Electrolysis and the Faraday Constant

A2 · 14 min

Electrolysis uses electrical energy to drive a redox reaction that would not happen on its own: a direct current passes through a molten or dissolved ionic compound and decomposes it. Paper 4 tests two things: predicting what is formed at each electrode (which depends on whether the electrolyte is molten or aqueous, on electrode potentials and on concentration), and calculating how much is formed from the current and time using the Faraday constant. You may also be asked to describe how electrolysis can be used to measure the Avogadro constant.

What happens in an electrolysis cell

An electrolyte is an ionic compound that conducts electricity when molten or in aqueous solution, because its ions are free to move. Two electrodes dip into it and are connected to a d.c. supply.

The cathode is connected to the negative terminal. Positive ions (cations) move towards it and gain electrons: reduction happens at the cathode.
The anode is connected to the positive terminal. Negative ions (anions) move towards it and lose electrons: oxidation happens at the anode.

In the external circuit, electrons flow through the wires from the anode to the cathode. Inside the electrolyte, charge is carried by moving ions, not electrons.

Tip

"Red cat, an ox": reduction at the cathode, anode oxidation. This is true in every cell, including the galvanic cells in the next notes. Only the signs of the electrodes change.

Predicting the products

Molten electrolytes

A molten salt contains only its own two ions, so the products are simple: the metal forms at the cathode and the non-metal at the anode.

\text{cathode: } \ce{Pb^{2+}(l) + 2e- -> Pb(l)} \qquad \text{anode: } \ce{2Br^-(l) -> Br2(g) + 2e-}

Aqueous electrolytes: water competes

In aqueous solution there are also water molecules (and tiny amounts of $\ce{H+}$ and $\ce{OH-}$ from water). At each electrode, the species that is most easily reduced (at the cathode) or most easily oxidised (at the anode) reacts. Standard electrode potentials, $E^{\ominus}$ , measure exactly this (see standard electrode potentials).

At the cathode, the choice is between the metal ion and water:

\ce{2H2O(l) + 2e- -> H2(g) + 2OH^-(aq)} \qquad E^{\ominus} = -0.83\ \text{V}

(In acidic solution, $\ce{2H+(aq) + 2e- -> H2(g)}$ , $E^{\ominus} = 0.00\ \text{V}$ .)

If the metal ion has a more positive $E^{\ominus}$ than water, the metal is deposited. Examples: $\ce{Cu^{2+}}$ ( $+0.34\ \text{V}$ ), $\ce{Ag+}$ ( $+0.80\ \text{V}$ ).
If the metal ion has a more negative $E^{\ominus}$ , as for reactive metals ( $\ce{Na+}$ $-2.71$ , $\ce{Mg^{2+}}$ $-2.38$ , $\ce{Al^{3+}}$ $-1.66\ \text{V}$ ), hydrogen is produced instead.

At the anode (with inert electrodes such as platinum or graphite), the choice is between the anion and water:

\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e-} \qquad (\text{reverse of } E^{\ominus} = +1.23\ \text{V})

(In alkaline solution: $\ce{4OH^-(aq) -> O2(g) + 2H2O(l) + 4e-}$ .)

Iodide ( $E^{\ominus}(\ce{I2/I-}) = +0.54\ \text{V}$ ) and bromide ( $+1.07\ \text{V}$ ) are oxidised more easily than water, so iodine or bromine forms.
Chloride ( $+1.36\ \text{V}$ ) is close to water ( $+1.23\ \text{V}$ ). In concentrated chloride solution, chlorine is the main product; in dilute solution, mainly oxygen forms.
Sulfate and nitrate ions are very hard to oxidise, so oxygen forms.

Why concentration matters

Electrode potentials depend on concentration (the Nernst equation makes this quantitative). In concentrated sodium chloride, the high $[\ce{Cl-}]$ makes the $\ce{Cl2/Cl-}$ electrode potential less positive, so chloride ions are oxidised in preference to water. As the solution is diluted, oxygen becomes the main product. The same idea can affect the cathode: at very high concentrations of some ions (for example $\ce{Zn^{2+}}$ ), the metal can be deposited even though its $E^{\ominus}$ is slightly more negative than that of water.

