Please fill in the rest of the table below. For theoretical potential, % error,
ID: 1073378 • Letter: P
Question
Please fill in the rest of the table below. For theoretical potential, % error, cell reactions for cathode, anode and net, and delta G in kJ.
Table 1
-0.605
Please fill in the rest of the table below. For theoretical potential, % error, cell reactions for cathode, anode and net, and delta G in kJ.
Table 1
Cell Measured TotalPotential from Multimeter (V)1 Individual Half-Cell Potentials Cell Reactions (anode, cathode, and net) Delta G (kJ) Cu Electrode
Standard Potential (V)2 Metal Electrode
Experimental Potential (V)3 Metal Electrode
Theoretical Potential (V)4 Metal Electrode
Potential % Error Cu Sn 0.419 0.34 -0.079 Cu Al 0.632 0.34 -0.292 Cu Fe 0.461 0.34 -0.121 Cu Zn 0.945 0.34
-0.605
Explanation / Answer
1) Cu + Sn system
Metal Electrode Theoretical Potential = Eo cathode - Eo anode
= 0.34 - (-0.14)
= 0.48 V
Metal Electrode Potential % Error = theoretical - experimental x 100 / experimental
= ( 0.48 - 0.419 / 0.419 ) x 100
= 14.6 %
cell reactions :
anode : Sn------------------> Sn+2 + 2e- , Eo = -0.14 V
cathode : Cu+2 + 2e- ---------------> Cu , Eo = +0.34 V
Delta G = - n F Eocell
= - 2 x 96485 x 0.48
= - 92626 J
= -92.6 kJ
1) Cu + Al system
Metal Electrode Theoretical Potential = Eo cathode - Eo anode
= 0.34 - (-1.66)
= 2.00 V
Metal Electrode Potential % Error = theoretical - experimental x 100 / experimental
= ( 2.00 - 0.632 / 0.632) x 100
= 216 %
cell reactions :
anode : Al -----------------> Al+3 + 3e- , Eo = -1.66 V
cathode : Cu+2 + 2e- ---------------> Cu , Eo = +0.34 V
Delta G = - n F Eocell
= - 2 x 96485 x 2
= - 385940 J
= -386 kJ