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EMF of Electrolyte Concentration Cell Reversible to Anion without Transference.ElectrochemistryPhysical Chemistry - Cepek Media

By Sunil Bhardwaj

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Consider a cell of the type $$\ominus Ag(s) | AgCl(s), \underset { ({ a }_{ 1 }) }{ HCl } \parallel \underset { ({ a }_{ 2 }) }{ HCl } AgCl(s), | Ag(s) \oplus$$ which consists of two metal-insoluble salt type electrodes (silver-silver chloride). The two HCl solutions of activities $${ a }_{ 1 }$$ and $${ a }_{ 2 }$$ are used. Let $${ a }_{ 1 }>{ a }_{ 2 }$$. Since the two solutions are connected by a salt bridge, there is negligible transfer of electrolyte at the liquid junction. The use of salt bridge is represented by the double line put between the two half cells. This cell is reversible with anions, $$Cl^{ - }$$ ions.

The cell reactions are

At L.H.E oxidation, $$\underset { { (s) } }{ { Ag } } + \underset { { ({ a }_{ 1 }) } }{ Cl^{ - } } \Longleftrightarrow \underset { { { (s) } } }{ { Ag }Cl } + { e }^{ - }$$ At R.H.E reduction, $$\underset { { { (s) } } }{ { Ag }Cl } + { e }^{ - } \Longleftrightarrow \underset { { s } }{ { Ag } } + \underset { { ({ a }_{ 2 }) } }{ Cl^{ - } }$$ therefore Net cell reaction $$\overline { \underset { { ({ a }_{ 1 }) } }{ Cl^{ - } } \Longleftrightarrow \underset { { ({ a }_{ 2 }) } }{ Cl^{ - } } }$$ The Nernst equation for the above cell may be written as $${ E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { RT }{ nF } \ln { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) } }$$ $${ E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { 2.303RT }{ F } \log { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) } }$$ (Here, the number of electrons involved in the electrode reactions. n=1) Since, for concentration cells, $${ E }_{ cell }^{ 0 } = 0V$$ $${ E }_{ cell } = - \frac { 2.303RT }{ F } \log { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) } }$$ At 298K, $$\frac { 2.303RT }{ F } =0.0591$$ $$\therefore { E }_{ cell } = - 0.0591 \log { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) } }$$ $$i.e. \boxed { { E }_{ cell } = 0.0591 \log { { \left( \frac { { a }_{ 1 } }{ a_{ 2 } } \right) } } }$$ Now the activity of single ion is given by $$a = m\gamma$$ where m = molality of ion and $$\gamma$$ = activity coefficient. $$\therefore { a }_{ 1 } = { m }_{ 1 }{ \gamma }_{ 1 } \ and \ { a }_{ 2 } = { m }_{ 2 }{ \gamma }_{ 2 }$$ $$i.e. \boxed { { E }_{ cell } = 0.0591 \log { { \left( \frac { { m }_{ 1 }{ \gamma }_{ 1 } }{ { m }_{ 2 }{ \gamma }_{ 2 } } \right) } } }$$ For spontaneous cell reaction, $${ a }_{ 2 } > { a }_{ 1 }$$.

Numerical problem: Calculate the e..m.f. of the following cell at 298K $$\ominus Ag(s) | AgCl(s), \underset { m = 0.1 \gamma = 0.90 }{ HCl } \parallel \underset { m = 0.05 \gamma = 0.98 }{ HCl } AgCl(s), | Ag(s) \oplus$$

Solution: The given cell is concentration cell without transference reversible to anions. Therefore, $$\boxed { { E }_{ cell } = 0.0591 \log { { \left( \frac { { m }_{ 1 }{ \gamma }_{ 1 } }{ { m }_{ 2 }{ \gamma }_{ 2 } } \right) } } }$$ $${ E }_{ cell } = 0.0591 \log { { \left( \frac { { 0.1 \times 0.90 } }{ 0.05 \times 0.98 } \right) } }$$ $${ E }_{ cell } = 0.0591 \log { { \left( \frac { 0.09 }{ 0.049 } \right) } }$$ $${ E }_{ cell } = 0.0591 \log { { \left( 1.837 \right) } }$$ $${ E }_{ cell } = 0.0591 \times 0.264$$ $${ E }_{ cell } = 0.0156V$$

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