ElectrochemistryHard
Question
The EMF of the cell: Hg(l)|Hg2Cl2(s), KCl sol. (1.0N)|Quinohydrone|Pt, is 0.210 V at 298 K. What is the pH of the quinohydrone solution, the potential of the normal calomel electrode is 0.279 V and E° for the quinohydrone electrode is 0.699 V, both at the same temperature. [2.303 RT/F = 0.06]
Options
A.3.5
B.7.0
C.1.85
D.−3.5
Solution
Ecell = Equinohydrone −Ecalomel
0.210 = Equinohydrone – 0.279
⇒ Equinohydrone = 0.489 V
Quinohydrone electrode is
$E = E^{o} - \frac{0.06}{2}.\log\frac{1}{\left\lbrack H^{+} \right\rbrack} = E^{o} - 0.06P^{H}$
or, 0.489 = 0.699 – 0.06.PH ⇒ PH = 3.5
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