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Date	November 2018	Marks available	2	Reference code	18N.3.hl.TZ0.15
Level	HL	Paper	3	Time zone	TZ0
Command term	Determine	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]

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

[2]

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) E^Θ_cell = 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]

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»

secondary cells need to be charged before use

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”.

Anode (negative electrode):

Li (graphite) → Li⁺ (electrolyte) + e^–

LiC6 (s) → 6C (s) + Li⁺ (electrolyte) + e^– ✔

Cathode (positive electrode):

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

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

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

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”.

«E = E^Θ− $\left( {\frac{{RT}}{{nF}}} \right)$ lnQ»

$0.19 = 0.14 - \left( {\frac{{8.31 \times 298}}{{2 \times 96500}}} \right)\,{\text{ln}}\,\left( {\frac{{\left[ {{\text{C}}{{\text{d}}^{2 + }}} \right]}}{{\left[ 1 \right]}}} \right)$