Question

If a current of 840.6 mA is passed through an electrochemical cell for 20.46 minutes, what charge, in coulombs, is passed through the cell?

Charge passed through the cell = coulombs

Suppose that the electrochemical cell consists of two copper electrodes immersed in a copper(II) sulfate solution. At which electrode, anode or cathode, does oxidationoccur?

The oxidation occurs at the

Do you expect that this electrode will increase or decrease in mass as the electrolysis proceeds?

The mass of the electrode will

Calculate the change in the mass of copper at the anode or cathode if the charge calculated above is passed through the cell described. Enter your answer as a positive quantity in mg.

Mass change at the electrode = mg

Answer 1

1 coulomb = 1A * 1sec

charge passed though the cell in in coulomb = (840.6 * 10^-3 A)* 20.46 *60 sec

                                                                  = 1031.92 coulomb

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oxidation occurs at anode

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mass will decrease as more and more Cu will convert to Cu²⁺ and will go into the solution

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According to Faraday's laws

where:

m = mass of the substance liberated at an electrode in grams

Q = total electric charge passed through the substance = 1031.92 C

F = 96485 C mol^?1 is the Faraday constant

M = molar mass of the substance = 63.5

z =valency number of ions of the substance (electrons transferred per ion) =2 (Cu--------> Cu²⁺)

subtituting the values in the above relationship

m = (1031.92 coulomb/96500 Cmole^-1 ) * (63.55 gmole^-1/ 2)

m = 0.3395g

   = 339.5 mg

Mass change at the electrode = 339.5 mg