In this set up a salt bridge provides ions to the two solutions to neutralize

In this set up a salt bridge provides ions to the two

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at both ends of the salt bridge. In this set-up, a salt bridge provides ions to the two solutions to neutralize its charges by adding negative ions to the positive half-cell (Zinc) and positive ions to the negative half-cell (Copper). This prevents the reactants to react directly with one another but still maintains electrical contact between the two half cells. The electrons produced from the Zinc rod passes through the wire, enter the copper rod and interact with the copper ions in solution. As the electrons travel through the wire, energy is produced and the bulb is lighted. In the experiment, a load like the light bulb was not connected to the wire, but a volt-meter. A volt- meter measures the voltage/ electric potential difference or the measure of electromotive force, the amount that can be extracted from the system. While doing this, the volt-meter does not consume the energy produced as it has an infinite resistance. The voltage also serves as a parameter for the tendency of the reaction to proceed to an equilibrium state. As the reaction proceeds, the potential continues to decrease until it reaches 0.000 V. However, absolute electrode potentials cannot be measured. This is because all volt-meters measure only differences in potential between two half cells. Thus, the values of electrode potential used in calculations are only those of the half cells when these are reacted with a standard hydrogen electrode which has an assigned value of potential of 0.00 V. In this reaction, the standard hydrogen electrode serves as the anode while the electrode to be tested is the cathode. From this reaction, the standard reduction potential ( ) is 2 3
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measured. This value can then be used to compute for the standard potential of the whole electrochemical cell using the equation: E cell = E cathode E anode [5] However, this standard potential only applies to solutions with the concentration of 1 M. In the experiment, it was assumed that the concentration of the halide ion did not significantly change with the electrolysis of the solution. But if the concentration was not 1 M, and other conditions such as the temperature and nature of the reactant were altered, the cell potential will be different from the standard potential. The effects of these factors are shown in the Nernst equation: E = RT nF lnQ [5] Where R is the gas constant with a value of 8.314 J/molK, T is the temperature in Kelvin, n= number of moles of electrons that appear in the half cell reaction, and F is Faraday which is equal to 96, 485 Coulombs. The Q in the equation is the reaction quotient, given by the formula Q = [ products ] m [ r eactants ] n [6] where m e the coefficients of the products and n are the coefficients of the reactants in the balanced equation. The values enclosed in brackets are the concentrations of the reactants and products. Simplifying the Nernst equation by substituting the constant values and indicating that the temperature is at 25⁰C, the cell potential can be calculated from the formula: E = O .O 592 n logQ [7] According to Faraday’s law, the mass of the product formed or the reactant used is proportional to the
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