Displacement reactions can be used to place the halogens in order of reactivity. A few drops of each halogen solution are added to solutions of potassium ha...
Question 1 Report
Displacement reactions can be used to place the halogens in order of reactivity. A few drops of each halogen solution are added to solutions of potassium halides. The results are recorded in Table 11.1, where 'no reaction' means no colour change was seen.
| chlorine water | bromine water | iodine solution | |
|---|---|---|---|
| potassium chloride solution | no reaction | no reaction | no reaction |
| potassium bromide solution | … | no reaction | no reaction |
| potassium iodide solution | … | … | no reaction |
(a) Complete the three shaded boxes, stating in each case whether a reaction occurs and, if so, the colour change seen. [3]
(b) Use the results to place chlorine, bromine and iodine in order of decreasing reactivity. [1]
(c) Explain why a more reactive halogen is able to displace a less reactive halogen from a solution of its salt. Refer to electron gain in your answer. [3]
(d) Write the ionic equation, with state symbols, for the reaction between bromine and potassium iodide solution. [2]
(e) In the reaction in (d), state which species is reduced and explain your choice in terms of electrons. [2]
(f) Predict whether astatine (below iodine in Group VII) would displace iodine from potassium iodide solution, and justify your prediction. [2]
Answer Details
Reactivity in Group VII decreases down the group, so chlorine is the most reactive of these three and iodine the least. A halogen can only displace another halogen that is below it (less reactive) in the group.
(a) The completed results, with the colour change seen when a halogen is displaced:
| chlorine water | bromine water | iodine solution | |
|---|---|---|---|
| potassium chloride | no reaction | no reaction | no reaction |
| potassium bromide | reaction: colourless to orange/brown | no reaction | no reaction |
| potassium iodide | reaction: colourless to brown | reaction: orange to brown | no reaction |
Marks: chlorine with potassium bromide, colourless to orange/brown (bromine set free) [1]; chlorine with potassium iodide, colourless to brown (iodine set free) [1]; bromine with potassium iodide, orange to brown (iodine set free) [1].
(b) Order of decreasing reactivity: chlorine, then bromine, then iodine [1].
(c) A halogen reacts by gaining an electron to become a halide ion. A more reactive halogen gains electrons more readily [1], so it pulls the electron away from the ion of the less reactive halogen [1]; the more reactive halogen becomes a halide ion while the less reactive halogen is set free as neutral molecules, that is, it is displaced [1].
(d) Ionic equation with state symbols [species 1; balancing and states 1]:
\[ \text{Br}_2(aq) + 2\text{I}^{-}(aq) \rightarrow 2\text{Br}^{-}(aq) + \text{I}_2(aq) \](e) Bromine, \(\text{Br}_2\), is reduced [1], because each bromine atom gains an electron to become \(\text{Br}^{-}\), and gain of electrons is reduction [1]. (The iodide ions are oxidised, losing electrons to form \(\text{I}_2\).)
(f) Astatine lies below iodine, so it is the least reactive halogen here and gains electrons less readily than iodine. It would therefore not displace iodine [1] from potassium iodide solution [1].
Download The App On Google Playstore
Everything you need to excel in your exams
