An electrochemical cell is constructed using two half-cells. an $\text{Sn}^{4+}/\text{Sn}^{2+}$ half-cell an $\text{Al}^{3+}/\text{Al}$ half-cell (a) State the material used for the electrode in each half-cell. $\text{Sn}^{4+}/\text{Sn}^{2+}$ half-cell $\text{Al}^{3+}/\text{Al}$ half-cell (b) The cell is operated at 298 K. The $\text{Al}^{3+}/\text{Al}$ half-cell has standard concentrations. The $\text{Sn}^{4+}/\text{Sn}^{2+}$ half-cell has $[\text{Sn}^{4+}] = 0.300\text{…

(a) Sn4+/Sn2+: platinum (Pt); Al3+/Al: aluminium (Al) · (b)(i) +0.16 V · (b)(ii) +1.82 V · (b)(iii) 2Al + 3Sn4+ -> 2Al3+ + 3Sn2+ · (c) 2420 kg · (d) Hydrogen gas is produced instead of chromium metal because H+/H2 (E° = 0.00 V) is more positive than Cr2+/Cr (E° = -0.91 V).

Chemistry 9701/43 — October/November 2019

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

100

marks

120

minutes

Topics Introduction to A Level Organic Chemistry · Equilibria · Hydrocarbons · Analytical Techniques · Nitrogen Compounds · Electrochemistry · +6 more

Q1ElectrochemistryFree sample

An electrochemical cell is constructed using two half-cells.

an $\text{Sn}^{4+}/\text{Sn}^{2+}$ half-cell
an $\text{Al}^{3+}/\text{Al}$ half-cell

(a)

State the material used for the electrode in each half-cell.

$\text{Sn}^{4+}/\text{Sn}^{2+}$ half-cell
$\text{Al}^{3+}/\text{Al}$ half-cell

DifficultyEasy

Worked solution

Answer

$\text{Sn}^{4+}/\text{Sn}^{2+}$ half-cell: platinum (Pt)
$\text{Al}^{3+}/\text{Al}$ half-cell: aluminium (Al)

Final answer

Sn4+/Sn2+: platinum (Pt); Al3+/Al: aluminium (Al)

Detailed explanation

Background Concept

In an electrochemical cell, each half-cell needs an electrode to allow electrons to be transferred to or from the species in solution. For a metal/metal-ion half-cell like $\text{Al}^{3+}/\text{Al}$ , the metal itself serves as the electrode — it is both the solid reactant/product and the electron conductor. For a redox couple where both species are in solution (like $\text{Sn}^{4+}/\text{Sn}^{2+}$ ), no solid metal is available, so an inert electrode must be used. Platinum is the standard choice because it is chemically unreactive and conducts electricity well.

Understanding the Question

The question asks for the electrode material in each half-cell. The $\text{Sn}^{4+}/\text{Sn}^{2+}$ couple involves two aqueous ions, so it needs an inert electrode. The $\text{Al}^{3+}/\text{Al}$ couple involves solid aluminium metal, so the metal itself acts as the electrode.

Approach

Identify whether each half-cell involves a solid metal. If yes, that metal is the electrode. If both species are aqueous ions, use an inert conductor — platinum is the standard.

Step-by-Step Reasoning

$\text{Sn}^{4+}/\text{Sn}^{2+}$ : Both tin species are aqueous ions. There is no solid tin metal in the half-cell. An inert electrode is needed — platinum (Pt) is the standard choice because it does not participate in the redox reaction and conducts electrons.
$\text{Al}^{3+}/\text{Al}$ : Aluminium is a solid metal. The aluminium metal itself acts as the electrode — it is the site where $\text{Al}^{3+}$ is reduced (or Al oxidised) and it conducts electrons.

Key Takeaways

Metal/metal-ion half-cells use the metal as the electrode.
Redox couples with only aqueous species need an inert electrode — platinum is standard.

Common Mistakes

Writing "graphite" instead of platinum — graphite is sometimes used but platinum is the standard answer for CIE.
Confusing the two — saying aluminium is used for the Sn half-cell.

Things to Be Careful About

The mark scheme requires BOTH answers for the single mark — missing one loses the mark.

Techniques used

identify the electrode materialrecognise inert vs reactive electrodes

(b)

The cell is operated at 298 K.

