What is the value of the equilibrium constant at 25 C for the reaction between t

Question

What is the value of the equilibrium constant at 25 C for the reaction between the pair: Mn(s) and Ag (aq) to give Ag(s) and Mn2(aq) Use the reduction potential values for Ag (aq) of +0.80 V and for Mn2 (aq) of -1.18 V Give your answer using E-notation with NO decimal places (e.g., 2 x 102 would be 2E-2; and 2.12 x 102 would also be 2E-2.). Do NOT include spaces, units, punctuation or anything else silly! [Blank1] Specified Answer for: Blank1 1.01e67 Correct Answers for: Blank1 Evaluation Method Case Sensitivity Correct Answer 8E66 Exact Match

Explanation / Answer

ANSWER:

Given,

T = 25 oC = 298 K

EoMn2+/Mn = -1.18 V (oxidation)

EoAg/Ag+ = +0.80 (reduction)

Eo = EoAg/Ag+- EoMn2+/Mn

Eo = + 0.80 - (-1.18) V = + 1.98 V

Now,

log Keq = {nF / 2.303 RT} Eocell

at 298 K

log Keq = {n / 0.06} Eocell

n=2, because total 2 electron transfer takes in a reaction.

log Keq = {2 / 0.06} x 1.98 = 66  

Keq = 1 x 1066

Hence, the value of the equilibrium constant at 25 oC for the pair of reaction is 1E66.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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