The ionic equation for reaction btwn chlorine and potassium iodide
Answers
Answer:
short answer: 2KI(aq) + Cl2(g) --> 2KCl(aq) + I2(aq) is the ionic equation. However if you want the net ionic equation without spectator ions. Is potassium iodide soluble in water?
pls mark me brainliest
Answer:
What is a balanced equation for the reaction between chlorine and potassium iodide?
Empower people to live healthier lives by sharing what you know.
This is a very interesting reaction, not only because it illustrates displacement of an anion (the overall reaction) but the “hidden” mechanism or reaction path is an example of redox and identifies a temporary intermediate compound.
The overall reaction is sufficient in most cases, but for those interested in reactions and how they come about, it hides a much more fascinating story...
There are “mega” more molecules of water than ions of I^(-) in the solution, therefore the Cl2 reacts with water in the first instance in a disproportionation reaction - simply because it is going to collide with water much more often than any I^(-) ions. This forms hypochlorous acid which actually oxidises the iodide ion to iodine.
By oxidising the iodide ion, the hypochlorous acid is used up, therefore it is essentially an intermediary in the reaction mechanism which is set out below…
The reaction is performed in aqueous solution. Adding chlorine to water gives…
Cl2 + H2O → HCl + HOCl ← hydrochloric and hypochlorous acids respectively. Call this Eqn (1)
(This is a disproportionation reaction since we start with Cl in oxidation state 0 (element) and produce Cl oxidation state -1 (Cl^(-) ion) and +1 (Hypochlorous ion OCl^(-))
OCl^(-) is a strong oxidant and oxidises the Iodide to iodine…
2KI + HOCl → I2 + KOH + KCl Call this Eqn (2)
Add Equations (1) and (2) so as to cancel out the intermediary, HOCl…
Cl2 + H2O + 2KI + HOCl → KCl + H2O + HOCl + I2 + KCl
Cl2 + H2O + 2KI → 2KCl +H2O + I2
H2O appears on both sides, therefore eliminate it…
Cl2 + 2KI → 2KCl + I2 ← Balanced stoichiometric equation.
Cl2 + 2I^(-) → I2 + 2Cl^(-) ← Balanced ionic equation.