Question

Voltaic Cells Electrode Potentials For all of the half-cells measured as well as the Ag', Ag reference cell, enter the reduction potentials in order of decreasing value in the first column. There should be seven half-cells. Smce standard reduction potentials are normally measured versus the standard hydrogen electrode (SHE), you will convert these by knowing that E = 0.80 V versus the H, H, electrode. Simply add 0.80 V to the values in the first column to obtain the values in the third column. (E = 0.00 V) 1.509 0.0919 0.4121 0.3714 0.1742 0.649 Reduction Half-Reaction Zn²+2ezn Te³++ le Te Cu2+ +2e-cu Ag++le Ag I2+2e21 Ag + le Ag (E = 0.00 V) PART B. Effect of Concentration on Cell Potentials 1. Complex ion formation: Potential, E" before adding 6 M NH3 Potential, E after Cu(NH3)42+ formed 426.3 1000 = 0.4263 0.4263 v 785.4 = 0.7854 0.7854 v 1000 Use Equation (10) to calculate the residual concentration of free Cu2+ ion in equilibrium with Cu(NH3)42+ in the solution. Take [Ag+] = 1 M. Show your calculations. 0.7854= 0.7854 0.4263 0.3967 95.50 M - 0.0592 \cdot log \frac{[Cu^{2+}]}{1} = 0.3967 \cdot 2 \cdot log[Cu^{2+}] 0.3967 log [Cu2+] = 1.980 [Cu2+] = 95.50M 101.980

          Voltaic Cells
Electrode Potentials
For all of the half-cells measured as well as the Ag', Ag reference cell, enter the reduction potentials
in order of decreasing value in the first column. There should be seven half-cells. Smce standard
reduction potentials are normally measured versus the standard hydrogen electrode (SHE), you
will convert these by knowing that E = 0.80 V versus the H, H, electrode. Simply add 0.80
V to the values in the first column to obtain the values in the third column.
(E = 0.00 V)
1.509
0.0919
0.4121
0.3714
0.1742
0.649
Reduction Half-Reaction
Zn²+2ezn
Te³++ le Te
Cu2+ +2e-cu
Ag++le Ag
I2+2e21
Ag + le Ag
(E = 0.00 V)
PART B. Effect of Concentration on Cell Potentials
1. Complex ion formation:
Potential, E" before adding 6 M NH3
Potential, E after Cu(NH3)42+ formed
426.3
1000
= 0.4263
0.4263 v
785.4
= 0.7854
0.7854 v
1000
Use Equation (10) to calculate the residual concentration of free Cu2+ ion in equilibrium with
Cu(NH3)42+ in the solution. Take [Ag+] = 1 M. Show your calculations.
0.7854=
0.7854
0.4263
0.3967
95.50 M
- 0.0592 \cdot log \frac{[Cu^{2+}]}{1}
= 0.3967 \cdot 2 \cdot log[Cu^{2+}]
0.3967
log [Cu2+] = 1.980
[Cu2+] = 95.50M
101.980

Voltaic Cells
Electrode Potentials
For all of the half-cells measured as well as the Ag', Ag reference cell, enter the reduction potentials
in order of decreasing value in the first column. There should be seven half-cells. Smce standard
reduction potentials are normally measured versus the standard hydrogen electrode (SHE), you
will convert these by knowing that E = 0.80 V versus the H, H, electrode. Simply add 0.80
V to the values in the first column to obtain the values in the third column.
(E = 0.00 V)
1.509
0.0919
0.4121
0.3714
0.1742
0.649
Reduction Half-Reaction
Zn²+2ezn
Te³++ le Te
Cu2+ +2e-cu
Ag++le Ag
I2+2e21
Ag + le Ag
(E = 0.00 V)
PART B. Effect of Concentration on Cell Potentials
1. Complex ion formation:
Potential, E" before adding 6 M NH3
Potential, E after Cu(NH3)42+ formed
426.3
1000
= 0.4263
0.4263 v
785.4
= 0.7854
0.7854 v
1000
Use Equation (10) to calculate the residual concentration of free Cu2+ ion in equilibrium with
Cu(NH3)42+ in the solution. Take [Ag+] = 1 M. Show your calculations.
0.7854=
0.7854
0.4263
0.3967
95.50 M
- 0.0592 ·log ([Cu^2+])/(1)
= 0.3967 ·2 ·log[Cu^2+]
0.3967
log [Cu2+] = 1.980
[Cu2+] = 95.50M
101.980

Added by Selena C.

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