A-LEVEL PAPER 1 PP8 MS

1.   (a)     Percentage of oxygen is 42.5% (M1)
Allow if shown clearly in the calculation.
1
Co 13.0 / 58.9 = 0.221, N 18.6 / 14 = 1.329,
K 25.9 / 39.1 = 0.662, O 42.5 / 16 = 2.656 (M2)
Allow alternative method if chemically correct.
If Ar has been divided by the percentage, chemical error, lose M2 and M3.
1
CoN6K3O12 (M3)
Allow in any order.
Correct answer without working scores this mark only.
1
(b)     Co(NO2)63−
Allow a correct diagram bonding through N or O
Do not allow CoN6O123−
Must have correct overall charge.
Allow consequential answer from part(a) if the charge on the anion is correct.
1
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2.             (i)      Fe + 2HCl → FeCl2 + H2 
(allow ionic formulae)
or Fe + 2H+ → Fe2+ + H2
1
(ii)     PV = nRT n = PV/RT
(allow either formula but penalise
contradiction)
1
n = 
1
= 4.53 × 10–3 (mol) 
(answer must have at least 3 sig. figs. Ignore units)
1
(iii)     Moles of iron = 4.5(3) × 10–3 mol 
(allow conseq on (a)(ii))
(or = 4.2(5) × 10–3 if candidate uses given moles of hydrogen)
1
Mass of iron = 4.53 × 10–3 × 55.8 = 0.253 g 
(mark is for method mass = moles × Ar)
(Mass of iron can be 56)
1
(iv)    0.253 × 100/0.263 = 96.1 % (mark is for answer to 2 sig. figs.)
(allow conseq on mass of iron. E.g. = 90% from
4.2(5) × 10–3 moles of H2 and Fe)
(Do not allow answers greater than or equal to 100%)
1
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3.      (a)     (i)

ONLY consider species involved in the step marked
(ii)     Cycling clockwise about (*)
CE if step missing
         ΔHaBa + 1st IEBa + 2nd IE Ba + 2ΔHaCl + 2EACl + LE – ΔHfBaCl2 = 0 (1)
+180 + 503 + 965 + 2 × 122 + 2EA –2056 + 859 = 0 (1)
EA = –695/2 = – (347 to 348) (1)
Ignore units
Calculation –1 for each error
Mark conseq.
Notes: –695 scores (2)
+(347 to 348) scores (2)
–(286 to 287) scores (2)
+(286 to 287) scores (1)
–573 scores (1)
+573 scores (0)
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(b)     ΔS = ΣS products – ΣS reactants
      = (63 +223) – 124 = 162 (1)
ΔG = ΔH – TΔS     or      ΔH = TΔS or T = ΔH/ΔS (1)
or used correctly
          ΔH = 859 × 103 (1) = T × 162
T = (5 300 to 5304) K (1)
Penalise if units °C
–1 for each error + mark conseq.
4
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4.   (a)     (i)      [H+][OH−] OR [H3O+][OH−]
Ignore (aq)
Must have [ ] not ( )
1
(ii)     3.46 × 10−14 (= 1.86 × 10−7)
If no square root, CE=0
1
pH = 6.73
Must be 2dp
1
(iii)    [H+] = 10−11.36 (= 4.365 × 10−12 OR 4.37 × 10−12)
Mark for working
1
Kw = [4.365 × 10−12 OR 4.37 × 10−12 × 0.047] = 2.05 × 10−13
Allow 2.05 × 10−13 − 2.1 × 10−13
Mark for answer
Ignore units
1
(b)     (i)      HCOOH    HCOO− +   H+
Must have  but ignore brackets.
OR   HCOOH + H2O     HCOO− +   H3O+
Allow HCO2− or CHOO− ie minus must be on oxygen, so penalise COOH−
1
(ii)      
 
Must have all brackets but allow ( )
Must be HCOOH etc.
Allow ecf in formulae from (b)(i)
1
(iii)    M1
 
Allow HA or HX etc.
Allow [H+] = (Ka × [HA]) for M1
1
M2    [H+] = 3.16 × 10−3
Mark for answer
1
M3    pH = 2.50     allow more than 2 dp but not fewer
Allow correct pH from their wrong [H+] here only If square root shown but not taken, pH = 5.00 can score max 2 for M1 and M3
1
(iv)    M1    Decrease     Mark M1 independently
1
M2    Eqm shifts / moves to RHS     OR    more H+    OR    Ka increases    
OR more dissociation
1
M3    To reduce temperature or oppose increase / change in temperature
Only award M3 following correct M2
1
(c)     (i)      M1    
 
If [HX]/[X-] upside down, no marks
1
M2    
         (= 2.27 × 10−4)
1
M3    pH = 3.64      allow more than 2 dp but not fewer
pH calc NOT allowed from their wrong [H+] here
1
(ii)     M1    Mol H+ added = 5.00 × 10−4
Mark on from AE in moles of HCl (eg 5 × 10−3 gives pH = 3.42 scores 3)
1
M2    Mol HCOOH = 2.40 × 10−2   and   Mol HCOO– = 1.79 × 10−2
If either wrong no further marks except AE (−1) OR if ECF in mol acid and / or mol salt from (c)(i), can score all 4
1
M3    
 
