\[\Delta_rG = -nFE,\qquad \Delta_rS = nF\left(\frac{\partial E}{\partial T}\right)_p,\qquad \Delta_rH = -nFE + nFT\left(\frac{\partial E}{\partial T}\right)_p\]
含义Meaning
可逆电池能做的最大电功等于反应吉布斯能的减少。测定电动势及其温度系数，即可求得反应的 ΔG、ΔS、ΔH。
The maximum electrical work of a reversible cell equals the decrease in Gibbs energy; measuring EMF and its temperature coefficient yields ΔG, ΔS and ΔH.
符号说明Symbols
| 符号Symbol | 含义 Meaning | 单位 Unit |
|---|---|---|
| \(n\) | 反应转移电子数electrons transferred | 1 |
| \((\partial E/\partial T)_p\) | 电动势温度系数temperature coefficient of EMF | V·K⁻¹ |
推导Derivation
恒温恒压下非体积功的最大值 \(W_\text{max} = -\Delta G\)。
At constant T and p the maximum non-expansion work is −ΔG.
电功 = 电量 × 电势差 = nFE，所以 \(\Delta G = -nFE\)。
Electrical work nFE gives ΔG = −nFE.
由 \((\partial\Delta G/\partial T)_p = -\Delta S\) 得 ΔS，再由 ΔH = ΔG + TΔS 得 ΔH。
Differentiate for ΔS; then ΔH = ΔG + TΔS.
适用条件与局限Assumptions & limitations
要求可逆（零电流）测量；实际放电时 E 降低，部分能量成为热。
Requires reversible (zero-current) conditions.
例题Worked example
丹尼尔电池 E° = 1.10 V：ΔG° = −2×96485×1.10 = −212 kJ/mol。
Daniell cell: ΔG° = −212 kJ/mol.