Faraday s Law. Including electrochemical reactions in stoichiometry

Transcription

1 Faraday s Law Including electrochemical reactions in stoichiometry copyright by James B. Condon (2013) all rights reserved Some of the graphics are supplied by Corel Corporation with their permission. Use of these materials by others is encouraged provided these notices are not altered. Donations to the Roane State Foundations would be appreciated in order to support the development of these presentations. 1

2 Faraday s Law Here is a question you should be able to answer: In the following reaction, how many grams of Ni will be plated out from 55.0 g of Nickel sulfamate? Ni(SO 3 NH 2 ) 2 + 2e! Ni + 2SO 3 NH 2 This is just a straight-forward stoichiometry problem. For the molar masses: MNi: = g mol 1 MNi(SO 3 NH 2 ) 2 : = g mol 1 for moles of Ni(SO 3 NH 2 ) 2 : n Ni(SO = 55.0 g / g mol 1 3 NH 2 ) 2 n Ni(SO 3 NH 2 ) 2 = mol Thus: m Ni = mol g mol 1 = 12.9 g Convert to moles of Ni: n Ni /1 = n Ni(SO 3 NH 2 ) 2 /1 n Ni = mol 2

3 Faraday s Law Here is another question you should be able to answer: In the following reaction, how many grams of Ni will be plated out using 1.50 mol of electrons? Ni(SO 3 NH 2 ) 2 + 2e! Ni + 2SO 3 NH 2 Again, this is just a straight-forward stoichiometry problem. Moles of electrons may be treated as moles of anything else. For the molar mass: M Ni : = g mol 1 (Obviously, the molar mass of electrons in not needed.) Convert to moles of Ni: n Ni /1 = n eg /2 n Ni = mol Thus: m Ni = mol g mol 1 m Ni = 44.0 g 3

4 Faraday s Law Here, however, is a question you should be able to answer without some additional information: In the following reaction, how many grams of Ni will be plated out using 13.5 A for 7.50 hr? Ni(SO 3 NH 2 ) 2 + 2e! Ni + 2SO 3 NH 2 What you need is to able to relate amps (A) and hours (hr) to moles of electrons. This is accomplished using the following relationships: Charge (Q) is related to amps (I) and time in seconds (t) by: Q = It Charge (Q) is related to moles of electrons by: Q = (96485 C mol 1 ) n eg 4

5 Faraday s Law Charge (Q) is related to amps (I) and time in seconds (t) by: Q = It Charge (Q) is related to moles of electrons by: Q = (96485 C mol 1) neg The units for charge (Q) are coulombs (C) The constant C mol 1 is called the Faraday s constant. 5

6 Faraday s Law Q = It and Q = (96485 C mol 1 ) n eg So, back to the original question now that you have the additional information: In the following reaction, how many grams of Ni will be plated out using 13.5 A for 7.50 hr? Ni(SO 3 NH 2 ) 2 + 2e! Ni + 2SO 3 NH 2 The charge used is: Q = (13.5 A)(7.50 hr)(3600 s hr 1 ) Q = C Moles of electrons is: n eg = C /(96485 C mol 1 ) n eg = 3.78 mol Thus moles of Ni is: n Ni /1 = ne 1 /2 n Ni /1 = 1.89 mol and mass of Ni is: m Ni = 1.89 mol g mol 1 m Ni = 111 g 6

7 Faraday s Law Q = It and Q = (96485 C mol 1 ) n eg Another example; How many grams of gold can be plated out from a bath of KAu(CN) 2 in 1.00 hr with 50.0 A? The ionic reaction is: Au(CN) 2 + 1e! Au + 2CN The charge used is: Q = (50.0 A)(1.00 hr)(3600 s hr 1 ) Q = C Moles of electrons is: n eg = C /(96485 C mol 1 ) n eg = 1.87 mol Thus moles of Au is: nau /1 = n eg /1 nau /1 = 1.87 mol and mass of Au is: m Au = 1.87 mol g mol 1 m Au = 368 g 7

8 Faraday s Law Q = It and Q = (96485 C mol 1 ) ne A third example: How long would it take to plate a mm layer of chromium onto a 1.00 m 2 part with a current of 1.20 A? The density of Chromium is 7.16 g ml 1. The reaction is: Cr e! Cr (s) In order to do this problem, the mass of the chromium is required. The volume in cm 3 ( the same as ml 1 ) is calculated from the thickness: x = mm [ m/1 mm] [1 cm/( m)] x = cm and the area: A = 1.00 m 2 [1 cm] 2 /[ m] 2 A = cm 2 8

9 Faraday s Law Q = It and Q = (96485 C mol 1 ) ne A third example: How long would it take to plate a mm layer of chromium onto a 1.00 m 2 part with a current of 1.20 A? The density of Chromium is 7.16 g ml 1. The reaction is: Cr e! Cr (s) Thus, the volume is: V = cm cm 2 V = 0.25 cm 3 (or ml) To get the mass, one uses the density formula: ρ = m/v 7.16 g ml 1 = m/ (0.25 ml) m = 1.79 g 9

10 Faraday s Law Q = It and Q = (96485 C mol 1 ) ne A third example: How long would it take to plate a mm layer of chromium onto a 1.00 m 2 part with a current of 1.20 A? The density of Chromium is 7.16 g ml 1. The reaction is: Cr e! Cr (s) So now let s reparse the question: I = 1.20 A m = 1.79 g t =? And the molar mass of Cr is needed: M = g mol 1 The equations needed are: n Cr n eg 1 = 2 Q = It F = C mol 1 Q = F n eg M = m n 10

11 Faraday s Law Q = It and Q = (96485 C mol 1 ) ne A third example: How long would it take to plate a mm layer of chromium onto a 1.00 m 2 part with a current of 1.20 A? The density of Chromium is 7.16 g ml 1. The reaction is: Cr e! Cr (s) Substituting the mass of Cr, obtained before as 1.79 g, to obtain the number of moles of Cr: g mol 1 = 1.79 g / n Cr yields: n Cr = mol n Cr n eg 1 = 2 Q = It F = C mol 1 Q = F n eg M = m n 11

12 Faraday s Law Q = It and Q = (96485 C mol 1 ) ne A third example: How long would it take to plate a mm layer of chromium onto a 1.00 m 2 part with a current of 1.20 A? The density of Chromium is 7.16 g ml 1. The reaction is: Cr e! Cr (s) Using the stoichiometric relationship to get moles of electrons: !2 n = eg 1 3 neg = 1.45 mol n Cr n eg 1 = 2 Q = It F = C mol 1 Q = F n eg M = m n 12

13 Faraday s Law Q = It and Q = (96485 C mol 1 ) ne A third example: How long would it take to plate a mm layer of chromium onto a 1.00 m 2 part with a current of 1.20 A? The density of Chromium is 7.16 g ml 1. The reaction is: Cr e! Cr (s) Converting the 1.45 mol of electrons to coulonbs: yields: Q = C mol! mol Q = C n n Cr e 1 = Q= It F = C mol Q= Fn - = e m n M 13

14 Faraday s Law Q = It and Q = (96485 C mol 1 ) ne A third example: How long would it take to plate a mm layer of chromium onto a 1.00 m 2 part with a current of 1.20 A? The density of Chromium is 7.16 g ml 1. The reaction is: Cr e! Cr (s) Using the current and Q = C, the time may be determined: C = (1.20 A)t so: t = 923 s (15.4 min) n n Cr e 1 = Q= It F = C mol Q= Fn - = e m n M 14

15 Faraday s Law Including electrochemical reactions in stoichiometry THE END 15