# Physical Chemistry: Calculating ΔG, ΔH, ΔS on a given equation

This question was brought in class, but I have no clue if my answer is correct or not.

I wonder if you guys can double check my answer.

Between $\pu{0 ^\circ C}$ and $90 ^\circ C$, the potential of a cell with $n=2$, is given by: $$\frac{E}{v} = 0.35510 - 0.3422\times 10^{-4} T - 3.2347\times 10^{-6}T^2$$

where $T$ is the Celsius temperature. Calculate $\Delta G, \Delta H, \Delta S$ for the cell at $50 ° C$.

\begin{align} E &= 0.35510 - 0.3422\times 10^{-4}\cdot 50 - 3.2347\times 10^{-6} (50)^2 \\ &= 0.35510 - 1.711\times 10^{-3} - 8.08675\times 10^{-3} \\ &= \pu{0.3453 V}\\[2ex] \left(\frac{\partial\Delta G}{\partial T} \right)_P &= -\Delta S\\ &= -nF\left( \frac{\partial E}{\partial T}\right)_P \\ &= -nF \cdot (-0.3422\cdot 10^{-4} -6.4694\cdot10^{-6}\cdot T) \\ \text{at $50 °C$,} \implies &= nF\cdot(3.5769\cdot10^{-4}) \\ &= 69.02 = -\Delta S\\ \therefore \Delta S &= \pu{-69.02 J//mol} \\[2ex] \Delta G &= nFE \\ &= -2\cdot \pu{96485 C//mol}\cdot \pu{0.3453 V}\\ &= \pu{-66633 kJ//mol}\\[2ex] \Delta H &= \Delta G + T\Delta S\\ &= \pu{-66633 kJ//mol} + (\pu{50 ^\circ C} + \pu{273.15 K})\cdot(\pu{-69.02 J//mol})\\ &= \pu{88.926 kJ} \end{align}
• Method/numbers seems to be ok except for little typo; it should be $\Delta G =-nFE$. – porphyrin May 9 '17 at 7:38