The electrode potential and the emf of the cell depend upon the nature of the electrode, temperature and the activities (concentrations) of the ions in solution. The variation of electrode and cell potentials with concentration of ions in solution can be obtained from thermodynamic considerations. For a general reaction such as
`M_1A + m_2B ..... n_1X + n_2Y + .... .......(i)`
Occurring in the cell, the Gibbs free energy change is given by the equation
`G = Delta G^o + 2.303 RT log _(10) ((a_x^(n1) xx a_y^(n2))/(a_A^(m1) xx a_B^(m2))) ......(ii)`
where 'a' represents the activities of reactants and products under a given set of conditions and
`DeltaG^o` refers to free energy change for the reaction when the various reactants and products are present at standard conditions.
The free energy change of a cell reaction is related to the electrical work that can be obtained from the cell, i.e.,
`DeltaG^o = -nFE_(cell)` and `DeltaG^o = -nFE^o`.
On substituting these values in Eq. (ii) we get
`nFE_(text(cell)) = -nFE^o + 2.303 RT log _(10) ((a_x^(n1) xx a_y^(n2))/(a_A^(m1) xx a_B^(m2))) ......(iii)`
`E_(text(cell)) = E^o + 2.303 RT log _(10) ((a_x^(n1) xx a_y^(n2))/(a_A^(m1) xx a_B^(m2))) ......(iv)`
This equation is known as Nernst equation.
Putting the values of
•R=8.314 J `K^(-1)` `text(mol)^(-1)` ,
•T = 298 K and
•F=96500 C,
Eq. (iv) reduces to
`E_(text(cell)) = E^o -(0.0591)/n log _(10) ((a_x^(n1) xx a_y^(n2))/(a_A^(m1) xx a_B^(m2))) ......(v)`
`E_(text(cell)) = E^o -(0.0591)/n log _(10) (([text(Product)])/([text(Reactant)]))...(vi)`
`text[Potential of single electrode (Anode)]` :
Consider the general oxidation reaction,
`M → M^(n+) + n e^(-)`
Applying Nernst equation
`E_(text(ox)) = E_(text(ox))^o - 0.059/n log_(10) (([M^(n+)])/[M])`
where `E_(text(ox))` is the oxidation potential of the electrode (anode), is the standard oxidation potential of the electrode.
Note : The concentration of pure solids and liquids are taken as unity.
`E_(text(ox)) = E_(text(ox))^o - 0.059/n log_(10) [M^(n+)]`
Let us consider a Daniell cell to explain the above equations. The concentrations of the electrolytes are not 1 M.
`Zn(s)+Cu^(2+)(aq) ⇋ Zn^(2+)(aq) + Cu(s)`
Cell Representation: `Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)|Cu`
`text[Potential at zinc electrode (Anode)]`
`E_(text(ox)) = E_(text(ox))^o - 0.059/n log_(10) [Zn^(2+)]`
`text[Potential at copper electrode (Cathode)]`
`E_(text(red)) = E_(text(red))^o - 0.059/n log_(10) [Cu^(2+)]`
Emf of the cell
`E_(text(cell)) = E_(text(ox)) + E_(text(red))`
`E_(text(cell)) = E_(text(ox))^o + E_(text(red))^o - 0.059/n ([Zn^(2+)])/([Cu^(2+)])`
The value of `n = 2` for both zinc and copper.
Let us consider an example, in which the values of n for the two ions in the two half-cells are not same. For example, in the cell
`Cu|Cu^(2+)||Ag^(+)|Ag`
The cell reaction is
`Cu(s) + 2Ag^(+) → Cu^(2+) + 2Ag`
The two half-cell reaction are:
`Cu → Cu^(2+) + 2e^(-)`
`Ag^(+) + e^(-) → Ag`
The second equation is multiplied by `2` to balance the number of electrons.
`2Ag^(+) + 2e^(-) → 2 Ag`
`E_(text(ox)) = E_(text(ox))^o - 0.059/2 log_(10) [Cu^(2+)]`
`=> E_(text(red)) = E_(text(red))^o - 0.059/2 log_(10)[Ag^(+)]^2`
`E_(text(cell)) = E_(text(ox)) + E_(text(red))`
`E_(text(cell)) = (0.0591)/2 log_(10) ([Cu^(2+)]/[Ag^(+)]^2) `
All the EMFs that have been used are reduction potentials in volts. Whatever be the value of reduction potentials, substitute them as such, taking into account the signs.
