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}}} $$$= \frac{0.059}{2}\left( P^{Ka_{1}} - P^{Ka_{2}} \right) = 0.059\text{ V}$$
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
What is the emf at 25oC for the cell, Ag Pt The standard reduction potentials for the half-reactions AgBr + e- → A...In the electrochemical cell :- Zn|ZnSO4(0.01M)| |CuSO4(1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentrat...Select the correct statement.(i) Delocalisation of σ-electron is hyperconjugation.(ii) Delocalisation of π-el...An acidic solution of copper (II) sulphate containing some contaminations of zinc and iron (II) ions was electrolysed ti...Chromium plating can involve the electrolysis of an electrolyte of an acidified mixture of chromic acid and chromium sul...