10.3 Reactions of halide ions Flashcards

1
Q

How are halide ions reducing agents?

A

. The halide ions lose/give away electrons and become halogen molecules

They themselves become oxidised, but they reduce other halogens

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2
Q

Describe the reducing ability down the group

A

.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

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3
Q

How do sodium halides react with concentrated sulphuric acid

A

Solid sodium halides react with concentrated sulphuric acid.

The products are different and reflect the reducing power of the halides

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4
Q

What is seen in the reaction between sodium chloride and concentrated sulphuric acid

A

. Drops of sulphuric acid are added to sodium chloride.
Steamy fumes of hydrogen chloride are seen .

Solid product is sodium hydrogensulfate.

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5
Q

Word and symbol equation for sodium chloride and sulphuric acid

A

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)

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6
Q

Why is the reaction between sodium chloride and sulphuric acid not a redox reaction

What type of reaction is it

A

. 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

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7
Q

How does sodium fluoride react with sulfuric acid

A

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

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8
Q

What forms from sodium bromide and concentrated sulphuric acid

A

You will see steamy fumes of hydrogen bromide, and brown fumes of bromine.

Colourless sulphur dioxide is also formed

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9
Q

First reaction equation for sodium bromide and sulphuric acid

(two reactions occur)

A

NaBr(s) + H2S04(l) —->
NaHS04(s) +HBr(g)

Sodium hydrogensulfate and hydrogen bromide are formed (in a similar acid base reaction to sodium chloride)

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10
Q

Why are there two reactions for sodium bromide and sulphuric acid

A

. 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

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11
Q

What is the redox reaction
(second reaction)for Sodium bromide and sulphuric acid

Exothermic reaction

A

-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

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12
Q

what can you see when reacting sodium iodide with concentrated sulphuric acid

A

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.

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13
Q

Acid base reaction of sodium iodide and sulfuric acid

A

NaI(s) + H2S04(l) —>
NaHS04(s) + HI (g)

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14
Q

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

A

2HI + H2SO4 —> SO2 + 2H2O + I2

SO2 + 6HI —> H2S + 3I2 + 2H2O

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15
Q

Why is the sulfur in sulphuric acid reduced more with iodide ions than with bromide ions

A

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.

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16
Q

Why can you see some solid yellow sulfer during the reaction of sodium iodide and sulfuric acid

A

The oxidation state of sulfur passes from +6 to -2 so goes through state 0 so you see the solid sulfur

17
Q

How do we identify metal halides

A

. Silver ions in aqueous solution eg silver nitrate, to form precipitate of silver halide

.

18
Q

How do we test for chloride ions

A

Cl- + Ag+ –> AgCl(s)

White precipitate is formed

19
Q

Why do we first add nitric acid to halide ions

A

. 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)

20
Q

Step by step to test for halide ions

A

. Add dilute nitric acid HN03

. then add a few drops of silver nitrate solution and halide precipitate forms

21
Q

What colour precipitate is formed when chloride ion is present

How can you further test for it

A

White precipitate of silver chloride

. It dissolves in dilute ammonia

22
Q

What colour precipitate is formed when bromide ion is present

How can you further test it

A

. cream precipitate of silver bromide

. It dissolves in concentrated ammonia

23
Q

What colour precipitate is formed when iodide ion is present

How can you further test it

A

. Pale yellow precipitate of silver iodide

.Insoluble in concentrated ammonia

24
Q

Why do you further test for halide ions with ammonia after silver nitrate

A

The precipitates formed from silver nitrate solution are too similar in colour to differentiate at times

25
Q

Equation of AgCl and dilute ammonia

A

AgCl + 2NH3 —> [Ag(NH3)2]+ + Br-