FARADAY\'S LAWS Name: Ca Partner\'s Name: seetion Day Hours: DATA. Time of elect

Question

FARADAY'S LAWS Name: Ca Partner's Name: seetion Day Hours: DATA. Time of electrolysis: 26: 31.21 duration in seconds 3lee Current in amperes: average lb3A Buret reading in mL: start l D520-end 35.SmL Volume of Ha in mL .2.4S Distance between the liquid levels in the beaker and gas buret (in mm): Mercury equivalent of the difference in liquid levels. 165 (Mercury is 13.6 times as dense as water, thus a column of water 13.6 mm high exerts as much pressure as a column of mercury 1.00 mm high) 1.00 mm Hg Torr) 13.6 mm H20 Atmospheric pressure in Torr: Gas temperature in °C: Vapor pressure of water in Torr Anode mass in g: start difference in mass LE-4 WS2016

Explanation / Answer

You have given most of the calculations correct. I am providing solutions to those questions which are left blank and for which you have given incorrect answers.

Moles of H2 formed :

In this, n = PV/RT , and P should be put in units of atm and not torr, because you are using R as 0.082. Thus P = 0.991 atm , and V = 0.03445 L

Solving we get :

n = 0.00141 moles H2

Since 1 mole H2 is produced when 2 moles of e- are passed, so moles of e- passed = 0.00282 moles e-

1 faraday = 96485 columbs, so columbs per faraday = 259.33 / 96485 = 0.00268

Number of faradys passed in this electrolysis is same as the number of moles of electrons passed.

Equivalent mass of Cu = mass of Cu oxidized / Faradays passed = 0.0869 / 0.00282 = 30.81

relation between atomic mass(AM) and equivalent mass(EM) is :

AM = EM * n , where n is no, of e- lost by metal during oxidation

For Cu, n = 2, AM = 63.54

So EM = 31.77

So, % error in experimental EM value = (31.77 - 30.81) / 31.77 * 100 = 3.02%

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