Question

$Pt\left(H_2\right)\left|\begin{array}{c}0.1Maceticacid+\\ 0.1Msodiumacetate\end{array}\right|\left|\begin{array}{c}0.1Mformicacid+\\ 0.1Msodiumformate\end{array}\right|Pt\left(H_2\right)at{25}^{0}C?$

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?

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

Answer 1

The correct option is B 0.063 volt

Ans $B$

For gives equation

$K=\frac{K_A\left(2abticAcid\right)}{K_a\left(formicAcid\right)}$

$V=1=\frac{18\times {10}^{-5}}{2.1\times {10}^{-4}}$

$=0.08571$

$\therefore \ln K=\frac{FE}{RT}$

$\Rightarrow F=\frac{RT}{F}\ln K$

$=\frac{8314\times 298}{1\times 96500}\ln \left(0.08571\right)$ volts

$Emf$ of the all $=-0.063$ volts

$\therefore$ Magnitude of $EMF$ of the all $=|-0.063|$ (moduloud) volts

$=0.063$ volts

Ans: $0.063V$