Post Lab Part 3 Why does an anode lose mass over the course of an experiment? .
ID: 867854 • Letter: P
Question
Post Lab
Part 3 Why does an anode lose mass over the course of an experiment?
. How long would it take to completely use up a lead anode with a mass of 8.000 g in a solution of Pb(NO3)2 at a current of 155 mA?
A 337 mA current is passed through a solution of cobalt (III) nitrate for 22.0 minutes in an electrolytic cell. How much cobalt will be deposited on the cathode?
Would it make a difference if Cu(NO3)2, instead of CuSO4, were used in the copper half cells in this experiment? Why or why not?
Look in your textbook to find the charge on one electron. What is the charge on a mole of electrons?
Explanation / Answer
1) At anode , oxidation occurs, i.e loss of electrons.Thus, the metal electrode looses electrons and forms anions which gets dissolved in the electrolytic solution.Hence in due coarse of time, the mass of the anode electrode decreases
Equivalent mass of Pb2+ = molar mass of Pb/charge on it = 207.2/2 = 103.6 g/equivalent
Thus, number of equivalents of Pb in 8 g of it = 8/103.6 = 0.07722
Now, As per Faraday's Law, charge required to deposit 1 equivalent of Pb = 96500 C.
Thus, charge required to deposit 0.07722 equivalents of Pb = 0.07722*96500 = 7451.74 C
Now, time required in seconds = charge/current = 7451.74/0.155 = 48075.73 sec
time required in hours = time in seconds/3600 = 13.354 hrs.
2) Equivalent mass of Co in Co(NO3)2 = Molar mass of Co/charge on it = 58.93/2 = 29.465 g
Now, total charge passed = urrent*time in seconds = 0.337*22*60 = 444.84 C
Now, As per Faraday's Law, charge required to deposit 1 equivalent of Co = 96500 C.
Thus, total Co deposited = (444.84)*96500)*equivalent mass of Co = 0.136 g
3) Yes , it would have made the differenece because the equivalent mas of Cu = 31.75 g/equivalent
4) Charge on an electron = 1.6*10-19 C
1 mole of electrons = 6.022*1023 electrons
Thus, charge on one mole of electrons = (6.022*1023)*(1.6*10-19) = 9.6352*104 C