TRANSITION METALS HW PART II MS

1.	Ligand substitution
	Suitable example e.g. reaction of thiocyanate ions with hexaaquairon(III)
to give [Fe(H2O)5(CNS)]2+ (1);
Observations e.g. formation of a blood-red colour (1)
Suitable equation e.g.
[Fe(H2O)6]3+ + CNS– → [Fe(H2O)5(CNS)]2+ + H2O (1)
Suitable example can be awarded from an equation
Equations do not need state symbols
	Precipitation
	Suitable example e.g. reaction between (aqueous) iron(II) chloride
with (aqueous) sodium hydroxide (1);
Observations e.g. formation of a green precipitate / formation of a green solid (1)
Suitable equation e.g. Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) (1)
Precipitate can be awarded state symbol in equation
	Redox
	Suitable example e.g. oxidation of iron(II) chloride by chlorine to
make iron(III) chloride (1)
Observation e.g. green solution becomes yellow / rust solution (1)
Suitable equation e.g. 2FeCl2 + Cl2 → 2FeCl3 (1)
Other examples could include iron and chlorine to make iron(III) chloride / iron and HCl to make FeCl2 / MnO4– and Fe2+ to make Fe3+
	And QWC
	One mark for correct spelling, punctuation and grammar in at least two sentences (1)	10
Answer must address the question
[10]

2.	Oxidation because oxidation state of Hg changes from 0 to +2 so oxidation (1)
Reduction because oxidation number of O changes from –1 to –2 (1)
	Or
	Correct identification of all the oxidation numbers (1)
Correct identification of oxidation and reduction (1)	2
Allow ecf for the identification of oxidation and reduction from wrong oxidation numbers
[2]

3.	(i)	Cu → Cu2+ + 2e- / Cu – 2e- → Cu2+ (1)	1
(ii)	2Cu + O2 + 4H+ → 2Cu2+ + 2H2O (1)	1
Allow any correct multiple
Allow ecf from (a)(i)
[2]

4.	Fe2O3 + 3Cl2 + 10OH– → 2FeO42– + 5H2O + 6Cl– (2)	2
Allow one mark if electrons shown
Allow one mark if correct reactants and products but not balanced
[2]

5.	Oxidation state of iron changes from +6 to +3 so is reduction (1)
	Oxidation state of oxygen changes from –2 to 0 so is oxidation (1)
To get the two marks for oxidation states marks any other oxidation state quoted must be correct.
Maximum one mark if any other oxidation number given is wrong
	OR
	Oxidation state of iron changes from +6 to +3 and
oxidation state of oxygen changes from –2 to 0 (1)
	Iron is reduced and oxygen is oxidised (1)	2
Allow ecf from wrong oxidation states
[2]

6.	Correct Mr for Fe2O3, 159.6, and of Na2FeO4,165.8 (1)
Moles of Fe2O3 = 0.00627 (1);
Mass of Na2FeO4 = 2.08 (1);
Percentage = 21.6 or 21.7 (%) (1)	4
Allow full marks for correct answer with some working
Answer must have 3 sig figs
Allow ecf from wrong moles or wrong mass
[4]

7.	(a)	Moles V2+ = 25.0 × 0.100 / 1000 = 0.0025 mols	1
Moles MnO4– = 30.0 × 0.0500 / 1000 = 0.00150 mols	1
1 mole of MnO4- changes its Oxidation State by 5 to change
the Oxidation State of 1.67 moles of V2+	1
Oxidation State of V2+ changes by 5 / 1.67 = 3	1
(b)	3MnO4– + 5V2+ + 3H2O → 3Mn2+ + 5VO3– + 6H+
(1 mark for correct species, 1 mark for balanced)	2
[6]

 8.	20 cm3 of 0.100 mol dm–3 VO2+ = 0.002 moles	1
0.002 moles VO2+ = 0.0004 moles MnO4–	1
0.0004 moles MnO4– are in 16.0 cm3	1
[3]

9.	(i)	MnO2 + 4H+ + 2Fe2+ → Mn2+ + 2H2O + 2Fe3+ (1)	1
Ignore state symbols
(ii)	Moles of Fe2+ that reacted with MnO2 = 0.02 – 0.0123 = 0.0077 (1)
Allow ecf within question
	Mass of MnO2 = 0.00385 × 86.9 = 0.335 (1)
% purity = 66.4% (1)
Allow 66.4 – 66.5
	Alternatively
	Moles of MnO2 in 0.504 = 0.00580
So moles of Fe2+ that should react with this is 0.0116 (1)
Moles of Fe2+ that reacted with MnO2 = 0.02 – 0.0123 = 0.0077 (1)
% purity = 66.4% (1)	3
[4]
 
10.	(i)	Decolorised / add starch which is decolorised
Allow blue/black → white or brown → white	1
Do not allow colourless
(ii)	moles S2O32– = 23.20 × 0.100/1000 = 0.00232 moles	1
Cu2+ ≡ S2O32– / moles Cu2+ = 0.00232 moles	1
But 25 cm3 of original = 10 × 0.00232 = 0.0232 moles	1
Concentration of original = 1000 × 0.0232 / 25	1
(iii)	Because concentration of Cu2+ is less than 1 mol dm–3 / less than standard	1
equilibrium moves to left (reducing +ve value of E)	1
[7]

 11.	(i)	Cr2O72– + 14H+ + 6I–  2Cr3+ + 3I2 + 7H2O
All species correct (ignore electrons for this mark)	1
Equation balanced (penalise if electrons not cancelled out)	1
(ii)	Brown colour disappears	1
S2O32– reacts with I2 (to form colourless I–)	1
Green colour remains due to Cr3+ (must say what gives green colour)	1
[5]
 
12.	(i)	Brown solution/brown precipitate/black solid
Add starch to get blue / black colour	1
(ii)	Titration / volumetric analysis	1
using sodium thiosulphate(with starch indicator)	1
(allow from equation)
	I2 + 2S2O32– → 2I– + S4O62–	1
	1 mol Cr2O72– = 6 mols S2O32–	1
[5]

13.	Moles of hydrogen = 3.17 × 10–3 / moles of zinc = 3.17 × 10-3 (1);
Not 3 × 10–3
	Mass of zinc = 0.207 g / moles of zinc × 65.4 (1);
Not 0.2
	Percentage of copper = 83.2 (1)	3
Allow ecf
Final answer must be to 3 or 4 sig figs
Penalise significant figures just once
Allow values between 82.9–83.2
[3]

14.	(i)	Cr2O72– + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
Correct reactants and products (1);
Correct balancing (electrons cancelled out) (1)	2
(ii)	Moles of dichromate(VI) = 3.53 × 10–4 (1);
Moles of iron(II) = 2.12 × 10–3 (1):
Mass of hydrated iron(II) sulphate = 0.588 g (1);
Percentage purity = 89.8 / 89.8 – 90.0 (1)	4
Allow alternative working out via mass instead of moles e.g. mass of iron in hydrated FeSO4 from percentage composition compared to mass of iron from moles of iron(II).
Allow ecf throughout unless percentage is above 100%
[6]
 
15.	(i)	(Oxidised to) iodine so a brown (solution) formed /
Fe3+ formed which is yellow or orange / Fe2+ formed which is green (1)	1
Allow red/brown or orange
(ii)	Nitrogen / N2 (1)	1
Allow any correctly named oxide of nitrogen / correct formulae / HNO3 etc.
[2]