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Brown's Chemistry  GCSE/IGCSE/GCE (basic A level)
O Level
Online Chemical Calculations
13.
Electrolysis product calculations (negative cathode and positive anode products)
Help for problem solving
in doing electrolysis calculations, using experiment data, making
predictions. Practice revision questions on quantitative calculations of
electrolysis electrode products (given electrode equations), mass of
products at cathode or anode electrodes, moles and volumes of gases
formed at the cathode and anode electrodes. This page describes and
explains, with fully worked out examples, methods of calculation
involving moles, masses or volumes of gases formed in an electrolysis
process. You need to understand electrode equations, interconvert mass
and moles and use the molar volume where gases are formed at the
electrodes in electrolytic processes.
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GCSE calculation? * Explaining electrolysis
and descriptions of experimental methods

13.
Electrolysis product calculations (negative cathode and positive anode products)
Relative atomic masses
needed: Na =
23, Cl = 35.5, H = 1, Cu = 63.5, Al = 27, O = 16 and the molar volume of
any gas is 24 dm^{3} at room temperature and pressure.
The common electrode equations you may come across are listed below.

electrode involved: () negative
cathode or (+) positive anode for the
Electrode Equation below

moles of electrons involved
(mass
of product formed) 
example of industrial process where this electrode
reaction happens 
sodium () Na^{+}_{(l)} + e^{}
==> Na_{(l)} 
1
(23g) =
1.0 mol Na metal per mol e^{}s 
electrolysis of molten chloride
salts to make chlorine and the metal 
chlorine (+) 2Cl^{}_{(l/aq)} 
2e^{}
==> Cl_{2(g)}

2
(71g) =
0.5 mol Cl_{2} gas (12 dm^{3})
released per mol e^{}s 
electrolysis of molten chloride
salts or their aqueous solution to make chlorine 
hydrogen () 2H^{+}_{(aq)} + 2e^{}
==> H_{2(g)} 
2
(2g) =
0.5 mol H_{2} gas (12 dm^{3})
released per mol e^{}s 
electrolysis of many salt
solutions to make hydrogen 
copper () Cu^{2+}_{(aq)} + 2e^{}
==> Cu_{(s)} 
2
(63.5g) =
0.5 mol Cu deposited per mol e^{}s 
deposition of copper in its
electrolytic purification or electroplating 
copper (+) Cu_{(s)} 
2e^{} ==> Cu^{2+}_{(aq)} 
2 (63.5g) =
0.5 mol Cu dissolves per mol e^{}s 
dissolving of copper in its
electrolytic purification or electroplating 
aluminium () Al^{3+}_{(l)} + 3e^{}
==> Al_{(l)} 
3 (27g) = 0.33 mol Al metal per mol e^{}s 
extraction of aluminium in the
electrolysis of its molten oxide ore 
oxygen (+) 2O^{2}_{(l)}  4e^{}
==> O_{2(g)} 
4 (32g) =
0.25 mol O_{2} (12 dm^{3})
gas released per mol e^{}s 
electrolysis of molten oxides 
oxygen (+) 4OH^{}_{(aq)}  4e^{}
==> 2H_{2}O_{(l)} + O_{2(g)} 
4 (32g) =
0.25 mol O_{2} gas (6 dm^{3})
released per mol e^{}s 
electrolysis of many salt
solutions such as sulphates, sulphuric acid etc. gives oxygen (but
chloride salts
==> chlorine) 
Explaining electrolysis
and descriptions of experimental methods
Part one:
The (+) anode and () cathode electrode product ratio

The amount of
material in moles formed at the electrode in electrolysis depends on three
factors.

The charge on the
ion. (compare the effect of one mole of electrons in the table above and
see examples 13.1.1 to 13.1.5 below in Part one)

The current flow.
(the current flowing in amperes, A, see examples in Part
two)

The time duration of
the electrolysis. (time in seconds, minutes or hours, see examples in
Part two)

If you know how much of a substance is made at one
electrode, you can theoretically calculate the amount of substance formed at
the other electrode.

The basis of these calculations is the ratio of the
electrons involved in both electrode reactions (hence the introductory table
of electrode equations above).

These
electrode equations in the table above are referred to in the examples below.

