By Sunil Bhardwaj

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Such a cell consists of two hydrogen gas electrodes operating at different pressures dipping in to a common solution of hydrochloric acid e.g. $$\underset { { P }_{ 1 } }{ Pt, { H }_{ 2 }\left( g \right) } | \underset { aq.soln }{ HCl } | \underset { { P }_{ 2 } }{ { H }_{ 2 }\left( g \right) , Pt }$$ This cell is reversible to cations $${ H }^{ + }$$ ions. Let the activity of $${ H }^{ + }$$ ions in solution be $${ a }_{ + }$$. The cell reactions are as follows:

At L.H.E oxidation, $$\frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 1 } \right) \Longleftrightarrow \underset { { a }_{ + } }{ H^{ + } } + { e }^{ - }$$ At R.H.E reduction, $$\underset { { a }_{ + } }{ { H }^{ + } } + { e }^{ - } \Longleftrightarrow \frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 2 } \right)$$ therefore Net cell reaction $$\overline { \frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 1 } \right) \Longleftrightarrow \frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 2 } \right) }$$ This reaction is independent of the concentration of the electrolyte. The Nernst equation for the above cell may be written as $${ E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { RT }{ nF } \ln { { \left( \frac { { P }_{ 2 } }{ P_{ 1 } } \right) }^{ \frac { 1 }{ 2 } } }$$ $${ E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { 2.303RT }{ F } \log { { \left( \frac { { P }_{ 2 } }{ P_{ 1 } } \right) }^{ \frac { 1 }{ 2 } } }$$ (Here, the number of electrons involved in the electrode reactions. n=1) Since, for concentration cells, $${ E }_{ cell }^{ 0 } = 0V$$ $${ E }_{ cell } = - \frac { 2.303RT }{ F } \log { { \left( \frac { { P }_{ 2 } }{ P_{ 1 } } \right) }^{ \frac { 1 }{ 2 } } }$$ $${ E }_{ cell } = - \frac { 2.303RT }{ 2F } \log { { \left( \frac { { P }_{ 2 } }{ P_{ 1 } } \right) } }$$ At 298K, $$\frac { 2.303RT }{ F } =0.0591$$ $$\therefore { E }_{ cell } = - \frac { 0.0591 }{ 2 } \log { { \left( \frac { { P }_{ 2 } }{ P_{ 1 } } \right) } }$$ $$i.e. \boxed { { E }_{ cell } = \frac { 0.0591 }{ 2 } \log { { \left( \frac { { P }_{ 1 } }{ P_{ 2 } } \right) } } }$$ For spontaneous cell reaction (i.e. to have positive cell e.m.f), $${ P }_{ 1 } > P_{ 2 }$$.

Numerical problem: Calculate the e.m.f. of the following electrode concentration cell at 298K. $$\underset { P_{ 1 } = 400torr }{ Pt, { H }_{ 2 } } | \underset { aq.soln }{ HCl } | \underset { { P }_{ 2 } = 200torr }{ { H }_{ 2 }, Pt }$$

Solution: The cell reactions are

At L.H.E oxidation, $$\frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 1 } \right) \Longleftrightarrow \underset { { a }_{ + } }{ H^{ + } } + { e }^{ - }$$ At R.H.E reduction,$$\underset { { a }_{ + } }{ { H }^{ + } } + { e }^{ - } \Longleftrightarrow \frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 2 } \right)$$ therefore Net cell reaction $$\overline { \frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 1 } \right) \Longleftrightarrow \frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 2 } \right) }$$ $$\boxed { { E }_{ cell } = \frac { 0.0591 }{ 2 } \log { { \left( \frac { { P }_{ 1 } }{ P_{ 2 } } \right) } } }$$ $${ E }_{ cell } = \frac { 0.0591 }{ 2 } \log { { \left( \frac { 400 }{ 200 } \right) } }$$ $${ E }_{ cell } = \frac { 0.0591 }{ 2 } \log { { \left( 2 \right) } }$$ $${ E }_{ cell } = 0.02955 \times 0.301 = 8.895 \times { 10 }^{ -3 }V$$

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