The $\text{Al}^{3+}/\text{Al}$ half-cell has standard concentrations.

The $\text{Sn}^{4+}/\text{Sn}^{2+}$ half-cell has $[\text{Sn}^{4+}] = 0.300\text{ mol dm}^{-3}$ and $[\text{Sn}^{2+}] = 0.150\text{ mol dm}^{-3}$ .

(i)

Use the Nernst equation to calculate the electrode potential, $E$ , of the $\text{Sn}^{4+}/\text{Sn}^{2+}$ half-cell under these conditions.

DifficultyMedium-Easy

Worked solution

Working

The Nernst equation for the $\text{Sn}^{4+}/\text{Sn}^{2+}$ half-cell is:

E = E^\ominus + \frac{0.0590}{z} \log \frac{[\text{Sn}^{4+}]}{[\text{Sn}^{2+}]}

where $z = 2$ and $E^\ominus = +0.15\text{ V}$ .

E = 0.15 + \frac{0.0590}{2} \log \frac{0.300}{0.150}

E = 0.15 + 0.0295 \times \log 2 = 0.15 + 0.0295 \times 0.301 = 0.15 + 0.00888

Answer

$E = +0.159\text{ V} \approx +0.16\text{ V}$

Final answer

+0.16 V

Detailed explanation

Background Concept

The Nernst equation relates the electrode potential of a half-cell to the standard electrode potential and the concentrations of the oxidised and reduced species:

E = E^\ominus + \frac{0.0590}{z} \log \frac{[\text{ox}]}{[\text{red}]}

where $z$ is the number of electrons transferred in the half-reaction and 0.0590 V is the value of $2.303RT/F$ at 298 K. The equation allows us to calculate the actual potential when concentrations are not standard (1 mol dm $^{-3}$ ).

For $\text{Sn}^{4+} + 2\text{e}^- \rightleftharpoons \text{Sn}^{2+}$ , $z = 2$ and $E^\ominus = +0.15\text{ V}$ .

Understanding the Question

We are given $[\text{Sn}^{4+}] = 0.300\text{ mol dm}^{-3}$ and $[\text{Sn}^{2+}] = 0.150\text{ mol dm}^{-3}$ , both non-standard. We need to calculate the actual electrode potential using the Nernst equation.

Approach

Substitute the given values into the Nernst equation. The ratio $[\text{ox}]/[\text{red}] = [\text{Sn}^{4+}]/[\text{Sn}^{2+}] = 0.300/0.150 = 2$ . Then compute $E$ .

Step-by-Step Reasoning

Write the Nernst equation: $E = E^\ominus + (0.0590/z) \log([\text{Sn}^{4+}]/[\text{Sn}^{2+}])$
Substitute: $E = 0.15 + (0.0590/2) \times \log(0.300/0.150)$
Compute the ratio: $0.300/0.150 = 2$
$\log 2 = 0.301$
$(0.0590/2) \times 0.301 = 0.0295 \times 0.301 = 0.00888\text{ V}$
$E = 0.15 + 0.00888 = 0.1589 \approx 0.159\text{ V} \approx 0.16\text{ V}$

The potential is slightly more positive than the standard value because $[\text{ox}] > [\text{red}]$ , which favours reduction and makes the potential more positive.

Key Takeaways

The Nernst equation corrects the standard potential for non-standard concentrations.
$z$ is the number of electrons in the half-reaction (here 2).
When $[\text{ox}] > [\text{red}]$ , $E > E^\ominus$ ; when $[\text{ox}] < [\text{red}]$ , $E < E^\ominus$ .

Common Mistakes

Using $z = 1$ instead of 2.
Using the wrong ratio — must be $[\text{oxidised}]/[\text{reduced}]$ .
Forgetting to take the logarithm of the concentration ratio.
Using $\ln$ instead of $\log_{10}$ — the 0.0590 factor already incorporates the conversion.

Things to Be Careful About

The mark scheme accepts 0.16 V (2 sig figs) or 0.159 V — the minimum is 2 significant figures.
State the units (V) in the final answer.
The sign is positive — a common error is to make it negative.

Techniques used

apply the Nernst equationcalculate electrode potential from concentrations

(ii)

Calculate the $E_{\text{cell}}$ under these conditions.

