The role of temperature in Nernst equation

Question

When measuring a non-standard voltaic cell’s electrode potential, regardless the standard conditions, as long as the concentration of product and reactant the same, a non-standard cell’s potential will be exactly equal to a standard cell’s potential.

$E= E^\circ -\frac{RT}{nF} \ln(Q)$

Q = 1 if the concentration of product equals to the concentration of reactant:

$E= E^\circ -\frac{RT}{nF} \ln(1)$

$E=E^\circ - 0$

In another word, no matter how temperature varies, as long as the concentration of both solution the same, temperature will have no effect on the voltage.

Is the understanding above valid? If it is, then my experimental result contradicted the Nernst equation. When I increase the temperature while having two half cell with the exact same concentration, the voltage increased as well. What is a possible cause for such result? (Assuming my voltemeter is funcotional one)

Answer 1

The standard cell potential is temperature-dependent itself, just like the Gibbs free energy. The easiest way to see this is to write the Gibbs free energy in terms of enthalpy, entropy and temperature:

$$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$

This shows you that $\Delta G^\circ$ is strongly temperature dependent (even if $ \Delta H^\circ$ and $\Delta S^\circ$ were independent of temperature, which they are not).

To relate this to the cell potential, remember that $\Delta G^\circ$ is a measure of the maximal work a system can do, and that the electrical work is electrical potential times charge moved across that potential.

$$\Delta G^\circ = w_\text{electrical} = -E^\circ n F$$

So $E^\circ$ is temperature dependent as well.