E = E˚ - log Q [Equation 1]
In the above equation, E˚ is equal to 0 because each cell is composed of the same element. The value of n, which is moles of electrons, is equal to 2. The values of F and R, which are constants, are 96485 C and 8.314 J/mol-K respectively. The value of Q is equal to the concentration of the diluted solution over the concentration of the more concentrated solution. Using this information, Equation 1 can be rearranged as shown below.
E = - log(0.9143)
E = 0.0389 J
The theoretical value of E of the copper concentrated cell was 0.0389 J, which is very close to our experimental value of 0.039 J. The difference between these two is either zero or immeasurable due to a lack of enough sig figs in our experimental value. Therefore, our theoretical and experimental values are incredibly precise in regard to one another.
U could be 0++ slightly due monitor, or concentrations that were not liom, cell putu 2....
use the nernst equation to calculate the theoretical value of E of the copper concentration cell and compare this value with the cell potential that you measured DATA TABLE Results of Parts I and II Cu/Pb X/Pb Y/Pb 0.431 0.172 0.581 Average cell potential (V) Results of Part III Cu concentration 0.042 Average cell potential (V)
Use the Nernst equation to calculate the theoretical value of E of th copper-concentration cell and compare this value with th cell potential you measured. E = E* - 0.0592 / n * logQ **So I believe this is the equation that I would use. However, i'm don't know what E* is suppose to be...** The my electrochemistry experiment the cell potential that i measured were: 0.130V, 0.115V, and 0.110V (average cell potential = 0.118V) The concentration of the copper...
I just want to double check my work for Q1, 2, 3 in the blank spaces and the Nernst equations. Concentration Cells 1) Concentration cell #1 : Which solution is being reduced, the IM Cu2 or the 0.001M Cu? VE 90 What was the measured cell potential for the 1M Cu/0.001 M Cu concentration cell? 14,S mV Measured Ecel= Use the Nernst equation to calculate the expected cell potential for the 1M Cu/0.001M Cu concentration cell; refer to the experimental...
First fill in your half cell and cell reactions. F in standard cell potentials as you calculate them. Oxidation at the Anode: The black (.) lead is attached to the electrode, which is the source of electrons. Write the anode half reaction: Reduction at the Cathode: The red lead (+) is attached to the electrode. Write the cathode half reaction: E degree - V. Overall Cell Reaction (Net Ionic equation): Write the overall cell reaction (balance electrons and add together):...
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If you could, a picture of the correct graph would be nice. Also the calculated slope is 0.0256, according to the book. 13.115 A silver concentration cell is constructed with the electrolyte at both electrodes being initially 0.10 M AgNO(aq) at 25°C. The electrolyte at one electrode is diluted by a factor of 10 five times and the cell potential measured each time. (a) Plot the potential of this cell on a graph as a function of InſAg 'Janode (b)...
Use the Nernst equation to calculate the expected voltage of the concentration cell before the addition of the 10 extra drops of 0.10 M copper(II) nitrate solution. Use your experimental temperature. How did the addition of 10 drops of 0.10 M cupric nitrate affect the concentration cells voltage? Why? Temp. 21C E=91V 10 drops 75.8V
pleas help me with lab assament Multimeter Consider a galvanic cell consisting of the following two redox couples: Ag+(0.010 M) + e-→ Ag(s) Eo = +0.80 V oi a. Write the equation for the half-reaction occurring at the Salt bridge C (0.010 M) Ag (0.010AM b. Write the equation for the half-reaction occurring at the anode. Cr Ag c. Write the equation for the cell reaction. d. What is the standard cell potential, Eelli for the cell? e. Realizing the...
Standard reduction potentials are listed for reactions under standard conditions. Standard conditions are 1 M concentrations of ions, 1 atm (or 1 bar) partial pressures for gases, and a temperature of 298 K a. In Part I, you were asked to compare your measured cell potential to a calculated standard cell 3. potential. The cell potential you measured was for a galvanic cell you prepared using 0.10 M solutions of Pb and Cu, not 1.0 M solutions. Write out the...
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