The table below list the products formed at the electrodes when various liquids and aqueous solutions are electrolysed.
Information taken from Chemistry counts by Graham Hill, Second Edition, published by Hodder & Stoughton in 1997/1998.
When acids and metal/non metal conduct electricity-
- Metals of hydrogen are formed at the cathode.
- Non- metals (except hydrogen are formed at the anode).
The compounds are decomposed by electrical energy and an element I produced at each electrode. This is different from conduction of electricity by which the metals do not decompose. The first two electrolyses in the table can be summarised in word equations-
The electrolysis of molten sodium chloride makes sodium and chloride. When an electrical current passes through molten sodium chloride a head of sodium (this head could be described as shiny) is produced at the cathode and chlorine gas is formed at the anode. This decomposition is caused by the electrical energy in the current. It occurs because sodium particles in the electrolytes are positive, they are attracted to the negative cathode. At the same time chlorine is produced at the anode, so chloride particles in the electrolyte are negative. Sodium chloride has the formula
NaCl → Na + and Cl-. Na Cl is neutral meaning the positive charge on the Na+ must be equal with the negative charge on the Cl-. These charged particles, which move to the electrodes during electrolysis, are called ions. During electrolysis Na+ ions near the cathode combine with negative electrons on the cathode forming neutral sodium atoms.
At the anode Cl- ions lose electrons to the positive anode leaving neutral chlorine atoms.
The Cl atoms then join up in pairs to form molecules of chlorine gas Cl2
These equations show that Na+ ions remove electrons from the cathode, and Cl- ions give up electrons to the anode during electrolysis. The electric current it being carried through the molten sodium chloride by ions. The electrolysis of other molten and aqueous substances can also be explained in term of ions.
When copper sulphate is electrolyses with copper electrodes copper is deposited in the cathode and the copper anode loses weight.
The aqueous copper sulphate contains copper ions (Cu 2+) and sulphate ions ( SO4 2-). During electrolysis Cu2+ ions are attracted to the cathode where they gain electrons and deposit on the cathode:
SO4 2+ ions are attracted to the anode, but they are not discharged. Instead, copper atoms, which make up the anode, give up two electrons each and go into solutions as Cu2+ ions.
The over all result on this electrolysis is that the anode loses weight and the cathode gains weight copper metal is transferred from the anode to the cathode.
Industrially this method is used to purify crude copper. The impure copper is the anode of the sell. The cathode is a thin sheet of pure copper. The electrolyte is copper sulphate solutions. The impure copper anode dissolved away, copper deposits on the cathode.
The famous scientist Michael Faraday was research electrolysis forming Faraday’s Law in 1843:
This is a physical law stating that the number of moles of substance produced at an electrode during electrolysis is directly proportional to the number of electrons transferred at the electrode. The amount of electric charge carried by one mole of electrons ( 6.02 x1023electrons) is called the faraday and is equal to 96,500 coulombs. The number of faradays required to produce one mole of substance at an electrode depends upon the way in which the substance is oxidised or reduced. For example in the electrolysis of molten sodium chloride, NaCl, one faraday, or one mole, of electrons is transferred at the cathode to one mole of sodium ions, Na+, to form one mole of sodium atoms, Na, while in the electrolysis of molten magnesium chloride MgCl2, two faradays of electrons must be transferred at the cathode to reduce one mole of magnesium ions, Mg+2 to one mole of magnesium atoms, Mg.
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Amount of substance deposited during electrolysis α to current passed and time taken.
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When 2 or more substances are deposited at different electrodes at the masses areαto the ionic charges of those substances.
Q=IT
Charge = Current x time
Equation for Faraday’s 1st Law
Faraday’s 2nd Law
The masses of different substances produced by the same current in the same time are α to their atomic masses divided by their ionic charges of valencies.
The aim of this experiment is to prove that by passing electric current through an aqueous copper sulphate solution changes the mass of each electrode. The anode’s mass will increase and the cathode’s mass will decrease. I will also attempt to investigate the relationship between the amount of change and the size of the voltage and current. If Faraday’s laws are correct the mass gained at the anode will equal the mass lost at the anode.
From my research I can predict that the Cathode will increase in mass and the Anode will decease in mass. I think that the higher the voltage the larger the change in mass will be, because with more volts passing through the electrons they move around more and therefore more electrons move to the cathode and more electrodes leave the anode. I can also predict that an increase in time and therefore and increase in the length of time the current flows will increase the change in mass. I also think that if the time doubles the amount of copper produced will also double baring mind that if the time doubles the charge doubles.
- Power Pack
- Two pieces of copper
- Stop Clock
- Copper sulphate solution
- Accurate balance
- I will be working with electricity for the main, which is as a high voltage so I will be careful in order to remain safe.
- I will also be working with a power pack so again I need to be careful to ensure my safety and that of my partners
- I will be using glass beakers so I need to be careful, in the event of breakages I will in sure that all glass I cleared away and placed in the “broken glass” container.
