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Derive an expression for the EMF of Electrode Concentration Cell Reversible to Cations (cell with gas electrode).ElectrochemistryPhysical Chemistry - Cepek Media

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 $$ \overline { \frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 1 } \right) \Longleftrightarrow \frac { 1 }{ 2 } { H }_{ 2 }\left( { P }_{ 2 } \right) } $$ $$ \boxed { { E }_{ cell } = \frac { 0.0591 }{ 2 } \log { { \left( \frac { { P }_{ 1 } }{ P_{ 2 } } \right) } } } $$ $$ { E }_{ cell } = \frac { 0.0591 }{ 2 } \log { { \left( \frac { 400 }{ 200 } \right) } } $$ $$ { E }_{ cell } = \frac { 0.0591 }{ 2 } \log { { \left( 2 \right) } } $$ $$ { E }_{ cell } = 0.02955 \times 0.301 = 8.895 \times { 10 }^{ -3 }V $$