10.3 Reactions of halide ions Flashcards
How are halide ions reducing agents?
. The halide ions lose/give away electrons and become halogen molecules
They themselves become oxidised, but they reduce other halogens
Describe the reducing ability down the group
.It increases down the group
. The size of ions increases down the group as there are more shells
. The outer shell gets further from the nucleus, so atomic radii increases
. So the attraction between the outer electron and nucleus is less
. So it can lose electrons more easily
How do sodium halides react with concentrated sulphuric acid
Solid sodium halides react with concentrated sulphuric acid.
The products are different and reflect the reducing power of the halides
What is seen in the reaction between sodium chloride and concentrated sulphuric acid
. Drops of sulphuric acid are added to sodium chloride.
Steamy fumes of hydrogen chloride are seen .
Solid product is sodium hydrogensulfate.
Word and symbol equation for sodium chloride and sulphuric acid
sodium chloride + sulfuric acid —>
sodium hydrogensulfate +
Hydrogen chloride
+1 -1 +1,+6,-2
NaCl (s) + H2SO4 (l) –>
+1,+1,+6,-2 +1,-1
NaHSO4(s) + HCL(g)
Why is the reaction between sodium chloride and sulphuric acid not a redox reaction
What type of reaction is it
. No oxidation state has changed.
The chloride ion is too weak a reducing agent to reduce the sulfur (oxidation state +6) in sulphuric acid.
Its an acid base reaction
How does sodium fluoride react with sulfuric acid
It produces hydrogen fluoride which is an extremely dangerous gas that can etch glass
The fluoride ion is even weaker a reducing agent than chloride ions
What forms from sodium bromide and concentrated sulphuric acid
You will see steamy fumes of hydrogen bromide, and brown fumes of bromine.
Colourless sulphur dioxide is also formed
First reaction equation for sodium bromide and sulphuric acid
(two reactions occur)
NaBr(s) + H2S04(l) —->
NaHS04(s) +HBr(g)
Sodium hydrogensulfate and hydrogen bromide are formed (in a similar acid base reaction to sodium chloride)
Why are there two reactions for sodium bromide and sulphuric acid
. Bromide ions are strong enough reducing agents to reduce the sulphuric acid to sulphur dioxide.
This means the oxidation state of the sulfur is reduced from +6 to +4 and that of the bromine increases from -1 to 0
What is the redox reaction
(second reaction)for Sodium bromide and sulphuric acid
Exothermic reaction
-1 +6
2HBr- + H2S04(l) —>
+4 0
S02(g) + 2H20(l) +Br2(l)
.Bromide ion reduces the sulphuric acid forming sulphur dioxide gas and bromine which is orange/brown
what can you see when reacting sodium iodide with concentrated sulphuric acid
Steamy fumes of hydrogen iodide
Black solid of iodine
Bad egg smell of hydrogen sulphide gas is present.
Yellow solid sulphur may be seen
Also colourless sulphur dioxide gases also evolved.
Acid base reaction of sodium iodide and sulfuric acid
NaI(s) + H2S04(l) —>
NaHS04(s) + HI (g)
What is the second reaction for sodium iodide and sulphuric acid, bring sulfur from +6 to +4
What about the final part of the reaction, bringing sulfur from +4 to -2
2HI + H2SO4 —> SO2 + 2H2O + I2
SO2 + 6HI —> H2S + 3I2 + 2H2O
Why is the sulfur in sulphuric acid reduced more with iodide ions than with bromide ions
Iodide ions are better reducing agents than bromide ions
They reduce the sulfur in sulphuric acid even further:
. from +6 (making S04)
. to 0 (making S8)
. and -2( forming H2S)
so sulpher dioxide, sulphur and hydrogen sulphide gas are produced.
Why can you see some solid yellow sulfer during the reaction of sodium iodide and sulfuric acid
The oxidation state of sulfur passes from +6 to -2 so goes through state 0 so you see the solid sulfur
How do we identify metal halides
. Silver ions in aqueous solution eg silver nitrate, to form precipitate of silver halide
.
How do we test for chloride ions
Cl- + Ag+ –> AgCl(s)
White precipitate is formed
Why do we first add nitric acid to halide ions
. Dilute nitric acid is first added to halide solution to get rid of any soluble carbonate, C03-(aq) or Hydroxide inequalities
.
Because..
It would interfere with the test by forming insoluble silver carbonate
2Ag+(aq) + C03 2- (aq)–> Ag2C03(s)
, or insoluble silver hydroxide
Ag+(aq) + 0H-(aq) –> Ag0H(s)
Step by step to test for halide ions
. Add dilute nitric acid HN03
. then add a few drops of silver nitrate solution and halide precipitate forms
What colour precipitate is formed when chloride ion is present
How can you further test for it
White precipitate of silver chloride
. It dissolves in dilute ammonia
What colour precipitate is formed when bromide ion is present
How can you further test it
. cream precipitate of silver bromide
. It dissolves in concentrated ammonia
What colour precipitate is formed when iodide ion is present
How can you further test it
. Pale yellow precipitate of silver iodide
.Insoluble in concentrated ammonia
Why do you further test for halide ions with ammonia after silver nitrate
The precipitates formed from silver nitrate solution are too similar in colour to differentiate at times
Equation of AgCl and dilute ammonia
AgCl + 2NH3 —> [Ag(NH3)2]+ + Br-
