e= 1.6 x 10^-19 e=MIt2 AMNA M= Molecular mass (Cu) (g) I = Current (A) Am= Change in mass (g) = M after-M before NA= Avo
Posted: Wed May 18, 2022 6:50 am
- E 1 6 X 10 19 E Mit2 Amna M Molecular Mass Cu G I Current A Am Change In Mass G M After M Before Na Avo 1 (61.87 KiB) Viewed 55 times
- E 1 6 X 10 19 E Mit2 Amna M Molecular Mass Cu G I Current A Am Change In Mass G M After M Before Na Avo 2 (58.57 KiB) Viewed 55 times
- E 1 6 X 10 19 E Mit2 Amna M Molecular Mass Cu G I Current A Am Change In Mass G M After M Before Na Avo 3 (35.66 KiB) Viewed 55 times
Calculations: Discussion of results and error analysis: 0/0 error = | Exp-Accpt|/Accpt x 100
deposit of one atom of copper requires the transfer of two electrons from the external circuit to neutralize the positive charge of the deposited ion, we can calculate the total charge transferred. This charge is a function of the electronic, charge (e) In order to derive a formula for the electronic charge () in terms of the quantities of the experiment we need to state Faradays Laws of electrolysis: Faradays Law 1. The mass of an element deposited at an electrode is directly proportional to the quantity of clectrical charge that passes through the cell. 2. The mass of the element deposited by a given quantity of charge is directly proportional to the chemical equivalent (atomic weight divided by valence) of the element. Q=the total charge through the cell mmass of each cu atom and 2e-charge carried by each Cu atom Am Then 9 20 (3) But the mass m of an atom equals its atomic weight (M) divided by Avogadro number (N1). Therefore by substituting m=M/N, equation (1) becomes Am MIN @ 2e (5) MO 2A But Q = It where I = current in amperes and t= time in seconds. Therefore, MI (6) 2A Equipment Ammeter, rheostat, switch. electrolytic cell power supply, clock, boalier (500 to 1000 ml.) and balance Experimental Procedure 1. Make the circuit shown in figure 1 and ask your instructor to check the circuit