CAIE A-Level Chemistry 24.1 Electrolysis

Practise selecting electrode products and completing quantitative electrolysis calculations from half-equations.

Exam points

choose cathode and anode products from molten or aqueous state, concentration and electrode potentials
write balanced reduction and oxidation half-equations with electrons and state symbols
use Q=It and electron stoichiometry with F to connect current and time to product amount

Covered syllabus points

24.1.1—The identities of substances liberated during 24.1.2—The relationship F = Le between the Faraday 24.1.3—Calculate 24.1.4—Determination of a value of the Avogadro

[Maximum number: 4]

Potassium iodide, KI, is used as a reagent in both inorganic and organic chemistry.

(a)

A student electrolyses a solution of KI(aq) for 8 minutes using a direct current. The half-equation for the reaction that occurs at the anode is given.

2 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{e}^{-}

[ 4 ]

(i)

Write a half-equation for the reaction that occurs at the cathode.

Include state symbols.

[ 1 ]

$2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{~g})$ state symbols required

(ii)

After the electrolysis, the $I_{2}(a q)$ produced requires $21.35 \mathrm{~cm}^{3}$ of $0.100 \mathrm{moldm}^{-3} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{aq})$ to react completely.

\mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow 2 \mathrm{NaI}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}(\mathrm{aq})

Calculate the average current used in 8 minutes during the electrolysis.
current = A

[ 3 ]

M1 moles of $\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}=\frac{21.35}{1000} \times 0.100=2.135 \times 10^{-3}$

M2 calc of Q $=2.135 \times 10^{-3} \times 0.5 \times 2 \times 96500=206.02 / 205.68$ ECF

M3 = answer I = $206 /(8 \times 60)=0.428$ OR 0.429 A min 2sf ECF

[Maximum number: 8]

(a)

Complete Table 5.1 to predict the substance liberated at each electrode during electrolysis of the indicated electrolyte with inert electrodes.

Table 5.1

[ 3 ]

electrolyte | substance liberated

at the anode

substance liberated

at the cathode

$\mathrm{PbBr}_{2}(\mathrm{l})$ $\mathrm{Br}_{2} /$ bromine

Pb / lead

concentrated NaCl( aq )

$\mathrm{Cl}_{2} /$ chlorine

$\mathrm{H}_{2} /$ hydrogen

$\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}($ aq )

$\mathrm{O}_{2} /$ oxygen $\left(+\mathrm{H}_{2} \mathrm{O}\right)$

Cu / copper

two for one mark, four for two marks, six for three marks

(b)

An electrolytic cell is set up to determine a value for the Avogadro constant, L. The electrolyte is dilute sulfuric acid and both electrodes are copper.

When a current of 0.600 A is passed through the acid for 30.0 minutes, the anode decreases in mass by 0.350 g .

[ 5 ]

(i)

State the relationship between the Faraday constant, F, and the Avogadro constant, L.

[ 1 ]

$F=\operatorname{Le} \mathbf{OR}$ F is directly proportional to L

(ii)

Use the experimental information in (b) and data from the table on page 23 to calculate a value for the Avogadro constant, L.

Show all working.

[ 4 ]

number of $\mathrm{Cu}^{2+}$ formed $=0.35 / 63.5=5.51 \times 10^{-3}$ $Q=I \times t=0.60 \times 30 \times 60=1080 C$

number of electrons $=1080 / 1.6 \times 10^{-19}=6.75 \times 10^{21}$ ecf

number of $\mathrm{Cu}^{2+}$ ions $=6.75 \times 10^{21} / 2=3.375 \times 10^{21}$ ecf

number of $\mathrm{Cu}^{2+}$ ions per mole $(L)=3.375 \times 10^{21} / 5.51 \times 10^{-3}=6.12 \times 10^{23}$ ecf min 2sf

all five points for four marks

Marking guidance:

ALLOW valid alternate calculations of L

All question bank results loaded

Covered syllabus points

24.1 Electrolysis question 1

Question (a)

Question (i)

Question (ii)

24.1 Electrolysis question 2

Question (a)

Question (b)

Question (i)

Question (ii)