1444 E 09524 V By using this method of calculation the percent error will

1444 e 09524 v by using this method of calculation

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0.1444 E ° = 0.9524 V By using this method of calculation, the percent error will decrease from 40.54% to 29.92%. These calculated reduction potentials can serve as a parameter for the spontaneity of a reaction. If the calculated values for the potentials of the half-cells are more positive than 0.34, the reactants of that half-cell are stronger oxidizing agents. If the potential difference is positive, the reaction follows a forward direction. Otherwise, the backward reaction is favored and work is needed to stimulate the forward reaction. An electrolytic cell set-up is used in this experiment to initiate a reaction like this, namely the halide half-cells. Consider the cathode iron half- cell in set-up 3 where the following reduction reaction occurred: 2 + ¿ ¿ →Fe ¿ 3 + ¿ + e ¿ Fe ¿ [13] When the chemically inert electrode, graphite, was inserted to the solution, a continuous exchange in 6
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electrons occurs between the electrode and the oxidizing and reducing substances that came in contact with it. At equilibrium, when Q = K eq , reaction [13] and its reverse reaction occur at the same rate, making the composition of the solution near the electrode, constant. The equilibrium potential, or the potential of the electrode with respect to the solution, can be calculated from the Nernst equation: 2 + ¿ Fe ¿ ¿ 3 + ¿ Fe ¿ ¿ ¿ E eq = O.O 592 n log ¿ [14] If the given potential of the electrode, E’, exceeds the E eq , the wire is not in equilibrium, thus electrons are still exchanged between the components of the solution until the potentials of the solution and the electrode are equal. The value of the ratio 2 + ¿ Fe ¿ ¿ 3 + ¿ Fe ¿ ¿ ¿ ¿ increases to lessen the value of E’ and the ferric iron tends to be reduced (reaction [13]). On the other hand, if E’ is less than E eq , the ratio decreases and reverse reaction predominates. Consider the electrolysis of a Chloride solution in part B: ¿ ¿ →Cl 2 + 2 e ¿ 2 Cl ¿ [15] In the electrolytic cell, voltage is applied to the two chemically inert graphite electrodes, one of which serves as the cathode, the other, the anode. The voltage applied can be calculated from the formula: V applied = E anode E cathode [16] If a potential applied to the cathode, E cathode , is less than E anode , the system is not in equilibrium. This causes the E cathode to increase and the E anode to decrease until E cell = 0, producing work instead of consuming it from the energy input (dry cell). So if the passage of current or electrolysis is aimed to occur, the E anode should be greater than E cathode . The potential at the anode of the cell and the reduction potential of the cathode can also be calculated from the Nernst equation: ¿ Cl ¿ ¿ ¿ ¿ E cat hode = O .O 592 n log [ Cl 2 ] ¿ [17] ¿ Cl ¿ ¿ ¿ 2 ¿ ¿ E anode = O .O 592 n log ¿ [18] Referring to the aforementioned concept, as the produced Cl 2 7
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increases, the oxidation potential decreases. Conversely, the reduction potential of the cathode increases with the increase of Cl 2 . This causes the potential needed by the system to decrease until it reaches equilibrium and making the reaction more spontaneous. After the electrolysis of
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