Products of electrolysis with inert electrodes

Electrolyte	Cathode	Anode
molten $\ce{NaCl}$	sodium	chlorine
molten $\ce{PbBr2}$	lead	bromine
$\ce{CuSO4(aq)}$	copper	oxygen
$\ce{AgNO3(aq)}$	silver	oxygen
concentrated $\ce{NaCl(aq)}$	hydrogen	chlorine
dilute $\ce{NaCl(aq)}$	hydrogen	mainly oxygen
$\ce{KI(aq)}$	hydrogen	iodine
$\ce{H2SO4(aq)}$	hydrogen	oxygen
$\ce{Na2SO4(aq)}$ or $\ce{NaOH(aq)}$	hydrogen	oxygen

Active electrodes

If the anode is made of a metal such as copper, the anode itself can be oxidised more easily than water or sulfate ions:

\ce{Cu(s) -> Cu^{2+}(aq) + 2e-}

Electrolysing copper(II) sulfate with copper electrodes therefore moves copper from the anode to the cathode: the anode loses mass, the cathode gains the same mass, and the solution concentration stays constant. This is used to purify copper and is the basis of the Avogadro constant experiment below.

Predicting products of aqueous electrolysis

List the species present: the cation, the anion, and water. Note whether the electrodes are inert.
Cathode: compare $E^{\ominus}$ of the metal ion with that of water ( $-0.83\ \text{V}$ , or $0.00\ \text{V}$ for $\ce{H+}$ in acid). The more positive one is reduced.
Anode: if the anode is an active metal, it dissolves. Otherwise compare the anion with water ( $+1.23\ \text{V}$ ). The species whose half-equation has the less positive $E^{\ominus}$ is oxidised; for chloride, use concentration to decide.
Write half-equations with state symbols and balanced electrons.

The Faraday constant

Each mole of electrons carries a fixed amount of charge, called the Faraday constant, $F$ .

Key result

F = Le

where $L$ is the Avogadro constant ( $6.022 \times 10^{23}\ \text{mol}^{-1}$ ) and $e$ is the charge on one electron ( $1.60 \times 10^{-19}\ \text{C}$ , magnitude). The Data Booklet value is $F = 9.65 \times 10^{4}\ \text{C mol}^{-1}$ .

The total charge passed is

Q = It

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

Check: $6.022 \times 10^{23} \times 1.602 \times 10^{-19} = 9.65 \times 10^{4}\ \text{C mol}^{-1}$ .

The charge needed to liberate one mole of a substance depends on how many electrons appear in its half-equation:

Half-equation	Electrons per mole of product	Charge per mole of product
$\ce{Ag+ + e- -> Ag}$	1	$F$
$\ce{Cu^{2+} + 2e- -> Cu}$	2	$2F$
$\ce{Al^{3+} + 3e- -> Al}$	3	$3F$
$\ce{2H+ + 2e- -> H2}$	2	$2F$
$\ce{2Cl- -> Cl2 + 2e-}$	2	$2F$
$\ce{2H2O -> O2 + 4H+ + 4e-}$	4	$4F$

Electrolysis calculations

Calculate the charge: $Q = It$ (convert minutes or hours to seconds).
Moles of electrons: $n(\ce{e-}) = Q / F$ .
Use the half-equation to convert to moles of product: divide by the number of electrons per mole of product.
Convert to mass ( $m = nM$ ) or to gas volume ( $V = n \times 24.0\ \text{dm}^3$ at room conditions).
To work backwards (find $I$ or $t$ ), reverse the steps.

Worked examples

Mass of copper deposited

A current of $2.00\ \text{A}$ is passed through aqueous copper(II) sulfate for 30.0 minutes. Calculate the mass of copper deposited at the cathode. ( $A_r$ : Cu 63.5)

Solution

Q = It = 2.00 \times 30.0 \times 60 = 3600\ \text{C}

n(\ce{e-}) = \frac{3600}{9.65 \times 10^{4}} = 0.03731\ \text{mol}

$\ce{Cu^{2+} + 2e- -> Cu}$ , so $n(\ce{Cu}) = 0.03731 / 2 = 0.01865\ \text{mol}$ .

m = 0.01865 \times 63.5 = 1.18\ \text{g}

Volumes of gases

Dilute sulfuric acid is electrolysed using platinum electrodes with a current of $0.500\ \text{A}$ for 1.00 hour. Calculate the volumes of hydrogen and oxygen produced at room conditions.

