Faraday's Laws of Electrolysis – Electrolysis (Senior 5 Chemistry) | Ganzaa

Electrolysis

Lessons

Senior 5 · Chemistry

Electrolysis

4 lessons

Lesson 4 of 4

Faraday's Laws of Electrolysis

25 min

Counting atoms with an ammeter

An electroplating workshop must decide how long to leave a car part in a nickel bath to give a coating exactly 10 micrometres thick. An aluminium smelter must know how much electricity each tonne of metal will cost. Both answers come from two simple laws found by Michael Faraday in 1833–1834, which connect the amount of electric charge passed to the amount of substance produced. Behind them is a single idea: electrons are particles, and each ion needs a fixed number of them.

Definition

Charge: $Q = I t$ , with $Q$ in coulombs (C), current $I$ in amperes (A) and time $t$ in seconds (s).

Faraday constant, $F$ : the charge on one mole of electrons, $F = N_A e = 6.022 \times 10^{23} \times 1.602 \times 10^{-19} = 96\,485\,\mathrm{C\,mol^{-1}}$ .

Faraday's first law: the mass of a substance produced at an electrode is directly proportional to the quantity of electricity passed, $m \propto Q$ .

Faraday's second law: when the same quantity of electricity passes through different electrolytes, the amounts (in moles) of the substances produced are inversely proportional to the number of electrons needed per particle (the charge $z$ on the ion).

From current to mass: the method

Combining both laws:

$n(e^{-}) = \frac{Q}{F} = \frac{It}{F} \qquad n(\text{product}) = \frac{n(e^{-})}{z} \qquad m = n(\text{product}) \times M = \frac{ItM}{zF}$

where $z$ is the number of electrons in the half-equation per particle of product: 1 for $\mathrm{Ag^{+} + e^{-} \rightarrow Ag}$ , 2 for $\mathrm{Cu^{2+} + 2e^{-} \rightarrow Cu}$ and for $\mathrm{2H^{+} + 2e^{-} \rightarrow H_2}$ , 3 for aluminium, 4 for $\mathrm{O_2}$ .

Figure 13.4: Cells in series. The same 0.0200 mol of electrons passes through all three, giving 0.0200 mol Ag (z = 1), 0.0100 mol Cu and 0.0100 mol $\mathrm{H_2}$ (z = 2), and 0.00500 mol $\mathrm{O_2}$ (z = 4). Volumes at 25 °C and 101.3 kPa.

Factors affecting the amount produced

Factor	Effect on the amount of product
current $I$	directly proportional
time $t$	directly proportional
charge on the ion, $z$	inversely proportional (moles)
molar mass $M$	directly proportional (mass)
side reactions	reduce the yield: current efficiency = actual ÷ theoretical × 100 %

The identity of the products depends on the factors in Lessons 1 and 2: the electrolyte (molten or aqueous), the position of the ions in the electrochemical series, their concentration, the electrode material and overpotential.

Worked example

Problem. A current of 2.00 A passes through copper(II) sulfate solution for 30.0 minutes. Calculate the mass of copper deposited. ( $A_{r}$ (Cu) = 63.55, $F = 96\,485\,\mathrm{C\,mol^{-1}}$ )

Solution.

$Q = It = 2.00 \times (30.0 \times 60) = 3600\,\mathrm{C} \qquad n(e^{-}) = \frac{3600}{96\,485} = 0.03731\,\mathrm{mol}$

$\mathrm{Cu^{2+} + 2e^{-} \rightarrow Cu}$ : $n(\mathrm{Cu}) = \frac{0.03731}{2} = 0.01866\,\mathrm{mol}$ ; $m = 0.01866 \times 63.55 = 1.19\,\mathrm{g}$ .

Worked example

Problem. A steel part with surface area 50.0 cm² is to be nickel-plated to a thickness of 10.0 µm using a current of 0.500 A. How long will it take? (Ni: density 8.91 g cm⁻³, $A_{r}$ = 58.69; $\mathrm{Ni^{2+} + 2e^{-} \rightarrow Ni}$ )

Solution. Thickness $10.0\,\mu\mathrm{m} = 1.00 \times 10^{-3}\,\mathrm{cm}$ .

$V = 50.0 \times 1.00 \times 10^{-3} = 0.0500\,\mathrm{cm^{3}} \qquad m = 0.0500 \times 8.91 = 0.4455\,\mathrm{g} \qquad n(\mathrm{Ni}) = \frac{0.4455}{58.69} = 7.59 \times 10^{-3}\,\mathrm{mol}$

$Q = 2 \times 7.59 \times 10^{-3} \times 96\,485 = 1465\,\mathrm{C} \qquad t = \frac{Q}{I} = \frac{1465}{0.500} = 2930\,\mathrm{s} = 48.8\,\mathrm{min}$

Worked example

Problem. (a) A charge of 965 C deposits 0.562 g of a metal X ( $A_{r}$ = 112.4) from a solution of its salt. Find the charge on the ion of X. (b) Estimate the electrical energy needed to make 1.00 tonne of aluminium if the cell runs at 4.5 V. ( $A_{r}$ (Al) = 26.98)

Solution. (a) $n(e^{-}) = \frac{965}{96\,485} = 0.0100\,\mathrm{mol}$ ; $n(\mathrm{X}) = \frac{0.562}{112.4} = 0.00500\,\mathrm{mol}$ . Electrons per atom $= \frac{0.0100}{0.00500} = 2$ : the ion is $\mathrm{X^{2+}}$ .

