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Electrolysis Experiment

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Electrolysis Experiment

Aim:

To find out how current affects the rate of electrolysis.

Pre-test:

Pre-test Apparatus:

1. Power Pack

2. Stop watch

3. 3 wires

4. Amp meter

5. A piece of Card

6. stop watch

7. Top pan balance

Pre-test Method:

Text Box:

1. Check to see if all the apparatus work by forming a small circuit
with a bulb

2. Collect apparatus and set them out like the diagram below. Ensure
that the positive wire leads to the anode and the negative leads to
the cathode.

3. Measure out 50 ml of copper sulphate

4. In to a small class beaker

5. Weigh cathode on top pan balance

6. Slide anode and cathode in to piece of card and place in
electrolyte

7. Turn on the power pack

8. Time until two minutes

9. Weigh cathode make sure you do not scrape off any copper

10. Repeat three times

11. Repeat for each current

Pre-test Findings:

By doing this pre-test I have learnt that:

* The analog amp meter is very unreliable it

* To be very accurate I need to use a multi meter

* To be careful not to scrape of copper when measuring weight

* Make sure that the anode and the cathode do not touch during the
experiment

* Make sure wires and power packs are working before you start

* To change the electrolyte after each experiment

* To use a stop watch instead of a clock as it is more accurate

* To use a variable resistor to get the exact current you need

Research:

There are many factors that affect the rate of electrolysis:

· The current

· The dept of the electrodes

· The distance between the electrodes

· Time allowed for electrolysis

· The concentration of the electrolyte

The current effects electrolysis. This is a straightforward
relationship. If you double the charge, then the rate of electrolysis
will double. This is because electrolysis involves ions taking
electrons from the cathode and depositing electrons at the anode.
Charge is the number of electrons. If there are twice as many
electrons available then twice as many ions can pick up electrons each
second.

The concentration of the solution does have a significant effect,
since the concentration of ions do not make it any easier for the
electrons to flow, nor does it affect any factors in the ionic
equation of reaction, so it should not have an effect. However, at
very low concentration, the resistivity of water increases due to the
lack of dissolved ions, and this would have an effect on the current
and therefore having an effect on the mass of copper deposited.

The time allowed for the electrolysis to run would have an effect on
the mass of copper deposited, since the deposition of copper is a
continuous process, and providing there is a fixed current it happens
at a fixed rate. Therefore, the longer the time, the more copper atoms
would be allowed to deposit and the bigger the combined mass. The mass
would be directly proportional to time since the rate is fixed.

If you increased the surface area by using larger pieces of electrode
(eg copper sheet) then this would make it easier for the electricity
to enter the electrolyte so there is more place for reactions to take
place so this might accelerate the electrolysis.

The shorter the distance between the two electrodes the faster the
rate of electrolysis. This is because if they are closer together the
current would flow through each electrode faster, therefore increasing
the making the rate of electrolysis faster.

[http://www.bbc.co.uk/schools/ks3bitesize/sosteacher/science/45200.shtml]

Faraday’s Law:

That the number of moles of substance produced at an electrode during
electrolysis is directly proportional to the number of moles of
electrons transferred at that electrode; the law is named for Michael
Faraday, who formulated it in 1834. The amount of electric charge
carried by one mole of electrons 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 oxidized or reduced

[http://www.encyclopedia.com/html/F/Faradays.asp]

Prediction:

I predict that copper in the solution would form outside the cathode.
This is because the electrolyte is a solution of copper sulphate and
water ( CuSO4 . H2O).

It consists of two positive ions, Copper Cu 2+ and the hydrogen ions
in the water H+. The copper ions are unstable, to become stable they
must lose 2 electrons therefore they would migrate towards the
negative electrode the Cathode in order to lose two electrons. To make
a balanced molecule the Cu2+ ion has to lose two electrons.

Cu2+ + 2e- = Cu

[www.bbc.co.uk/physics/electrolysis/sos.12]

I also predict that the electrolyte would turn from blue to
colourless. This is because the copper is the substance which gives
the solution the blue colour. If the copper is extracted the solution
would turn clear.

I predict that the hydroxide would form on the anode. This is because
the negative ions in the solution are the sulphate ion SO42-, and the
hydroxide ion OH-. These will migrate to the positive anode. The
hydroxide would lose it’s oxygen. The evidence to this is that during
the pre-test you could see fizzing and bubbles coming off the anode.
In this case the hydroxide ion is discharged by loosing its electron
to form water and oxygen, 4OH- - 4e- = 2O + 2H2O.

I also predict that carbon dioxide would be formed, as oxygen is being
released fro the hydroxide the oxygen will react with the carbon anode
to form carbon dioxide this, attacks the carbon anode and produces
carbon dioxide. C + O2 =CO2..

[www.bbc.co.uk/physics/electrolysis/sos.12]

I predict that the rate of electrolysis would be higher as the current
increases.

This is because electrolysis involves ions taking electrons from the
cathode and depositing electrons at the cathode. If you increase the
current you increases the flow of electricity, thus you increase the
overall charge of the electrodes so ions can pick up more electrons
each second. I also predict that the results will be directly
proportional because if you double the current you double the charge
so you double the amount of copper deposited.

Faraday’s law:

Q (the amount of charge in Coulombs) = I (the current in Amps) x t
(time in seconds).

1 mole of electrons is called a Faraday and is equal to 96500
coulombs.

