ElectrochemistryMedium
Question
The dissociation constant for CH3COOH is 1.8 × 10−5 at 298 K. The electrode potential for the half-cell:
Pt|H2 (1 bar)|0.5 M – CH3COOH, at 298 K is (log 2 = 0.3; log 3 = 0.48; 2.303 RT/F = 0.06)
Options
A.−0.3024 V
B.−0.1512 V
C.+0.3024 V
D.+0.1512 V
Solution
$H_{2}(g) \rightleftharpoons 2H(aq) + 2e^{-}$
$E_{H_{2}|CH_{3}COOH} = E_{H_{2}|H^{+}} = E_{H_{2}|H^{+}}^{o} - \frac{0.06}{2}.\log\frac{\left( H^{+} \right)^{2}}{P_{H_{2}}}$
$= 0 - \frac{0.06}{2}.\log\frac{\left( \sqrt{Ka \times C} \right)^{2}}{P_{H_{2}}} $$$= - \frac{0.06}{2}.\log\frac{1.8 \times 10^{- 5} \times 0.5}{1} = + 0.1512\text{ V}$$
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
At 25°C, the solubility product of CuCl is 2.0 × 10−7 and $E_{Cl^{-}|CuCl|Cu}^{o}$ is 0.128 V. The value of $E_{Cu^{+}|C...In the electrolytic cell, flow of electrons is from...100 mL of buffer of 1 M NH3(aq) and 1 M NH4+(aq) are placed in two comparments of a voltaic cell separately. A current o...The chemical reaction2AgCl(s) + H2 (g) → 2HCl (aq) + 2Ag (s)taking place in a galvanic cell is represented by the ...When silver chloride is dissolved in a large excess of ammonia, practically all silver ion can be assumed to exist in fo...