Chemistry, Third edition

ANSWERS TO EXERCISES AND REVISION QUESTIONS

Revision questions

6.1 (i)Pure nitric ‘acid’ is a covalent liquid and does not

contain any H+(aq) ions.

(ii)It produces OH–(aq) ions when it dissolves in water.

(iii)The water present in the ammonia would react with

the concentrated acid producing dilute sulfuric acid

which would neutralize the ammonia.

6.2 (i)CuSO 4 ·5H 2 O(s)Cu^2 +(aq)SO 42 – (aq)

5H 2 O(l)

(ii)Li+,OH–(s)Li+(aq)OH–(aq)

6.3 (i)(COOH) 2 (aq)2KOH(aq)(COOK) 2 (aq)

2H 2 O(l),

or H+(aq)OH–(aq)H 2 O(l)

(ii)(COOH) 2 (aq)Zn(s)(COO) 2 Zn(aq)H 2 (l)

or 2H+(aq)Zn(s)H 2 (g)Zn^2 +(aq)

6.4 (i)MgCO 3 (aq)H 2 SO 4 (aq)

MgSO 4 (aq)H 2 O(l)CO 2 (g),

or 2H+(aq)CO 32 – (aq)CO 2 (g)H 2 O(l)

(ii)(NH 4 ) 2 SO 4 (aq)2NaOH(aq)

2NH 3 (g)2H 2 O(l)Na 2 SO 4 (aq),

or NH 4 +(aq)OH–(aq)H 2 O(l)NH 3 (g)

(iii)Na 2 S(aq)2HNO 3 (aq)2NaNO 3 (aq)H 2 S(g),

or S^2 – (s)2H+(aq)H 2 S(g)

(iv)Na 2 SO 3 (aq)2CH 3 CO 2 H(aq)

2CH 3 CO 2 Na(aq)SO 2 (g)H 2 O(l)

or SO 32 – (aq)2H+(aq)SO 2 (g)H 2 O(l)

6.5 (i)Both gases react with water, producing acidic

solutions. SO 2 forms a more strongly acidic solution than

CO 2 , i.e. it produces more H+(aq) ions per mole of

dissolved gas in aqueous solution. (SO 2 is also more

soluble than CO 2 .) The presence of greater numbers of

H+(aq) ions means that more limestone can be

destroyed.

(ii)The OH–(aq) ions force any dissolved CO 2 to produce

carbonate ions, CO 32 – (aq), which react with Ca^2 +

producing insoluble calcium carbonate:

Ca^2 +(aq)CO 32 – (aq)CaCO 3 (s)

6.6The blue colour suggests that the substance

contains copper(II) ions (K Cu^2 +). The fizzing with acid

suggests the presence of a carbonate ion (J CO 32 – ). In

the fizzing reaction, the carbonate is converted to a salt

and to CO 2 , while the copper(II) ion is unaffected. The

copper(II) ion forms a blue precipitate of copper(II)

hydroxide with OH–(aq).

6.7GFe^3 +(aq) (confirmed by precipitate of Fe(OH) 3 );

HNO 3 – (confirmed by brown ring test).

6.8SS^2 – and T Pb^2 +:

S^2 – (s)2H+(aq)H 2 S(g)

This reaction causes Pb^2 +ions in the solid PbS to enter

aqueous solution where they may now be identified by

the addition of sulfate ions which precipitate white

lead(II) sulfate:

Pb^2 +(aq)SO 42 – (aq)PbSO 4 (s)

and by the addition of hydroxide ions:

Pb^2 +(aq)2OH–(aq)Pb(OH) 2 (s)

Pb(OH) 2 (s)2OH–(aq)[Pb(OH) 4 ]2–(aq)

6.9Under alkaline conditions, any free aluminium ions

are precipitated as Al(OH) 3 or as higher hydroxides:

Al^3 +(aq)3OH–(aq)Al(OH) 3 (s).

The precipitate ‘locks up’ Al^3 +(aq). In acidic conditions,

free aluminium ions are produced:

Al(OH) 3 (s)3H+(aq)Al^3 +(aq)3H 2 O(l)

toxic

Unit 7

Exercises

7A

(i)Cu^2 +(aq)2e–Cu(s)

(ii)Zn(s)Zn^2 +(aq)2e–

(iii)Cu^2 +(aq)Zn(s)Cu(s)Zn^2 +(aq)

(iv)The sulfate ions remain the same throughout the

reaction – they do not take part in the redox reaction. The

sulfate ions arespectatorions.