As copper is in a pair ( Cu2+) the formula would be:

96500 x 2 =193000, 2 Faradays

One Faraday will deposit ½ a mole of copper metal

So 1 mole of copper metal would be deposited

Q= I x T

0.2 x 120 = 24

0.4 x 120 = 48

0.6 x 120= 72

0.8 x 120 =96

1 x 120=120

[IMAGE]Atomic weight =64g

Charge = 193000

Current x time=

0.2 x 300 = 60

60 x atomic weight of cathode = 60 x 64 = 0.02 g

Time 193000

Current x time=

0.4x 300= 120

120atomic weight of cathode = 120 x 64 = 0.04g

Time 193000

Current x time=

0.6 x 300= 180

180 x atomic weight of cathode = 180 x 64 = 0.06g

Time 193000

Current x time=

0.08 x 300= 240

240 x atomic weight of cathode = 240 x 64 = 0.08g

Time 193000

Current x time=

1 x 300 = 300

300 x atomic weight of cathode = 300 x 64 = 0.1g

Time 193000

Safety:

ü Wear goggles for each experiment

ü Avoid contact from eyes

ü Wipe and spills immediately

ü Do not touch circuit with wet hands

ü Make sure power pack is off when it is not in use

ü Make sure there is no break in circuit

ü Tuck all chairs under tables

ü Put bags out of the way

ü Avoid contact from skin, cuts and grazes

ü Do not touch the electrodes while the power pack is on

ü Start of with your highest current and work your

ü Way down wards so you do not blow a fuse.

Fair test:

In each experiment make sure that the following are the same:

ü The starting temperature of the electrolyte

ü The concentration of the ions in the electrolyte

ü The surface area of the of the electrodes

ü The distance between the electrodes

ü Make sure that the two electrodes do not touch

For accurate results repeat each test three times.

Method:

1. Text Box: Check to see if all the apparatus work by forming a
small circuit with a bulb

2. Collect apparatus and set them out like the diagram Ensure that the
positive wire leads to the anode and the negative leads to the
cathode.

3. Measure out 50 ml of copper sulphate In to a small class beaker

4. Weigh cathode on top pan balance

5. Slide anode and cathode in to piece of card and place in
electrolyte

6. Turn on the power pack

7. Time until two minutes

8. Weigh cathode make sure you do not scrape off any copper

9. Repeat three times

10. Repeat for each current

Observation:

Observations made during the experiment:

· You could see fizzing and bubbles coming off the anode. In this
case the hydroxide ion is discharged by loosing its electron to form
water and oxygen, 4OH- - 4e- = 2O + 2H2O.

· You could see the colour of the electrolyte turning pale as the
copper is the substance which gives the solution the blue colour.

· A type of sludge formed outside the cathode

Analysis:

From this experiment I have found out that the current does affect the
rate of electrolysis. As you can see from my graph and results the
current 1 amp had the most amount of copper deposited on the
electrode. This is because electrolysis involves ions taking electrons
from the cathode and depositing electrons at the cathode. If you
increase the current you increases the flow of electricity, thus you
increase the overall charge of the electrodes so ions can pick up more
electrons each second. The current 0.2 amp had the least amount of
copper deposited on the electrode. My graph is a strong positive
correlation and proves that as the current increases the amount of
copper deposited on the electrode increases. This proves that my
prediction was right as I predicted that the copper in the solution
would form outside the cathode. This is because the electrolyte is a
solution of copper sulphate and water ( CuSO4. H2O).

It consists of two positive ions, Copper Cu 2+ and the hydrogen ions
in the water H+. The copper ions are unstable, to become stable they
must lose 2 electrons therefore they would migrate towards the
negative electrode the Cathode in order to lose two electrons.

Cu2+ + 2e- = Cu

I also predicted that the electrolyte would turn from blue to
colourless. As you can see from the observation it has.

I predicted that the hydroxide would form on the anode. This is
because the negative ions in the solution are the sulphate ion SO42-,
and the hydroxide ion OH-. These will migrate to the positive anode.
The hydroxide would lose it’s oxygen. The evidence to this is that
during the experiment I could see fizzing and bubbles coming off the
anode. In this case the hydroxide ion is discharged by loosing its
electron to form water and oxygen, 4OH- - 4e- = 2O + 2H2O.

However my results do not match the results that I had predicted.

Predicted

Actual

0.02

0.03

0.04

0.08

0.06

0.09

0.08

0.1

0.1

0.2

However they do show the pattern that I predicted. I had also
predicted that the results would turn out to be directly proportional
however they aren’t.

Conclusion:

My results are consistent but not very accurate. The percentage
accuracy for 0.2 amps is 66%, 0.4 amps 53%, 0.6amps 63%, 0.8 amps 60%
and 1 amps 52%.They outline my theory that “the higher the current the
more copper is deposited on the cathode”. I predicted that the
results would turn out to be directly proportional however they
aren’t. However I do not have any anomile results. I think that the
reason for this error is that the current kept on fluctuating. Even
when you are using the variable resistor the current would not stay
the same you had to keep on adjusting it during the experiment making
the test unfair.

Another error was that the two electrodes may have touched. This was
because there was no effective way of keeping them is place. If I was
to redo this experiment I would use a clamp stand to hold the
electrodes in place.

How to Cite this Page

MLA Citation:
"Electrolysis Experiment." 123HelpMe.com. 30 Oct 2014
    <http://www.123HelpMe.com/view.asp?id=150108>.




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