Gibbs-Helmholtz Equation for the emf of a Reversible Cell

A reversible cell is one which may be restored to its original condition, after discharge, by passing a reverse current through it. The energy required for charging would be exactly equal to that obtained from the cell during discharge. Helmhomtz said that the emf of a cell should depend on the heat produced during the chemical reactions taking place within it and also on the heating effects produced in the cell. He applied the laws of thermodynamics to a reversible cell and obtained an expression for its emf.

Consider a cell of emf E at an absolute temperature T. Let a small charge q be drawn from the cell. The line AB on the emf-charge graph represents its passage occurring isothermally at temperature T. During this process emf remains constant.

Let a further small quantity of charge dq pass through the cell adiabatically. A certain amount of energy is absorbed from the cell. Hence, there is a fall of temperature δT.  The new emf of the cell is (E – dE/dT Dt). This adiabatic change is represented by the line BC.

Let a charge q now pass in the revere direction isothermally at temperature T – dT. This is shown by Cd.

Finally, let the cell be brought back to its original state by passing a charge dq adiabatically through the cell. The line DA represents this adiabatic change.

The work done along AB = Eq
The work done long CD = (E – dT dE/dT)q
The work done during the adiabatic changes BC and DA are neglected.
The net work done by the cell during a complete cycle is
    W = Eq – (E – dT dE/dT)q = q dT dE/dT

Let h be the heat absorbed from the source at the higher temperature T.
Hence     q dE/dT dT = dT            (from thermo-dynamics)
             _____ h             T
or        h = qTdE/dT

Now h is the energy drawn from the cell in the form of heat. Let H represent the energy liberated chemically by the passage of one coulomb.
Then the energy liberated corresponding to a charge q is Hq.
The total energy provided by the cell is
        Eq    = Hq + h
            = Hq + qTdE/dT
        E    = H + T dE/dT
This is the Gibbs-Helmholtz equation,
In the case of Daniel Cell (dE/dT) is 0. Hence, E = H.
If (dE/dT) is + ve, E increases as T increases. The increase is caused by drawing heat form the cell. The will result in cooling of the cell.
Similarly if dE/dT is –ve, the cell will heat up on delivering a current.