CAIE A Level Chemistry 24 Electrochemistry

Practise predicting electrolysis products and using charge, current, time, Faraday constant and electron moles, linking quantitative calculations to molten or aqueous electrolytes.

Covered syllabus points

24.1 Electrolysis 24.2 Standard electrode potentials E ⦵ , standard cell potentials E ⦵ cell and the Nernst

[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: 13]

(a)

Define the term standard cell potential.

[ 2 ]

the potential difference between two half-cells/two electrodes (in a cell)

under standard conditions of $1 \mathrm{~atm} ., 298 \mathrm{~K}$ , (all) solutions being $1 \mathrm{~mol} \mathrm{dm}^{-3}$

(b)

Draw a fully labelled diagram of the experimental set-up you could use to measure the standard electrode potential of the $\mathrm{Pb}^{2+}(\mathrm{aq}) / \mathrm{Pb}(\mathrm{s})$ electrode. Include the necessary chemicals.

[ 4 ]

Figure for Answer to Question (b) — CAIE A-Level Chemistry A2

8 marking points, any 2 points for each mark

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

correct delivery system for $\mathrm{H}_{2}$ $\mathrm{Pb}^{2+}$ (aq)

Pb electrode

Pt electrode

$\mathrm{H}^{+}(\mathrm{aq})$ solution

salt bridge

voltmeter/V labelled

(c)

The $E^{\ominus}$ for a $\mathrm{Pb}^{2+}(\mathrm{aq}) / \mathrm{Pb}(\mathrm{s})$ electrode is -0.13 V .

Suggest how the E for this electrode would differ from its $E^{\ominus}$ value if the concentration of $\mathrm{Pb}^{2+}(\mathrm{aq})$ ions is reduced. Indicate this by placing a tick $(\checkmark)$ in the appropriate box in the table.

Table for Question (c) — CAIE A-Level Chemistry A2

Explain your answer.

[ 2 ]

more negative

shifts $\mathrm{Pb}^{2+}\left(+2 \mathrm{e}^{-}\right) \leftrightharpoons \mathrm{Pb}$ equilibrium / reaction to the left

(d)

Car batteries are made up of rechargeable lead-acid cells. Each cell consists of a negative electrode made of Pb metal and a positive electrode made of $\mathrm{PbO}_{2}$ . The electrolyte is $\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})$ .

When a lead-acid cell is in use, $\mathrm{Pb}^{2+}$ ions are precipitated out as $\mathrm{PbSO}_{4}(\mathrm{~s})$ at the negative electrode.

\mathrm{Pb}(\mathrm{~s})+\mathrm{SO}_{4}{ }^{2-}(\mathrm{aq}) \rightarrow \mathrm{PbSO}_{4}(\mathrm{~s})+2 \mathrm{e}^{-}

[ 3 ]

(i)

Calculate the mass of Pb that is converted to $\mathrm{PbSO}_{4}$ when a current of 0.40 A is delivered by the cell for 80 minutes.

\text { mass of } \mathrm{Pb}=

[ 2 ]

$\begin{aligned} & Q=0.4 \times 80 \times 60=1920 \mathrm{C} \text { and use of } 96500 / 193000 & \text { Moles of } \mathrm{Pb}=1920 / 193000=9.95 \times 10^{-3} & \text { Mass of } \mathrm{Pb}=207.2 \times 9.95 \times 10^{-3}=\mathbf{2 . 1 ~ g} \end{aligned}$

OR
$\begin{aligned} & \mathrm{Q}=0.4 \times 80 \times 60=1920 \mathrm{C} \text { and use of } 1.6 \times 10^{-19} / 1.2 \times 10^{22} & \text { atoms } \mathrm{Pb}=6 \times 10^{21} ; \text { moles of } \mathrm{Pb}=6 \times 10^{21} / 6 \times 10^{23}=0.01 & \text { Mass of } \mathrm{Pb}=207.2 \times 0.01=\mathbf{2 . 1 ~ g} \end{aligned}$

(ii)

Complete the half-equation for the reaction taking place at the positive electrode.

\mathrm{PbO}_{2}(\mathrm{~s})+\mathrm{SO}_{4}{ }^{2-}(\mathrm{aq})+\ldots \ldots \ldots \ldots . .+\ldots \ldots \ldots \ldots . \rightarrow \mathrm{PbSO}_{4}(\mathrm{~s})+\ldots \ldots \ldots \ldots .

[ 1 ]

$\mathrm{PbO}_{2}(\mathrm{~s})+\mathrm{SO}_{4}{ }^{2-}(\mathrm{aq})+\mathbf{4 H}^{+}+\mathbf{2 e}^{-} \rightarrow \mathrm{PbSO}_{4}(\mathrm{~s})+\mathbf{2 H}_{2} \mathbf{O}$

(e)

The diagrams show how the voltage across two different cells changes with time when each cell is used to provide an electric current.

Figure for Question (e) — CAIE A-Level Chemistry A2

Suggest a reason why
- the voltage of the lead-acid cell changes after several hours,
- the voltage of the fuel cell remains constant.

[ 2 ]

reagents/ $\mathrm{PbO}_{2} / \mathrm{H}_{2} \mathrm{SO}_{4}$ and used up/concentration decreases

as fuel/hydrogen is being continuously supplied/fuel has not run out

[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

[Maximum number: 17]

Some electrode potentials are shown in Table 3.1.

Table 3.1

(a)

Complete the diagram to show a standard hydrogen electrode.

Label your diagram. Identify all substances. You do not need to state standard conditions.

