19.1 Electrochemical Cells

Question 1

Marks: 1

Which of the following can be used for a standard hydrogen electrode (SHE)?

	Electrode	Electrolyte solution
A.	Graphite	1 mol dm^-3 H₂SO₄
B.	Graphite	1 mol dm^-3 HCl
C.	Platinum	0.5 mol dm^-3 H₂SO₄
D.	Platinum	0.5 mol dm^-3 HCl

Key Concepts

The Standard Hydrogen Electrode

Question 2

Marks: 1

What are the ratios of gases produced at the electrodes in the electrolysis of dilute vs concentrated sodium chloride solution?

	Ratio of gas produced at cathode : anode in the electrolysis of dilute NaCl	Ratio of gas produced at cathode : anode in the electrolysis of concentrated NaCl
A.	1:1	2:1
B.	1:1	1:2
C.	1:2	1:1
D.	2:1	1:1

Key Concepts

Electrolysis of Aqueous Solutions

Question 3

Marks: 1

Which of the following could be used to electroplate a zinc medal?

1.0 mol dm^-3 AgNO₃ solution with a silver anode
1.0 mol dm^-3 AgNO₃ with a silver cathode
1.0 mol dm^-3 CuSO₄ solution with a copper cathode
1.0 mol dm^-3 solution of SnCl₂ and a tin anode

Key Concepts

Electroplating

Question 4

Marks: 1

The diagram below shows the set-up of aluminium and silver cells in series:

silver-and-aluminium-cells-in-series-ib-hl-mcq-19-1-q4

What calculation shows the loss in mass of the aluminium electrode if the silver electrode gains 0.25 g?

Al mass lost = $\frac{107.87 \times 3 \times 0.25}{26.98}$
Al mass lost = $\frac{107.87 \times 0.33 \times 0.25}{26.98}$
Al mass lost = $\frac{26.98 \times 0.33 \times 0.25}{107.87}$
Al mass lost = $\frac{26.98 \times 3 \times 0.25}{107.87}$

Key Concepts

Quantitative Electrolysis

Question 5

Marks: 1

The oxidation of iron is a spontaneous process described by the overall equation:

Fe (s) + $\frac{1}{2}$ O₂(g) + H₂O (l) → Fe(OH)₂ (s)

(ΔG^θ = -164 kJ mol^-1at 298 K)

The two half equations for the process are:

Fe²⁺ (s) + 2e^- → Fe (s)

$\frac{1}{2}$ O₂(g) + H₂O (l) + 2e^- → 2OH^- (s)

(ΔG^θ= -nFE^θ, F = 9.65 x 10⁴ C mol^-1)

Which is the correct calculation to work out E^θin V?

E^θ= $\frac{- 164}{- 2 \times 9.65}$
E^θ= $\frac{- 164 000}{- 2 \times 96 500}$
E^θ= $\frac{- 164 000}{- 0.5 \times 96 500}$
E^θ= $\frac{- 164}{- 0.5 \times 9.65}$

Key Concepts

Free Energy & Eº

DP IB Chemistry: HL

Topic Questions

19.1 Electrochemical Cells

Question 1

Key Concepts

Question 2

Key Concepts

Question 3

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Question 4

Key Concepts

Question 5

Key Concepts

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