The electrode potential and the emf of the cell depend upon the nature of the electrode, temperature and the activities (concentrations) of the ions in solution. The variation of electrode and cell potentials with concentration of ions in solution can be obtained from thermodynamic considerations. For a general reaction such as
`M_1A + m_2B ..... n_1X + n_2Y + .... .......(i)`
Occurring in the cell, the Gibbs free energy change is given by the equation
`G = Delta G^o + 2.303 RT log _(10) ((a_x^(n1) xx a_y^(n2))/(a_A^(m1) xx a_B^(m2))) ......(ii)`
where 'a' represents the activities of reactants and products under a given set of conditions and
`DeltaG^o` refers to free energy change for the reaction when the various reactants and products are present at standard conditions.
The free energy change of a cell reaction is related to the electrical work that can be obtained from the cell, i.e.,
`DeltaG^o = -nFE_(cell)` and `DeltaG^o = -nFE^o`.
On substituting these values in Eq. (ii) we get
`nFE_(text(cell)) = -nFE^o + 2.303 RT log _(10) ((a_x^(n1) xx a_y^(n2))/(a_A^(m1) xx a_B^(m2))) ......(iii)`
`E_(text(cell)) = E^o + 2.303 RT log _(10) ((a_x^(n1) xx a_y^(n2))/(a_A^(m1) xx a_B^(m2))) ......(iv)`
This equation is known as Nernst equation.
Putting the values of
•R=8.314 J `K^(-1)` `text(mol)^(-1)` ,
•T = 298 K and
•F=96500 C,
Eq. (iv) reduces to
`E_(text(cell)) = E^o -(0.0591)/n log _(10) ((a_x^(n1) xx a_y^(n2))/(a_A^(m1) xx a_B^(m2))) ......(v)`
`E_(text(cell)) = E^o -(0.0591)/n log _(10) (([text(Product)])/([text(Reactant)]))...(vi)`
`text[Potential of single electrode (Anode)]` :
Consider the general oxidation reaction,
`M → M^(n+) + n e^(-)`
Applying Nernst equation
`E_(text(ox)) = E_(text(ox))^o - 0.059/n log_(10) (([M^(n+)])/[M])`
where `E_(text(ox))` is the oxidation potential of the electrode (anode), is the standard oxidation potential of the electrode.
Note : The concentration of pure solids and liquids are taken as unity.
`E_(text(ox)) = E_(text(ox))^o - 0.059/n log_(10) [M^(n+)]`
Let us consider a Daniell cell to explain the above equations. The concentrations of the electrolytes are not 1 M.
`Zn(s)+Cu^(2+)(aq) ⇋ Zn^(2+)(aq) + Cu(s)`
Cell Representation: `Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)|Cu`
`text[Potential at zinc electrode (Anode)]`
`E_(text(ox)) = E_(text(ox))^o - 0.059/n log_(10) [Zn^(2+)]`
`text[Potential at copper electrode (Cathode)]`
`E_(text(red)) = E_(text(red))^o - 0.059/n log_(10) [Cu^(2+)]`
Emf of the cell
`E_(text(cell)) = E_(text(ox)) + E_(text(red))`
`E_(text(cell)) = E_(text(ox))^o + E_(text(red))^o - 0.059/n ([Zn^(2+)])/([Cu^(2+)])`
The value of `n = 2` for both zinc and copper.
Let us consider an example, in which the values of n for the two ions in the two half-cells are not same. For example, in the cell
`Cu|Cu^(2+)||Ag^(+)|Ag`
The cell reaction is
`Cu(s) + 2Ag^(+) → Cu^(2+) + 2Ag`
The two half-cell reaction are:
`Cu → Cu^(2+) + 2e^(-)`
`Ag^(+) + e^(-) → Ag`
The second equation is multiplied by `2` to balance the number of electrons.
`2Ag^(+) + 2e^(-) → 2 Ag`
`E_(text(ox)) = E_(text(ox))^o - 0.059/2 log_(10) [Cu^(2+)]`
`=> E_(text(red)) = E_(text(red))^o - 0.059/2 log_(10)[Ag^(+)]^2`
`E_(text(cell)) = E_(text(ox)) + E_(text(red))`
`E_(text(cell)) = (0.0591)/2 log_(10) ([Cu^(2+)]/[Ag^(+)]^2) `
All the EMFs that have been used are reduction potentials in volts. Whatever be the value of reduction potentials, substitute them as such, taking into account the signs.