In studying the examples
below you must refer to the electrode equations in the table above,

Electrolysis calculation Example 13.1.1

The electrolysis of brine, aqueous sodium
chloride solution, NaCl_{(aq)} produces hydrogen gas, H_{2(g)}
at the ve electrode and chlorine gas, Cl_{2(g)} at the positive
electrode. Atomic masses: H = 1, Cl = 35.5

2H^{+}_{(aq)} + 2e^{}
==> H_{2(g)} and 2Cl^{}_{(l/aq)}  2e^{}
==> Cl_{2}_{(g)}

2
electrons are involved in both the formation of a hydrogen molecule [M_{r}(H_{2})
= 2] or a chlorine molecule [M_{r}(Cl_{2}) = 71].

The ratio of the products
for H_{2(g)} : Cl_{2(g)}
is 1 mol : 1 mol or 24dm^{3} : 24 dm^{3}
or 2g : 71g

If during the electrolysis of
sodium chloride solution, 25 cm^{3} of hydrogen were produced, what
volume of chlorine is theoretically formed?

Since the mole ratio is 1 : 1
for H_{2} : Cl_{2} for every 25 cm3 of hydrogen
formed, 25 cm^{3} of chlorine will be formed.

Electrolysis calculation Example 13.1.2

The electrolysis of molten aluminium oxide
Al_{2}O_{3} is a more complicated affair.

Its best to think of the
ratio effect of a current of 12 moles
of electrons passing through the electrolyte.

4Al^{3+}_{(l)} +
12e^{} ==> 4Al_{(l)} and 6O^{2}_{(l)} 
12e^{} ==>
3O_{2(g)}

It takes 3 moles
of electrons to form 1 mole of Al from 1 mole of Al^{3+} ions.

and 4 moles of
electrons to form 1 mole of O_{2} molecules from 2 moles of O^{2} ions.

Atomic masses:
Al = 27, O = 16

The ratio of the products
from 12 moles of electrons is therefore

If 0.1 mol of molten
aluminium oxide is completely electrolysed (i) what mass of aluminium is
formed and (ii) what volume of oxygen is formed (at RTP)

atomic mass Al = 27, molar
volume of any gas at RTP = 24 dm^{3}

(i) From 1 mole of Al_{2}O_{3}
you get 2 moles of Al

therefore from 0.1 mol of Al_{2}O_{3}
you get 0.2 mol of Al

mass Al = moles Al x atomic mass

mass Al = 0.2 x 27 = 0.54 g
Al

(ii) From 1 mole of Al_{2}O_{3}
you get the equivalent of 3 moles of O atoms,

BUT you must treat this as 1.5
mol O_{2} molecules

therefore from 1 mol Al_{2}O_{3}
you get 1.5 mol of O_{2} molecules

so from 0.1 mol Al_{2}O_{3}
you get 0.15 mol of O_{2} gas

volume of oxygen gas = moles of
oxygen gas x 24

volume of oxygen = 0.15 x
24 = 3.6 dm^{3} (3600 cm3)

Electrolysis calculation
Example 13.1.3

In
the electrolysis of dilute sulphuric acid, 36 cm^{3} of hydrogen, H_{2}
was formed at the negative electrode (cathode).

What volume of oxygen, O_{2}
would be formed at the positive electrode (anode)?

2H^{+}_{(aq)} +
2e^{} ==> H_{2(g)} and 4OH^{}_{(aq)}  4e^{}
==> 2H_{2}O_{(l)} + O_{2(g)}

It takes an electron
transfer of 2 electrons to form each hydrogen molecule from 2 hydrogen,
H^{+} ions and the transfer of 4 electrons to make 1 molecule of
oxygen from 4 hydroxide, OH^{} ions.

Therefore, from the
same amount of electrons (current), the ratio of hydrogen : oxygen
formed is 2 : 1

so the volume of
oxygen formed is 18 cm^{3}. (36 : 18 have the ratio 2 : 1)

Electrolysis calculation
Example 13.1.4

In
the electrolysis of copper sulphate solution using carbon electrodes, what
mass and volume of oxygen would be formed at the positive electrode if 254g
of copper was deposited on the negative electrode? Atomic masses: Cu = 63.5,
O = 16.

Cu^{2+}_{(aq)} + 2e^{}
==> Cu_{(s)} and 4OH^{}_{(aq)}  4e^{}
==> 2H_{2}O_{(l)} + O_{2(g)}

It takes a transfer
of 2 moles of electrons to form 1 mole of solid copper (63.5g) from 1
mole of copper(II) ions, Cu^{2+}

and a transfer of 4
moles of electrons to form 1 mole of oxygen from 4 moles of hydroxide,
OH^{} ions.

