ElectrochemistryHard
Question
Two weak acid solutions HA1 and HA2 each with the same concentration and having pKa values 3 and 5 are placed in contact with hydrogen electrodes (1 atm, 25°C) and are interconnected through a salt bridge. EMF of the cell is
Options
A.0.0295 V
B.0.118 V
C.0.0885 V
D.0.059 V
Solution
Cell reaction: $H^{+}\left( C_{1},HA_{1} \right) \rightarrow H^{+}\left( C_{2},HA_{2} \right)$
$E_{cell} = 0 - \frac{0.059}{1}.\log\left( \frac{C_{2}}{C_{1}} \right) = 0.059.\log\sqrt{\frac{Ka_{2}}{Ka_{1}}} $$$= \frac{0.059}{2}\left( P^{Ka_{1}} - P^{Ka_{2}} \right) = 0.059\text{ V}$$
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