82%-- Thu Jul 12 1:45:02 PM ople Window Help nline teaching and x Grades for Wen

Question

82%-- Thu Jul 12 1:45:02 PM ople Window Help nline teaching and x Grades for Wendell Soto CHIX * Syllabus for CHM2046 603U akeAssignment/takeCovalentActivity.do?locato-assignment-take&takeAssignmentSessi; ) ? E Use the References to access important values if needed for this question. When the Hg concentration is 1.40 M, the observed cell potential at 298K for an electrochemical cel with the following reaction is 3.328V. What is the Mg concentration? Hg(a) + Mg() Hg)+Mg(a) Answer: Submit Answer Retry Entire Group 9 more group attempts remalning Previous Next

Explanation / Answer

The two half cell reactions are

Reduction reaction

Hg2+ + 2e- = Hg

Ered = 0.85 V

Oxidation reaction

Mg = Mg2+ + 2e-

Eox = 2.37 V

Overall cell reaction

Hg2+ + Mg = Hg + Mg2+

E° = Eox + Ered

= 0.85 + 2.37

= 3.22 V

From the Nernst equation

E = E° - (0.0592/2) log [Mg2+] / [Hg2+]

3.328 = 3.22 - (0.0592/2) log [Mg2+] / [1.40]

0.108 = - 0.0296 log [Mg2+] / [1.40]

log [Mg2+] / [1.40] = - 3.648

[Mg2+] / [1.40] = 0.00022490

[Mg2+] = 3.15 x 10^-4 M

Related Questions

Navigate

• Browse (All)
• Browse 8
• Subjects

---

---

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