Please answer all and please answer if you\'re 100% sure the answer is right! 5:

Question

Please answer all and please answer if you're 100% sure the answer is right! 5:33 PM -107 AT&T; 25%, D webassign net My Notes A solution contains 0.301 M 3+ 3.01 x 10 M Ce4+, 3.01x10 M Mn Ce 2+ 0.301 M Mno4 and 3.01 MHCIO4. The Faraday's constant is 9.649x 104 Coulombs/mole.) Reduction (volts) (a) Write a balanced net reaction that can occur between species in this solution. (Use the lowest possible coefficients. Omit states of matter.) HINT: make permanganate a product. (b) Calculate AG, and K for the reaction (at 298 K, two sig figs). (c) Calculate E for the conditions given. (d) Calculate AG for the conditions given (two sig figs). (e) At what pH would the given concentrations of Ce4+, Ce Mn' 3+ and Mno4 be in equilibrium at 298 K? (two sig figs) The quinhydrone electrode was introduced in 1921 as a means of measuring

Explanation / Answer

a) Ce4+ + e– Ce3+ Eo = + 1.7 V
  Mn2+ + 4 H2O MnO4– + 8 H+ + 5 e–   Eo = -1.57 V
Balance the equation before writing the complete equation,
5Ce4++ 5e– 5Ce3+   Eo = + 1.7 V
Mn2+ + 4 H2O   MnO4– + 8 H+ + 5 e–   Eo = -1.57 V
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5Ce4+ + Mn+2 + 4H2O 5 Ce3+ + MnO4- + 8H+ E cell = 0.13 V

b) G = - nFEcell = - 5 * 96.49 kJ/mol e- V * 0.13 = -62.7185 kJ = -63 kJ
    G = -RT ln K
   -62.7185 * 103 J = - 8.314 J/ mol K * 298 K ln K
   ln K = 25.3145
   K = 9.8616*1010 = 1011

c) E = Ecell - (0.0592 /n) * log( [products] / [reactants] )
   consider the overall reaction,
   E = 0.13 - ( 0.0592/5) * log ( [Ce3+]5 [MnO4-] [H+]8 / [ Ce+4]5 [Mn+2] )
   E = 0.13 - (0.0592/5) * log ( [0.301]5 [0.301] [3.01]8  / [3.01*10-4]5 [3.01*10-4] )
   E = 0.13 - (0.0592/5) log ( 5.011075 / 7.43702*10-22)
E = 0.13 - 0.2584498 = -0.12844 V
but according to your question the equation is MnO4- / Mn+2
E = + 0.1284 V

d) G = -nFE
   G = - 5 * 96.49 kJ /mol e- V * 0.12844 V = -61.9706 kJ = -62 kJ

e) At equilibrium E becomes 0
   K = ( [products] / [reactants] )
1011 = (  [0.301]5 [0.301] [ H+]8  / [3.01*10-4]5 [3.01*10-4] )
1*10-7 = [ H+]8   
   [H+] = 0.133352 M
  PH = - log10[H+] = 0.875

     

   

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