7B

(i)Zn oxidized (gains O), O 2 reduced (loses O!)

(ii)Mg oxidized (loses electrons), Cu^2 +reduced (gains

electrons)

(iii)Cl 2 reduced (gains electrons), I–oxidized (loses

electrons)

(iv)Since sodium chloride is Na+Cl–, Na oxidized (loses

electron), Cl 2 reduced (gains electrons)

(v)Cl 2 reduced (gains H), H 2 oxidised (loses H!).

7C

(i) 4 (ii) 3 (iii)O (iv) 5 (v) 5

(vi) 7 (vii) 2 (viii)5 (contains VO 3 – )

(ix) 5 (x)6.

7D

(i)

(ii)No, the oxidation number of sulfur does not change.

(iii)SO 2 is reduced to S because the oxidation number

of sulfur decreases 4 to 0. H 2 S is oxidised to S

because the oxidation number of sulfur increases from

2 to 0.

7E

7F

(i)Those with oxidation numbers 4 and 0, because

they can be oxidized (increase their oxidation number)

and reduced (decrease their oxidation number).

(ii)Those with oxidation number 6, because they can

be reduced only.

(iii)H 2 S, because it can be oxidized only.

7G

1. (i)I–3H 2 OIO 3 – 6H+6e–
   (ii)NO 3 – 4H+3e–NO2H 2 O
   (iii)Cr 2 O 72 – 14H+6e–2Cr^3 +7H 2 O
   (iv)H 2 O 2 2H+2e–2H 2 O
   (v)To balance S^2 – SO 42 –
   Balance the Os with H 2 O:
   S^2 – 4H 2 OSO 42 –
   Balance the hydrogens by adding 8H 2 O to the opposite
   side and 8OH–to the same side:
   S^2 – 4H 2 O8OH–SO 42 – 8H 2 O
   Balance the charges:
   S^2 – 4H 2 O8OH–SO 42 – 8H 2 O8e–
   Simplify the equation:
   S^2 – 8OH–SO 42 – 4H 2 O8e–
   2. (i)Cu2Ag+Cu^2 +2Ag
   (ii)Cu2NO 3 – 4H+Cu^2 +2NO 2 2H 2 O
   (iii)I 2 2S 2 O 32 – S 4 O 62 –  2I–
   (iv)2I–H 2 O 2 2H+I 2 2H 2 O
   (v)2H 2 O 2 ClO 2 –  2O 2 Cl–2H 2 O
   7H
   (i)
   Pt | H 2 (g) | H+(aq) || Cu^2 +(aq) | Cu(s)
   (ii)0.34 V E^ —(Cu^2 +(aq)/Cu(s))0, therefore E—^
   (Cu^2 +(aq)/Cu(s))0.34 V
   (iii)H 2 (g)Cu^2 +(aq)2H+(aq)Cu(s)
   7I
   (i)yes

I 2 (s)2e– 2I–(aq) 0.54

Br 2 (l)2e– 2Br–(aq) 1.09

(ii)no

2H+(aq)2e– H 2 (g) 0.00

Cu^2 +(aq)2e– Cu(s) 0.34

(iii)yes

Zn^2 +(aq)2e– Zn(s 0.76

Fe^3 +(aq)e– Fe^2 +(aq) 0.77

(iv)yes

I 2 (s)2e– 2I–(aq) 0.54

Fe^3 +(aq)e– Fe^2 +(aq) 0.77

7J

Work out the reactions by using anticlockwise arrows:

(i)Mg(s)2Ag+(aq)Mg^2 +(aq)2Ag(s)

(ii)no reaction

(iii)Ni(s)Cu^2 +(aq)Ni^2 +(aq)Cu(s)

(iv)Zn(s)2H+(aq)Zn^2 +(aq)H 2 (g)

(v)no reaction.

7K

(i)There is no dissolved oxygen in boiled water.

(ii)Water is more highly conducting if it carries dissolved

salt and rusting occurs more rapidly. A great deal of salt

is around at coastal locations.

(iii)Acid rain provides H+ions, making rusting occur

more readily (see the equations for rusting).

7L

The standard electrode potential of iron is more negative

than that for zinc, showing that iron loses electrons to

form Fe^2 +(aq) more readily than tin forms Sn^2 +(aq).

Tin plating is not as effective as galvanizing. Provided that

the coating is undamaged, the tin protects the iron under-

neath from water and oxygen. Once the surface is

scratched, however, iron loses electrons more readily

than tin to form Fe^2 +ions, which then are oxidized to

rust.

7M

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