1.
(a)
What do you understand by the term standard electrode potential?
potential / emf / voltage / half cell of electrode relative (½) to (standard)
hydrogen electrode (½) solutions of unit concentration (½) and gases
at 1 atm / standard temperature (allow from (b)(ii) (½) round up half marks
(2)
(b)
If a metal is immersed in a solution of its ions, the potential set up between the metal and
the solution cannot be measured without using a reference electrode. Explain why this is
so.
voltmeter requires two connection / measure potential difference (1)
and so another electrode is introduced into system (1)
(1) for Redox has to have a source / sink of electrons
(2)
(i)
What is meant by disproportionation?
m
(c)
Use the following data to deduce whether MnO2 will disproportionate in acidic
solution.
MnO4– + 4H+ + 3e–1
MnO2 + 4H+ + 2e–
MnO2 + 2H2O
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(ii)
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reaction where a given (chemical) species / substance / molecule / ion not
atom is simultaneously / both oxidised and reduced (1)
Mn2+ + 2H2O
E
= +1.70 V
E
= +1.23 V
2.
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E
for disproportionation is (+1.23) – (+1.70)V = –0.47V (1)
If equivalent in words (1)
(this is negative) so reaction does not occur (1)
second mark must be consequential on the first statement
(3)
(Total 7 marks)
What do you understand by the term standard electrode potential?
If a metal is immersed in a solution of its ions, the potential set up between the metal and the
solution cannot be measured without using a reference electrode. Explain why this is so.
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3.
(Total 2 marks)
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potential / emf / voltage / half cell of electrode relative (½) to (standard)
hydrogen electrode (½) solutions of unit concentration (½) and gases
at 1 atm / standard temperature (allow from (b)(ii) (½) round up half marks
voltmeter requires two connection / measure potential difference (1)
and so another electrode is introduced into system (1)
(1) for Redox has to have a source / sink of electrons
4.
(a)
(Total 3 marks)
What is meant by disproportionation?
reaction where a given (chemical) species / substance / molecule / ion not
atom is simultaneously / both oxidised and reduced (1)
NT Exampro
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(b)
Use the following data to deduce whether MnO2 will disproportionate in acidic solution.
MnO4– + 4H+ + 3e–1
MnO2 + 4H+ + 2e–
MnO2 + 2H2O
Mn2+ + 2H2O
E
= +1.70 V
E
= +1.23 V
E
for disproportionation is (+1.23) – (+1.70)V = –0.47V (1)
If equivalent in words (1)
(this is negative) so reaction does not occur (1)
second mark must be consequential on the first statement
(Total 3 marks)
Use the data below to explain concisely why zinc is used in preference to tin for coating steel
which is used to manufacture cars.
m
5.
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Sn2+ (aq) + 2e– = Sn(s)
Fe2+ (aq) + 2e– = Fe(s)
Zn2+ (aq) + 2e– = Zn(s)
E /V
–0.14
0.44
0.76
6.
(a)
em
ac
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ve
not just prevents corrosion
Zn oxidises in preference to iron / to form Zn2+ / iron (oxidises) in preference to tin (1)
since E more negative (1) not lower
hence prevents corrosion starting by preventing first stage Fe → Fe2+ + 2e– / oxidation
of Fe (1)
not electrochemical series argument
OR
similar arguments for tin
(Total 3 marks)
The standard reduction potential for the MnO4– |Mn2+ electrode is given by
Mn2+(aq) + 4H2O(1)
E
= + 1.51 V
ch
MnO4– (aq) + 8H+(aq) + 5e–
why is a reference electrode necessary to measure E ?
(b)
ww
w.
potential of single electrode can’t be measured/
second electrode has a potential/
second electrode needed (1)|
have to measure potential difference (1)
need a source or sink of electrons (1) (max 2)
(2)
The standard reduction potential for the Cl2 |Cl– electrode is given by
½Cl2(g) + e–
(i)
Cl– (aq)
E
= 1.36 V
Combine this half equation with the one in (a) to give the reaction of
manganate(VII) ions with chloride ions.
