Anode And Cathode Reaction Lab Report
b) Write the anode and cathode half-reactions, and the net cell reaction for the above diagram. [ T/I / 2 ]
Oxidation: Zn (s) Zn2+ (aq) + 2e-
Reduction: Cu2+ (aq) +2e- Cu(s) Zn (s) + Cu2+ (aq) +2e- Zn2+ (aq) + 2e- + Cu(s) Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)
c) Write the cell notation for the above galvanic cell drawn. [ T/I / 1 ]
Cu(s) | Cu2+(aq) || Zn 2+(aq) | Zn (s)
d) Calculate the percent error of the cell potential for the above galvanic cell. Show all work. [ T/I / 1 ]
(Theoretical Value-Experimental Value )/(Theoretical Value) x100%
=(1.10-0.91)/1.10 x100%
=17.3%
Therefore, the percent error of the cell potential for the above galvanic cell is approximately 17.3%.
Compare your measured …show more content…
Table 2: Sample data collected by your teacher.
Cathode Anode Measured Potential Difference (V)
Pb(NO3)2 Cu(NO3)2 - 0.50 V
Zn(NO3)2 Pb(NO3)2 - 0.34 V
Using the knowledge that you have obtained from the experiment, speculate as to why the values obtained were negative. [ C / 1 ]
The values obtained were negative due to the teacher connecting the alligator clips to the incorrect metal. The issue can be solved by switching the clips around or by the teacher referring to the table to know which metal is the cathode and which metal is the anode in the cell.
You were asked to make a battery for a new prototype car that was being designed by Ford. Based on your experimental data, which galvanic cell would you pick and why? [ C / 2 ]
Based on my experimental data, I would pick the Cu(NO3)2 and Zn(NO3)2 galvanic cell because it has not only the highest theoretical cell potential but the highest measured cell potential which would have the greatest output and be most beneficial to the car and to the company as it will likely guarantee the highest customer satisfaction possible and result in better
Oxidation: Zn (s) Zn2+ (aq) + 2e-
Reduction: Cu2+ (aq) +2e- Cu(s) Zn (s) + Cu2+ (aq) +2e- Zn2+ (aq) + 2e- + Cu(s) Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)
c) Write the cell notation for the above galvanic cell drawn. [ T/I / 1 ]
Cu(s) | Cu2+(aq) || Zn 2+(aq) | Zn (s)
d) Calculate the percent error of the cell potential for the above galvanic cell. Show all work. [ T/I / 1 ]
(Theoretical Value-Experimental Value )/(Theoretical Value) x100%
=(1.10-0.91)/1.10 x100%
=17.3%
Therefore, the percent error of the cell potential for the above galvanic cell is approximately 17.3%.
Compare your measured …show more content…
Table 2: Sample data collected by your teacher.
Cathode Anode Measured Potential Difference (V)
Pb(NO3)2 Cu(NO3)2 - 0.50 V
Zn(NO3)2 Pb(NO3)2 - 0.34 V
Using the knowledge that you have obtained from the experiment, speculate as to why the values obtained were negative. [ C / 1 ]
The values obtained were negative due to the teacher connecting the alligator clips to the incorrect metal. The issue can be solved by switching the clips around or by the teacher referring to the table to know which metal is the cathode and which metal is the anode in the cell.
You were asked to make a battery for a new prototype car that was being designed by Ford. Based on your experimental data, which galvanic cell would you pick and why? [ C / 2 ]
Based on my experimental data, I would pick the Cu(NO3)2 and Zn(NO3)2 galvanic cell because it has not only the highest theoretical cell potential but the highest measured cell potential which would have the greatest output and be most beneficial to the car and to the company as it will likely guarantee the highest customer satisfaction possible and result in better