Question

# Copper from a 200.0 mL solution of copper(II) sulfate is plated on the cathode of an...

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.

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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