• Displacement

• HI
• Misty white fumes

•. Oxidation

• I2
• Dark grey solid / purple vapour

•. Reduction

• SO2 - choking gas
• S - yellow solid
• H2S - bad egg smell

Question 21

What products are formed when I- reduces H2SO4? Do equations for all

Answer 21

• H2SO4 + 2I- → SO42- + 2HI
• H2SO4 + 2H+ + 2I- → SO2 + I2 + 2H2O
• H2SO4 + 6H+ + 6I- → S + 3I2 + 4H2O
• H2SO4 + 8H+ + 8I- → H2S + 4I2 + 4H2O