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Investigation into Electrolysis

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Practical Experiment: Electrolysis Introduction: During electrolysis, electrical energy is converted into chemical energy, forcing chemical reactions to take place. The negative cell emf of an electrolytic cell indicates that the overall reaction does not occur spontaneously but rather needs electricity for the reaction to take place. In an electrolytic cell, two electrodes are immersed in a common electrolyte. At the cathode, one of the electrodes, reduction takes place, but as it is attached to the negative terminal, it has a negative charge. The anode, the positively charged electrode is where oxidation takes place. To predict the products of electrolysis, all possible reactions must be considered. If aqueous electrolytes are present or reactive electrodes used, all possible half reactions for each electrode should be considered, with the half reaction with the most positive Eo value being the most likely to occur. Electrodes such as carbon and platinum are inert and simply provide a surface for electron transfer. The concentration of the electrolyte solution can make the discharge of certain ions, such as chloride, become more favourable. Even though a particular reaction has a higher Eo value, a reaction a second ion with a lower Eo value will become more likely if the concentration of the second ion is higher. Electroplating is the method of applying a metallic coating to another material. This is done by putting a negative charge onto the object to be plated and immersing it into a solution which contains a salt of the metal to be deposited. The metallic ions of the salt carry a positive charge and are attracted to the cathode, which is the object to be plated. ...read more.


+ 2e- Eo = +0.26V 2H2O(l)--> O2 (g) + 4H+(aq) + 4e- Eo= -1.23V Ni2+(aq) + 2e- --> Ni(s) Eo = -0.26V 2H2O(l) + 2e- --> H2 (g) + 2OH-(aq) Eo = -0.83V K+(aq) + 2e- -->K(s) Eo = -2.93V 3a. Zn(s) --> Zn2+(aq) + 2e- Eo= +0.76 2H2O(l)--> O2 (g) + 4H+(aq) + 4e- Eo= -1.23V Zn2+(aq) + 2e- --> Zn(s) Eo= -0.76V 2H2O(l) + 2e- --> H2 (g) + 2OH-(aq) Eo = -0.83V SO42-(aq) + 4H+(aq) + 2e- --> SO2 (aq) + 2H2O (l) Eo = + 0.16 3b. ZnSO4 (aq) + 2KNO3 (aq) --> Zn(NO3)2 (aq) + K2SO4 (aq) Therefore, Ions do not precipitate out, there are simply K+, NO3-, Zn2+ and SO42- ions in solution Zn(s) --> Zn2+(aq) + 2e- Eo= +0.76 2H2O(l)--> O2 (g) + 4H+(aq) + 4e- Eo= -1.23V 4a. If pure aluminium: Al(s) --> Al3+ + 3e- Eo= -1.66V 2H2O(l)--> O2 (g) + 4H+(aq) + 4e- Eo= -1.23V If aluminium coated with aluminium oxide: (Al2O3) 2H2O(l)--> O2 (g) + 4H+(aq) + 4e- Eo= -1.23V Al3+ +3e- --> Al(s) Eo = 1.66V 2H2O(l) + 2e- --> H2 (g) + 2OH-(aq) Eo = -0.83V Discussion: This experiment demonstrated the competing electrode reactions in an electrolytic cell and how the introduction of further ions, in the potassium nitrate, changes these reactions further. In doing so, in every part b, where Potassium Nitrate is added to solution, a series of reactions takes place in conjunction with the reduction and oxidation reactions at the electrodes to produce both a precipitate and the continuation of the normal electroplating reaction. Our experiment was designed to be both qualitative and quantitative, based on both observation and measurements so as to provide more accurate analysis of the resulting reactions. ...read more.


To confirm Faraday's laws qualitatively, one could observe the quality of the electroplating on the cathode, as metals plated at a certain current density might form a durable and shiny coating on the cathode, while some other current density might form an excessively grainy, dull coating. A higher potential difference (voltage) applied to the cell means the cathode will have more energy to bring about reduction, and the anode will have more energy to bring about oxidation. Higher potential difference would enable the electrolytic cell to oxidize and reduce energetically more "difficult" compounds. This may drastically change what products will form in this experiment. These could be qualitatively observed. The distance between the electrodes of the experiment was kept constant, however to investigate the affect of this one could have a constant such as stage 1a and then vary the distance between electrodes and calculate the difference in rate of reaction by measuring the amount of product formed in the same period of time. It could be hypothesized that the greater the distance between electrodes means that the average time for ions to reach them is greater so the electron flow through the wires is less. This experiment demonstrated the competing electrode reactions and how the concentration of ions influences the redox half reactions that take place at the anode or cathode. It also demonstrated that less active metals, according to the reactivity series, have higher rates of reactions in electroplating electrolytic cells. Further investigation of variables influencing electrolytic cells through suggested experiments, greater care taken in measurements and greater amount of reactants used would make the experiment more successful. 1 ...read more.

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