Question

Acetic acid has $K_a=1.8\times {10}^{-5}$ while formic acid has $K_a=2.1\times {10}^{-4}.$ What would be the magnitude of the emf of the cell?
$Pt\left(H_2\right)\left|\begin{array}{c}0.1Maceticacid+\\ 0.1Msodiumacetate\end{array}\right|\left|\begin{array}{c}0.1Mformicacid\\ 0.1Msodiumformate\end{array}\right|Pt\left(H_2\right)at{25}^{o}C?$

• A. 0.032 vol
• B. 0.063 volt
• C. 0.0456 volt
• D. 0.055 volt

Answer 1

The correct option is B 0.063 volt

$Reaction:H_c^{+}\stackrel{1e^{-}}{\to }{H}_A^{+}$
$\therefore E=\frac{0.059}{1}log\frac{2.1\times {10}^{-4}}{1.8\times {10}^{-5}}=0.0629V=0.063V$