Thursday, August 22, 2019

Electrochemical Cell Essay Example for Free

Electrochemical Cell Essay Introduction To investigate the effect of change in temperature of the solutions on the voltage of an electrochemical cell. How does increase in temperature of the electrolytes in en electrochemical cell affect the voltage? An electrochemical cell produces electrical energy from chemical energy, where the chemical energy comes from the reactions in the cell. An electrochemical cell consists of two half-cells. Each half-cell consists of an electrode, and an electrolyte (salt solution). A salt bridge is used that connects the two solutions in the containers to allow flow of ions so there is no charge build up in either solution. Charge build-up would shift the equilibrium and the reaction would cease. When the two half cells, consisting of a metal electrode and a conducting solution are connected with an external wire, the strongest oxidizing agent will undergo a reduction in one half cell and the strongest reducing agent will undergo an oxidation in the other half cell. An electrochemical cell makes use of a redox reaction and uses the chemical reaction to produce an electric current. The two metals that will be used in this experiment are Copper and Zinc, and their salts are used in an aqueous form in the container, being CuSO4 and ZnSO4. A salt bridge that is dipped in KNO3 will be used. The following reaction takes place in each half cell: Cu (s) Cu2+ (aq) + 2e- Zn (s) Zn2- (aq) + 2e- Independent Variable: Temperature of the electrolytes (à ¯Ã‚ ¿Ã‚ ½C) Dependant Variable: Voltage (V) Controlled Variables: Nature of electrodes: The electrodes will be of the same elements i.e. Copper and Zinc. Changing the nature of the electrodes will change the voltage since different elements produce different forms of ions which then affects the voltage of the cell. Size of electrodes: The size of electrodes will be kept constant to 6cm x 1cm. Increasing the size of electrodes means that the electrodes will react produce more ions hence increasing the voltage of the cell. Nature of ion transfer: A salt bridge (dipped in KNO3) will be used as a pathway for the ion transfer between the two half cells. Changing the nature of ion transfer will change the amount of ions are transferred between the half cells hence affecting the voltage. The concentration of the electrolytes: is kept constant. A constant amount of metal salt will be added to make each electrolyte. Having a varying concentration will affect the amount of ions produced hence varying the voltage accordingly. Voltmeter: The voltmeter is kept constant since changing the voltmeter can affect the voltage since different voltmeters may have different resistance which may affect the readings. Materials required * Copper strip, 6cm x 1cm * Zinc strip, 6cm x 1cm * CuSO4 solution, 0.99M à ¯Ã‚ ¿Ã‚ ½ 0.01M * ZnSO4 solution, 0.99M à ¯Ã‚ ¿Ã‚ ½ 0.01M * KNO3 solution, 1.99M à ¯Ã‚ ¿Ã‚ ½ 0.01M * Distilled water, 300mL à ¯Ã‚ ¿Ã‚ ½ 1mL * Beakers (3), 250mL à ¯Ã‚ ¿Ã‚ ½ 50mL * Graduated cylinder (100mL à ¯Ã‚ ¿Ã‚ ½ 1mL) * Digital Voltmeter with crocodile wires * Hot plate * Thermometers (2), à ¯Ã‚ ¿Ã‚ ½ 1à ¯Ã‚ ¿Ã‚ ½C * Paper strips, 10cm x 1cm * Magnetic stirrer * Rubber gloves * Apron * Safety Goggles Method Method to prepare solutions 1. Concentration required (CuSO4): 0.99M AMU of CuSO4: 159.61 So, 1.00M of CuSO4 has 159.61g of CuSO4 Hence 0.99M has: 159.61 x 0.99 = 158.01g à ¯Ã‚ ¿Ã‚ ½ 0.01g Mix 158.01g à ¯Ã‚ ¿Ã‚ ½ 0.01g of CuSO4 in 100mL of distilled water. Use a magnetic stirrer if necessary == Use 100mL à ¯Ã‚ ¿Ã‚ ½ 1mL of 0.99M à ¯Ã‚ ¿Ã‚ ½ 0.01M of CuSO4 solution. 2. Concentration required (ZnSO4): 0.99M AMU of ZnSO4: 161.44 So, 1.00M of CuSO4 has 161.44g of ZnSO4 Hence 0.99M has: 161.44 x 0.99 = 159.83g à ¯Ã‚ ¿Ã‚ ½ 0.01g Mix 159.83g à ¯Ã‚ ¿Ã‚ ½ 0.01g of ZnSO4 in 100mL of distilled water. Use a magnetic stirrer if necessary == Use 100mL à ¯Ã‚ ¿Ã‚ ½ 1mL of 0.99M à ¯Ã‚ ¿Ã‚ ½ 0.01M of ZnSO4 solution. 3. Concentration required (KNO3): 1.99M AMU of CuSO4: 101.11 So, 1.0M of CuSO4 has 101.11g of KNO3 Hence 1.99M has: 101.11 x 1.99 = 201.21g à ¯Ã‚ ¿Ã‚ ½ 0.01g Mix 201.21g à ¯Ã‚ ¿Ã‚ ½ 0.01g of KNO3 in 100mL of distilled water. Use a magnetic stirrer if necessary == Use 30mL à ¯Ã‚ ¿Ã‚ ½ 1mL of 1.99M à ¯Ã‚ ¿Ã‚ ½ 0.01M of KNO3 solution. 4. Now put the paper strip (salt bridge) in the KNO3 solution and leave it for 30 seconds. Method for investigation 1. Before starting the experiments, all the materials should be acquired with all safety precautions 2. Put the two beakers that have electrolytes on the hot plate. 3. Now put the salt bridge between the beakers. Also, put the two thermometers in each container 4. Connect the crocodile wires from the voltmeter to each electrode, but dont put it in the half cells yet. 5. Now turn on the hot plate and let the temperature increase from room temperature to 30à ¯Ã‚ ¿Ã‚ ½C 6. As soon as the temperature increases to 30à ¯Ã‚ ¿Ã‚ ½C, quickly insert the electrodes in the respective solutions (Cu in CuSO4 and Zn in ZnSO4) and record the voltage 7. Remove the electrodes and clean and dry them. 8. Now increase the temperature by a difference of 10à ¯Ã‚ ¿Ã‚ ½C i.e. 40à ¯Ã‚ ¿Ã‚ ½C, 50à ¯Ã‚ ¿Ã‚ ½C, 60à ¯Ã‚ ¿Ã‚ ½C, 70à ¯Ã‚ ¿Ã‚ ½C, 80à ¯Ã‚ ¿Ã‚ ½C and 90à ¯Ã‚ ¿Ã‚ ½C and repeat steps 5-7. 9. After the experiment has been completed, put the cleaned materials back to their original place. Raw Data Table Volume of each electrolyte (mL à ¯Ã‚ ¿Ã‚ ½ 1mL) Length of salt bridge (cm à ¯Ã‚ ¿Ã‚ ½ 0.01 cm) Length of each electrode (cm à ¯Ã‚ ¿Ã‚ ½ 0.01 cm)

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