Figure for Question (a) — CAIE A-Level Chemistry A2

[ 1 ]

hydrogen, delivery system, $\mathrm{H}^{+}$ , platinum, [1]

(b)

An electrochemical cell is set up using an $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ electrode and a standard hydrogen electrode.

Identify the positive electrode in the electrochemical cell and the direction of electron flow in the external circuit.
positive electrode
Electrons flow from the electrode to the electrode.

[ 1 ]

(c)

The vanadium-containing species in the electrode reactions given in Table 3.1 are $\mathrm{V}, \mathrm{V}^{2+}$ , $\mathrm{V}^{3+}, \mathrm{VO}^{2+}$ and $\mathrm{VO}_{2}{ }^{+}$ .

[ 3 ]

(i)

Identify one vanadium-containing species that does not react with $\mathrm{Fe}^{2+}$ ions under standard conditions.

Use data from Table 3.1 to explain your answer.

[ 1 ]

(for specified $\mathrm{V}^{2+}, \mathrm{V}^{3+}$ or $\mathrm{VO}^{2+}$ ) $\mathrm{E}^{\ominus}$ is more positive than / above -0.44 AND more negative than / below 0.77 V [1]

(ii)

Identify all the vanadium-containing species that will react with $\mathrm{Fe}^{2+}$ ions under standard conditions.

[ 1 ]

V and $\mathrm{VO}_{2}{ }^{+}$ [1]

(iii)

Write an equation for one of the possible reactions identified in (ii).

[ 1 ]

$\begin{aligned} \mathrm{V}+\mathrm{Fe}^{2+} \rightarrow \mathrm{V}^{2+}+\mathrm{Fe} \mathrm{VO}_{2}{ }^{+}+2 \mathrm{H}^{+}+\mathrm{Fe}^{2+} \rightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}+\mathrm{Fe}^{3+}[1] \end{aligned}$

(d)

Another electrochemical cell is set up using an $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ electrode and an alkaline $\mathrm{ClO}^{-} / \mathrm{Cl}^{-}$ electrode.

The concentration of $\mathrm{Fe}^{3+}$ is 1000 times greater than the concentration of $\mathrm{Fe}^{2+}$ in the $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ electrode. All other conditions are standard.

[ 3 ]

(i)

Use the Nernst equation to calculate the E value of the $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ electrode.

Show your working.

[ 2 ]

Nernst: $\mathrm{E}=0.77+(0.059 / \mathrm{z}) \log [\mathrm{ox}] /[\mathrm{red}][1]$
0.947 [1]

(ii)

Write an equation for the reaction that occurs in the cell, under these conditions.

[ 1 ]

$2 \mathrm{Fe}^{3+}+\mathrm{Cl}^{-}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}[1]$

(e)

Another electrochemical cell is set up using an $\mathrm{Fe}^{2+} / \mathrm{Fe}$ electrode and an alkaline $\mathrm{ClO}^{-} / \mathrm{Cl}^{-}$ electrode under standard conditions.

Calculate the value of $\Delta G^{\ominus}$ for the cell.
$\Delta G^{\ominus}=$ $\mathrm{kJ} \mathrm{mol}^{-1}$

[ 3 ]

$\begin{aligned} \mathrm{E}^{\ominus}{ }_{\text {cell }}=1.33 \mathrm{~V}[1] \Delta \mathrm{G}^{\ominus}=-\mathrm{nE}^{\ominus}{ }_{\text {cell }} F[1] -257[1] \end{aligned}$

(f)

A solution of iron(II) sulfate, $\mathrm{FeSO}_{4}(\mathrm{aq})$ is electrolysed with iron electrodes. Under the conditions used, no gas is evolved at the cathode.

A current of 0.640 A is passed for 17.0 minutes. The mass of the cathode increases by 0.185 g .
Use these results to calculate an experimental value for the Avogadro constant, L.
Show your working.

[ 3 ]

$0.64 \times 17 \times 60=653 / 652.8$ Coulombs [1]

$652.8 \div 1.6 \times 10^{-19}=4.08 \times 10^{21}$ (number of electrons)

$4.08 \times 10^{21} \div 2=2.04 \times 10^{21}$ (number of atoms Fe) [1]

$0.185 \div 55.8=3.31 \times 10^{-3}$ (number of moles Fe atoms

$2.04 \times 10^{21} \div 3.31 \times 10^{-3}=L=6.153 \times 10^{23}$ [1]

(g)

Iron(II) chloride, $\mathrm{FeCl}_{2}$ , is oxidised by chlorine to form iron(III) chloride, $\mathrm{FeCl}_{3}$ , under standard conditions.

2 \mathrm{FeCl}_{2}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{FeCl}_{3}(\mathrm{~s})

\Delta H^{\ominus}=-128 \mathrm{~kJ} \mathrm{~mol}^{-1}

Table 3.2

[ 3 ]

(i)

Use Table 3.2 and other data to calculate the Gibbs free energy change, $\Delta G^{\ominus}$ , for this reaction.

Show your working.

\Delta G^{\ominus}=

\mathrm{kJ} \mathrm{~mol}^{-1}

[ 3 ]

$\Delta \mathrm{S}=-179[1]$ $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}[1]$

-74.7

All question bank results loaded

Covered syllabus points

24. Electrochemistry question 1

Question (a)

Question (i)

Question (ii)

24. Electrochemistry question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (i)

Question (ii)

Question (e)

24. Electrochemistry question 3

Question (a)

Question (b)

Question (i)

Question (ii)

24. Electrochemistry question 4

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (iii)

Question (d)

Question (i)

Question (ii)

Question (e)

Question (f)

Question (g)

Question (i)