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Date November 2018 Marks available 1 Reference code 18N.3.hl.TZ0.15
Level HL Paper 3 Time zone TZ0
Command term Identify Question number 15 Adapted from N/A

Question

Chemical energy from redox reactions can be used as a source of electrical energy.

The chemical structure of a photosensitive dye found in blueberries and a schematic diagram of a solar cell are shown.

Outline how a rechargeable battery differs from a primary cell.

[1]
a.

Formulate half-equations for the reactions at the anode (negative electrode) and cathode (positive electrode) during discharge of a lithium-ion battery.

[2]
b.

A voltaic cell consists of a nickel electrode in 1.0 mol dm−3 Ni2+ (aq) solution and a cadmium electrode in a Cd2+ (aq) solution of unknown concentration.

Cd (s) + Ni2+ (aq) → Cd2+ (aq) + Ni (s)               EΘcell = 0.14 V

Determine the concentration of the Cd2+ (aq) solution if the cell voltage, E, is 0.19 V at 298 K. Use section 1 of the data booklet.

[2]
c.

Identify the structural feature of the dye that allows the conversion of solar energy into electrical energy.

[1]
d.i.

Outline the effect of sunlight on the dye in the solar cell.

[1]
d.ii.

State the purpose of TiO2.

[1]
d.iii.

Deduce the reduction half-equation at the cathode.

[1]
d.iv.

Markscheme

«redox» reaction in rechargeable battery is reversible «but not in a primary cell»

OR

secondary cells need to be charged before use

OR

secondary cells have greater rate of self-discharge ✔

 

Accept “primary cells cannot be recharged/reused”, “primary cells can be used only once” OR “lithium batteries may explode”.

a.

Anode (negative electrode):

Li (graphite) → Li+ (electrolyte) + e

OR

LiC6 (s) → 6C (s) + Li+ (electrolyte) + e

 

Cathode (positive electrode):

Li+ (electrolyte) + e + MnO2 (s) → LiMnO2 (s)

OR

Li+ (electrolyte) + e + NiO2 (s) → LiNiO2 (s)

OR

Li+ (electrolyte) + e + CoO2 (s) → LiCoO2 (s)

OR

2Li+ (electrolyte) + 2e + 2CoO2 (s) → Co2O3 (s) + Li2O (s) ✔

 

Accept “polymer” for “electrolyte”.

Award [1 max] if electrodes are reversed.

Do not accept “CO” for “Co”.

b.

«E = EΘ −  ( R T n F ) ln

0.19 = 0.14 ( 8.31 × 298 2 × 96500 ) ln ( [ C d 2 + ] [ 1 ] )

OR

0.05 = –0.01283 ln [Cd2+]

OR

ln[Cd2+] = – 3.897 ✔

[Cd2+] = 0.020 «mol dm–3» ✔

 

Award [2] for correct final answer.

c.

«extensive» conjugation

OR

«extensive» alternate single and double bonds ✔

 

Accept “delocalization”.

d.i.

electrons excited/released «from dye» ✔

 

Accept “photooxidation/oxidizes dye”.

d.ii.

transfers e to external circuit ✔

 

Accept “provides large surface area”.

d.iii.

I3 (aq) + 2e → 3I (aq) ✔

 

Accept “3I2 (aq) + 2e → 2I3 (aq)”.

d.iv.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
d.iii.
[N/A]
d.iv.

Syllabus sections

Options » C: Energy » C.8 Photovoltaic and dye-sensitized solar cells (HL only)
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