If [HX]/[X-] upside down here after correct expression in (c)(i), no further marks
                
If [HX]/[X-] upside down here and is repeat error from (c)(i), max 3 (pH = 3.88 after 3.86 in (c)(i))
1
M4    pH = 3.62      allow more than 2 dp but not fewer
pH calc NOT allowed from their wrong [H+] here
1
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5.	(a)     Electron acceptor / gains electrons / takes electrons away
Do not allow electron pair acceptor / gain of electrons / definition of redox (QWC)
1
(b)     Cd(OH)2
Do not allow ‘Cd(OH)2/Cd’
1
Species (on LHS) with the least positive/most negative electrode potential / lowest E
/ smallest E
Only allow this mark if M1 answer given correctly or blank
Do not allow negative emf
1

(c)     (i)      1.5 (V) / 1.50
1
(ii)     2MnO2 + 2H2O + Zn →2MnO(OH) + 2OH– + Zn2+
Ignore state symbols
e– must be cancelled
(take care that Zn2+ is on RHS)
1
(iii)    Allows ions to pass (through it) or words to that effect
Penalise passage of electrons
Allow mention of particular ions
1
(iv)     Allows electrons to flow / makes electrical contact / conductor
Allow acts as an (inert) electrode / anode / cathode
1
(v)      Zn is ‘used up’ / has reacted / oxidised
Allow idea that zinc reacts
Do not allow just zinc corrodes
1
(d)     (i)      3 / +3 / III
1
2Ni(OH)2 + Cd(OH)2 → 2NiO(OH) + Cd + 2H2O
For correct nickel and cadmium species in correct order (allow H2O missing and OH– not cancelled)
1
For balanced equation (also scores M2)
Allow max 1 for M2 and M3 if correct balanced equation but reversed.
Ignore state symbols
1
(ii)     Metal / metal compounds are re-used / supplies are not depleted / It (the cell)
can be re-used
Allow does not leak / no landfill problems / less mining / less energy to extract metals / less waste
Do not allow less CO2 unless explained
1
(e)     (i)      C2H5OH + 3O2 → 2CO2 + 3H2O
Allow C2H6O
1
(ii)     C2H5OH + 3H2O → 2CO2 + 12H+ + 12e–
Allow C2H6O
1
(iii)    (+)0.23 (V)
1
(iv)     CO2 released by combustion / fermentation / fuel cell / reaction with water
Can be answered with the aid of equations
1
(atmospheric) CO2 taken up in photosynthesis
1
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6.     (a)     Ligand: -
atom, ion or molecules which can donate a pair of electrons to a metal ion.
1
co-ordinate bond:- a covalent bond
1
in which both electrons are donate by one atom
1

(b)     (i)      Two correct complex ions
1
Balanced equation
1
Two correct colours
2
(ii)     Complex with a bidentate ligand
1
Balanced equation
NB en not allowed as a ligand unless structure also given
1
         More molecules/ions formed
1
Increase in entropy
1
more stable complex formed
1
Max 2
(c)     ΔE; energy absorbed by electron, ground to excited state (QoL)
1
h; Planck’s constant or a constant
1
Change in
Oxidation state
1
Ligand
1
Co-ordination number
Apply list principle to incorrect additional answers
1
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7.       (a)     Alternative route
Allow mechanism outlined
allow forms intermediate species
1
Lower activation energy
1
(b)     Variable oxidation state
allow changes oxidation states
1
(c)     (i)      SO2 + V2O5 → SO3 + V2O4
allow 2VO2 instead of V2O4
1
O2 + 2V2O4 → 2V2O5
1
(ii)     Poison attaches to surface
Allow blocks active site/surface
Decreases surface area
1
(iii)     Purify reactants
Allow remove impurities
1
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8.   (a)     Water in the gaseous state from the precipitate absorbed by drying agent
OR
Water vapour from the precipitate absorbed by drying agent
Allow ‘water vapour reacts with drying agent’.
Do not allow ‘absorb water’ without qualification.
1
(b)     (Blue to) pink / pink colour observed
1
[2]

9.    (a)     To reduce any Fe3+ ions to Fe2+ ions
Allow ‘to ensure that all of the iron present is in the form of Fe2+ ions’ or ‘to ensure that no Fe3+ ions are present’.
1
(b)     Zinc would react with MnO4− / Fe3+ produced in titration
Do not allow ‘would increase titre value’.
Do not allow ‘zinc would react’ without further qualification.
1
[2]

10.    (a)     Species             [Co(H2O)6]2+
1
          Precipitate         Co(H2O)4(OH)2
1
(b)     [Co(NH3)6]2+
1
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11.	(a)		
	(b)		
	(c)		
	(d)		
	(e)		
	(f)		

	(g)		
	(h)		
	(i)		
	(j)	
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