Therefore the
expected mole ratio of Cu_{(s)} : O_{2(g)} from the
electrolysis is 2 : 1

The moles of Cu
deposited = 254/63.5 = 4 moles

so moles oxygen
formed = 2 moles, since M_{r}(O_{2}) = 2 x 16 = 32

mass of oxygen
formed = 2 x 32 = 64g, volume of oxygen = 2 x 24 = 48 dm^{3}
Electrolysis calculation
Example 13.1.5

In the industrial manufacture of aluminium by electrolysis of the molten
oxide (plus cryolite) 250kg of aluminium are formed.

What volume of oxygen
would be theoretically formed at room temperature and pressure?

[ A_{r}(Al)
= 27 and 1 mole of gas at RTP = 24 dm^{3} (litres) ]

Aluminium oxide is
Al_{2}O_{3}, so on splitting in electrolysis the atomic
ratio for Al : O is 2 : 3,

and a mole ratio
of Al : O_{2} or 4 : 3

4Al^{3+}_{(l)} +
12e^{} ==> 4Al_{(l)} and
6O^{2}_{(l)} 
12e^{} ==> 3O_{2(g)}

Note: It
takes 12 electrons added to four Al^{3+} ions to make four atoms
of Al, and 12 electrons removed from six oxide ions, O^{2},
to form six oxygen atoms, which combine to form three O_{2}
molecules (see next line).

BUT, oxygen exists
as O_{2} molecules, so the mole ratio of Al atoms : O_{2}
molecules is 4 :
3

250kg Al = 250000g,
Al = 250000/27 moles = 9259.26 moles Al metal.

Therefore scaling
for moles O_{2} = 9259.6 x 3/ 24 = 6944.44 moles O_{2}
molecules.

Since volume of 1
mole of gas at RTP = 24 dm^{3} (litres)

Volume of oxygen
formed = 6944.44 x 24 = 166667 dm^{3}
Selfassessment Quizzes [pec, based on part 1 only]
type in answer
Honly or
multiple choice
Honly
Explaining electrolysis
and descriptions of experimental methods
Part Two:
The relationship between current and the quantity of electrode product

1 Faraday (F)
= 96 500 Coulombs (C) = 1 mole of electrons.

This can be
expressed as the Faraday Constant = 96500 Cmol^{1}

A current of 1A = 1C/s, 1 mole
of any gas = 24 dm^{3} or 24000cm^{3} at 25^{o}C/1
atmosphere.

Quantity of electricity in
coulombs = current in amps x time in
seconds

Example 13.2.1:
A current was passed through an electrolysis circuit of silver nitrate
solution and O.54g of silver was formed.

A_{r}(Ag)= 108 and the
electrode equation is Ag^{+} (aq) + e^{} ==> Ag_{(s)}

A_{r}(Ag)= 64 and the
electrode equation is Cu^{2+}_{ (aq)} + 2e^{}
==> Cu_{(s)}

If in the same circuit a
copper(II) sulphate and copper electrodes cell was connected, how much
copper is deposited at the negative () cathode?

0.54g Ag = 0.54 / 108 = 0.005 mol
Ag

now 1 mole electrons deposits 1
mol of silver, but only 0.5 mol of copper for the same electrons.

so mol copper deposited = 0.005 /
2 = 0.0025 mol Cu, mass Cu = 0.0025 x 64 = 0.16g Cu

Example 13.2.2:
How much copper is deposited if a current of 0.2 Amps is passed for 2 hours
through a copper(II) sulphate solution ?

Example 13.2.3:
In the electrolysis of molten sodium chloride 60 cm^{3} of chlorine
was produced.

Electrode equations:

Calculate ...

(a) how many moles of were
chlorine produced?

(b) what mass of sodium would be
formed?

from the electrode equations 2
mol sodium will be made for every mole of chlorine

so 0.0025 x 2 = 0.005 mol
sodium will be formed. A_{r}(Na) = 23

mass = mol x atomic or formula
mass = 0.005 x 23 = 0.115g Na

(c) for how long would a current
of 3 A in the electrolysis circuit have to flow to produce the 60cm^{3}
of chlorine?

To produce 0.0025 mol of Cl_{2}
you need 0.005 mol of electrons

0.005 mol electrons = 0.005 x
96500 coulombs = 482.5 C

Q = I x t, so 482.5 = 2 x t,
therefore t = 482.5 / 3 = 161 s (to nearest second)

Example 13.2.4:
In an electrolysis of sodium chloride solution experiment a current of 2 A
was passed for 2 minutes.

