DP Chemistry (last assessment 2024)

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Question 22N.2.hl.TZ0.3

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Date November 2022 Marks available [Maximum mark: 15] Reference code 22N.2.hl.TZ0.3
Level hl Paper 2 Time zone TZ0
Command term Calculate, Determine, Justify, Label, Outline, Predict, Show, State, Write Question number 3 Adapted from N/A
3.
[Maximum mark: 15]
22N.2.hl.TZ0.3

Consider the following reaction:

Cu2+ (aq) + Fe (s) → Fe2+ (aq) + Cu (s)

(a)

State the ground-state electron configuration for Fe2+.

[1]

Markscheme

1s22s22p63s23p63d6

OR

[Ar]3d6

Examiners report

Over half the candidates gave the correct electronic structure of the Fe2+ ion, however quite a few did not read the question carefully so gave that of the parent atom and a few gave electron configurations that retained 4s electrons.

(b)

The mass spectrum for copper is shown:

Source: WebElements, n.d. Copper: isotope data [online] Available at:
https://www.webelements.com/copper/isotopes.html [Accessed 6 October 2021].


Show how a relative atomic mass of copper of 63.62 can be obtained from this mass spectrum.

[1]

Markscheme

63×69+65×31100

OR

65x+1-x63=63.62  AND  x=0.31/31«%»  AND  1-x=0.69 / 69«%» ✔

Examiners report

Almost every student managed to illustrate how isotopic abundances, from the given mass spectrum, could be used to calculate the relative atomic mass of the element.

(c)

Predict, with a reason, whether Cu or Cu2+ has the greater ionization energy.

[1]

Markscheme

Cu2+ AND fewer shielding electrons/less electron-electron repulsion «from same nuclear charge»

OR

Cu2+ AND larger effective nuclear charge

OR

Cu2+ AND more energy required to remove electron from positive ion than neutral parent atom

OR

Cu2+ AND smaller radius

OR

Cu2+ AND electron is being lost from a lower energy/inner/3d orbital ✔

Examiners report

Whilst a few students considered the ionisation energy of copper atom to be greater than that of Cu2+, the majority of students, who instinctively realised the subsequent ionisation energies of an atom would be greater than the first, found it difficult to construct a clear argument to support this, hence the average mark was only 0.6.

(d)

Determine the frequency, in s−1, of a photon that will cause the first ionization of copper. Use sections 1, 2 and 8 of the data booklet.

[2]

Markscheme

Alternative 1

«E=745kJmol-1×1mol6.02×1023atoms=» 1.24×10-21 «kJatom-1» ✔

 «E=hv»

«1.24×10-21kJ×1000J1kJ=6.63×10-34Js×v»

 «v=» 1.87×1015  «s-1» ✔

 

Alternative 2

 «E=hv»

 «745×103Jmol-1=6.63×10-34Js×v»

 «v=» 1.12×1039  «s-1mol-1» ✔

 «=1.12×1039s-16.02×1023» =1.87×1015  «s-1» ✔

 

Award [2] for correct final answer.

Examiners report

Whilst many students managed to identify the correct formula and constants to use to convert energy into frequency, quite a few failed to convert the ionization enthalpy from kJ to J, then even more were unable to convert from a molar to an atomic scale.

(e)

Outline the magnetic properties of iron by referring to its electron configuration.

[2]

Markscheme

«iron atoms have 4» unpaired electrons ✔


aligns with a magnetic field/paramagnetic

OR

has a magnetic moment

OR

ferromagnetic ✔

 

For M1 accept diagrams showing unpaired electrons.

Examiners report

Many candidates realised that iron had some unusual magnetic properties (paramagnetic or ferromagnetic). A few correctly correlated these to the presence of unpaired electrons, however quite a number seemed to think they just arose from having partially filled d-orbitals.

The diagram shows an unlabelled voltaic cell for the reaction:

Cu2+ (aq) + Fe (s) → Fe2+ (aq) + Cu (s)

(f.i)

Label the diagram with the species from the equation and the direction of electron flow.

[2]

Markscheme

all 4 species correctly labelled ✔

arrow showing electron flow from anode to cathode in the external circuit ✔

 

Accept any soluble salt of copper(II) for Cu2+ and any soluble salt of iron(II) for Fe2+.

Do not apply ECF for M2.

Examiners report

Most students correctly annotated the voltaic cell and usually also marked the correct direction of electron flow. Some however had the electrodes reversed, whilst others displayed a higher level of confusion with ionic species comprising the electrodes.

(f.ii)

Write the half-equation for the reaction occurring at the anode (negative electrode).

[1]

Markscheme

Fe (s) → Fe2+ (aq) + 2e

 

Accept equilibrium arrow.

Do not award ECF for Cu (s) → Cu2+ (aq) + 2e.

Examiners report

Most, almost irrespective of the annotation of the diagram, managed to write the correct half equation for the reaction at the iron anode.

(f.iii)

The diagram includes a salt bridge that is filled with a saturated solution of KNO3. Outline the function of the salt bridge.

[1]

Markscheme

«keep» each half-cell/electrolyte «electrically» neutral ✔

 

Accept balance charges/ions.

Accept allow ion flow «between cells».

Examiners report

Most students were aware of the role of the salt bridge, even if only as completing the circuit.

(f.iv)

Predict the movement of all ionic species through the salt bridge.

[2]

Markscheme

NO3 to anode/Fe/left ✔

K+ «and Fe2+» to cathode/Cu/right ✔

 

Accept other specific anions in addition to nitrate for M1.

Award [1 max] for “anions/negative ions to anode AND cations/positive ions to cathode”.

Examiners report

In spite of appearing to be aware of the role of the salt bridge, over half the candidates lost these marks by framing their answers in terms of the ions in the electrolyte, rather than those in the salt bridge.

(f.v)

Calculate the standard cell potential, in V, for this cell. Use section 24 of the data booklet.

[1]

Markscheme

«E = + 0.34 V – (–0.45 V) = +» 0.79 «V» ✔

Examiners report

About two thirds of the students correctly identified the electrode potentials and combined these to deduce the cell potential, though a number used Fe3+/Fe2+ rather than Fe2+/Fe.

(f.vi)

Calculate the standard free energy change, in kJ, for the cell. Use your answer in (f)(v) and sections 1 and 2 of the data booklet.

If you did not obtain an answer in (f)(v), use 0.68 V, although this is not the correct answer.

[1]

Markscheme

«ΔG = – nFE = – 2 mol × (9.65 x 104 C mol–1) × (0.79 V) =» –152 «kJ» ✔

 

Accept answers in the range 150 – 153.

Examiners report

Most correctly calculated the free energy change from the cell potential, however the value of "n" (the number of electrons transferred) and converting the result to kJ, as was required by the question, created difficulties for some.