MnO4– + 8H+ + 5e–
Mn2+ + 4H2O
–
–
5/2CI2 + 5e (1) for correct multiplication
5Cl
Mn2+ + 5/2CI2 + 4H2O (1) or doubled
MnO4– + 5CI– + 8H+
if the final equation is given with no working then (2)
if the equation is the wrong way round but everything else is
correct then (1) only)
(2)
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(ii)
State, giving a reason whether you would expect this reaction to occur under
standard conditions.
E = 1.51 – 1.36 V = 0.15 V (1) or cell potential is positive
(some valid comparison of the 2 E values required)
E > 0 so reaction feasible (1)
a clear statement that the cell has a positive potential is required
the 2nd mark is consequential on the 1st ie if Ecell is –ve then ‘no reaction’ or
‘equilibrium far to the left’ should be given for (max 1)
if ‘no restriction because E<0.3V’ given then (max 1)
(2)
(iii)
Given that
MnO2 (s) + 4H+(aq) + 2e–
Mn2+(aq) + 2H2O(1)
E
= + 1.23V
m
comment on the fact that the reaction of MnO2 with hot concentrated HCI is a
standard laboratory method for the preparation of chlorine.
7.
(a)
(i)
(ii)
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ac
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ve
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E = 1.23 – 1.36 V = .0.3V ∴ not possible (1)
some quantitative evidence is needed but calculation not necessary
reaction conds. are not standard (1)
eg conc HCl / or temperature/concentration of CI– ions/removal of CI2
(. E > 1.23) (1)
Al2O3 has very high melting point or it is very expensive to melt
aluminium oxide (1)
(3)
(Total 9 marks)
1
2O2– → O2 + 4e– or 4AlO33– → 2Al2O3 + 3O2 +12e– (1)
O2 reacts (to form oxides of carbon) or C + O2 → CO2 (1)
(C + 2O2– → CO2 + 4e– overall equation (2))
ch
which are gaseous products or anodes get smaller or anodes burnt away or anodes
used up or corrode or oxidised or combust (1)
Anodes erode or anodes disintegrate or anodes worn (0)
3
maintenance of electrolyte temperature or keeping electrolyte molten or
reaction endothermic (1) not high temperature needs to be reached.
electrolytic or electrolysis reactions at electrode or aluminium ion needs three
electrons (1)
2
(iv)
List A
low density ( NOT light) or high strength or resistance to corrosion (or stable oxide
layer)
ww
w.
(iii)
List B
Alternative – less good – answers include good conductivity or good conductor of
heat or very malleable or very ductile or attractive appearance or not toxic
An answer with two from list A scores 2 marks
An answer with one from list A and one from list B scores 2 marks
An answer with two from list B scores 1 mark
high strength to weight ratio(2)
2
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(b)
(i)
oxygen or air or inert porous material or equation 2 or II (1)
1
(ii)
4Al(s) + 3O2 (g) + 6H2O(l) → 4Al(OH)3(s) (1)
State symbols not needed; ignore OH— if this is on both sides
1
hydroxide or oxide layer building up on the surface of the anode
or use up oxygen or cathode materials
or dissolve aluminium or use up all the aluminium
or low on water
or any sensible chemical comment (1)
NOT aluminium hydroxide is insoluble
1
(iii)
(c)
nothing (1)
explanation relating to emf of cell or electrodes /
reference to relative strength reducing agent(1)
(i)
2
2
2Fe + O2 + 2H2O → 2Fe(OH)2
/
2+
2Fe(s) + O2 (g) + 2H2O(l) → 2Fe (aq) + 4OH–(aq) (2)
correct species (1) balancing (1)
2
reacts preferentially / reacts as sacrificial anode (1)
Mg more negative redox potential / stronger reducing agent (1)
2
[8]
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w.
ch
(ii)
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(b)
the potential difference between a metal and a
1 mol dm–3 solution of its ions (under standard
conditions) / the half-cell with 1 mol dm–3 solution(1)
and the standard hydrogen electrode (1)
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(a)
em
ac
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8.
m
[11]
NT Exampro
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