Electrode equations:

(a) Calculate the volume of
chlorine gas produced.

Q = I x t, so Q = 2 x 2 x 60 =
240 C

240 C = 240 / 96500 = 0.002487
mol electrons

this will produce 0.002487 / 2
= 0.001244 mol Cl_{2} (two electrons/molecule)

vol = mol x molar volume =
0.001244 x 24000 = 29.8 cm^{3} of Cl_{2}

(b) What volume of hydrogen would
be formed?

(c) In practice the measured
volume of chlorine can be less than the theoretical value. Why?

Example 13.2.5:
In a copper(II) sulphate electrolysis experiment ...

Electrode equation: () cathode Cu^{2+}_{(aq)}
+ 2e^{} ==> Cu_{(s)} and A_{r}(Cu) = 64

(a) how much copper is deposited
on the cathode by a 0.2A current flowing for 10 minutes?

Q = I x t, Q = 0.2 x 10 x 60 =
120 C, mole electrons = 120 / 96500 = 0.001244 mol e^{}

2 mole electrons deposits 1
mol of Cu, so mol Cu deposited = 0.001244 / 2 = 0.000622

mass = mol x atomic or formula
mass = 0.000622 x 64 = 0.0398g Cu

(b) how long must a 0.1 A current
be passed to deposit 1g of copper on the cathode?

1g Cu = 1 / 64 = 0.015625 mol,
needs 0.015625 x 2 mol electrons = 0.03125 mol e^{}

0.03125 mol e^{} =
0.03125 x 96500 = 3016 C

Q = I x t, 3016 = 0.1 x t, t =
3016 / 0.1 = 30160s, 30160 / 3600 = 8.38 hours.

Example 13.2.6:
What volume of oxygen is formed by passing a current of 5A through acidified
water for 25 minutes at a temperature of 25oC and 101kPa (1 atmosphere
pressure)

Electrode equations:

Quantity of electricity in
Coulombs = current in A x time in seconds

Q = I x t = 5 x 25 x 60 = 7500 C,
now 1 mole of electrons = 96500 C

so moles of electrons = 7500 /
96500 = 0.07772 moles

it takes 4 moles of electrons to
form 1 mole of oxygen gas

therefore moles of oxygen formed =
0.07772 / 4 = 0.01943

1 mole of gas = 24000 cm3,
therefore volume of gas = 0.01943 x 24000 = 466.3 cm^{3} of O_{2}

Example 13.2.7:
How long will it take to produce 2 dm^{3} of chlorine gas by passing
a 6A current through concentrated sodium chloride solution at 25C and 101kPa
(1 atmosphere pressure)

(+) anode 2Cl^{} 2e^{}
==> Cl_{2}

therefore chlorine to be produced
= 2/24 = 0.08333 moles of chlorine

2 moles of electrons must be
removed from 2 moles of chloride ions to produce 1 mole of chlorine gas,

therefore moles of electrons
required = 0.08333 x 2 = 0.1666

1 mole of electrons = 96500
Coulombs, therefore quantity of electricity required

= 0.1666 x 96500 = 16077 Coulombs

quantity of electricity in
Coulombs = current in A x time in seconds

16077 = 6 x time in seconds, so
time in seconds = 16077 / 6 = 2679.5 seconds

or 2679.5 / 60 = 44.66 minutes
to produce 2 dm^{3} of chlorine gas.


Explaining electrolysis
and descriptions of experimental methods
OTHER CALCULATION PAGES

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relative isotopic mass and calculating relative atomic mass

Calculating relative
formula/molecular mass of a compound or element molecule

Law of Conservation of Mass and simple reacting mass calculations

Composition by percentage mass of elements
in a compound

Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)

Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination

Introducing moles: The connection between moles, mass and formula mass  the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)

Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)

Moles and the molar volume of a gas, Avogadro's Law

Reacting gas volume
ratios, Avogadro's Law
and GayLussac's Law (ratio of gaseous
reactantsproducts)

Molarity, volumes and solution
concentrations (and diagrams of apparatus)

How to
do volumetric titration calculations e.g. acidalkali titrations
(and diagrams of apparatus)

Electrolysis products calculations (negative cathode and positive anode products)
(this page)

Other calculations
e.g. % purity, % percentage & theoretical yield, volumetric titration
apparatus, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy

Energy transfers in physical/chemical changes,
exothermic/endothermic reactions

Gas calculations involving PVT relationships,
Boyle's and Charles Laws

Radioactivity & halflife calculations including
dating materials
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