# Plotting ΔG vs Temp for Electrochemical Reaction

I’m doing an electrochemistry lab that requires me to construct various galvanic cells in an attempt to solve for $\Delta H$, $\Delta G$, and $\Delta S$ values of an unknown. During one part of the experiment, I had to heat a cell to a temperature of 70 °C and record its voltage in 15 degree increments until the temperature reached 10 °C.

The cell I am using to perform this step on is represented by the equation:

$\ce{Zn + Sn^2+ -> Zn^2+ + Sn}$

After performing the procedure, I’m getting concerned that my data may be inaccurate. Theoretically, I was using 0.1 M concentrations of each solution in the cells. According to the Nernst equation, it seems that the voltage or potential of the cell should not be affected by a change in temperature if the concentrations are equal because $\ln(0.1/0.1) = 0$.

However, according to my professor, I should be seeing a linear relationship when I plot $\Delta G$, calculated from the voltage measurements I took, vs the temperature in K.

My question is, theoretically, what should happen to the voltage as I heat the cell, assuming that the concentrations of the solutions were off? I’ve calculated the $\Delta S$ value based on the standard values listed for these entities and I’m getting a value of −85.18 J/(K mol).

Calculation: $\Delta S = (-112.1 + 51.55) - (41.63 + -17.0) = -85.18\ \mathrm{J/(mol\ K)}$

Based off of this result, it would appear that the slope of this plot must come out to be a positive value because $m = - \Delta S$. Does this mean that I should be seeing an increase in voltage as temperature decreases? Regarding the $\Delta H$ of this reaction, which is represented by the Y-intercept (b) of the equation of the best fit line for my data, it appears I must also have a negative value of −145090 (in units of J) based upon the calculation I did for this, also using standard values.

Calculation: $\Delta H = (-153.89 + 0) - (0 + -8.8) = -145.09\ \mathrm{kJ/mol}$

I’ve tried to be as detailed as possible in describing by question, and hopefully it makes sense. I’m just getting confused on what the data should look like. Unfortunately, its not possible for me to redo the experiment. Please note that this is for my own understanding, I will not submit the assignment with fixed data. I appreciate any assistance that anyone can offer me.

My question is, theoretically, what should happen to the voltage as I heat the cell, assuming that the concentrations of the solutions were off?

This question is tough to answer because I don't know what you mean by "off". So let me ask (and answer) a few related questions that I think will help you.

## What should happen to voltage as a function of temperature in an electrochemical cell

The electrical potential of electrochemical cells varies with temperature, even when the cell is at standard conditions. A 1985 NIST document explains the situation well. Only the standard hydrogen electrode has a standard potential that does not vary with temperature. (The SHE potential is defined as 0 volts at every temperature.) For other cells, usually the assumption of a linear variation with temperature is usually very good. That is,

$${\cal E}_T^\circ = {\cal E}_{298}^\circ + (T-298.15)\cdot \frac{d{\cal E}_{298}^\circ}{dT}$$

Note that this variation is completely unrelated to the Nernst equation. The same reference gives values for $\frac{d{\cal E}_{298}^\circ}{dT}$ for zinc(II)/zinc metal of 0.119 milliVolts per Kelvin and for tin(II)/tin metal -0.32 milliVolts per Kelvin.

## Don't be confused by the $T$ in the Nernst equation

The Nernst equation is

$${\cal E}= {\cal E}^\circ + \frac{RT}{nF}\ln Q$$

It does contain a $T$ in there, but don't be mislead into thinking that the $T$ you see is the sole means by which a cell potential ${\cal E}$ can vary with temperature. ${\cal E}^\circ$ also varies with temperature (see above).

According to the Nernst equation, it seems that the voltage or potential of the cell should not be affected by a change in temperature if the concentrations are equal because ln(0.1/0.1)=0.

Thus, hopefully my answer makes clear that this statement is wrong, because ${\cal E}^\circ$ varies with temperature.