Solution

Q = 0.500 \times 3600 = 1800\ \text{C} \qquad n(\ce{e-}) = \frac{1800}{96500} = 0.01865\ \text{mol}

Cathode: $\ce{2H+ + 2e- -> H2}$ , so $n(\ce{H2}) = 0.01865/2 = 0.009326\ \text{mol}$ ; $V = 0.009326 \times 24.0 = 0.224\ \text{dm}^3$ .

Anode: $\ce{2H2O -> O2 + 4H+ + 4e-}$ , so $n(\ce{O2}) = 0.01865/4 = 0.004663\ \text{mol}$ ; $V = 0.112\ \text{dm}^3$ .

The 2 : 1 volume ratio follows from the electrons: 2 per $\ce{H2}$ , 4 per $\ce{O2}$ .

Finding the time

How long, in minutes, must a current of $1.50\ \text{A}$ flow to deposit $5.00\ \text{g}$ of silver from silver nitrate solution? ( $A_r$ : Ag 107.9)

Solution

n(\ce{Ag}) = \frac{5.00}{107.9} = 0.04634\ \text{mol} = n(\ce{e-}) \quad (\ce{Ag+ + e- -> Ag})

Q = 0.04634 \times 96500 = 4472\ \text{C} \qquad t = \frac{Q}{I} = \frac{4472}{1.50} = 2981\ \text{s} = 49.7\ \text{min}

Deducing the charge on an ion

A current of $1.00\ \text{A}$ for exactly 1 hour deposits $0.647\ \text{g}$ of chromium from a solution of a chromium salt. Deduce the charge on the chromium ion. ( $A_r$ : Cr 52.0)

Solution

n(\ce{e-}) = \frac{1.00 \times 3600}{96500} = 0.03731\ \text{mol} \qquad n(\ce{Cr}) = \frac{0.647}{52.0} = 0.01244\ \text{mol}

\frac{n(\ce{e-})}{n(\ce{Cr})} = \frac{0.03731}{0.01244} = 3.00

Three moles of electrons per mole of chromium, so the ion is $\ce{Cr^{3+}}$ .

Determining the Avogadro constant by electrolysis

Because $F = Le$ , measuring the charge needed to deposit a known amount of a metal gives $L$ , provided the charge on the electron is known.

Copper electrodes in copper(II) sulfate solution, in series with an ammeter and a variable resistor. The shorter line of the cell symbol is the negative terminal.

Measuring the Avogadro constant

Method

Clean two copper electrodes with fine emery paper, rinse with propanone, dry, and weigh each one accurately.
Set up the circuit shown, with the electrodes in aqueous copper(II) sulfate.
Switch on and start a stopwatch. Use the variable resistor to keep the current constant (for example at $0.300\ \text{A}$ ) throughout.
After a measured time (for example 60 minutes), switch off. Remove the electrodes, rinse gently with distilled water then propanone, dry and reweigh.

Calculation

$Q = It$ , so the number of electrons passed is $Q/e$ .
The mass change of an electrode gives moles of copper; each $\ce{Cu}$ needs 2 electrons, so moles of electrons $= 2 \times n(\ce{Cu})$ .
$L = \dfrac{\text{number of electrons}}{\text{moles of electrons}}$ .

Sources of error and improvements

The current may drift: use a variable resistor and record the current frequently, or use a constant-current supply.
Copper may flake off the cathode or not all be deposited firmly: use a low current density (large electrodes, small current).
The anode mass loss is often slightly larger than the cathode gain, because impurities and loose copper fall from the anode. The cathode gain is usually the more reliable measurement.
Electrodes must be completely dry before weighing; a balance reading to $0.001\ \text{g}$ or better is needed because the mass changes are small.
Copper oxidises if heated to dry it; rinse with propanone and air-dry instead.

Calculating L from experimental data

In the experiment above, a constant current of $0.300\ \text{A}$ was passed for 60.0 minutes. The cathode gained $0.350\ \text{g}$ . Calculate a value for the Avogadro constant. ( $A_r$ : Cu 63.5; $e = 1.60 \times 10^{-19}\ \text{C}$ )

Solution

Q = 0.300 \times 3600 = 1080\ \text{C}

\text{number of electrons} = \frac{1080}{1.60 \times 10^{-19}} = 6.75 \times 10^{21}

n(\ce{Cu}) = \frac{0.350}{63.5} = 5.512 \times 10^{-3}\ \text{mol} \quad\Rightarrow\quad n(\ce{e-}) = 1.102 \times 10^{-2}\ \text{mol}

L = \frac{6.75 \times 10^{21}}{1.102 \times 10^{-2}} = 6.12 \times 10^{23}\ \text{mol}^{-1}

This is about 2% above the accepted value; the expected mass gain was $0.355\ \text{g}$ , so a little copper was probably lost from the cathode.

