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Date	May 2019	Marks available	2	Reference code	19M.3.hl.TZ2.19
Level	HL	Paper	3	Time zone	TZ2
Command term	Determine	Question number	19	Adapted from	N/A

Question

Electricity can be generated in a variety of ways.

Outline how a microbial fuel cell produces an electric current from glucose.

C₆H₁₂O₆ (aq) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l)

[3]

The cell potential for the spontaneous reaction when standard magnesium and silver half-cells are connected is +3.17 V.

Determine the cell potential at 298 K when:

[Mg²⁺] = 0.0500 mol dm⁻³
[Ag⁺] = 0.100 mol dm⁻³

Use sections 1 and 2 of the data booklet.

[2]

Outline one difference between a primary and a secondary cell.

[1]

Markscheme

Any three of:
C₆H₁₂O₆ (aq) + 6H₂O (l) → 6CO₂ (g) + 24H⁺ (aq) + 24e^–
OR
electrons released by oxidation of glucose [✔]

enzymes «in bacteria» oxidize glucose
OR
«bacteria» transfer «released» electrons directly to anode [✔]

24H⁺ (aq) + 6O₂ (g) + 24e^– → 12H₂O (l)
OR
electrons consumed by reduction of oxygen [✔]

PEM/membrane separates two half reactions
OR
PEM/membrane allows proton/H⁺ transfer from anode to cathode [✔]

electrons flow though external circuit [✔]

Note: Accept 4H⁺ (aq) + O₂ (g) + 4e^– → 2H₂O (l).

«E = Eᶱ $\frac{{RT}}{{nF}}$ × lnQ»

lnQ = « $\frac{{[{\text{M}}{{\text{g}}^{2 + }}]}}{{{{[{\text{A}}{{\text{g}}^ + }]}^2}}} = \frac{{0.0500}}{{{{0.100}^2}}} = \ln 5.00 =$ » 1.61 [✔]

E = «3.17 V $- \frac{{8.31{\text{ J }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \times 298{\text{ K}}}}{{2 \times 96500{\text{ J }}{{\text{V}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}}} \times \ln \frac{{0.0500}}{{{{0.100}^2}}} = 3.17 - 0.021 = +$ »3.15 «V» [✔]

Note: Award [2] for correct final answer.

Any one of:

Note: Accept “primary cannot be recharged AND “secondary can be recharged”.

Examiners report

The question on microbial cell invited varied responses. Half equations for the oxidation of glucose or reduction of oxygen were rarely written. PEM/membrane separates two half- reactions and allows proton transfer from anode to cathode was missed by most of the candidates.

The cell potential was correctly calculated by several candidates with some candidates managed an ECF mark for an error in the calculation. Unfortunately, the ln Q part was frequently wrong due to candidates forgetting to square the denominator.

Most candidates were able to state one difference between a primary and a secondary cell.

Syllabus sections

Options » C: Energy » C.6 Electrochemistry, rechargeable batteries and fuel cells (HL only)

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