One drop has a volume of about 0.05 mL, thus the concentration of Cu2+ in the...
One drop has a volume of about 0.05 mL, thus the concentration of Cu2+ in the dilute half-cell is about (0.05 mL/25 mL)(0.1 M)=2x10-4 M. At 298K use this equation:
E= -(0.0592/n)xlog(An+ Mdil/ An+ Mconc) to calculate it.
Homework Answers
Molarity of Cu+2 in dilute solution = 2 x 10^-4 M
molarity of Cu+2 in concentration solution = 0.1 M
number of electrons trnasfered = 2 = n
E= -(0.0592/n) x log(An+ Mdil/ An+ Mconc)
E = (-0.0592 / 2 ) x log ( 2 x 10^-4 / 0.1)
E = 0.0799 V
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To determine the solubility product of copper(II) carbonate,
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of the lab handout is constructed. The temperature of the Galvanic
cell is measured to be 22.5°C, and the cell potential 282 mV (0.282
V). Using this data and Equation 8 in the lab manual, calculate the
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