Common mistakes

Time in minutes. $Q = It$ needs seconds.
Forgetting the electron ratio. Copper needs 2 moles of electrons per mole, aluminium 3, oxygen 4.
Products of aqueous electrolysis. Sodium, potassium, magnesium and aluminium are never deposited from aqueous solution; hydrogen forms instead.
"Chloride gives chlorine" always. Only in concentrated solution; dilute chloride solutions give mainly oxygen.
Mixing up electrodes. Reduction at the cathode (negative in electrolysis), oxidation at the anode (positive).
Using $e$ with its sign. In $L = (Q/e)/n(\ce{e-})$ use the magnitude $1.60 \times 10^{-19}\ \text{C}$ .

Exam technique

Exam tip

To "predict the products", name both and give a brief reason based on $E^{\ominus}$ values or concentration. Half-equations are often worth a separate mark.
Show $Q = It$ , then moles of electrons, then moles of product: each is often a separate method mark.
Keep 3 or 4 significant figures in intermediate values and round only the final answer.
"Describe how the Avogadro constant could be determined" (4 to 6 marks): copper electrodes, copper(II) sulfate, weigh before and after, constant measured current and time, then the calculation using $Q/e$ and $2n(\ce{Cu})$ .
When two cells are in series, the same charge passes through both: use moles of electrons to link them.

Summary

Electrolysis: reduction at the cathode (−), oxidation at the anode (+); ions carry the charge in the electrolyte.
Molten salts give the metal and the non-metal. In water, less reactive metals ( $\ce{Cu}$ , $\ce{Ag}$ ) are deposited; otherwise hydrogen forms.
At an inert anode, $\ce{I-}$ and $\ce{Br-}$ give halogens, concentrated $\ce{Cl-}$ gives chlorine, dilute $\ce{Cl-}$ , sulfate and nitrate give oxygen. A copper anode dissolves.
$F = Le = 9.65 \times 10^{4}\ \text{C mol}^{-1}$ ; $Q = It$ .
Moles of product $= Q / (zF)$ , where $z$ is the number of electrons per mole of product.
$L$ can be found by weighing a copper cathode before and after electrolysis at a measured constant current for a measured time.

Practice questions

Question

Predict the products at each electrode, using inert electrodes, for: (a) molten lead(II) bromide; (b) aqueous copper(II) sulfate; (c) concentrated aqueous sodium chloride; (d) aqueous potassium iodide; (e) aqueous sodium sulfate.
Write the half-equations, with state symbols, for the reactions at each electrode when aqueous copper(II) sulfate is electrolysed with inert electrodes.
Use $L = 6.022 \times 10^{23}\ \text{mol}^{-1}$ and $e = 1.60 \times 10^{-19}\ \text{C}$ to calculate a value for $F$ .
Calculate the mass of aluminium produced when a current of $10.0\ \text{A}$ passes through molten aluminium oxide for 2.00 hours. ( $A_r$ : Al 27.0)
Calculate the volume of chlorine, in cm³ at room conditions, produced when a current of $0.250\ \text{A}$ passes through concentrated sodium chloride solution for 40.0 minutes.
Calculate the time, in minutes, needed to deposit $2.00\ \text{g}$ of copper using a current of $0.800\ \text{A}$ .
Two cells are connected in series. One contains aqueous silver nitrate and the other aqueous copper(II) sulfate. When $1.08\ \text{g}$ of silver has been deposited, what mass of copper has been deposited?
In an Avogadro constant experiment, a copper anode lost $0.240\ \text{g}$ when a current of $0.200\ \text{A}$ was passed for 1.00 hour. Calculate a value for $L$ , and suggest why using the anode mass loss tends to give a value that is too low.
A current of $0.600\ \text{A}$ for 45.0 minutes deposits $0.996\ \text{g}$ of tin. Deduce the charge on the tin ion. ( $A_r$ : Sn 118.7)
When dilute aqueous sodium chloride is electrolysed using platinum electrodes, the gas collected at the anode is mainly oxygen, but when the solution is concentrated it is mainly chlorine. Use the $E^{\ominus}$ values $\ce{Cl2 + 2e- <=> 2Cl-}$ $+1.36\ \text{V}$ and $\ce{O2 + 4H+ + 4e- <=> 2H2O}$ $+1.23\ \text{V}$ to explain these observations.