I will do my up most to obtain accurate measurements, to do this I will-
- Tare the balance before weighing
- Ensure the stop clock it started and stopped at the correct time.
- There is no specific amount of Copper Sulphate although the measure meant will be around 250ml because that is the size of the glass beaker I will be using
- Ensure the molarity of the copper sulphate solution is kept at a constant.
I will ensure that both electrodes are clean before starting the experiment. I will mark one of the electrode with a +, this will become the anode and one of the electrode with a (-) this will become the cathode. I will weigh each electrode on an accurate balance Tareing the balance before hand. I will pour the copper sulphate solution with a molarity of 2M into a 250ml glass beaker. The copper electrode will then be placed into the solution and stuck to each end of the beaker; the electrodes are connected to a cell along with an ammeter and a voltmeter. At the correct time the electrode will be re-weighed. In order to do this the cell will be switched off for safety and accuracy and then the electrodes will be removed from the solution. The copper will stick to the electrode permanently. Once the electrodes are weighed using an accurate scale they can be reused to record another result so that the relationship between the difference in mass and the amount of voltage and current can be recorded.
I will record my results on a table with the following headings:
I thought it would be a good idea if I had a rough idea of what kind of results I should be obtaining. It was possible to do this with the using a series of simple equations.
Charge (C) = Current (A) x Time (sec.)
Moles of Electrons or Faradays = Charge (C) / 96500
Moles of Copper = Moles of electrons or Faradays / ratio=2
Mass = moles x RAM
If the Current is 0.2A and the time taken 5 minutes
Ø Charge = 0.2 x (5x60)
Ø Faradays = 60/96500
Ø Moles of Copper = 0.0006217/2
Ø Mass = 0.0003108 x 64
Ø Mass = 0.0199 grams
If the Current is 0.2A and the time taken 10 minutes
Ø Charge = 0.2 x (10 x 60)
Ø Faradays = 120/96500
Ø Moles Copper = 0.0012435/2
Ø Mass = 0.0006217 x 64
Ø Mass = 0.0398 grams
If the current is 0.2A and the time taken 15 minutes
Ø Charge = 0.2 x (15 x 60)
Ø Faradays = 180/96500
Ø Moles Copper = 0.0018652/2
Ø Mass = 0.0009326 x 64
Ø Mass = 0.0597 grams
If the current is 0.2A and the time taken 20 minutes
Ø Charge = 0.2 x (20 x 60)
Ø Faradays = 240/96500
Ø Moles Copper = 0.002487/2
Ø Mass = 0.0012435 x 64
Ø Mass = 0.0759 grams
If the current is 0.2A and the time taken 25 minutes
Ø Charge = 0.2 x (25 x 60)
Ø Faradays = 300/96500
Ø Moles Copper = 0.0031088/2
Ø Mass = 0.0015544 x 64
Ø Mass = 0.0995 grams
These equations will help to support my predictions, as from these equations a “theoretical” table of values can be produced and those can be plotted against the actual results obtained. From this comparison, it will be possible to spot any anomalies in the results and from this explain why these may have occurred
THEORETICAL RESULTS
Information taken from the Internet
On reflection, although my results were accurate I did not record the information to prove what I intended to prove. The main factor I wanted to investigate was the effect of time on the change of mass. However I only did this for 5 minutes when I investigated the effect of different voltages and different currents. The remaining results do not reflect what I wanted to achieve in this experiment. However I have left them in as experimental proof. To investigate the effect of time on the change in mass I will use results collected form a secondary resource.
My Results
ANODE (+)
CATHODE (-)
AVERAGE
Results taken from a secondary source
The results I obtained from a secondary resource support my prediction that if the time increases the so does the amount of copper deposited on the cathode. It is also possible to say that if the time is doubled and therefore the charge is doubled then so is the amount of copper produced. This can be seen in the results:
In 10 minutes 0.0405 grams of Copper is produced.
In 20 minutes 0.0810 grams of Copper is produced.
0.0810 grams is exactly double 0.0405 grams. This proves the prediction that the longer the experiment lasts, the higher the charge and therefore, the higher the amount of Copper produced.
The actual results produce an almost straight-line graph. Therefore, it has now been proved, through this experiment, that both of Faraday’s Laws Of Electrolysis are correct.
Faraday’s First Law of electrolysis states that:
“The mass of any element deposited during electrolysis is directly proportional to the number of coulombs of electricity passed”
Although my results were accurate I failed to record the correct information and I changed too many variables. Once I realised this there was only enough laboratory time left to obtain the results for 5 minutes and repeat this. These results were accurate; they proved my prediction and Faradays law. In order to improve this investigation I should continue to record results in the same way for varying times repeating each time at least once. The experimental results I obtain from a secondary source appeared to be very accurate and therefore they product of a successful experiment; however the following improvement are true for both experiments. The electrodes, which, even after a thorough cleaning were still fairly dirty and seemed to have irremovable substances from previous experiments still attached to them. If this experiment were to be repeated for a second time, to a greater degree of accuracy, it would be important to use a new pair of electrodes, which had never been used before.