24 % -Thu 1 1 :03 AM a Safari File Edit View History Bookmarks Window Help sapli

Question

24 % -Thu 1 1 :03 AM a Safari File Edit View History Bookmarks Window Help saplinglearning.com 1.What Is The Oxidation State Of Each Element Chegg.com University of Missouri, Columbia CHEM 1330 - Fall15 PERRY: HW 18 Announcements - CHEM 1330: College Chemistry II, Sec. 01 Avai Due Poin Gra Des Poli Periodic Table Periadic Table Print Calculator Question 3 of 15 Map O sapling learning Calculate the cell potential for the following reaction as written at 25.00 , given that [Cr2+1-0.815 M and Fe2]-0.0140 M. Standard reduction potentials can be found here Cr(s) + Fe2+(aq) Crs) +Fe a Cr2 .(aq) + Fe(s) You You You Ther Number Te 2

Explanation / Answer

Half cell reactions are:

Cr (s) ----> Cr2+ (aq) + 2e- .......... E = 0.91 V

Fe2+ (aq) + 2e- ------> Fe (s) ............ E = - 0.44 V

Overall Reaction:

Cr (s) + Fe2+ (aq) -----> Cr2+ (aq) + Fe (s)

Eo cell = 0.91 + (-0.44) = 0.47 V

We know that,

E cell = Eo cell - (0.0591 / n) log (Q)

Eo cell = 0.47 V

n = number of electrons involved = 2

Q = [Cr2+] / [Fe2+] = 0.815 / 0.014 = 58.2

=> E cell = 0.47 - (0.0591 / 2) log (58.2) = 0.418 V = Cell Potential

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.