Question

Post-Laboratory Report (Page 1/1) Group Number Date: Day Time 1. The salt bridge in Goal #1 was soaked with a saturated ammonium nitrate solution. In which direction do the ammonium ions move when this salt bridge completed the electric circuit? Toward the zinc metal or toward the copper metal? In which direction do the nitrate ions move? 2. In a concentration cell, what would be the potential if we studied two solutions that had the same concentration? Show that the answer derived from common sense is the same as the one derived mathematically from your plot of Ecell vs log(Cu2+ Conc. Ration 3. The Nernst equation is used to predict the impact of concentration on the cell potential. What value of slope does the Nemst equation predict for a plot of Ecell vs log(Cu2. Conc. Ratio)? How does this value (magnitude only) compare with the value observed for your Ecell vs log(Cu Conc. Ratio) plot? Show all of your work. Table 1. Half-Cell Connected to “+" Terminal te 6 Table 2. Cell Potentials Measured Relative to the Tin Halt-Cell (Take Data From Table 1 Measured Cu vs. Sn Zn vs. Sn Al vs. Sn - nom Mg vs. Sn 43m Table 3. Data A Comparison of Measured Ecell VS. The Values Predicted With The Tin Half Cell Ecel Measured Ecen Predicted (Use data from Table 2) (taken from Table 1) Cu vs. Zn Cu vs. A 130 V-132 m 1541m 11m Zn vs. Mg Al vs. Mg 12 Goal #1. Zinc-Copper Galvanic Cell 1. Magnitude of voltage: 2. Species oxidized: 2n 3. Anode: Zn 4. Direction of flow of electrons (check correct answer): Cu--Zn Zn-Cu 91n m V Species reduced. Co. Cathode:C Part Il. Cu Concentration Cell Cu/'T of the stock solution 0.SO M Potential difference for various concentration cells 0.0005 M 0.005 M 0.5 M 0.05 M 0.5 M 58.sv 50.5 v 24.S V 0.05M 0.005 M 0.0005 M Concentration tronog (Conc Rato) Measured Ratio)Potential Difference Ratio 0.5 M/0.05 M 0.45M 0: I 0.5M,0.005 M İ o.49SM|100.1 0 5 M /0909 M 0.45M1000 ら 3 120.aV Attach 3 graphs: (1.) Ecell vs. (3) Ecell vs. log(Conc. Ratio) Difference: (2) Ecell vs. Concentration Ratio; Plot That Provides The Clearest Linear Presentation of the Data - Slope of Plot 14

Answer 1

Table 1.

Cathode Cu, Anode Al, Ecell °= 0.337 - (-1.662) = 1.992V

2Al(s) + 3Cu2+(aq) ------> 3Cu(s) + 2Al3+(aq)

Cathode Zn, Anode Al, Ecell° = -0.7618 -(-1.662) = 0.9002V

2Al(s) + 3Zn2+(aq) -----> 3Zn(s) + 2Al3+(aq)

Sn anode, Cu cathode, Ecell°= 0.337- (-0.13) = 0.467V

Sn(s) + Cu2+(aq) ------> Sn2+(aq) + Cu(s)

Slope of the graph is -2.303RT/ nF

= - 2.303× 8.314×298/2×96485 = - 0.029