DifficultyMedium-Easy

Worked solution

Working

The more positive potential is the cathode (reduction); the more negative is the anode (oxidation).

E_{\text{cell}} = E_{\text{Sn}} - E_{\text{Al}} = 0.16 - (-1.66) = +1.82\text{ V}

Answer

$E_{\text{cell}} = +1.82\text{ V}$

Final answer

+1.82 V

Detailed explanation

Background Concept

The cell potential is the difference between the two electrode potentials. By convention, $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$ , where the cathode has the more positive potential (reduction occurs) and the anode has the more negative potential (oxidation occurs).

Understanding the Question

We have calculated the Sn electrode potential as +0.16 V. The $\text{Al}^{3+}/\text{Al}$ half-cell has standard conditions, so $E^\ominus(\text{Al}^{3+}/\text{Al}) = -1.66\text{ V}$ . We need $E_{\text{cell}}$ .

Approach

Identify which is the cathode (more positive) and which is the anode (more negative), then subtract.

Step-by-Step Reasoning

$E(\text{Sn}^{4+}/\text{Sn}^{2+}) = +0.16\text{ V}$ (from part i)
$E(\text{Al}^{3+}/\text{Al}) = -1.66\text{ V}$ (standard value from Data Booklet)
Sn is more positive → cathode (reduction: $\text{Sn}^{4+} \rightarrow \text{Sn}^{2+}$ )
Al is more negative → anode (oxidation: $\text{Al} \rightarrow \text{Al}^{3+}$ )
$E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 0.16 - (-1.66) = +1.82\text{ V}$

Key Takeaways

$E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$ .
The more positive potential is always the cathode.
A positive $E_{\text{cell}}$ indicates a spontaneous reaction.

Common Mistakes

Adding instead of subtracting: $0.16 + (-1.66) = -1.50\text{ V}$ (wrong).
Forgetting the double negative when subtracting $-1.66$ .
Using the standard value +0.15 V instead of the Nernst-corrected +0.16 V.

Things to Be Careful About

Use the value from part (i), not the standard value, for the Sn half-cell.
The answer must be to at least 3 significant figures: +1.82 V.
Include the positive sign and the unit.

Techniques used

combine electrode potentials to find cell potentialidentify cathode and anode

(iii)

Write an equation for the overall cell reaction that occurs.

DifficultyMedium-Easy

Worked solution

Answer

Oxidation (anode): $2\text{Al} \rightarrow 2\text{Al}^{3+} + 6\text{e}^-$

Reduction (cathode): $3\text{Sn}^{4+} + 6\text{e}^- \rightarrow 3\text{Sn}^{2+}$

Overall: $2\text{Al} + 3\text{Sn}^{4+} \rightarrow 2\text{Al}^{3+} + 3\text{Sn}^{2+}$

Final answer

2Al + 3Sn4+ -> 2Al3+ + 3Sn2+

Detailed explanation

Background Concept

In a cell, the half-reaction with the more positive potential runs as reduction (cathode), and the one with the more negative potential runs as oxidation (anode). The overall cell reaction is the sum of the two half-reactions, balanced so that the number of electrons cancels.

Understanding the Question

We need the overall equation for the spontaneous cell reaction. From part (ii), $\text{Sn}^{4+}/\text{Sn}^{2+}$ is the cathode, $\text{Al}^{3+}/\text{Al}$ is the anode.

Approach

Write both half-reactions in the correct direction, balance electrons, and add.

Step-by-Step Reasoning

Cathode (reduction): $\text{Sn}^{4+} + 2\text{e}^- \rightarrow \text{Sn}^{2+}$
Anode (oxidation): $\text{Al} \rightarrow \text{Al}^{3+} + 3\text{e}^-$
Balance electrons: LCM of 2 and 3 is 6.
- $3 \times (\text{Sn}^{4+} + 2\text{e}^- \rightarrow \text{Sn}^{2+})$ : $3\text{Sn}^{4+} + 6\text{e}^- \rightarrow 3\text{Sn}^{2+}$
- $2 \times (\text{Al} \rightarrow \text{Al}^{3+} + 3\text{e}^-)$ : $2\text{Al} \rightarrow 2\text{Al}^{3+} + 6\text{e}^-$
Add: $2\text{Al} + 3\text{Sn}^{4+} + 6\text{e}^- \rightarrow 2\text{Al}^{3+} + 3\text{Sn}^{2+} + 6\text{e}^-$
Cancel electrons: $2\text{Al} + 3\text{Sn}^{4+} \rightarrow 2\text{Al}^{3+} + 3\text{Sn}^{2+}$

