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  • A hydrogen electrode placed in a solution containing sodium acetate and acetic acid in the ratio of x : y and y : x has an electrode potential value E1 and E2n

A hydrogen electrode placed in a solution containing sodium acetate and acetic acid in the ratio of x : y and y : x has an electrode potential value E1 and E2 volts respectively at 25oC. The pKa value of acetic acid is:

Option: 1

\frac{-\left(E_{1}+E_{2}\right)}{2 \times 0.059}


Option: 2

\frac{\left(E_{1}+E_{2}\right)}{2 \times 0.059}


Option: 3

\frac{E_{2}-E_{1}}{2 \times 0.059}


Option: 4

\frac{-\left(E_{1}+E_{2}\right)}{0.059}


Answers (1)

best_answer

The reaction occuring in the hydrogen electrode is given below:

\mathrm{H^++e^-\longrightarrow\frac{1}{2}H_2}

Using the Nernst Equation, the electrode potential can be written as 

\mathrm{E= -0.059 ~log\frac{1}{[H^+]}=-0.059~pH}

In the two given cases, the Hydrogen ions are obtained by buffer solution having the ratio of \mathrm{Salt:Acid} as \mathrm{x:y} and \mathrm{y:x}

The pH of an acidic buffer can be calculated as 
\mathrm{\mathrm{pH} =\mathrm{p} K_{a}+\log \left(\frac{\mathrm{Salt}}{\mathrm{Acid}}\right)}
\mathrm{\therefore \mathrm{\: pH}_{1}=\mathrm{p} K_{\mathrm{a}}+\log \frac{x}{y}}
Similarly,
\mathrm{\mathrm{pH}_{2}=\mathrm{p} K_{\mathrm{a}}+\log \frac{y}{x}}

Thus, the respective electrode potentials can be written as 

\begin{aligned}{\therefore \text{E}_{1}=-0.059 \mathrm{\: pH}_{1}} \\ {\: \: \: \text{E}_{2}=-0.059 \mathrm{\: pH}_{2}}\end{aligned}

Now, adding the two potentials give us 

\mathrm{{\left(E_{1}+E_{2}\right)=-0.059\left(\mathrm{p} K_{\mathrm{a}}+\log \frac{x}{y}\right)} \ {-\: 0.059\left(\mathrm{pK}_{\mathrm{a}}+\log \frac{y}{x}\right)}}

\mathrm{(E_{1}+E_{2})=-0.059\times 2 \times pKa}

\mathrm{\mathrm{p} K_{\mathrm{a}}=\frac{-\left(E_{1}+E_{2}\right)}{2 \times 0.059}}

Therefore, option(1) is correct

Posted by

vinayak

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