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Safari File Edit View History Bookmarks Window Help # +9 % 78961, Wed 11:07 AM elaineschultz QE saplinglearning.ca University of Manitoba - CHEM 1310 Lecture & Lab - Winter15 - J. Cullen: Assignment 5 Home | Chegg.com 0 4/10/2015 11:00 PM 82.9/1004/5/2015 03:35 PM Assignment Information Available From: 1/6/2015 12:00 A Due Date: Points Possible: 100 Grade Category: Homework Description Policies Attempts Score Gradebook Print Calculator Periadic Table 4 95 Question 32 of 36 4/10/2015 11:00 P 23 100 Mapoob sapling learning 24 100 Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg21-0792 M and [Fe2] 0.0200 M. Standard reduction potentials can be found here 25 100 CHEM 1310 assig Mg(s) + Fe2+(aq) Mg2+(aq) + Fe(s) 26 4 85 You can check your answers. 27 90 Number You can view solutions when you com up on any question. 28 4 70 Tools You have ten attempts per question. 29 90 You lose 5% of the points available to in your question for each incorrect att answer 30 100 31 100 For multiple-choice questions, the per on the number of choices available. 32 Help With This Topic 34 Web Help & Video:s 35 Technical Support and Bug Re Previous Give Up & View Solution Check Answer Next Exit Hint 36 Copyright © 2011-2015 Sapling Learning, Inc.-40 about uscareerspartnersprivacy policy terms of use contact us help 8

Explanation / Answer

the cell reaction is

Mg(s) + Fe+2 (aq) ----> Mg+2 (aq) + Fe (s)

Cathode : ( reduction )

Fe+2 (aq) + 2e- ----> Fe (s)

anode (oxidation )

Mg(s) -----> Mg+2 (aq) + 2e-

Eo cell = Eo cathode - Eo anode

Eo cell = Eo Fe+2/Fe - Eo Mg+2/Mg

Eo cell =   -0.44 + 2.372

Eo cell = 1.932

now

using nenrst equation

E= Eo - ( 0.05916 / n ) log [Mg+2 / Fe+2]

here n = 2 as two electrons are transferred

E= 1.932 - ( 0.05916/2) log [ 0.792 / 0.02)

E= 1.88

so the cell potential is 1.88 V

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