Group 7 Halogens and Chlorine Chemistry

Flashcards covering Group 7 halogen reactions, redox trends of halogens and halides, and their specific chemical reactions with water, sodium hydroxide, and sulfuric acid.

Chlorine and cold water reaction

$$Cl_2 + H_2O \rightarrow HCl + HClO$$

Chlorine and sodium hydroxide reaction

$$Cl_2 + 2NaOH \rightarrow NaCl + NaClO + H_2O$$

Chlorine and water reaction in sunlight

$$2Cl_2 + 2H_2O \rightarrow 4HCl + O_2$$

$$ClO^-$$

The active ingredient responsible for killing harmful microbes and bacteria in swimming pools or bleach.

Safety of Chlorine in water

Even though $Cl_2$ is toxic, it is used in low enough concentrations that the benefits of clean water outweigh the risks.

Oxidising ability trend (Halogens)

Decreases down the group ($F_2$, $Cl_2$, $Br_2$, $I_2$) because larger molecules have a weaker attraction between the positive nucleus and outer electrons, making it harder to accept an electron.

Oxidising power demonstration: Chlorine and Potassium Bromide

Reaction: $Cl_2 + 2KBr \rightarrow 2KCl + Br_2$; Observation: formation of an orange solution.

Ionic equation: Chlorine and Bromide ions

$$Cl_2(aq) + 2Br^-(aq) \rightarrow 2Cl^-(aq) + Br_2(aq)$$

Oxidising power demonstration: Chlorine and Potassium Iodide

Reaction: $Cl_2 + 2KI \rightarrow 2KCl + I_2$; Observation: formation of a brown solution.

Oxidising power demonstration: Bromine and Potassium Iodide

Reaction: $Br_2 + 2KI \rightarrow 2KBr + I_2$; Observation: formation of a brown solution.

Reducing ability trend (Halide ions)

Increases down the group ($F^-$, $Cl^-$, $Br^-$, $I^-$) because the ion gets bigger, reducing the attraction between the nucleus and outer electrons, making it easier to donate an electron.

Acid-Base reaction of sodium halides with sulfuric acid

$NaX + H_2SO_4 \rightarrow NaHSO_4 + HX$; characterized by the transfer of a proton with no change in oxidation state.

Role of halide ion in Acid-Base step

A base (proton acceptor).

Observation of HX production

Misty white fumes.

Reactivity of $NaF$ and $NaCl$ with $H_2SO_4$

These stop at the acid-base reaction step because fluoride and chloride ions are not strong enough reducing agents to proceed to redox.

Redox step: Bromide and Sulfuric acid

$$2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O$$

Overall reaction: Sodium Bromide and Sulfuric acid

$$3H_2SO_4 + 2NaBr \rightarrow 2NaHSO_4 + SO_2 + Br_2 + 2H_2O$$

Oxidation half equation for Bromide

$$2Br^- \rightarrow Br_2 + 2e^-$$

Reduction half equation for Sulfuric acid with Bromide

$$H_2SO_4 + 2H^+ + 2e^- \rightarrow SO_2 + 2H_2O$$

Roles in Bromide and Sulfuric acid redox reaction

Halide ion acts as the reducing agent; Sulfuric acid acts as the oxidising agent.

Observation for Bromide redox step

Orange fumes or solution due to the formation of $Br_2$.

Iodide

Strongest reducing agent and reduces sulfuric acid to three different sulfur-based substances.

Reaction to produce colorless gas of sulfur dioxide (SO₂)

Oxidation half equation: 2I ➡ I2 + 2e-

Reduction half equation: H2SO4 + 2H+ + 2e- ➡ SO2 + 2H2

Overall equation: H2SO4 + 2H+ + 2I- ➡ I2 + SO2 + 2H2O.

Foul smelling/rotten egg smell of hydrogen sulfide gas (H₂S)

Oxidation half equation: 2I ➡ I2 + 2e-

Reduction half equation: H2SO4 + 6H+ + 6e- ➡ H2S + 4H2O

Overall equation: H2SO4 + 6H+ + 6I- ➡ 3I2 + H2S + 4H2O

Yellow solid of sulfur (S)

Oxidation half equation: 2I- ➡ I2 + 2e-

Reduction half equation: H2SO4 + 8H+ + 8e- ➡ S + 4H2O

Overall equation: 2SO4 + 8H+ + 8I- ➡ 4I2 + S + 4H2O

Role of halide ion

Reducing agent.

Role of Sulfuric acid

Oxidizing agent.

Observation of I₂ production

I₂ is a black solid or brown solution or purple fumes.