By Sunil Bhardwaj

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Consider a cell of the type $$ \ominus Ag | \underset { ({ a }_{ 1 }) }{ AgN{ O }_{ 3 } } \parallel \underset { ({ a }_{ 2 }) }{ AgN{ O }_{ 3 } } | Ag \oplus $$ which consists of two silver electrodes immersed in two solutions of \(AgN{ O }_{ 3 }\) of activities \({ a }_{ 1 }\) and \({ a }_{ 2 }\). Let \({ a }_{ 2 } > { a }_{ 1 }\). 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 cations, \({ Ag }^{ + }\) ions. The cell reactions are

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

Numerical Problem: Calculate the e.m.f. of the following cell at 298K $$ \ominus Ag | \underset { (0.01m) }{ AgN{ O }_{ 3 } } \parallel \underset { (0.1m) }{ AgN{ O }_{ 3 } } | Ag \oplus $$ Given that \(0.1m AgN{ O }_{ 3 }\) and \(0.01m AgN{ O }_{ 3 }\) solutions are 80% and 90% dissociated.

Solution: From the percentage dissociation of \(AgN{ O }_{ 3 }\) we can first calculate the dissociation per mole.

\(\therefore\) Conc. of \({ Ag }^{ + }\) ion in 0.01m \(AgN{ O }_{ 3 } =\frac { 0.01 \times 90 }{ 100 } = 0.009m \)

Similarly, Conc. of \({ Ag }^{ + }\) ion in 0.1m \(AgN{ O }_{ 3 } =\frac { 0.1 \times 80 }{ 100 } = 0.08m\)

For concentration cell without transference reversible to cations, we have $$ \boxed { { E }_{ cell } = 0.0591 \log { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) } } } $$ $$ { E }_{ cell } = 0.0591 \log { { \left( \frac { 0.08 }{ 0.009 } \right) } } $$ $$ { E }_{ cell } = 0.0591 \log { { \left( 8.89 \right) } } $$ $$ { E }_{ cell } = 0.0591 \times 0.949 $$ $$ { E }_{ cell } = 0.056V $$