Investigate the factors that affect the mass of Copper deposited on the Copper Cathode during the Electrolysis of Copper (11) Sulphate Solution using Copper Electrodes.

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Georgina Leach

An Experiment to investigate the factors that affect the mass of Copper deposited on the Copper Cathode during the Electrolysis of Copper (11) Sulphate Solution using Copper Electrodes

Background Information

Electrolysis is the decomposition of a molten or aqueous compound by electricity.  Electrolysis occurs only in liquids called electrolytes.  Electrolytes are compounds, which conduct electricity when molten or dissolved in water, but not when they are in a solid state as the oppositely charged ions are held together by strong ionic bonds in a giant lattice and this means electricity cannot pass through it and the ions are unable to move.

If electrolytes are molten or dissolved in water their ions are free to move, and the movement of ions in liquids is key in electrolysis.  Electrolysis takes place in an electrolyte cell.

The diagram below shows the apparatus used in the electrolysis of molten or aqueous ionic compounds.

The electrodes (anode and cathode) are known as conducting rods.  Normally they are made out of platinum or graphite as these substances are chemically unreactive and will not interfere with the experiment. They are inert electrodes. However in this experiment we are using copper electrodes.  Copper electrodes are active electrodes and these metal electrodes take part in electrolysis.  Copper electrodes are used in industry to obtain pure copper.

The anode is the positive electrode, electrons flow from the anode to the battery. The anions, which are negatively charged non-metal ions (except for some complex transition metal ions) are attracted to the anode.

The cathode is the negative electrode.  Electrons flow into the cathode from the battery.  The cations, which are positively charged metal ions (except for hydrogen ions and ammonium ions) are attracted to the cathode.

In the diagram copper (11) sulphate solution is being electrolysed using graphite electrodes, one would expect the following to happen.

At the cathode

The ions Cu   and H   are both attracted to the cathode, but because copper is lower down in the reactivity series than hydrogen, copper ions migrate to the cathode forming a layer of copper at the cathode.

Cu   + 2e   ⇒   Cu atom

At the anode

The ions OH   (hydroxide ions) and So     (sulphate ions) are attracted here.  As So      is not a halogen, OH   ions give up electrons more readily and migrate to the anode forming oxygen there.

OH   - 4e   ⇒   2H  O + O

Electrolysis involves both Oxidation and Reduction reactions at the electrodes.  They are chemical opposites, oxidation is loss of electrons and reduction is the gain of electrons.

At the cathode, cations gain electrons and are reduced and at the anode, anions lose electrons and are oxidised

As we know there are two different types of electrolytes, aqueous and molten.  I am now going to discuss what happens when the compounds are electrolysed.

The electrolysis of molten electrolytes

When a molten ionic compound is used in electrolysis a metal is always formed at the cathode and a non-metal is always formed at the anode.  There is only one exception to this and that is hydrogen, the only non-metal that forms at the cathode e.g. electrolysis of molten sodium chloride.

The electrolysis of aqueous electrolytes

When an aqueous ionic compound is used in electrolysis the following occurs.  Metals above hydrogen in the reactivity series are considered more reactive.  At the cathode the more reactive the metal, the more eager it is to exist as ions and remain in the solution so we can say if you have a more reactive metal, the H+ ions accept electrons and hydrogen molecules are formed at the cathode and the metal stays in solution.

The ions of less reactive metals (below hydrogen on the reactivity series) will accept electrons and form metal atoms at the cathode, leaving hydrogen ions in solution.

At the anode the following occurs.  If ions of a halogen are present (chlorine ions, bromine ions and iodine ions) they will give up electrons more eagerly than hydroxide ions (OH  ) do.  This means molecules of chlorine, bromine and iodine will form at the cathode.  If a halogen is not present hydroxide ions will give up electrons more readily than other non-metal ions do and oxygen will form at the anode.

Factors that affect the mass of copper deposited on the cathode:

  • Surface area of electrodes:

The greater the surface area of the electrodes, the greater the amount of Cu 2+ ions lost at the anode and gained at the cathode

  • Concentration of Copper (11) Sulphate Solution:

The greater the concentration of copper (11) sulphate solution, the more copper deposited at the cathode

  • Time

The more time the copper electrodes are left in the copper (11) sulphate solution, the more time there will be for cu 2+ ions to be deposited.

  • Current

If the current is larger, there will be more electrons flowing around the circuit and more copper will be deposited.

  • Temperature

The greater the temperature the more copper deposited at the cathode as particles will obtain more energy and collide more if placed in a warmer temperature.

The factor I will investigate is current. This is because it is easy to obtain results and make predictions, calculating the theoretical value.  Also using scientific knowledge I can make a qualitative prediction .  The experiment will also be easy to set up and to achieve accurate results.

Qualitative prediction

In this investigation I am investigating the electrolysis of Copper (11) sulphate solution using copper electrodes.  When this experiment is undertaken I predict that the anode will dissolve into the copper (11) sulphate solution and this is predicted because it’s atoms give up electrons to form ions in the solution

Cu atom  -  2e     Cu    ion

I also predict that the opposite occurs at the cathode, the copper ions pick up electrons and become copper atoms

Cu     +     2e        Cu atom

Here we can see that as copper ions leave the copper (11) sulphate solution to form copper atoms at the cathode, they are replaced in the solution at the anode.  This means it can be predicted that the mass of copper gained at the cathode equals the mass of copper lost at the anode.  This is what one expects to happen in the electrolysis of copper(11) sulphate solution using copper cathodes, now the factor which is going to be investigated, the current will be observed.

        It is predicted that the current will affect the amount of copper deposited on the cathode in the following ways.  The larger the current the more copper will be deposited and so from this we can say if you double the current you double the amount of copper deposited at the cathode.  This is because as the current increases, so does the amount of electrons flowing around the circuit, so in a given time (20 minutes) more ions will be formed at the anode by losing two electrons (oxidation) and therefore more copper atoms will be formed at the cathode as ions will gain two electrons (reduction).  So as we can see the greater the amount of electrons (charge) flowing around the circuit, the greater the mass of copper deposited at the cathode.  This means it can be predicted that the amount of copper deposited at the cathode is directly proportional to the size of the current.  The currents that shall be used are 0.2A, 0.4A, 0.6A, 0.8A and 1A.  I predict that as the current increases more copper will be deposited on the cathode and the amount of copper deposited on the cathode when using the current 0.2A will be half the amount of copper deposited on the copper cathode using the current 0.4A as 0.4A is twice 0.2A.  This also applies to my other currents such as 0.4A and 0.8A.

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Quantitative Prediction

        Since how the current affects the mass of copper deposited on the copper cathode is being investigated it is sensible to pick five different readings, 0.2 Amps, 0.4 Amps, 0.6 Amps, 0.8 Amps and 1.0 Amps.  This range and number of currents were chosen because they cover a wide scale and will therefore produce a broad range of results, which hopefully support my prediction.  All the readings were left for a period of 20 minutes.  I will now calculate the theoretical values for each reading in order to predict my results.

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