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1.11 Rearrangement of equation 1.4.16 yields the following expression for / as a function of E, which is
convenient for calculating i¡ªE curves for nernstian reactions:
ilii = {1 + exp[(nF/RT)(E - El/2)]}-1
(a) Derive this expression, (b) Consider the half-reaction Ru(NH3)£+ + e *¡À Ru(NH3)^+. The E¡ã
for this reaction is given in Appendix C. A steady-state i-E curve is obtained with a solution containing
10 mM Ru(NH3)6
+ and 1 M KC1 (as supporting electrolyte). The working electrode is a Pt
disk of area 0.10 cm2 operating under conditions where m = 10~3 cm/s
for both Ru species. Use a spreadsheet program to calculate and plot the expected i-E curve.

1.12 (a) Derive an expression for / as a function of E, analogous to that in Problem 1.11, from equation
1.4.20, using (1.4.15) as the definition of §¦§è2, for use in solutions that contain both components
of a redox couple, (b) Consider the same system as in Problem 1.11, but for a solution containing
10 mM Ru(NH3)^+ and 5.0 mM Ru(NH3)^+ in 1M KC1. Use a spreadsheet program to calculate
the i-E curve and plot the results, (c) What is 77conc at a cathodic current density of 0.48 mA/cm2?
(d) Estimate Rmt.

[ Last edited by wolfghost on 2013-6-6 at 10:02 ]
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2Â¥2013-05-18 10:10:12
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hqsy

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wolfghost: °ïÄã¸ÄÁËϱêÌ⣬ÈÃÓ¦ÖúµÄ³æ×ÓÄÜÖ±½ÓÃ÷°×ÄãÇóÖúµÄÄÚÈÝ£¡ 2013-06-06 10:03:09
ÒýÓûØÌû:
2Â¥: Originally posted by wolfghost at 2013-05-18 10:10:12
Â¥Ö÷´ËÌùÊÇÇóÖúʲô£¿·­Ò룿ÇëÃ÷ʾ£¬·½±ã³æ×ÓÃǸúÌù£¡

½â´ðÕâЩϰÌâ¡£1. 11 ÕûÀí¹«Ê½(1. 4. 16 ) ¿ÉµÃµ½ÏÂÃæ¼ÆËãÄÜË¹ÌØ·´Ó¦µÄi-E ÇúÏߵļò±ã±í´ïʽ:
i/ il = {1 + exp [ ( nF/ R T) ( E - E1/ 2 ) ]} - 1
( a ) ÍÆµ¼´Ë±í´ïʽ¡£( b) ¿¼Âǰ뷴Ӧ: Ru ( NH3 )3 +
6 + e Ru ( NH3 )2 +
6 , ÆäE􊯟 ¿ÉÓɸ½Â¼C Öвé
µ½¡£ÔÚº¬ÓÐ10mmol/ L Ru (NH3 )3 +
6 ºÍ1mol/ L KCl ×÷Ϊ֧³Öµç½âÖʵÄÈÜÒºÖÐ, ¿ÉµÃµ½Ò»¸öÎÈ̬µÄi-E
ÇúÏß¡£ÔÚÁ½ÖÖRu »¯ºÏÎïm = 10 - 3 cm/ s Ìõ¼þÏÂ, ²ÉÓÃÃæ»ýΪ0. 1cm2 µÄ²¬µç¼«½øÐÐʵÑé, ÀûÓóÌÐò½ø
ÐмÆËã²¢»­³öÔ¤ÆÚµÄi-E ÇúÏß¡£
1. 12 ( a ) ÀàËÆÓÚϰÌâ1. 11 , ²ÉÓÃʽ( 1. 4. 15 ) ¶ÔÓÚE1/ 2 µÄ¶¨Òå, ´Ó¹«Ê½( 1. 4. 20) ÍÆµ¼³ö¿ÉÓÃÓÚÈÜÒºº¬
ÓÐÑõ»¯»¹Ô­Á½ÖÖ×é·ÖÇé¿öϵĵçÁ÷×÷ΪµçÊÆº¯ÊýµÄ±í´ïʽ; ( b) ¿¼ÂÇÓëϰÌâ1. 11 ÏàͬµÄÌåϵ, µ«ÈÜ
ÒºÖк¬ÓÐ10mmol/ L Ru(N H3 )3 +
6 ¡¢5mmol/ L Ru( NH3 )2 +
6 ºÍ1mol/ L KCl, Ó¦ÓÃÒ»¸ö³ÌÐò½øÐмÆËã²¢
»­³öÆäi-E ÇúÏß¡£( c ) ÔÚÒõ¼«µçÁ÷ÃܶÈΪ0. 48mA/ cm2 µÄÇé¿öÏÂ, ¦Çconc ÊǶàÉÙ? ( d) ¹ÀËãRmt ¡£
ÓëÊÀÖð¶ø²»ÔðÓÚÈË¡£
3Â¥2013-06-06 10:00:24
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