To test how the amount of electricity effects the liberation of copper ions, during electrolysis.

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Chemistry course-work

Electrochemistry

Aim:

To test how the amount of electricity effects the liberation of copper ions, during electrolysis.

Scientific Ideas

The prediction of this experiment should be easily done.  After gathering all the information about electrochemistry and researching about redox reactions the predictions should be as accurate as possible.

Two laws which need to be taken into account are that discovered by Michael Faraday.  The first laws states that the quantity of material transformed at each electrode is proportional to the amount of electricity passed through. The second  law states that the weight of the elements transformed is proportional to the equivalent  weights of the elements, that is, to the atomic weights of the elements divided by their valences (the effective charge)

i.e. the equivalent weight         = ATOMIC WEIGHT
                                         CHARGE

My prediction will be base on these two laws.

Other background ideas which will help my attempt at a prediction is my knowledge on electrolysis and my being able to implement half - equations.

Electrolysis is the passing of an electric current (flow of e-) through a liquid or molten compound dissociates (splits) in ion (charged atoms). An atom which loses an electron becomes positive (CATION) and an atom which gains an electron becomes negative (ANION) -

Reduction and Oxidation occur together as opposites -: As metal copper will lose electrons at the  at the same time as gaining electrons at the Cathode. E.g.

        Cu  → Cu2 → + 2e- (at the Anode)

        Cu 2 + 2e- → Cu (at the Cathode)

Both of the above are half equations.

Predictions:

The current of 1 AMP (A) = 1 coulomb (c) per second

One mole of electrons charge = 96500 coulombs

(6.02 X 1023 X the charge of an electron)

Hypothesis:

300        X        0.5        = 150         (c)

600        X        0.5        = 300         (c)

900        X        0.5        = 450         (c)

1200        X        0.5        = 600         (c)

1500        X        0.5        = 750         (c)

1800        X        0.5        = 900         (c)

2100        X        0.5        = 1050 (c)

2400        X        0.5        = 1200 (c)

To work out Coulombs

Work out Cu (g) Lost at Anode

1 mole of Cu needs 193000 Coulombs

=        150        X 64        = 0.05g of copper
        193000

=        300        X 64        = 0.10g of copper
        193000

=        450        X 64        = 0.15g of copper
        193000

=        600        X 64        = 0.20g of copper
        193000

=        750        X 64        = 0.25g of copper
        193000

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=        900        X 64        = 0.30g of copper
        193000

=        1050        X 64        = 0.35g of copper

         193000

=        1200        X 64        = 0.40g of copper

         193000

        It is clear from my table, that the longer the electricity is distributed, the more copper is liberated from the  to the Cathode. It is lost from  and gained to the Cathode.

Apparatus

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