these data the experimental value of potential of the different half cells

# These data the experimental value of potential of the

• 9

This preview shows page 5 - 7 out of 9 pages.

these data, the experimental value of potential of the different half cells connected to the copper half-cell can be calculated using equation [5]. If the copper served as the anode in the voltaic cell, the formula for the potential of the variable half- cell is calculated from: E cathode = E cell + 0.34 [10] Otherwise, the potential of the half cell will be E anode = 0.34 E cell [11] For set-ups 1, 2, 4, 5 and 6, the reference electrode, Cu, served as the anode. Thus, given the E red of Cu as 0.34, the standard potential of the cathode was calculated from equation [10] . The first set-up should have yielded a voltage reading of 0.00 V because both half-cells supposedly have the same content and concentration. Because of this, the E anode and E cathode should have the same value and the E cell should be equal to 0.00 V. However, a difference in concentration can also add to the potential difference, so even a slight error in the volume measurements can add to the error of the results. Set-ups 4, 5 and 6 were affected by the concentration. However, in the experiment, it was assumed that the change in concentration due to the hydrolysis was negligible and was not taken into account in the calculations. Because in set-up 3, copper served as the cathode, the value calculated for was the standard potential of the anode. This was computed from equation [11]. Using these equations, the resulting experimental value for the standard reduction potentials of the variable half-cells are as follows: Table 2. Calculated standard potential Set- up # Variable Half-Cell Notation Calcula ted Volts Boo k Valu e % Error 1 Cu|Cu 2+ 0.2504 0.00 26 % 2 Zn|Zn 2+ - 0.74 - 0.76 3 3.01 % 3 Fe 3+ |Fe 0.754 0.77 1 2.2 % 4 C|Cl - ,Cl 2 0.808 1.35 9 40.54 % 5 C|Br - ,Br 2 0.609 1.08 7 43.97 % 6 C|I - ,I 2 0.461 0.61 5 25.04 % These values were compared to the book value and yielded percentages of error within the range of 2.2%-43.97%. The percent error were highest for the last three set-ups for these were calculated without taking into account the changes in concentration due to the electrolysis done prior to setting up the voltaic cell. 5

Subscribe to view the full document.

If the change in concentration was taken into account, the current (in Amperes) and time length of the electrolysis should be measured. For example, in the experiment, Cl - is converted into Cl 2 . which follows the reaction:ly ¿ + Cl 2 ( s ) ¿ 2 e ¿ 2 Cl ( aq ) ¿ [12] The Cl- solution was electrolyzed for 4 minutes and 30 seconds and the measured current was 0.065 Amperes. Substituting these values to equation 8, the calculated amount of electricity is 17.55 Coulombs. This is then converted to mole e - by dividing 17.55 by 96,500. Using stoichiometry, the number of moles of Cl - is calculated, hence, also the concentration of Cl- in the solution. This new concentration can be substituted to equation [7] and since ¿ Cl ¿ ¿ ¿ ¿ Q = [ Cl 2 ] ¿ and the activity of Cl 2 is taken as 1, Q is only equal to ¿ Cl ¿ ¿ ¿ ¿ 1 ¿ . The measured voltage of the voltaic cell, which is 0.468 V, was then used to calculate the potential of the Chloride half-cell: E cathode = 0.468 + 0.34 = 0.808 V The n, calculated Q, and potential of the cathode was substituted in the Nernst equation to get the standard reduction potential of the cathode: E = 0.0592 2 log 1 [ 0.00364 ] 2 0.808 = 0.1444

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask 0 bonus questions You can ask 0 questions (0 expire soon) You can ask 0 questions (will expire )
Answers in as fast as 15 minutes