(b) $n(\mathrm{Al}) = \frac{1.00 \times 10^{6}}{26.98} = 3.706 \times 10^{4}\,\mathrm{mol}$ ; $\mathrm{Al^{3+} + 3e^{-} \rightarrow Al}$ : $Q = 3 \times 3.706 \times 10^{4} \times 96\,485 = 1.073 \times 10^{10}\,\mathrm{C}$ .

$E = QV = 1.073 \times 10^{10} \times 4.5 = 4.83 \times 10^{10}\,\mathrm{J} = \frac{4.83 \times 10^{10}}{3.6 \times 10^{6}}\,\mathrm{kWh} = 1.34 \times 10^{4}\,\mathrm{kWh}$

Answer: (a) 2+; (b) about 13 400 kWh per tonne, which is why aluminium smelters are built where electricity is cheap, and why recycling aluminium saves so much energy.

Warning

Typical errors: using minutes instead of seconds in $Q = I t$ ; forgetting to divide by $z$ ; using $z = 2$ for oxygen (it is 4 electrons per $\mathrm{O_2}$ ); and assuming 100 % current efficiency when some charge is wasted on side reactions such as hydrogen release at a plating cathode.

Safety: plating solutions can contain toxic metal salts (nickel salts are skin sensitisers). Wear gloves and goggles, and wash hands after practical work. If a solution gets in the eyes, rinse with water for at least 15 minutes (CDC/NIOSH first aid procedures).

Tip

Use the chain I·t → Q → mol e⁻ → mol product → mass or volume, and write the half-equation first: it tells you how many electrons each particle of product needs.

Jason Goudie: Calculations involving Faraday's Laws of Electrolysis. Follow the worked example step by step: charge from current and time, moles of electrons, then moles and mass of product.

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Practice

State Faraday's first and second laws of electrolysis.
Calculate the mass of silver deposited when 0.250 A flows for 1.00 hour through silver nitrate solution ( $A_{r}$ (Ag) = 107.87).
Acidified water is electrolysed with 1.50 A for 20.0 minutes. Calculate the volumes of hydrogen and oxygen at 25 °C and 101.3 kPa (24.5 dm³ mol⁻¹).
The same charge deposits 1.079 g of silver in one cell and copper in a second cell in series. What mass of copper is deposited? What mass of aluminium would the same charge give from molten $\mathrm{Al_2O_3}$ ? ( $A_{r}$ : Ag 107.87, Cu 63.55, Al 26.98)
How long must a current of 3.00 A flow to deposit 2.00 g of chromium from a solution containing $\mathrm{Cr^{3+}}$ ? ( $A_{r}$ (Cr) = 52.00)
In an experiment, 0.300 A for 30.0 min deposited 0.165 g of copper. Calculate the current efficiency.
A student calculates $F$ from a copper-plating experiment and obtains 1.07 × 10⁵ C mol⁻¹. Suggest and evaluate two possible reasons for the high value.

Show answersHide answers

First: the mass of product is proportional to the quantity of electricity passed. Second: for the same quantity of electricity, the moles of different products are inversely proportional to the charges on their ions (electrons needed per particle).
$Q = 0.250 \times 3600 = 900\,\mathrm{C}$ ; $n = \frac{900}{96\,485} = 9.33 \times 10^{-3}$ ; $m = 1.01\,\mathrm{g}$ .
$Q = 1800\,\mathrm{C}$ ; $n(e^{-}) = 0.01866$ ; $\mathrm{H_2}$ : 0.00933 mol = 229 cm³; $\mathrm{O_2}$ : 0.00466 mol = 114 cm³.
$n(e^{-}) = n(\mathrm{Ag}) = 0.0100$ ; Cu: $0.00500 \times 63.55 = 0.318\,\mathrm{g}$ ; Al: $\frac{0.0100}{3} \times 26.98 = 0.0899\,\mathrm{g}$ .
$n(\mathrm{Cr}) = 0.03846$ ; $Q = 3 \times 0.03846 \times 96\,485 = 1.113 \times 10^{4}\,\mathrm{C}$ ; $t = 3711\,\mathrm{s} \approx 61.8\,\mathrm{min}$ .
Theoretical: $Q = 540\,\mathrm{C}$ , $n(\mathrm{Cu}) = 2.798 \times 10^{-3}$ , $m = 0.178\,\mathrm{g}$ ; efficiency $= \frac{0.165}{0.178} \times 100 = 92.8\,\%$ .
$F = \frac{Q}{n(e^{-})}$ : a high value means too little copper was weighed for the charge. Possible causes: some current was used in a side reaction (hydrogen formed at the cathode), or loosely attached copper fell off and was lost during washing. A current reading that was too high would also do it. Each is plausible; checking for gas at the cathode and drying the cathode gently would help decide.

1Counting atoms with an ammeter

2From current to mass: the method

3Factors affecting the amount produced

Counting atoms with an ammeter

From current to mass: the method

Factors affecting the amount produced