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  • Consider the following cell with hydrogen electrodes at different pressure p1 and p2. <img alt="\operatorname{Pt}, \mathrm{H}_{2}\left(p_{1}\right)\left|\begin{arr

Consider the following cell with hydrogen electrodes at different pressure p1 and p2.
\operatorname{Pt}, \mathrm{H}_{2}\left(p_{1}\right)\left|\begin{array}{c}{\mathrm{H}^{+}(\mathrm{aq})} \\ {1 \mathrm{M}}\end{array}\right| \mathrm{H}_{2}\left(p_{2}\right), \mathrm{Pt}
The EMF of the cell is given by:

Option: 1

\mathrm{\frac{R T}{F} \ln \frac{p_{1}}{p_{2}}}


Option: 2

\mathrm{\frac{R T}{2F} \ln \frac{p_{1}}{p_{2}}}


Option: 3

\mathrm{\frac{R T}{F} \ln \frac{p_{2}}{p_{1}}}


Option: 4

\mathrm{\frac{R T}{2F} \ln \frac{p_{2}}{p_{1}}}


Answers (1)

best_answer

For the given concentration cell, the Nernst equation can be written as 

\begin{aligned}\mathrm E_{\text {cell }} &=0.059\left[\mathrm{pH}_{\mathrm{a}}-\mathrm{pH}_{\mathrm{c}}+\frac{1}{2} \log \frac{\left(\mathrm{p}_{\mathrm{H}_{2}}\right)_{\mathrm{a}}}{\left(\mathrm{p}_{\mathrm{H}_{2}}\right)_{\mathrm{c}}}\right] \\ &=0.059\left[0-0+\frac{1}{2} \log \frac{p_{1}}{p_{2}}\right] \\ &=\frac{0.059}{2} \log \frac{p_{1}}{p_{2}}=\frac{R T}{2 F} \ln \frac{p_{1}}{p_{2}} \end{aligned}

Therefore,option(2) is correct

Posted by

vishal kumar

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