Question

ADVANCED MATERIAL Recognizing consistency among equilibrium constant, free energ... A student made measurements on some electrochemical cells and calculated three quantities: • The standard reaction free energy AG". • The equilibrium constant K at 25.0°C. . The cell potential under standard conditions E. His results are listed below. Unfortunately, the student may have made some mistes. Examine his results carefully and tick the box next to the incorrect quantity in each row, if any. Note: If there is a mistake in a row, only one of the three quantities listed is wrong. Also, you may assume the number of significant digits in each quantity is correct. Also note: for each cell, the number of electrons transferred per redox reaction is 1. calculated quantities (Check the box next to any that are wrong.) cell A1 AGK - 31. kJ/mol o 3.71 x 100 -48. kJ/mol of 390 x 10" o -134. klimot o 3.34 × 10 20 X E -0,32 V -0.5o v 1.39 V 5 0 o ol C ?

Answer 1

For Case A and B sign of ∆G^o should be opposite. And for Case C K value is wrong.

4U SHOU Cell n -0-32 V AG - 31 ka/mol -48 kamol -134 kg/mol 3.71X10-6 3.90X107 3.34X1024 © -0.50 V 1.39 v e 46°. - R Teu Kopa - NF E&am Kreqm Caracterie at 25e AGO ARTIM (3.71X106) 7-8-314 X 298 x 10( 3.71X10 E-RT luban Cell n k A 1. 3.71X 166 9462-31 KT . -FE', -31% 103 I - E ---31 X103 E - +032 V = + 30.98 KT. 46°2-IXFXE F = -1X96458x - 0-32 964589) malce (2) E's 0.324 AG = 30.9 or 31 kI/mole s+30.86 KI an fk Then 4G3-RT lu (3.90x109) B 1 23.90X10-91 2-8-314x298 x In (3.71810) 2 +47 98 KT at Ea -0.50V -> 4Gº 2 -1%96458 X-0.50 49.229 Kamole 4G2 + 48 X103 I 2-RT luk If luk a -488103. 8.314x298 of a fue} 4 then ko t 3.85X10-9 given 3.90x10-9 AG2748X10%-MFEO. Here also 46 then Eº > 48X103 7. Sou Should be tve) 96458 Then RT @ nat. If 4G .+134.Kā mol. oo luk » +134X103 I As 462-134. KJ/mol 9. luk2 -134x103 :, - HEE°3-134X103 7/md 134 x 103 AFE02-134 X10 Ilma 8.314 X 298 k 13.24x10-24 given 3.34x20²24 here kvalue is different to 96458 11.39 o