Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 4

Table of contents
- a 4
  - Solution a
- b 4
  - Solution b
- c 4
  - Solution c
- d 4
  - i d
    - Solution i
  - ii d
    - Solution ii
- e 4
  - Solution e

Solution

Write both electrode reactions first as reductions:

C H_{3} CO O^{-} + 5 H^{+} + 4 e^{-} ⟶ C_{2} H_{5} OH + H_{2} O,

(1)

Ce O_{2} + 4 H^{+} + e^{-} ⟶ C e^{3 +} + 2 H_{2} O .

(2)

The conventional cell reaction runs with oxidation on the left and reduction on the right. Reverse the first equation and multiply the second by four:

C_{2} H_{5} OH + 4Ce O_{2} + 11 H^{+} ⟶ C H_{3} CO O^{-} + 4C e^{3 +} + 7 H_{2} O .

(3)

Four electrons pass through the external circuit per reaction as written. This direction consumes the oxide; deposition requires the reverse electrolysis direction.

Solution

 0  0

E^{\circ} = E_{right}^{\circ} - E_{left}^{\circ} = 1.660 - (- 0.285) = 1.945 V .

(1)

With

n = 4

Δ_{r} G^{\circ} = - n F E^{\circ} = - 4 (96485) (1.945) = - 7.51 \times 1 0^{5} J mo l^{- 1} .

(2)

Thus the conventional forward reaction is spontaneous under standard conditions, dissolving CeO

_{2}

while oxidizing ethanol. Oxide deposition is its reverse and needs external electrical work; its intention does not change the sign convention.

Solution

 0  0

Measure the reversible cell potential at several temperatures while maintaining standard-state thermodynamic activities and the same phases. From

Δ_{r} G^{\circ} = - n F E^{\circ}

and

Δ_{r} S^{\circ} = - (\partial Δ_{r} G^{\circ} / \partial T)_{p}

Δ_{r} S^{\circ} = n F (\frac{\partial E ^{\circ}}{\partial T})_{p} .

(1)

The sign of the slope gives the sign of the standard reaction entropy. Measuring a loaded cell voltage without correcting polarization would not give this reversible thermodynamic derivative.

Solution

 0  0

Pure solid CeO

_{2}

and liquid water have unit thermodynamic activity in the stated approximation, so

Q = \frac{a _{C e^{3 +}}^{4} a _{C H_{3} CO O^{-}}}{a _{C_{2} H_{5} OH} a _{H^{+}}^{11}}, E = E^{\circ} - \frac{RT}{4 F} ln Q .

(1)

The power 11 and the four-electron denominator follow from the balanced conventional reaction; using the deposition reaction without reversing the potential would give inconsistent signs.

At pH 7, $E ≃ 0.792 V$ . This is the formal fixed-thermodynamic activity calculation requested; actual cerium speciation and thermodynamic activities would need to remain compatible with the chosen pH to realize it experimentally.

e

 0  0

Solution

 0  0

Passivation by the oxide film is a likely cause. As poorly conducting CeO

_{2}

covers palladium and thickens, electron transport through the deposit becomes difficult and the interfacial resistance rises. Transport of ions/defects through the film can also become limiting.

Thus abundant cerium in bulk solution does not guarantee continued interfacial deposition. The film can block the electronic or ionic path required to grow more oxide.

This is an oxide-film transport limitation during cerium dioxide electrodeposition.

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 4

a

Solution

b

Solution

c

Solution

d

i

Solution

ii

Solution

e

Solution

 Ancestors (11)

 Discussion (0)

 Articles by others on the same topic (0)

Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 4

 Discussion (0)  Subscribe (1)

 Discussion (0)