Section 1i Electrolysis

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    Section 1i: electrolysisElectrolytes

    1.48 understand an electric current as a flow of electrons or ions1.49 understand why covalent compounds do not conduct electricity1.50 understand why ionic compounds conduct electricity only when molten or in solution1.51 describe experiments to distinguish between electrolytes and non-electrolyte.

    n this topic we will study the effect of passing an electric current through a compound.

    !n electric current is a flow of electrons. "or a substance to conduct electricity it must have mobile

    charged particles i.e. electrons or ions. "or instance metals can conduct electricity #a current passes

    through them$ because metals have delocali%ed electrons.

    &ovalent compounds do not have any charged particles. 'hey consist of atoms or molecules which are

    electrically neutral. 'hey cannot conduct electricity even if we melt or dissolve them in water #if they are

    soluble that is$.

    (n the other hand ionic compounds have charged particles #metal positive ions and negative non-metal

    ions$. )owever* ionic compounds can only conduct when the ionic compound is molten or dissolved in

    water. (nly in the li+uid or a+ueous state can the ions #charged particles$ move to the oppositely charged

    electrode.! li+uid #molten or solution$ that conducts electricity is called an electrolyte.

    ,e can tests substances if they are electrolytes or non-electrolytes #li+uids which do not conduct

    electricity$ by doing simple tests.

    ! circuit is set up as shown below. t consists of a battery or power supply* wires* carbon electrodes and

    an indicator of a current i.e. a bulb or an ammeter. &arbon is piced as a material for the electrodes

    because it is inerti.e. it does not react and does therefore not interfere in the electrolysis reaction. 'he

    electrode which is connected to the positive terminal of the battery is named the anode the electrode

    connected to the negative one is the cathode.

    'he substance to be tested is either molten or dissolved in water. f the bulb lights or a current is

    measured by an ammeter* the substance tested is an electrolyte. n the diagram below the compound onthe left is ionic / the bulb is on - whilst the one on the right is covalent as the bulb does not light.

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    Electrolysis

    1.52 understand that electrolysis involves the formation of new substances when ionic compoundsconduct electricity

    1.5 describe simple experiments for the electrolysis* using inert electrodes* of molten salts such as lead#$ bromide1.54 describe simple experiments for the electrolysis, using inert electrodes, of aqueous

    solutions of sodium chloride, copper(! sulphate and dilute sulphuric acid and predict the products1.55 write ionic half-e+uations representing the reactions at the electrodes during electrolysis.

    6ection 1i electrolysis 1 |P a g e

    molten or

    dissolved

    ionic

    compound

    (electrolyte)

    covalent

    compound

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    ,hen a solution is electroly%ed predicting the products

    becomes more complicated as also the water #the solvent$

    also becomes involved. ,ater* although a covalent

    compound* ioni%es a little bit producing hydrogen and

    hydroxide ions which are also attracted to the electrodes. n

    most cases a gas will be produced which needs to be

    collected and tested. ,hen carrying out electrolysis during

    which a gas will be produced the following set up ca be used

    to collect the gas#es$.

    6tarting

    chemicals

    concentrated sodium chloride solution or brine* a&l #a+$

    ions present

    at cathode a ) at the anode &l- ()-

    only 1 ion can react or be discharged at each electrode* the other remains in

    solution

    observations anode

    bubbles* yellow green gas* nearly 11 ratio with gas at cathode

    test gas bleaches damp litmus papercathode

    bubbles* colourless gas

    test :pop; sound with lit splint* more than 11 ratio

    red litmus turns blue

    products anode

    chlorine is discharged and not the hydroxide ions which remain in solution

    ionic half e+uation 2&l-#a+$ &l2#g$ 2e- < oxidation

    cathode

    hydrogen is discharged

    o ionic half e+uation 2)#a+$ 2e- )2#g$ < reduction

    sodium hydroxide a and ()- are left in the solution as sodium hydroxide which can

    be detected using an indicator

    overalle+uation

    at anode 2&l-#a+$ &l2#g$ 2e-

    at cathode 2)#a+$ 2e- )2#g$

    2)#a+$ 2&l-#a+$ &l2#g$ )2#g$

    'rend for any concentrated solution of a salt made from a group 1 metal and a group = non-metal there

    will always be three products #1$ hydrogen* #2$ the halogen* and #$ the hydroxide of the group 1 metal.

    Electrolysis of copper (! sulphate solution using inert electrodes

    starting

    chemicals

    copper sulphate solution* &u6(4#a+$

    ions present at cathode &u2 ) at the anode 6(42- ()-

    observations anode

    bubbles* colourless gas

    test relights glowing splint

    cathodered deposit on electrode

    products anode

    oxygen is produced ionic half e+uation 4()-#a+$ (2#g$ 2)2( #l$ 4e

    - < oxidation

    sulphate* 6(42-remains in solution

    6ection 1i electrolysis " |P a g e

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    cathode

    copper

    o is discharged and not the hydrogen ions as they remain in the solution

    o ionic half e+uation &u2 #a+$ 2e- &u #s$ < reduction

    hydrogen ions remain in solution

    overall e+uation at anode 4()-#a+$ (2#g$ 2)2( #l$ 4e-

    at cathode 2&u2 #a+$ 4e- 2&u #s$

    4()-#a+$ 2&u2#a+$ (2#g$ 2)2( #l$ 2&u #s$

    Electrolysis of dilute sulphuric acid

    'he two previous solutions are consideredconcentrated solutions. 'he electrolyte*sulphuric acid* in this example is diluted and thisalso has an effect on ind of products of theelectrolysis as none of the ions of the electrolytewill be discharged. nstead they remain in thesolution. 'he ions which are discharged are thehydrogen and hydroxide ions which formhydrogen gas and oxygen gas.

