a 1.0M CuSO4 solution was electrolyzed for 28 minutes and 22 seconds using coppe

Question

a 1.0M CuSO4 solution was electrolyzed for 28 minutes and 22 seconds using copper electrodes. The average current flowing through the cell over the time period was 0.622 A. The mass of the copper anode before electrolysis was 2.4852g and 2.1335g afterward. a) Calculate the number of moles of copper oxidized and moles of electrons that passed through the cell. b) Calculate the total charge (coulombs) that passed through the cell. c) How many electrons passed through the cell during the 28.0 minute and 22 second time period. d) From the data ,calculate Avogadro's number.

Explanation / Answer

The amount of Copper "dissolved" during the electrolysis was: m(Cu) = 2.4852 - 2.1335 = 0.3517 g This corresponds to n(Cu) = 0.3517 / 63.5 = 5.54 x 10^-03 moles of Cu electrolyzed, since the atomic weight of Cu is 63.5 g/mol [ Answer (a.1) ] For each mole of Cu oxidized to Cu2+ 2 moles of electrons were needed. Therefore, the number of moles of electrons that passed through the cell was: n(e-) = 2 x 5.54 x 10^-03 = 1.1 x 10^-02 moles of electrons [ Answer (a.2) ] The amount of charge passed through the cell was: Q = I(A) x t(secs) = 0.622 x 1,702 =1,058.6 Coulombs [ Answer (b) ] If 1.1 x 10^-02 moles of electrons passed through the cell during the electrolysis time, the number of electrons may be calculated by direct proportionality: 1 mol of electrons ====== > 6.02 x 10^23 electrons 1.1 x 10^-02 moles ====== > X electrons and by cross-multiplication: X = (1.1 x 10^-02) x (6.02 x 10^23) = 6.62 x 10^21 electrons [ Answer ( c ) ] For the Avogadro number, it was determined that: 1,058.6 Coulombs ===== > 1.1 x 10^-02 moles of electrons 96,500 Coulombs ====== > Y moles of electrons (should correspond to 1 mol = 6,02 x 10^23) Y = 1.003 moles of electrons ~ 1.003 x 6.02 x 10^23 ~ 6.04 x 10^23 electrons or N(Avogadro) ~ 6.04 x 10^23

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