Q1.
In which one of the following reactions does the metal species undergo reduction?
A       MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
B       [Cu(H2O)6]2++ 4Cl− → [CuCl4]2− + 6H2O
C       CrO + 2OH−  → 2CrO + H2O
D       TiO2 + 2C + 2Cl2 → TiCl4 + 2CO
(Total 1 mark)
Q2.
Where appropriate, use the standard electrode potential data in the table below to answer the questions which follow.
 
 	 	 	 	 	 	 	Eο/V	
Zn2+(aq)	+	2e–	→	Zn(s)	 	 	–0.76 	
V3+(aq)	+	e–	→	V2+(aq)	 	 	–0.26	
+ 2H+(aq)	+	2e-	→		+	H2O(l) 	+0.17 	
VO2+(aq) +2H+(aq)	+	e–	→	V3+(aq)	+	H2O(l)	+0.34	
Fe3+(aq)	+	e–	→	Fe2+(aq)	 	 	+0.77	
+ 2H+(aq)	+	e–	→	VO2+(aq)	+	H2O(l)	+1.00	
Cl2(aq)	+	2e–	→	2Cl–(aq)	 	 	+1.36 	
(a)     From the table above select the species which is the most powerful reducing agent.
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(1)
(b)     From the table above select
(i)      a species which, in acidic solution, will reduce  to VO2+(aq) but will not reduce VO2+(aq) to V3+(aq),
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(ii)     a species which, in acidic solution, will oxidise VO2+(aq) to .
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(2)

 (c)     After acidification, 25.0 cm3 of a solution of hydrogen peroxide reacted exactly with 
16.2 cm3 of a 0.0200 mol dm–3 solution of potassium manganate(VII). The overall equation for the reaction is given below.
 + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2
(i)      Use the equation for this reaction to determine the concentration, in g dm–3, of the hydrogen peroxide solution.
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(ii)     Calculate the maximum volume of oxygen, measured at a pressure of 98 kPa and a temperature of 298 K, which would be evolved in this reaction.
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(8)
Q3.
The table below shows some successive ionisation energy data for atoms of three different elements X, Y and Z.
Elements X, Y and Z are Ca, Sc and V but not in that order.
 
 	First	Second	Third	Fourth	Fifth	Sixth	
X	648	1370	2870	4600	6280	12 400	
Y	590	1150	4940	6480	8120	10 496	
Z	632	1240	2390	7110	8870	10 720	
(a)     Which element is calcium?
X       
Y       
Z       
(1)
(b)     Which element is vanadium?
X       
Y       
Z       
(1)
(c)     Justify your choice of vanadium in part (b)
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(1)
(d)     An acidified solution of NH4VO3 reacts with zinc.
Explain how observations from this reaction show that vanadium exists in at least two different oxidation states.
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(2)
(e)     The vanadium in 50.0 cm3 of a 0.800 mol dm−3 solution of NH4VO3 reacts with 506 cm3 of sulfur(IV) oxide gas measured at 20.0 °C and 98.0 kPa.
Use this information to calculate the oxidation state of the vanadium in the solution after the reduction reaction with sulfur(IV) oxide.
Explain your working.
The gas constant R = 8.31 J K−1 mol−1.

  

 Oxidation state = _______________
(6)
(Total 11 marks)

Mark schemes
Q1.
A
[1]
Q2.
(a)     most powerful reducing agent:   Zn;
1
(b)     (i)      reducing species:                            Fe2+
1
(ii)     oxidising species:                            Cl2;
1
 (c)     (i)      moles KMnO4 = 16.2 × 0.0200 ×10–3 = 3.24 ×10–4;
1
moles H2O2 = Moles KMnO4 × 5 / 2 = 8.10 × –4;
1
8.10 × 10–4 moles H2O2 in 25 cm3
8.10 × 10–4 × 1000 / 25 in 1000 cm3 = 0.0324 mol dm–3;
1
hence g dm–3 = mol dm–3 × Mr = 0.0324 × 34 = 1.10;
(penalise use of an incorrect H2O2 to KMnO4 ratio by two marks)
1
(ii)     PV = nRT;
1
hence V = nRT / P
= 8.10 × 10–4× 8.31 × 298/98000;
1
= 2.05 × 10–5;
1
units m3;
(mark consequentially to answers in (c)(i))
(allow correct answers with other units)
(answers to (c)(i) and (ii) must be to 3 significant figures; penalise once only)
1
[14]
Q3.
(a)     Y
1
(b)     X
1
(c)     Jump in trend of ionisation energies after removal of fifth electron
Fits with an element with 5 outer electrons (4s23d3) like V
1
(d)     Explanation: Two different colours of solution are observed
1
Because each colour is due to vanadium in a different oxidation state
1
(e)     Stage 1: mole calculations in either order
Moles of vanadium = 50.0 × 0.800 / 1000 = 4.00 × 10–2
Extended response
Maximum of 5 marks for answers which do not show a sustained line of reasoning which is coherent, relevant, substantiated and logically structured.
1
Moles of SO2 = pV / RT = (98 000 × 506 × 10–6 ) / (8.31 × 293)
= 2.04 × 10–2
1
Stage 2: moles of electrons added to NH4VO3
When SO2 (sulfur(IV) oxide) acts as a reducing agent, it is oxidised to sulfate(VI) ions so this is a two electron change
1
Moles of electrons released when SO2 is oxidised = 2.04 × 10–2 × 2
= 4.08 × 10–2
1
Stage 3: conclusion
But in NH4VO3 vanadium is in oxidation state 5
1
4.00 × 10–2 mol vanadium has gained 4.08 × 10–2 mol of electrons 
therefore 1 mol vanadium has gained 4.08 × 10–2 / 4.00 × 10 – 2 = 1 mol
of electrons to the nearest integer, so new oxidation state is 5 – 1 = 4
1
[11]