Answers

(a) Cathode lead, anode bromine. (b) Cathode copper, anode oxygen. (c) Cathode hydrogen, anode chlorine. (d) Cathode hydrogen, anode iodine. (e) Cathode hydrogen, anode oxygen.
Cathode: $\ce{Cu^{2+}(aq) + 2e- -> Cu(s)}$ . Anode: $\ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e-}$ .
$F = 6.022 \times 10^{23} \times 1.60 \times 10^{-19} = 9.64 \times 10^{4}\ \text{C mol}^{-1}$ .
$Q = 10.0 \times 7200 = 72\,000\ \text{C}$ ; $n(\ce{e-}) = 72\,000/96\,500 = 0.7461\ \text{mol}$ ; $n(\ce{Al}) = 0.7461/3 = 0.2487\ \text{mol}$ ; mass $= 0.2487 \times 27.0 = 6.71\ \text{g}$ .
$Q = 0.250 \times 2400 = 600\ \text{C}$ ; $n(\ce{e-}) = 6.218 \times 10^{-3}\ \text{mol}$ ; $n(\ce{Cl2}) = 3.109 \times 10^{-3}\ \text{mol}$ ; $V = 3.109 \times 10^{-3} \times 24\,000 = 74.6\ \text{cm}^3$ .
$n(\ce{Cu}) = 2.00/63.5 = 0.03150\ \text{mol}$ ; $n(\ce{e-}) = 0.06299\ \text{mol}$ ; $Q = 0.06299 \times 96\,500 = 6079\ \text{C}$ ; $t = 6079/0.800 = 7598\ \text{s} = 127\ \text{min}$ .
$n(\ce{Ag}) = 1.08/107.9 = 0.01001\ \text{mol} = n(\ce{e-})$ . The same charge passes through both cells, so $n(\ce{Cu}) = 0.01001/2 = 0.005005\ \text{mol}$ ; mass $= 0.005005 \times 63.5 = 0.318\ \text{g}$ .
$Q = 0.200 \times 3600 = 720\ \text{C}$ ; electrons $= 720/1.60 \times 10^{-19} = 4.50 \times 10^{21}$ . $n(\ce{Cu}) = 0.240/63.5 = 3.780 \times 10^{-3}\ \text{mol}$ ; $n(\ce{e-}) = 7.559 \times 10^{-3}\ \text{mol}$ . $L = 4.50 \times 10^{21} / 7.559 \times 10^{-3} = 5.95 \times 10^{23}\ \text{mol}^{-1}$ . The anode can lose extra mass as loose copper and impurities fall off without being oxidised, so the measured mass loss overstates the copper oxidised; dividing by too large a number of moles gives $L$ too low.
$Q = 0.600 \times 2700 = 1620\ \text{C}$ ; $n(\ce{e-}) = 0.01679\ \text{mol}$ ; $n(\ce{Sn}) = 0.996/118.7 = 0.008391\ \text{mol}$ ; ratio $= 2.00$ , so the ion is $\ce{Sn^{2+}}$ .
At the anode the species oxidised is the one whose half-equation has the less positive electrode potential. Under standard conditions, $E$ for $\ce{O2/H2O}$ ( $+1.23\ \text{V}$ ) is less positive than for $\ce{Cl2/Cl-}$ ( $+1.36\ \text{V}$ ), so water is oxidised to oxygen in dilute solution. The two values are close. Increasing $[\ce{Cl-}]$ (the reduced species) makes $E(\ce{Cl2/Cl-})$ less positive, so in concentrated solution chloride ions are oxidised in preference and chlorine is the main product.

What happens in an electrolysis cell

Predicting the products

Molten electrolytes

Aqueous electrolytes: water competes

Why concentration matters

Active electrodes

The Faraday constant

Worked examples

Determining the Avogadro constant by electrolysis

Exam technique

Summary

Practice questions

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