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Investigating the rate of electrolysis Essay

It is known that by passing a constant electric current through a copper sulphate solution the passage of ions through this solution results in copper atoms being dissolved into the solution from the anode, which has a positive charge while positive copper, ions (cations) are also being discharged at the cathode which has a negative charge. Normally anions, which have a negative charge, are discharged at the anode.

The experiment I will be carrying out is aimed to observe the amount of Copper (Cu) metal deposited during the electrolysis of Copper Sulphate solution (CuSo4) using Copper electrodes, when certain variables are changed

In this investigation I will change variables within the experiment, which will hopefully change the rate of reaction and also the deposit of copper metal at the cathode.

These variables could include:

· Voltage

· Concentration of solution/ Quantity of Solution

· Surface area/ Size of Electrodes

· Temperature

· Molarity/Concentration of Solution

· Distance between the electrodes

These variables all have a way of changing the rate of reaction.


Changing the voltage of the circuit would affect the rate of reaction because as Ohm’s law states, As charged particles try to make their way round a circuit they encounter resistance to their flow which means that they collide with atoms in the conductor. More resistance means that more energy is needed to push the same number of electrons through part of the circuit. So by increasing the voltage more electrons will flow through the circuit, which means there would be a lot more electrons flowing in the circuit therefore there will be a lot more energy being produced. This means that the more voltage flowing through the circuit the faster the reaction will take place.

Concentration of a solution:

For many reactions involving liquids or gases, increasing the concentration of the reactants increases the rate of reaction. In a few cases, increasing the concentration of one of the reactants may have little effect of the rate so changing the concentration could speed up the reaction or it could have no affect at all. Although, it is very common that when you increase the concentration the reaction does speed up.

If we were to say in this reaction increasing the concentration would speed up the reaction I shouldn’t assume that if I double the concentration of one of the reactants that I will double the rate of the reaction. It may happen like that, but the relationship may be more complicated. In order for any reaction to happen the particles must first collide. This is true whether both particles are in solution, or whether one is in solution and the other a solid. If the concentration is higher, the chances of collision are greater as there are more of the particles that react in the solution therefore they can collide more frequently which means they will react more frequently, speeding up the reaction. These same rules would also apply if you increased the quantity of the solution.

Below is a diagram of the molecules of the solutions in their regular state (left) and one where the concentration is higher (on the right)

As you can see when the concentration is higher the ratio of molecules in the solution to the reactant is higher therefore there will be more collisions and the reaction will happen faster.

Surface Area:

A solid in a solution can only react when particles collide with the surface.

The bigger the area of the solid surface,

the more particles can collide with it per second,

and the faster the reaction rate is. You can increase the surface area by breaking it up into smaller pieces.

A powder has the largest surface area and will have the fastest reaction rate. This is why catalysts are often used as powders as they speed up the reaction anyway but as they are also a powder they do it even more rapidly.


Raising the temperature of a reaction makes the particles move faster. This means that more particles collide with each other per second. Making the rate of the reaction increase raising the temperature of a reaction by 10 °C will double the rate of the reaction.

Therefore gradient of the plot of a graph will be twice as steep.

If you were given the value for a collision shown as an example on the diagram below. You can see that the area shaded shows the amount of particles that react; the value of collision shows the amount energy the particles need before a reaction can actually take place. When you raise the temperature of the reaction this then gives the particles more energy therefore a higher proportion of successful collisions the particles have the amount of energy needed so that they can react. This means there will be a lot more collisions and the reaction will take place sooner.

With all these factors in mind I predict that:

· if the current used to electrolyse the copper sulphate solution is increased, then the mass of copper deposited on the copper cathode will increase.

· as the current is increased the mass of the anode will decrease.

This is because of what happens during electrolysis. All electrolytes contain ions, which are both positively and negatively charged ions. In copper sulphate the solution I will be using there are positively charged copper ions (Cu2+) and negatively charged sulphate ions (S2-). When the copper sulphate is a solid, the ions in the copper sulphate solution are held tightly together in a regular lattice structure, and they cannot move. However, the copper sulpate I will be using in my investigation is a solution an the ions are not in a structure at all the ions are free from their lattice structure and can move around freely making it easier for them to react.

During the electrolysis of copper sulphate solution, the positively charged copper ions are attracted to the negatively charged cathode, because as in my previous years of schooling when using magnets I have found that opposite charges attract. Electricity is the flow of negatively charged electrons, when the copper ions come within contact of the cathode, they attract electrons from the cathode. The electrons then cancel out the positive charge of the copper ions, leaving it as a neutral copper atom. The positively charged copper ion has been discharged, and it forms copper metal, and this metal is deposited on the cathode as copper, so the mass of the cathode will increase during the experiment.

