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  4. E text cell of the following cell is P t(s) mid H2(g), bar | H +( l M ) | H +(0.1 M )| H 2(g), 1 bar mid Pt (s)

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Q. $E_{\text {cell }}$ of the following cell is $P t(s) \mid H_{2}(g)$, bar $\left| H ^{+}( l M ) \| H ^{+}(0.1 M )\right| H _{2}(g), 1$ bar $\mid Pt (s)$

KEAMKEAM 2019Electrochemistry

Solution:

$(a) \operatorname{Pt}(s) \mid H _{2}(g), 1$ bar $\left| H ^{+}(1\, M ) \| H ^{+}(0.1\, M )\right|$

$H _{2}(g), 1$ bar $\mid \operatorname{Pt}(s)$

Half cell corresponding to the reaction are At cathode,

$H ^{+}(a q)+e^{-} \longrightarrow \frac{1}{2} H_{2}$

At anode, $\frac{1}{2} H_{2} \longrightarrow H ^{+}(a q)+e^{-}$

Nernst equation, $E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2303\, R T}{n F} \log \left[\frac{1}{ H ^{+}}\right]$

$E_{\text {cell }}=0-\frac{2303\, R T}{F} \log \left[\frac{1}{0.1}\right]$

$=-\frac{2303\, R T}{F} \log\, 10$

$E_{\text {cell }}=-\frac{2303\, R T}{F}$