Answers to questions on photovoltaic and dye-sensitized solar cells

1. (a) Metals have relatively low ionization energies and their outer electrons are delocalized so can move freely as the metal conducts electricity. In non-metals the outer electrons are normally held in fixed positions forming covalent bonds between atoms so are unable to move.

(b) Adding gallium (a group 13 element) causes a ‘hole’ in the silicon layer as it has one outer electron less than silicon. This ‘hole’ can be regarded as a positive carrier so it is known as a p-type. Arsenic (a group 15 element) contains an extra outer electron, which can move more easily through the silicon lattice making it a better conductor than pure silicon. This makes it a negative carrier or n-type semiconductor

2. (a) Ethene 1, benzene 4 and β-carotene 13.

(b) Ethene contains just one double bond so requires high energy (shorter wavelength) light to excite an electron to an anti-bonding orbital. As the amount of conjugation increases the energy required is less so the absorbance occurs at longer wavelength. Benzene has three conjugated double bonds and in β-carotene there is extensive conjugation (so much so that the absorption occurs in the visible region).

(c) In 1,3-pentadiene there is conjugation (as the two C=C double bonds are separated by just one single C-C bond) so it will absorb at a longer wavelength (the actual value is 225 nm) than 1,4-pentadiene (178 nm) where there is no conjugation as the two C=C double bonds are separated by two C-C single bonds.

3. (a) (i) The dye absorbs the visible light and releases an electron as it becomes oxidized.

(ii) The tin(IV) oxide acts as the anode from which electrons are conducted through the external circuit.

(iii) The titanium dioxide provides a large surface area upon which the photosensitive dye is covalently bonded and it acts as a semiconductor.

(iv) The platinum acts as the cathode receiving electrons form the external circuit and feeding them into the electrolyte so a redox reaction can take place.

(b) Iodide ions in the electrolyte reduce the dye back to its normal state (after it has been oxidized by the light). In this process iodide ions are converted into triiodide ions (which can be thought of as iodine and iodide ions).

3I⁻(aq) → I₃⁻ (aq) + 2e⁻

The electrolyte receives electrons from the platinum cathode which reduce I₃⁻(aq) ions back to iodide ions.

I₃⁻ (aq) + 2e⁻ → 3I⁻(aq)

(c) Any two from:
Simpler to mIB Docs (2) Teamfacture
Cheaper
Absorb a larger range of wavelengths
Semi-flexible

Answers to Photovoltaic and dye-sensitized solar cells questions