Given the following reaction half reaction: Cd2+ + 2e ï‚® Cd Eï‚° = ï€ 0.403 V Pb2+ + 2eï€ï‚®Pb Eï‚° = ï€ 0.126 V where [Cd2+] = 0.020 M and [Pb2+] = 0.200 M construct a galvanic cell using half cell reactions and find
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i) Which half cell –reaction is the cathode? ii) Write the cell reaction. iii) Calculate Eï‚°cell of this galvanic cell. iv) Calculate Ecell of this galvanic cell v) Calculate equilibrium constant using Eï‚°cell potentials of this galvanic cell based on the construction of ions
The first step is to determine the cell reaction and total cell potential.
In order for the cell to be galvanic, E0cell > 0.
For this reaction to be galvanic, the cadmium reaction must be the oxidation reaction. Cd → Cd2+ + 2 e- E0 = +0.403 V
Pb2+ + 2 e- → Pb E0 = -0.126 V
The total cell reaction is:
Pb2+(aq) + Cd(s) → Cd2+(aq) +
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