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?
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
= 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 = -
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,
Hope it helps!
Dec 18th, 2014
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