For the reactions in parts A and B, calculate the value of the equilibrium const

Question

For the reactions in parts A and B, calculate the value of the equilibrium constant (K_eq). In addition, calculate the electrode potential of the system at the equivalence point (E_eq) of a hypothetical titration. Finally, choose the indicator from the list below that can be used to signal the end point of the hypothetical titration. Methylene blue (E_In = +0.53 V, n=2) 2,3'-Diphenylamine dicarboxylic acid (E_In = +1.12V. n=2) Erioglaucin A (E_In = +0.98 V, n=2) Diphenylamine sulfonic acid (E_In = +0.85 V, n=2) Diphenylamine (E_In = +0.76 V, n=2) Indigo tetrasulfonate (E_In = +0.36 V, n=2) Phenosafranine (E_In = +0.28 V, n=1) B) 2Cu^+ + SeO_4^2- + 4H^+ 2Cu^2+ + H_2SeO_3 + H_2O Methylene blue (E_In = +0.53 V, n=2) Diphenylamine (E_In = +0.76 V, n= 2) Phenosafranine (E_In = +0.28 V, n=1) Indigo tetrasulfonate (E_In = +0.36 V, n=2) 2,3'-Diphenylamine dicarboxylic acid (E_In = +1.12V, n=2) Erioglaucin A (E_In = +0.98 V, n=2) Diphenylamine sulfonic acid (E_In = +0.85 V, n=2)

Explanation / Answer

(A) G° = R T ln Keq and Go = nFEocell

Thus, RTlnKeq = nFE0cell OR   RTlnKeq = nFE0cell

ln Keq = (nFE0cell ) / RT -------------(1)

where, number of electron transferred = n = 1

Universal gas constant R = 8.314 J/mol.K

Temperature T = 298 K (standard)

E0 cell = cell potential = Ecathode + Eanode or Ereduction + Eoxidation

= 1.9V + (-0.37V)

= 1.53 V

Faraday's constant F = 96485 C/mol

Now put the values in eq---(1)

ln Keq = (1 * 96485 C/mol * 1.53V) / (8.314 J/mol.K * 298K)

= 147622 C*V / 2477.572 J (because V = J/C)

ln Keq = 59.6

Keq = e59.6= ( 7.65 * 1025)

Eeq = Eeq = (1.9 - 0.37)/2 = 0.765 V

Correct answer is (diphenylamine)

(B)

ln Keq = (nFE0cell ) / RT -------------(1)

E0 cell = cell potential = Ecathode + Eanode or Ereduction + Eoxidation

= 1.2V + 0.15 V

= 1.35 V

Faraday's constant F = 96485 C/mol

Now put the values in eq---(1)

ln Keq = (2 * 96485 C/mol * 1.35V) / (8.314 J/mol.K * 298K)

= 260509.5 C*V / 2477.572 J (because V = J/C)

ln Keq = 105.15

Keq = e105.15= ( 4.63* 1045)

Eeq = Eeq = (1.2 - 0.15)/2 = 0.525 V

Correct answer is (METHYLENE BLUE)

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