Write the balanced equation for the reaction in this voltaic cell.

Calculate the cell potential for $0.0100 \mathrm{mol} {\mathrm{dm}}^{-3} {\mathrm{Mg}}^{2+}\left(\mathrm{aq}\right)$ and $0.800 \mathrm{mol} {\mathrm{dm}}^{-3} {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)$ at $298 \mathrm{K}$. Use sections 1, 2 and 24 of the data booklet.

Predict, giving a reason, how an increase in temperature affects the potential of this cell.

$\mathrm{Mg}\left(\mathrm{s}\right)+{\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)\to {\mathrm{Mg}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Ni}\left(\mathrm{s}\right)$ ✔

Accept a balanced molecular equation such as “$Mg\mathit{+}NiS{O}_{\mathit{4}}\mathit{\to }MgS{O}_{\mathit{4}}\mathit{+}Ni$”.

$E^{\mathrm{Ɵ}}=«2.37-0.26=»(+)2.11«\mathrm{V}»$ ✔

$«Q=\left(\frac{0.0100}{0.800}\right)=»0.0125$ AND $«n=»2$ ✔

$«E=E^{Ɵ}-\left(\frac{RT}{nF}\right) \ln Q=2.11-\left(\frac{8.31\times 298}{2\times 96 500}\right) \ln 0.0125=»(+)2.17«\mathrm{V}»$ ✔

Award [3] for correct final answer.

cell potential/$E$ increases AND increasing temperature favours forward reaction
OR
cell potential/$E$ increases AND $∆G$ becomes more negative
OR
cell potential/$E$ increases AND $RT/nF \ln Q$ becomes more negative ✔

Accept any correct mathematical explanation using the Nernst equation.

Nearly all were able to write the balanced equation for the reaction occurring in the voltaic cell.

The cell potential numerical problem was well executed and many scored all three marks, by calculating the final answer as +2.17 V. The most common errors related to either incorrectly calculating Q or identifying n = 2 in the intermediate, numerical step.

This was often gained as an easy subsequent mark for those that scored full marks in (b). The majority approached this problem by applying a mathematical explanation using the Nernst equation.

The Geobacter species of bacteria can be used in microbial fuel cells to oxidise aqueous ethanoate ions,
CH₃COO⁻(aq), to carbon dioxide gas.

State the half-equations for the reactions at both electrodes.

A concentration cell is an example of an electrochemical cell.

(i) State the difference between a concentration cell and a standard voltaic cell.

(ii) The overall redox equation and the standard cell potential for a voltaic cell are:

Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)     E^θ_cell = +1.10 V

Determine the cell potential E at 298 K to three significant figures given the following concentrations in mol dm⁻³:

[Zn²⁺] = 1.00 × 10⁻⁴       [Cu²⁺] = 1.00 × 10⁻¹

Use sections 1 and 2 of the data booklet.

(iii) Deduce, giving your reason, whether the reaction in (b) (ii) is more or less spontaneous than in the standard cell.

Dye-sensitized solar cells (DSSC) convert solar energy into electrical energy.

(i) Describe how a DSSC converts sunlight into electrical energy.

(ii) Explain the role of the electrolyte solution containing iodide ions, I⁻, and triiodide ions, I₃⁻, in the DSSC.

Negative electrode (anode): CH₃COO⁻ (aq) + 2H₂O (l) → 2CO₂ (g) + 7H⁺ (aq) + 8e⁻

Positive electrode (cathode): O₂ (g) + 4H⁺ (aq) + 4e⁻ → 2H₂O (l)

Accept equilibrium signs in equations.
Award [1 max] if correct equations are given at wrong electrodes.

i
concentration cell has different concentrations of electrolyte «solutions» «but same electrodes and electrolytes»
OR
standard voltaic cell has different electrodes/electrolytes «but same concentration of electrolytes»
Accept “both half-cells in concentration cell made from same materials”.

ii
«$E=1.10-\left(\frac{RT}{nF}\right)\ln \frac{\left[\text{Z}{\text{n}}^{2+}\right]}{\left[\text{C}{\text{u}}^{2+}\right]}=1.10-\left(\frac{8.31\times 298}{2\times 96500}\right)\ln \frac{{10}^{-4}}{{10}^{-1}}=1.10+0.0886=$»

(+) 1.19 «V»
3 significant figures needed for mark.

iii
more spontaneous because E > E^θ_cell

i

photon/«sun»light absorbed by the dye/photosensitizer/«transition» metal complex
OR
dye/photosensitizer/«transition» metal complex excited by photon/«sun»light 

electron«s» move«s» to conduction band
OR
electron«s» transferred to semiconductor/TiO₂     

ii

I₃⁻ + 2e⁻ → 3I⁻ «at cathode»
OR
triiodide ions/I₃⁻ reduced into/produce iodide ions/I⁻ «at cathode»

iodide ions/I⁻ reduce dye/act as reducing agent AND oxidized into/produce triiodide ions/I₃⁻ 
OR
dye⁺ + e⁻ → dye AND 3I^- → I₃⁻ + 2e⁻

Deduce the half-equations for the reactions occurring at the electrodes.

Anode (negative electrode):

Cathode (positive electrode):

Calculate the cell potential, E^θ, in V, using section 24 of the data booklet.

Suggest how PEM fuel cells can be used to produce a larger voltage than that calculated in (b)(i).

Suggest an advantage of the PEM fuel cell over the lead-acid battery for use in cars.

Outline the functions of the dye, TiO₂ and the electrolyte in the operation of the DSSC.

Dye: 

TiO₂:

Electrolyte:

Suggest an advantage of the DSSC over silicon-based photovoltaic cells.

Anode (negative electrode):
H₂ (g) → 2H⁺ (aq) + 2e⁻ ✔

Cathode (positive electrode):
O₂ (g) + 4H⁺ (aq) + 4e⁻ → 2H₂O (l) ✔

NOTE: Accept any correct integer or fractional coefficients. Award [1 max] for M1 and M2 if correct half-equations are given at the wrong electrodes OR if incorrect reversed half-equations are given at the correct electrodes.

(+)1.23 «V» ✔

NOTE: Do not accept “-1.23 «V»”.

connect several fuel cells in series
OR
increase pressure/concentration of reactant/hydrogen/oxygen ✔

NOTE: Do not accept changes in [H⁺]/pH as they do not affect cell potential in this case.
Do not accept reference to quantity for “concentration”.

liquid in cell is less/not corrosive
OR
does not contain lead/toxic chemicals
OR
larger energy density/charge capacity/current per unit mass
OR
does not have to be charged prior to use / is always ready for use «as long as fuel is available» ✔

Dye:
absorbs photons/light
OR
releases electrons ✔

TiO2:
conducts current/electricity
OR
semiconductor ✔

Electrolyte:
reduces/regenerates «the oxidized» dye ✔

Any one of:
cheaper/ease of manufacture
OR
plentiful and renewable resources «to construct DSSC cells» ✔

use light of lower energy/lower frequency/longer wavelength
OR
use of nanoparticles provides large surface area for exposure to sunlight/sun/light
OR
can absorb better under cloudy conditions ✔

operate at lower «internal» temperatures
OR
better at radiating heat away «since constructed with thin front layer of conductive plastic compared to glass box in photovoltaic cells» ✔

better conductivity ✔

more flexible/durable ✔

NOTE: Accept “lower mass/lighter «so greater flexibility to integrate into windows etc.»” OR “greater power-conversion efficiency «with latest DSSC models»”.