By Sunil Bhardwaj

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Consider a cell of the type $$ \underset { \overset { { t }_{ { H }^{ + } } }{ \underset { { t }_{ Cl^{ - } } }{ \Longleftrightarrow } } }{ \ominus \underset { (s) }{ Ag } | AgCl(s) \underset { ({ a }_{ 1 }) }{ HCl } | \underset { ({ a }_{ 2 }) }{ HCl } AgCl(s) | \underset { (s) }{ { Ag } } \oplus } $$ this cell consists of two metal-insoluble salt type electrodes (i.e. silver-silver chloride) in contact with HCI solutions of different concentrations or activities. The two solutions are in direct contact with each other through porous partition. This cell is reversible to anions \(Cl^{ - }\) ions. Let \({ a }_{ 1 }>{ a }_{ 2 }\) As the two solutions are in direct contact, the transfer of electrolyte takes place from one compartment to another. Hence, there will be a liquid junction potential. When one faraday of electricity passes through the cell. The following reactions occure at the respective electrodes of the cell,

At L.H.E oxidation, $$ Ag(s) + \underset { { a }_{ 1 } }{ { Cl }^{ - } } \Longleftrightarrow AgCl(s) + { e }^{ - } $$ At R.H.E reduction,$$ AgCl(s) + { e }^{ - } \Longleftrightarrow Ag(s) + \underset { { a }_{ 2 } }{ { Cl }^{ - } } $$ therefore Net cell reaction$$ \overline { \underset { { a }_{ 1 } }{ { Cl }^{ - } } \Longleftrightarrow \underset { { a }_{ 2 } }{ { Cl }^{ - } } \qquad ....(1) } $$ The electrons liberated move in the external circuit from L.H.E. to R.H.E. To complete the circuit similar charges must pass from right to left through the liquid junction. These charges are carried by the ions of the solution. As \({ Cl }^{ - }\) ions carry negative charge they move in the direction of electrons. The \({ H }^{ + }\) ions carry positive charge and thus they move in the opposite direction. Thus, while the current is passing \({ t }_{ + }\) gm-ions of \({ H }^{ + }\) ions will migrate across the junction between the two solutions from left to right and \({ t }_{ - }\) gm-ions of \({ Cl }^{ - }\) ions will move in the opposite direction. Here\( { t }_{ + }\) and \({ t }_{ - }\) are the transference numbers of\( { H }^{ + }\) and \({ Cl }^{ - }\) ions respectively. Therefore, \({ t }_{ + }\) faraday of charge will be carried by \({ t }_{ + }\) gm-ion of \({ H }^{ + }\) from left to right and \({ t }_{ - }\) faraday of charge will be carried by \({ t }_{ - }\) gm-ion of \({ Cl }^{ - } \)from right to left. These reactions can be written as follows: For the transfer of \({ H }^{ + }\) ions $$ \boxed { { t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } \Longleftrightarrow { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } \qquad ....(2) } $$ For the transfer of \({ Cl }^{ - }\) ions. $$ { t }_{ - } \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow { t }_{ - } \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } $$ Since \({ t }_{ + } + { t }_{ - } = 1\) $$ \therefore { t }_{ - } = (1 - { t }_{ + })$$ Hence, $$ \boxed { (1 - { t }_{ + }) \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow (1 - { t }_{ + }) \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } \qquad ....(3) } $$ The net cell reaction is given by addition of equations (1), (2) and (3) $$ i.e. \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } + { t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + (1 - { t }_{ + }) \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } + { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + (1 - { t }_{ + }) \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } $$ $$ \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } + { t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } - { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } + { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } - { t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } $$ $$ { t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } - { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } - { t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } $$ $$ { t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + { t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } \Longleftrightarrow { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } $$ $$ { t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { { HCl } } } \Longleftrightarrow { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { { HCl } } } $$ From this net cell reaction, the Nernst equation for the above cell may be written as, $$ { E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { RT }{ nF } \ln { { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) }^{ { t }_{ + } } } } $$ $$ { E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { 2.303RT }{ F } \log { { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) }^{ { t }_{ + } } } } $$ (Here , n= 1) Since, for concentration cells, \({ E }_{ cell }^{ 0 } = 0V\) $$ { E }_{ cell } = - \frac { 2.303RT }{ F } \log { { { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) }^{ { t }_{ + } } } } } $$ At 298K, \(\frac { 2.303RT }{ F } =0.0591\) $$ \therefore { E }_{ cell } = - 0.0591 \log { { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) }^{ { t }_{ + } } } } $$ $$ { E }_{ cell } = - 0.0591 \times { t }_{ + } \log { { { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) } } } } $$ $$ i.e. \boxed { { E }_{ cell } = 0.0591 \times { t }_{ + } \log { { { \left( \frac { { a }_{ 1 } }{ a_{ 2 } } \right) } } } } $$ Since, HCl is uni-univalent electrolyte, $$ { a }_{ 1 } = { { m }_{ 1 } }^{ 2 }{ { \gamma }_{ 1 } }^{ 2 } and { a }_{ 2 } = { { m }_{ 2 } }^{ 2 }{ { \gamma }_{ 2 } }^{ 2 }$$ where m and \(\gamma\) represent molality and activity coefficient. Thus, $${ E }_{ cell } = 0.0591 \times { t }_{ + } \log { { { \left( \frac { { { m }_{ 1 } }^{ 2 }{ { \gamma }_{ 1 } }^{ 2 } }{ { { m }_{ 2 } }^{ 2 }{ { \gamma }_{ 2 } }^{ 2 } } \right) } } } $$ $$ { E }_{ cell } = 0.0591 \times { t }_{ + } \log { { { \left( \frac { { { m }_{ 1 } }{ { \gamma }_{ 1 } } }{ { { m }_{ 2 } }{ { \gamma }_{ 2 } } } \right) } }^{ 2 } } $$ $$ \boxed { { E }_{ cell } = 0.0591 \times 2{ t }_{ + } \log { { { \left( \frac { { { m }_{ 1 } }{ { \gamma }_{ 1 } } }{ { { m }_{ 2 } }{ { \gamma }_{ 2 } } } \right) } } } } $$ For spontaneous cell reaction, \({ a }_{ 1 } > { a }_{ 2 }.\) It should be noted that if the electrodes of the concentration cell are reversible with respect to anions, the transference number of the cations gets involved in the equation. Note: If the electrolyte used in the above cell is bi-univalent or uni-bivalent, then $$ { a }_{ 1 } = 4{ { m }_{ 1 } }^{ 3 }{ { \gamma }_{ 1 } }^{ 3 } \ and \ { a }_{ 2 } = 4{ { m }_{ 2 } }^{ 3 }{ { \gamma }_{ 2 } }^{ 3 }$$ Hence the expression for e.mf. of this cell becomes $$ \boxed { { E }_{ cell } = 0.0591 \times 3{ t }_{ + } \log { { { \left( \frac { { { m }_{ 1 } }{ { \gamma }_{ 1 } } }{ { { m }_{ 2 } }{ { \gamma }_{ 2 } } } \right) } } } } $$