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 (Keq). In addition, calculate the electrode potential of the system at the equivalence point (Eea) 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 Number Number 1.900 V Co Co V eq eq Ti,, ,,Ti:'=-0.370 Choose the correct indicator. Shown in parenthesis is the reduction potential and the number of electrons transferred in the oxidation or reduction of the indicator O Indigo tetrasulfonate (En +0.36 V, n-2) O Erioglaucin A (En +0.98 V, -2) O Diphenylamine sulfonic acid (En +0.85 V, n-2) +1.12V. n-2) 2,3'-Diphenylamine dicarboxylic acid (Ein Phenosafranine (En = +0.28 V, n=1) Scroll down to view the rest of the question. O Diphenylamine (Ein+0.76 V, n-2) Methylene blue (En = +0.53 V, n=2) ssume the H concentration is 0.080 M Cu2+/Cu. =0.150 V Seoj./H,seOs:= 1.200 V Number Number eq Choose the correct indicator. Shown in parenthesis is the reduction potential and the number of electrons transferred in the oxidation or reduction of the indicator safranine (Ein = +0.28 V, n=1) 2,3-Diphenylamine dicarboxylic acid (Ein+1.12V, n-2) O Erioglaucin A (En+0.98 V, n-2) O Methylene blue (Ein = +0.53 V, n-2) O Diphenylamine sulfonic acid (En+0.85 V, n-2) O Indigo tetrasulfonate (Ein+0.36 V, n-2) Diphenylamine (Eln_ +0.76 V,n=2)

Explanation / Answer

A

Let us first write the half cell reactions

Co3+ +e- -------> Co2+

Ti3+ + e- ------------> Ti2+

So, n=1

E-cell = E-cathode - E -anode = 1.9 - (- 0.37) V = 2.27 V

At equlibrium point, Ecell = 0, therfore applying Nernst equation,

0 = 2.27 -2.303 RT log K/ F

or, log K = 38.47

Keq = 3 X 1038 at 298 K

For electrode potential at equivalenece point,

Eeq = 1.9+(-0.37)/ (1+1) = 0.76 V

So, the correct indicator will be Diphenylamine

* The given option maybe wrong as n=1 for this reaction

B.

As in above let us first write the half cell reactions

2Cu2+ + 2e- ----------------->2Cu+

SeO-24 + 4H+ + 2 e--------------> H2SeO3 + H2O

So, n=2

Using above equations we get

Keq = 3.91 X 1035 at 298 K

Now,

Eeq = (2) (1.2) + (0.15)(1)/ (2+1)

= 0.85 V

SO, the indicator would be, Diphenylamine sulphonic acid

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