Question

Copper from a 200.0 mL solution of copper(II) sulfate is plated on the cathode of an electrolytic cell. Hydronium ions are generated at one of the electrodes. Is this electrode an anode or a cathode? How many moles of hydronium are produced if a current of 0.120 A is applied to the cell for 30.0 hr? If the pH of the solution was initially at 7.0, what will be the pH at the end of the electrolysis? Assume the solution volume stays constant.

Answer #1

**consider the plating**

**Cu+2 + 2e- ---> Cu (s)**

**we know that**

**reduction ---> cathode**

**oxidation ---> anode**

**so**

**this electrode is a cathode**

**2)**

**we know that**

**moles of metal plated = I x t / F x z**

**so**

**moles of Cu plated = 0.12 x 30 x 60 x 60 / 96485 x
2**

**moles of Cu plated = 0.06716**

**now**

**H2 ---> 2H+ + 2e-**

**Cu+2 + 2e- --> Cu**

**so**

**H2 + Cu+2 ---> 2H+ + Cu**

**we can see that**

**moles of H+ produced = 2 x moles of Cu
produced**

**so**

**mooles of H+ produced = 2 x 0.06716**

**moles of H+ produced = 0.13432**

**now**

**concentration = moles x 1000 / volume (ml)**

**so**

**conc of [H+] added = 0.13432 x 1000 / 200 =
0.6716**

**inital [H+] = 10-7**

**so**

**final [H+] = 0.6716**

**now**

**pH = -log [0.6716]**

**pH = -0.1728**

**so**

**the pH at the end is -0.1728**

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