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The quinhydrone electrode can be used for the potentiometric determination of pH

Chemistry
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Q + 2H+ + 2e–     H2Q     E0 = 0.699 V What is the pH of a solution saturated with quinhydrone if the potential of a platinum electrode  in the solution, measured against a saturated calomel electrode, is −0.205 V? 

Dec 9th, 2014

The electrode reaction is 

C6H4O2 + 2H+ + 2e- → C6H4(OH)2

Quinone is oxidant, and hydroquinone is reductant.

For the potentiometric measurements, it is a combination of the quinhydrone electrode with suitable reference electrode to create electrochemical cell. Thus,

pH = (E0(Q/H2Q) – Ecalomel – E) / 2.303 RT/F

E0(Q/H2Q) = 0.699 V

E calomel is a standard table data that was supposed to be given to you in a problem conditions along with the Temperature. I suppose it is room T conditions or that would be stated otherwise. Thus, E calomel at T room = 0.2410 V

E is measured = - 0.205 V

T = 25 C

R = 8.314 J kmol-

F = 96485 C mol-

Plugging in numbers: pH = (0.699 V - 0.2410 V + 0.205 V) / 0.05916 V =  0.663 / 0.05916 = 11.2

Wasn't easy problem, huh? :)

Hope it helps!


Dec 18th, 2014

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