Question

In: Chemistry

Please fill in the rest of the table below. For theoretical potential, % error, cell reactions...

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 Total
Potential 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

Solutions

Expert Solution

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


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