Key Takeaways

The cathode half-reaction is written as reduction, the anode as oxidation.
Electrons must cancel — multiply half-reactions by appropriate factors.
The overall equation must be balanced in both atoms and charge.

Common Mistakes

Writing the reaction in the wrong direction (e.g., $\text{Sn}^{2+} \rightarrow \text{Sn}^{4+}$ ).
Not balancing the electrons — e.g., $\text{Al} + \text{Sn}^{4+} \rightarrow \text{Al}^{3+} + \text{Sn}^{2+}$ .
Writing the electrons in the overall equation.

Things to Be Careful About

The mark scheme gives one mark for the correct species and one for correct balancing.
State symbols are not required by the mark scheme here, but are good practice.

Techniques used

balance redox half-equationswrite the overall cell reaction

(c)

Aluminium is produced industrially by electrolysis of a melt containing large amounts of $\text{Al}^{3+}$ ions.

Calculate the mass of aluminium that is obtained when a current of 300 000 A is passed for 24 hours. Give your answer to three significant figures.

DifficultyMedium

Worked solution

Working

Charge passed:

Q = It = 300\,000 \times 24 \times 60 \times 60 = 2.59 \times 10^{10}\text{ C}

Moles of electrons:

n(\text{e}^-) = \frac{Q}{F} = \frac{2.59 \times 10^{10}}{9.65 \times 10^4} = 2.69 \times 10^5\text{ mol}

Moles of aluminium ( $\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}$ ):

n(\text{Al}) = \frac{2.69 \times 10^5}{3} = 8.95 \times 10^4\text{ mol}

Mass of aluminium:

m = n \times M_r = 8.95 \times 10^4 \times 27 = 2.42 \times 10^6\text{ g} = 2420\text{ kg}

Answer

$2420\text{ kg (3 s.f.)}$

Final answer

2420 kg

Detailed explanation

Background Concept

Faraday's laws of electrolysis: the amount of substance liberated at an electrode is proportional to the quantity of charge passed. $Q = It$ (charge = current × time). The number of moles of electrons is $n(\text{e}^-) = Q/F$ , where $F = 9.65 \times 10^4\text{ C mol}^{-1}$ is the Faraday constant. For $\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}$ , three moles of electrons are needed per mole of aluminium.

Understanding the Question

A current of 300,000 A is passed for 24 hours through a melt containing $\text{Al}^{3+}$ ions. We need the mass of aluminium produced, to 3 significant figures.

Approach

Calculate the total charge: $Q = It$ (convert hours to seconds).
Convert charge to moles of electrons: $n(\text{e}^-) = Q/F$ .
Use stoichiometry: 3 mol e $^-$ → 1 mol Al.
Convert moles to mass: $m = n \times M_r$ ( $M_r(\text{Al}) = 27$ ).

Step-by-Step Reasoning

$Q = 300\,000 \times 24 \times 60 \times 60 = 2.592 \times 10^{10}\text{ C}$
$n(\text{e}^-) = 2.592 \times 10^{10} / 9.65 \times 10^4 = 2.686 \times 10^5\text{ mol}$
$n(\text{Al}) = 2.686 \times 10^5 / 3 = 8.95 \times 10^4\text{ mol}$
$m(\text{Al}) = 8.95 \times 10^4 \times 27 = 2.417 \times 10^6\text{ g} = 2417\text{ kg} \approx 2420\text{ kg}$

Key Takeaways

$Q = It$ is the first step in any Faraday calculation.
The stoichiometry of the electrode half-reaction (3 e $^-$ per Al) is crucial.
The Faraday constant $F = 9.65 \times 10^4\text{ C mol}^{-1}$ .

