c13 JWBS043-Rogers September 13, 2010 11:27 Printer Name: Yet to Come
216 COULOMETRY AND CONDUCTIVITYbe within our measurement range. If the amount of current we can maintain is about
50 milliamperes (mA), we know everything in the equation except the time. Trials
show that it takes about an hour (1 h=3600 s) to deposit 0.2 g of Ag.F=
I×t
moles Ag=
0. 050 × 3600
0. 2
107. 9
=
180
0. 00185
= 97 , 110
Now in our precise one-hour experiment, the real amperostat reading is 49.883 mA
over the time period of 1.000 h= 3600 .0 s, and the weight of silver gained by the
(carefully dried) silver cathode is 0.20149 gF=
coulombs
mol=
I×t
mol=
0. 049883 ( 3600. 00 )
0. 20101
107. 868
=
179. 5788
0. 00186534
= 96 , 367
which is a little more than 0.1% in error.Example 13.2
Suppose you are in the unlikely circumstance of needing to know the faraday to nine
significant figures. Propose a way of calculating F from fundamental constants.Solution 13.2 Unknown in Faraday’s time, of course, unit charge is 1. 602176487 ×
10 −^19 C and Avogadro’s number isNA= 6. 02214179 × 1023. Since the faraday is
the charge per mole of (singly charged) particles, we haveF=
(
6. 02214179 × 1023
)
1. 602176487 × 10 −^19 = 96485 .3398 C mol−^1These data are from the handbook, but you will get some argument on the last two
digits.Example 13.3
(a) A Hittorf cell having 0.1000 mol of AgNO 3 (aq) in each of its three compart-
ments had 0.0100 faradays of charge passed through it. A reduction reaction
depositing solid silver, Ag(s), took place at the cathode:Ag+(aq)+e−→Ag(s)Analysis of the contents of the cathode compartment after the experiment
showed that 0.00947 mol of Ag+(aq) remained. What are the transport num-
bersofAg+(aq) and NO− 3 (aq)?