This also means the negatively charged sulphate ions are then attracted to the positively charged anode using the same rules applied before saying that opposite charges attract. Also the sulphate ions have two too many electrons, whereas the anode has a shortage of electrons, that’s why it is positively charged. When the negatively charged sulphate ions reach the anode, the anode attracts their electrons. Resulting in the sulphate ions losing their two extra electrons, leaving them as neutral sulphur atoms. This sulphur remains in the solution, and may sink to the bottom of the beaker during the experiment. However the reason that the anode loses mass is because of what the power supply is doing to the electrons in the copper as the electrons are being ripped away by the cell.

And this is why the anode loses mass, and the cathode gains mass.

My reasoning behind my prediction is that the mass loss of the anode is the same as the mass gained at the cathode, which is clearly stated in Faraday’s Law.

This law states that the amount of copper deposited on the cathode and lost from the anode depends on the number of electrons passing through the circuit, i.e. upon the charge passed through the power supply. The charge passed, q in Coulombs. A Coulomb is the quantity of electricity that flows when one amp is allowed to flow for one secon. 96,500 Coloumbs is 1 faraday which is one mole of electricity. Is related to the current, I (in amps) and time, t (in seconds). The formula is,

q = It

From this equation I would be able to predict the results I will be getting from my investigation.

As the time for this experiment is constant, then as the current flowing round the circuit is increased, then the charge passed will increase, and therefore more copper will be deposited on the cathode, and more copper will be lost at the anode.


Two copper electrodes will be chosen so that they had the same approximate surface area and length. One of these electrodes will be chosen as the cathode being negatively charged, and one the anode being positively charged. The anode will then to be weighed, and the mass noted. The circuit will then be set up as shown below.

My apparatus will consist of:

Ø Two Copper Electrodes (Anode(+)Cathode(-) )

Ø Beaker

Ø Ammeter

Ø Copper Sulphate Solution

Ø Battery Power Pack

Ø Crocodile Clips

Ø Circuit Wires

Ø Stopclock

Ø Scales

The power supply will be set to different volt settings in my experiment first at 2 volts, then 4, then 6, then8 then 12. I will then switch on the power supply for five minutes at two volts. After the five minutes I will then take out the anode dry it and then weigh the anode and note the mass. I will then subtract this mass from the mass I recorded at the start of my experiment to see how much weight it has lost. I will then move onto 4 volts and so on. After successfully carrying out this experiment to 12 volts I will then repeat this experiment twice more to make sure I get accurate results using different electrodes each time. I will then take an average of the results. Throughout this experiment safety glasses must be warn at all times. This is because copper sulphate is an irritant and could cause damage to the eyes.

To make sure that this test is fair at all times, the only variable I will be altering during the experiment, is the current. Therefore, everything else that is used during the experiment must remain the same. These things are, , the resistance in the circuit, the surface area of the two electrodes, and the time that the solution is electrolysed for. Most of these things can be monitored fairly easily. The time that the solution is electrolysed for can also be kept at roughly five minutes throughout the experiment. This is only rough because I am stopping the timer myself and therefore there is the factor of human error, so the time will not be exactly two minutes all of the time.

The surface area of the two electrodes can also be kept the same, because the same two electrodes will be used throughout the experiment. The one variable that may not be very easy to control will be the resistance. This is because when the circuit is connected, the wires used create resistance. However, if the same wires are used throughout the experiment, then this will not be a problem, because the resistance will be the same throughout the experiment, and the current can be adjusted, using the variable resistor, at the start of the experiment, so that it is always the same.

The measurements in this experiment will be as accurate as they can be with the equipment that is available to me. The mass of the anode will be kept accurate to three decimal places, because this is the accuracy of the scales that are available to me. The time, as I have mentioned before, may be able to be measured to one hundredth of a second using the electronic stop-clock, but as I will be stopping the timer, the human error factor only makes this time correct to one second. To make sure that my results are as reliable as possible, I will repeat each reading three times.

This is to make sure that even if one, or two of the results go wrong, then the other two or three will be accurate. In addition to these factors I will also be weighing only the anode as this is the elctrode that loses mass whereas the cathode gains because of the copper that gets deposited on it. I will be weighing the anode becausee if I weighed the cathode the reasults may become irregular as some of the copper deposited on the catode may fall of when moving it to the scales or even drying the cathode. Therefore I would not be weighing the true mass that has been gained.

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