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EMF of Electrolyte Concentration Cell Reversible to Cation with Transference.ElectrochemistryPhysical Chemistry - Cepek Media

By Sunil Bhardwaj

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Consider a cell of the type $$\underset { \overset { { t }_{ { H }^{ + } } }{ \underset { { t }_{ Cl^{ - } } }{ \Longleftrightarrow } } }{ \ominus \underset { (1 atm) }{ Pt, { H }_{ 2 } } | \underset { ({ a }_{ 1 }) }{ HCl } | \underset { ({ a }_{ 2 }) }{ HCl } | \underset { (1 atm) }{ { H }_{ 2 }, Pt } \oplus }$$ which consists of two hydrogen gas electrodes 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 cations $${ H }^{ + }$$ ions. Let $${ a }_{ 2 }>{ a }_{ 1 }$$

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. 1gm atom of hydrogen dissolves at the L.H.E. to yield 1gm-ion of $${ H }^{ + }$$ and the same amount of $${ H }^{ + }$$ ions is reduced to form 1gm-atom of hydrogen gas which is liberated at the R.H.E.

Therefore, the reactions at the respective electrodes of the cell can be written as,

At L.H.E oxidation,$$\frac { 1 }{ 2 } { H }_{ 2 }(g) \Longleftrightarrow \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + { e }^{ - }$$ At R.H.E reduction, $$\underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + { e }^{ - } \Longleftrightarrow \frac { 1 }{ 2 } { H }_{ 2 }(g)$$ therefore Net cell reaction $$\overline { \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } \Longleftrightarrow \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } \qquad ....(1) }$$ 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 $${ t }_{ + } \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } \Longleftrightarrow { t }_{ + } \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } }$$ Since $${ t }_{ + } + { t }_{ - } = 1$$ $$\therefore { t }_{ + } = (1 - { t }_{ - })$$ Hence, $$\boxed { (1 - { t }_{ - }) \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } \Longleftrightarrow (1 - { t }_{ - }) \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } \qquad ....(2) }$$ For the transfer of $${ Cl }^{ - }$$ ions. $$\boxed { { t }_{ - } \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow { t }_{ - } \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } } \qquad ....(3) }$$ The net cell reaction is given by addition of equations (1), (2) and (3) i.e. $$\underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + (1 - { t }_{ - }) \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + { t }_{ - } \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + (1 - { t }_{ - }) \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + { t }_{ - } \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } }$$ $$\underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } - { t }_{ - }\underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + { t }_{ - } \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow \underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + \underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } - { t }_{ - }\underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + { t }_{ - } \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } }$$ $$- { t }_{ - }\underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + { t }_{ - } \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow - { t }_{ - }\underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + { t }_{ - } \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } }$$ $${ t }_{ - }\underset { { { ({ a }_{ 2 }) } } }{ { { H }^{ + } } } + { t }_{ - } \underset { { { ({ a }_{ 2 }) } } }{ { Cl^{ - } } } \Longleftrightarrow { t }_{ - }\underset { { { ({ a }_{ 1 }) } } }{ { { H }^{ + } } } + { t }_{ - } \underset { { { ({ a }_{ 1 }) } } }{ { Cl^{ - } } }$$ $$\therefore { t }_{ - }\underset { { { ({ a }_{ 2 }) } } }{ { { HCl } } } \Longleftrightarrow { t }_{ - }\underset { { { ({ a }_{ 1 }) } } }{ { { HCl } } }$$ From this net cell reaction, the Nernst equation for the above cell may be written as, $${ E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { RT }{ nF } \ln { { { \left( \frac { { a }_{ 1 } }{ a_{ 2 } } \right) }^{ { t }_{ - } } } }$$ $${ E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { 2.303RT }{ F } \log { { { \left( \frac { { a }_{ 1 } }{ a_{ 2 } } \right) }^{ { t }_{ - } } } }$$ (Here , n= 1) Since, for concentration cells, $${ E }_{ cell }^{ 0 } = 0V$$ $${ E }_{ cell } = - \frac { 2.303RT }{ F } \log { { { { \left( \frac { { a }_{ 1 } }{ a_{ 2 } } \right) }^{ { t }_{ - } } } } }$$ At 298K, $$\frac { 2.303RT }{ F } =0.0591$$ $$\therefore { E }_{ cell } = - 0.0591 \log { { { \left( \frac { { a }_{ 1 } }{ a_{ 2 } } \right) }^{ { t }_{ - } } } }$$ $${ E }_{ cell } = - 0.0591 \times { t }_{ - } \log { { { \left( \frac { { a }_{ 1 } }{ a_{ 2 } } \right) } } }$$ $$i.e. \boxed { { E }_{ cell } = 0.0591 \times { t }_{ - } \log { { { \left( \frac { { a }_{ 2 } }{ a_{ 1 } } \right) } } } }$$ 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, $${ E }_{ cell } = 0.0591 \times t- \log { { { \left( \frac { { { m }_{ 2 } }^{ 2 }{ { \gamma }_{ 2 } }^{ 2 } }{ { { m }_{ 1 } }^{ 2 }{ { \gamma }_{ 1 } }^{ 2 } } \right) } } }$$ $${ E }_{ cell } = 0.0591 \times t- \log { { { \left( \frac { { { m }_{ 2 } }{ { \gamma }_{ 2 } } }{ { { m }_{ 1 } }{ { \gamma }_{ 1 } } } \right) } }^{ 2 } }$$ $$\boxed { { E }_{ cell } = 0.0591 \times 2t- \log { { { \left( \frac { { { m }_{ 2 } }{ { \gamma }_{ 2 } } }{ { { m }_{ 1 } }{ { \gamma }_{ 1 } } } \right) } } } }$$ For spontaneous cell reaction, $${ a }_{ 2 } > { a }_{ 1 }.$$ It should be noted that if the electrodes of the concentration cell are reversible with respect to cation, the transference number of the anion 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 $$\boxed { { E }_{ cell } = 0.0591 \times 3t- \log { { { \left( \frac { { { m }_{ 2 } }{ { \gamma }_{ 2 } } }{ { { m }_{ 1 } }{ { \gamma }_{ 1 } } } \right) } } } }$$

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