    n essence when a dilute solution is electroly%edwater is decomposed. 'he identity and natureof the electrolyte does not really matter

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    starting

    chemicals

    dilute sulphuric acid* )26(4#a+$

    ions present at cathode ) at the anode 6(42- ()-

    observations anode

    bubbles* colourless gas*

    volume approximately half of the gas produced at the cathode

    test relights glowing splint

    cathode

    bubbles* colourless gas volume twice as much as gas at the anode

    test :pop; sound with lit splint*

    products anode

    oxygen is produced

    ionic half e+uation 4()-#a+$ (2#g$ 2)2( #l$ 4e- < oxidation

    cathode

    hydrogen is discharged

    ionic half e+uation 2)#a+$ 2e- )2#g$ < reduction

    overall e+uation at anode 4()-#a+$ (2#g$ 2)2( #l$ 4e-

    at cathode 4) #a+$ 4e- 2)2#g$

    6ection 1i electrolysis # |P a g e

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    4()-#a+$ 4)#a+$ (2#g$ 2)2( #l$ 2)2#g$

    Electrolysis of copper solution using copper electrodes as used in the purification of copper.

    cathode

    &u2#a+$ 2e-

    &u# s$

    anode

    &u #s$ &u2#a+$

    2e-

    (bservation&opper ions fromsolution aredeposited onto piece ofpure copper.>ass of electrodeincreases

    (bservation&opper from impurecopper goesinto solution impuritiessin.>ass of electrodedecreases

    Electrolysis calculations

    1.5 recall that one faraday represents one mole of electrons1.5 calculate the amounts of the products of the electrolysis of molten salts and aqueous solutions

    'he amount of product of an electrolysis reaction can be calculated and depends on

    'he amount of charge passed which is expressed in a number of faraday.

    1 faraday = the charge of 1 mole of electrons = 96 500 coulombs (C)

    'he amount of charge that passes in a circuit depends on the current #amperes or !$ of the circuitand the amount of time #in seconds$ the current is switch on.

    'he number of faraday can be calculated byo multiplying the amperes by seconds which gives coulombs #&$

    o dividing the number of coulombs by 93 500 &

    charge of the ions the greater the charge of the ion* the greater the number of electrons needed or

    the greater the number of faraday

    a e- a this means that to produce 1 mole of sodium atoms #i.e.2 grams$ one mole of electrons or 1 faraday is needed.

    &a2 2e- &a this means that to obtain 1 mole of calcium atoms twomoles of electrons or 2 faraday are needed.

    2&l-#a+$ &l2#g$ 2e- this means that to obtain 1 mole of chlorine molecules two

    moles of electrons or 2 faraday are needed.

    4()-#a+$ (2#g$ 2)2( #l$ 4e- this means that to obtain 1 mole of oxygen molecules 4

    moles of electrons or 2 faraday are needed.

    ,ored example

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    ! solution of copper #$ sulphate is electroly%ed. )ow much copper will be deposited by a current of 2 !flowing for 20 minutes?

    6tep 1 calculate the amount of coulombs #g2>g2 2e- >g

    0.24

    @r- 2@r-#a+$ @r2#g$ 2e- 10

    ()-4()-#a+$ (2#g$ 2)2( #l$ 4e-

    0.025

    &l-2&l-#a+$ &l2#g$ 2e

    -

    0.5

    2. &opper #$ sulphate solution is electroly%ed for 15 minutes using copper electrodes and a 4 ! current.&alculate the mass of copper produced.

    . !luminium oxide is extracted by electrolysis of molten aluminium oxide. 'he aluminium ion is !l

    .,hat mass of aluminium will be obtained if a current of 25000 ! flows for 24 hours?

    4. ,hen sodium chloride is electroly%ed* the gases hydrogen and chlorine are obtained.

    #a$ ,rite the e+uation for the formation of chlorine gas at the electrode#b$ ! current of 2! is passed for 40 minutes. &alculate the volume of chlorine released at room

    temperature and pressure.

    5. 1g of sodium is produced by electroly%ing sodium chloride. ,hat volume of chlorine is produced atthe same time?

    3. Auring the electrolysis of lead #$ nitrate solution* lead is deposited at the cathode and oxygen at theanode according to the e+uations below

    f a current of 0.50 ! flows for 1000 seconds calculate the mass of lead and volume of oxygen

    6ection 1i electrolysis % |P a g e

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    obtained at B'C.

    cathode reaction Cb2 2e- Cb

    anode reaction 4()-#a+$ (2#g$ 2)2( #l$ 4e-

    6ection 1i electrolysis & |P a g e