A solution contains 0.181 M Ce3+, 1.81104 M Ce4+, 1.81104 M Mn2+, 0.181 M MnO4,

Question

A solution contains 0.181 M Ce3+, 1.81104 M Ce4+, 1.81104 M Mn2+, 0.181 M MnO4, and 1.81 M HClO4. (The Faraday's constant is 9.649104 Coulombs/mole.) substance E (volts) Ce3+ 2.336 Ce4+ +1.700 Mn2+ 1.182 MnO4 +1.507

(a) Write a balanced net reaction that can occur between species in this solution. (Use the lowest possible coefficients. Omit states-of-matter.) chemPadHelp 5Ce4+ + Mn2+ + 4H2O 5Ce3+ + MnO4- + 8H+ 5Ce^4+ + Mn^2+ + 4H_2O <=> 5Ce^3+ + MnO_4^- + 8H^+ Correct.

(b) Calculate G and K for the reaction. G WebAssign will check your answer for the correct number of significant figures. kJ K WebAssign will check your answer for the correct number of significant figures.

(c) Calculate E for the conditions given. WebAssign will check your answer for the correct number of significant figures. V

(d) Calculate G for the conditions given. WebAssign will check your answer for the correct number of significant figures. kJ (e) At what pH would the given concentrations of Ce4+, Ce3+, Mn2+, and MnO4 be in equilibrium at 298 K

Explanation / Answer

(a) Balanced net chemical equation :

5Ce4+ + Mn2+ + 4H2O ----> 5Ce3+ + MnO4- + 8HClO4

(b) Using the table for standard reduction potential,

Ecell = 1.61-1.51

         = 0.1 V

dG = -nFEcell = - 5 x 96485 x 0.1 = 48242.5 = -48.24 kJ

dG = -RTlnK

48240 = -8.314 x 298 ln K

K = 2.86 x 10^8

(c) E = 0.1 - 0.0592/5 log([Ce4+]^5[Mn2+]/[Ce3+]^5[MnO4-][H+]^8)

= 0.1 - 0.0592/5 log[(1.81 x 10^-4)^5(0.181 x 10^-4)/(0.181)^5(0.181)(1.81)^8]

= -0.14 V

(d) dG = -nFE

    = 67.54 kJ

(e) 0 = 0.1 - 0.0592/5 logK

K = 2.80 x 10^8

[H+] = 9.38 x 10^-3 M

pH = 2.03

At pH 2.03 the equilibrium will be established.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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