a You have a concentration cell with Cu electrodes and [Cu2+] = 1.00 M (right side) and 0.0085 M (left side). Calculate the potential for this cell at 25°C. Ecell = 0.061 Correct Since the concentration cell is given, E°cell = 0. 0.0591 8.5 x 10-3 E = 0- - log( =0.061V 1.00 b You have a concentration cell with Cu electrodes and [Cu2+] Cu2+ ion reacts with NH3 to form Cu(NH3)4* = 1.00 M (right side) and 0.0085 M (left side). The 2+ by the following equation: Cu2+(aq) + 4NH3(aq) → Cu(NH3)42+(aq) K = 1.0x1013 Calculate the new cell potential after enough NH3 is added to the left cell compartment such that at equilibrium [NH3] = 1.3 M. Ecell = V

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a You have a concentration cell with Cu electrodes and [Cu2+] = 1.00 M (right side) and 0.0085 M (left side).
Calculate the potential for this cell at 25°C.
%3D
Ecell
= 0.061
V
Correct
Since the concentration cell is given, E°cell
0.
0.0591
8.5 x 10-3
E = 0- -
- log(
- ) =0.061V
2
1.00
bYou have a concentration cell with Cu electrodes and [Cu2+] = 1.00 M (right side) and 0.0085 M (left side). The
Cu2+ ion reacts with NH3 to form Cu(NH3)4°
2+
by the following equation:
Cu2+(aq) + 4NH3(aq) → Cu(NH3)4²+(aq)
K = 1.0×1013
Calculate the new cell potential after enough NH3 is added to the left cell compartment such that at equilibrium
[NH3] = 1.3 M.
Ecell
V
Transcribed Image Text:a You have a concentration cell with Cu electrodes and [Cu2+] = 1.00 M (right side) and 0.0085 M (left side). Calculate the potential for this cell at 25°C. %3D Ecell = 0.061 V Correct Since the concentration cell is given, E°cell 0. 0.0591 8.5 x 10-3 E = 0- - - log( - ) =0.061V 2 1.00 bYou have a concentration cell with Cu electrodes and [Cu2+] = 1.00 M (right side) and 0.0085 M (left side). The Cu2+ ion reacts with NH3 to form Cu(NH3)4° 2+ by the following equation: Cu2+(aq) + 4NH3(aq) → Cu(NH3)4²+(aq) K = 1.0×1013 Calculate the new cell potential after enough NH3 is added to the left cell compartment such that at equilibrium [NH3] = 1.3 M. Ecell V
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