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o2 Mail -kiohnson10 x y Indiana State Univ x wwws aplinglearni x VG wou are titrating 1 x www.saplinglearni x chegg study Gu x V D chem.321.Modul x twinsofcosplay Vibiscm mod/ibis/view.php?id 3224083 sapling learning.com Apps YouTube N Netflix f Facebook secure cloud Access D Tower of God.seanla lng postsecret Acomplete guide to nMaltese dog breeders Kids Rides& Fides fo her bookmark tr Print Calculator Periodic Table Available From 4/3/2017 09:00 AM Question 1 of 4/19/2017 09:00 AM Due Date: Map Points Possible You are titrating 100.0 mL of 0.0200 M Fe n 1 M HCIO4 with o.100 M cu to give Fe2 and cu?" using Pt and saturated Ag IAgCl electrodes to find the endpoint. Grade Category: Default Description: (a) Write the balanced titration reaction, Policies SFW Custom (b) Complete the two half reactions for E -0.161 V Cu e the Pt electrode. You can check your answers. E -0.767 V Fe e You can view solutions when you complete or give up on any questio (c From the list in the column at the right, select the E-0.767-0.05916log A two comect Nernst equations for the cell voltage. Fe 0.197 You can keep trying to answer each question until Fe you get it right or give up. (Each applying at different points in the titration E of the Ag IAgol electrode is 0.197 V You lose 10% of the points available to each Fe (B) E-0.767-0.05916 log 0.197 answer in your question for each incorrect attemp at that answer. Fe C) E 0.16 0.05916 log 0.197 O eTextbook Fe Fe O Help With This Topic D) E 0.16 0.05916 0.197 Fe OWeb Help & Videos Cu n.197 OTechnical Support and Bug Reports Previous 8 Give Up & View salution Check Answer O d Exit Next D 2017 Sapling Learning Inc. privacy policy hel 3:14 AM B 4/19/2017Explanation / Answer
1)
a) balanced equation is
Fe+3 + Cu+ = Fe+2 + Cu+2
b) half cell reactions-
Cu+2 + e- = Cu+, E=0.161v
Fe+3 + e- = Fe+2 , E= -0.767v
C) correct options from the list are as follows
Option- B & H
(B) E= 0.767 - 0.05916 * [log Fe+3/Fe+2] - 0.197
(H) E= 0.161- 0.05916 * [ log Cu+2/Cu+] - 0.197
Fe electrode acts as ANODE(due to less E value) and Cu electrode acts as CATHODE(more E value)
overall equation is
E= [0.161- (-0.767)] - 0.05916 * [log Fe+3 / Cu+2] -0.197
E = 0.928 - 0.05916 * [log Fe+3 / Cu+2] - 0.197
d)
(1) for 2ml of Cu+ solution, E is as follows-
E = 0.928 - 0.05916 * [log Fe+3 / Cu+2] - 0.197...given that [Fe]= 0.02M of 100 mL & [Cu] = 0.100M
E = 0.928 - 0.05916 * log[100 * 0.02/ 0.1 * 2] -0.197
= 0.928 - 0.05916 * log [20/2] -0.197
= 0.928 - 0.05916 * log 10 -0.197.....log 10 = 1
= 0.928 -0.05916 -0.197
E =0.6718 v (for 2 mL)
(2) for 10ml of Cu+ solution, E is as follows-
E = 0.928 - 0.05916 * [log Fe+3 / Cu+2] - 0.197
=0.928 - 0.05916 * log[100 * 0.02/ 0.1 * 10 ] -0.197
= 0.928 - 0.05916 * log[2/ 1] -0.197
= 0.928 - 0.05916 * 0.3010 -0.197......log 2 = 0.3010
= 0.928 - 0.01780-0.197 = 0.7132
E = 0.7132 v (for 10 mL)
(3) for 18.5 ml of Cu+ solution, E is as follows-
E = 0.928 - 0.05916 * [log Fe+3 / Cu+2] - 0.197
=0.928 - 0.05916 * log[100 * 0.02/ 0.1 * 18.5 ] -0.197
= 0.928 - 0.05916 * log[ 2/1.85] - 0.197
= 0.928 - 0.05916 * [log 2- log 1.85] - 0.197
= 0.928 - 0.05916 * [0.3010 - 0.2671] - 0.197
= 0.928 - 0.05916 * 0.0339 - 0.197 = 0.729v
E = 0.729v (for 18.5 mL)
(4) for 20 ml of Cu+ solution, E is as follows-
E = 0.928 - 0.05916 * [log Fe+3 / Cu+2] - 0.197
=0.928 - 0.05916 * log[100 * 0.02/ 0.1 * 20 ] -0.197
= 0.928- 0.05916 * log [ 2/2] - 0.197........log1=0
= 0.928 - 0.197= 0.731v
E = 0.731v (for 20 mL)
(5) for 21 ml of Cu+ solution, E is as follows-
E = 0.928 - 0.05916 * [log Fe+3 / Cu+2] - 0.197
=0.928 - 0.05916 * log[100 * 0.02/ 0.1 * 21] -0.197
= 0.928 - 0.05916 * [log 2- log 2.1] -0.197..........log 2= 0.3010, log 2.1= 0.3222
= 0.928 - 0.05916 * (-0.0212)- 0.197
= 0.928 + 0.00125 - 0.197 = 0.7322 v
E= 0.7322 v( for 21 mL)
(6) for 40 ml of Cu+ solution, E is as follows-
E = 0.928 - 0.05916 * [log Fe+3 / Cu+2] - 0.197
=0.928 - 0.05916 * log[100 * 0.02/ 0.1 * 40] -0.197
= 0.928 - 0.05916 * log [1/2] -0.197
= 0.928 - 0.05916 * [log 1-log 2] -0.197.......log 1 = 0, log 2 = 0.3010
= 0.928 +0.01780 -0.197= 0.7488v
E = 0.7488 v (for 40mL)
40 mL 0.7488v
volume of Cu+ solution E value 2 mL 0.6718v 10 mL 0.7132v 18.5 mL 0.729 v 20 mL 0.731v 21.0 mL 0.7322vRelated Questions
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