Common Mistakes

Forgetting to convert hours to seconds.
Using the wrong number of electrons per Al (using 1 or 2 instead of 3).
Using $M_r = 26.98$ instead of 27 (the syllabus uses 27).
Not converting grams to kg (though the answer can be given in g).

Things to Be Careful About

The answer must be to 3 significant figures: 2420 kg or $2.42 \times 10^6$ g.
The mark scheme's answer is 2420 kg — the unit matters.
Error carried forward: if you get the charge wrong but use it correctly afterwards, you can still gain method marks.

Techniques used

calculate charge from current and timeapply Faraday's lawconvert moles to mass

(d)

Explain why chromium metal cannot be obtained by the electrolysis of dilute aqueous chromium(II) sulfate. Your answer should include data from the Data Booklet.

DifficultyMedium-Easy

Worked solution

Answer

$\text{Cr}^{2+} + 2\text{e}^- \rightleftharpoons \text{Cr}$ , $E^\ominus = -0.91\text{ V}$

$2\text{H}^+ + 2\text{e}^- \rightleftharpoons \text{H}_2$ , $E^\ominus = 0.00\text{ V}$

Since $E^\ominus$ for $2\text{H}^+/\text{H}_2$ is more positive than for $\text{Cr}^{2+}/\text{Cr}$ , hydrogen ions are reduced in preference to $\text{Cr}^{2+}$ ions. Hydrogen gas is produced instead of chromium metal.

Final answer

Hydrogen gas is produced instead of chromium metal because H+/H2 (E° = 0.00 V) is more positive than Cr2+/Cr (E° = -0.91 V).

Detailed explanation

Background Concept

In aqueous electrolysis, there is competition between the reduction of metal ions and the reduction of water (or H $^+$ ions). The species with the more positive (less negative) electrode potential is reduced preferentially. At standard conditions:

$\text{Cr}^{2+} + 2\text{e}^- \rightleftharpoons \text{Cr}$ , $E^\ominus = -0.91\text{ V}$
$2\text{H}^+ + 2\text{e}^- \rightleftharpoons \text{H}_2$ , $E^\ominus = 0.00\text{ V}$

Since $0.00\text{ V} > -0.91\text{ V}$ , H $^+$ is reduced in preference to $\text{Cr}^{2+}$ , so hydrogen gas is evolved at the cathode instead of chromium metal being deposited.

Understanding the Question

We need to explain why electrolysing dilute aqueous chromium(II) sulfate does not produce chromium metal. The answer must include the relevant $E^\ominus$ values from the Data Booklet.

Approach

Compare the $E^\ominus$ values of the $\text{Cr}^{2+}/\text{Cr}$ and $\text{H}^+/\text{H}_2$ couples. The more positive one is reduced preferentially.

Step-by-Step Reasoning

Write the two competing reduction half-reactions with their $E^\ominus$ $E^{⊖}$ values:
- $\text{Cr}^{2+} + 2\text{e}^- \rightleftharpoons \text{Cr}$ , $E^\ominus = -0.91\text{ V}$
- $2\text{H}^+ + 2\text{e}^- \rightleftharpoons \text{H}_2$ , $E^\ominus = 0.00\text{ V}$
The $\text{H}^+/\text{H}_2$ couple has the more positive $E^\ominus$ , so H $^+$ ions are reduced in preference to $\text{Cr}^{2+}$ ions.
Therefore hydrogen gas is produced at the cathode, not chromium metal.

Key Takeaways

In aqueous electrolysis, compare $E^\ominus$ values to decide which species is reduced.
The more positive $E^\ominus$ is reduced preferentially.
Water (or H $^+$ ) is often reduced instead of metals with very negative $E^\ominus$ values.

Common Mistakes

Not quoting the $E^\ominus$ values (the question explicitly asks for data from the Data Booklet).
Saying "water is reduced" without the $E^\ominus$ comparison.
Confusing which is more positive — $0.00 > -0.91$ .

Things to Be Careful About

The mark scheme requires both $E^\ominus$ values to be seen (M1) and the conclusion that hydrogen is formed (M2).
The Data Booklet values are: $\text{Cr}^{2+}/\text{Cr} = -0.91\text{ V}$ , $\text{H}^+/\text{H}_2 = 0.00\text{ V}$ .

Techniques used

compare standard electrode potentialspredict electrolysis products

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