TASC TCE 2023-style practice: How many grams of chromium are obtained from the electrolysis of Cr(NO_3)_3(aq) if 10.0\ A is passed through the cell for 12.0\ hours? (F = 96\,500\ C mol^-1, M(Cr) = 52.00\ g mol^-1.)

Charge passed: convert time to seconds, 12.0 × 3600 = 4.32 × 10^4\ s, so Q = It = 10.0 × 4.32 × 10^4 = 4.32 × 10^5\ C. (1 mark) Moles of electrons: n(e^-) = Q/F = 4.32 × 10^5 / 96\,500 = 4.477\ mol. (1 mark) Cathode: Cr^3+ + 3e^- → Cr, so n(Cr) = 4.477/3 = 1.492\ mol. Mass = 1.492 × 52.00 = 77.6\ g. (1 mark)

§-Syllabus dot point

TASChemistrySyllabus dot point

How much product does electrolysis make for a given charge?

Apply Faraday's laws to calculate masses and volumes produced in electrolysis.

Quantitative electrolysis using Q = It, the Faraday constant, cells in series, and calculating mass or gas volume from charge passed, with fully worked TASC-style examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

TASC expects you to perform quantitative electrolysis calculations linking charge, moles of electrons and mass or volume of product, including cells connected in series.

Charge and the Faraday constant

One mole of electrons carries a charge equal to the Faraday constant, $F = 96\,500\ \text{C mol}^{-1}$ . So the moles of electrons passed is $n(e^-) = Q/F$ .

Linking electrons to product

The electrode half-equation tells you how many electrons are needed per mole of product. The general routine is:

Find charge $Q = It$ (time in seconds).
Find moles of electrons $n(e^-) = Q/F$ .
Use the half-equation ratio to find moles of product.
Convert moles of product to mass ( $m = nM$ ) or gas volume.

Mass of copper deposited

Calculate the mass of copper deposited when a current of $2.00\ \text{A}$ flows through copper(II) sulfate solution for $30.0$ minutes.

Step 1: Convert time and calculate the charge

The formula $Q = It$ requires time in seconds. Convert the minutes first, then multiply by the current to find the total charge that passed through the cell:

t = 30.0 \times 60 = 1800\ \text{s}, \qquad Q = It = 2.00 \times 1800 = 3600\ \text{C}.

Step 2: Convert charge to moles of electrons

The Faraday constant $F = 96\,500\ \text{C mol}^{-1}$ tells us that one mole of electrons carries $96\,500$ coulombs. Dividing the total charge by $F$ gives the moles of electrons delivered to the cathode:

n(e^-) = \frac{Q}{F} = \frac{3600}{96\,500} = 0.0373\ \text{mol}.

Step 3: Use the half-equation to find moles of copper

The cathode reaction is $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$ . Each copper atom requires two electrons, so the mole ratio of copper to electrons is $1:2$ . Divide the moles of electrons by $2$ :

n(\text{Cu}) = \frac{0.0373}{2} = 0.01865\ \text{mol}.

Step 4: Convert moles of copper to mass

Multiply the moles of copper by its molar mass ( $M(\text{Cu}) = 63.55\ \text{g mol}^{-1}$ ) to find the deposited mass:

m = nM = 0.01865 \times 63.55 = 1.19\ \text{g}.

Final answer: 1.19 g of copper is deposited.

Gas volumes

If the product is a gas (for example $\text{O}_2$ at the anode, $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ ), convert moles to volume using the molar gas volume at the stated conditions, or the ideal gas equation $pV = nRT$ . The same electron bookkeeping applies; just watch the electrons per mole of gas, which is often more than for a metal.

Predicting the products of electrolysis

When a solution (rather than a molten salt) is electrolysed, water can also react, so you must decide which species reacts at each electrode. At the cathode, the species most easily reduced wins: a metal ion below hydrogen in reactivity (such as $\text{Cu}^{2+}$ or $\text{Ag}^+$ ) is deposited, but for very reactive metal ions (such as $\text{Na}^+$ or $\text{K}^+$ ) water is reduced instead, releasing hydrogen, $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$ . At the anode, the species most easily oxidised wins: halide ions are usually oxidised to the halogen, but with oxoanions such as sulfate, water is oxidised instead, releasing oxygen, $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ . Concentration and overpotential can override the simple potential order, so the predicted product is sometimes only the most likely one.

Industrial electrolysis

Faraday's laws underpin large-scale electrochemical industry. Aluminium is extracted by electrolysing molten aluminium oxide dissolved in cryolite, with the cathode half-equation $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ requiring three moles of electrons per mole of metal, which is why the process is so energy intensive. Electroplating deposits a thin even layer of metal (chromium, silver, gold) onto an object made the cathode, with the plating mass controlled precisely through $Q = It$ .

In the exam, convert time to seconds, compute $Q = It$ , divide by the Faraday constant for moles of electrons, then apply the half-equation ratio before converting to mass or volume; for series cells, reuse the same moles of electrons in each cell.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20233 marksHow many grams of chromium are obtained from the electrolysis of

\text{Cr}(\text{NO}_3)_3(aq)

10.0\ \text{A}

is passed through the cell for

12.0\ \text{hours}

? (

F = 96\,500\ \text{C mol}^{-1}

M(\text{Cr}) = 52.00\ \text{g mol}^{-1}

Show worked answer →

Charge passed: convert time to seconds, $12.0 \times 3600 = 4.32 \times 10^{4}\ \text{s}$ , so $Q = It = 10.0 \times 4.32 \times 10^{4} = 4.32 \times 10^{5}\ \text{C}$ . (1 mark)

Moles of electrons: $n(e^-) = Q/F = 4.32 \times 10^{5} / 96\,500 = 4.477\ \text{mol}$ . (1 mark)

Cathode: $\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}$ , so $n(\text{Cr}) = 4.477/3 = 1.492\ \text{mol}$ . Mass $= 1.492 \times 52.00 = 77.6\ \text{g}$ . (1 mark)

TCE 20225 marksTwo electrolytic cells are connected in series with a current of

1.55\ \text{A}

through each. The first contains

\text{Cu}^{2+}

and the second contains

\text{Au}^{3+}

(both nitrates). (a) Calculate the time the current flowed if

1.87\ \text{g}

of copper was deposited in the first cell. (b) Calculate the mass of gold produced. (

M(\text{Cu}) = 63.55

M(\text{Au}) = 197.0\ \text{g mol}^{-1}

Show worked answer →

(a) $n(\text{Cu}) = 1.87/63.55 = 0.02943\ \text{mol}$ . Cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ , so $n(e^-) = 2 \times 0.02943 = 0.05886\ \text{mol}$ . $Q = n(e^-) \times F = 0.05886 \times 96\,500 = 5680\ \text{C}$ . $t = Q/I = 5680/1.55 = 3665\ \text{s}$ (about $61\ \text{min}$ ). (3 marks)

(b) The cells are in series, so the same $0.05886\ \text{mol}$ of electrons passes through the gold cell. Cathode: $\text{Au}^{3+} + 3e^- \rightarrow \text{Au}$ , so $n(\text{Au}) = 0.05886/3 = 0.01962\ \text{mol}$ . Mass $= 0.01962 \times 197.0 = 3.87\ \text{g}$ . (2 marks)