热力学与电化学

Entropy 熵 $S$ measures the dispersal 分散 of energy and matter – loosely, the number of ways to arrange a system. Entropy usually increases on melting or vaporisation 汽化 and when accessible gas states increase. Dissolving a solid can increase or decrease the system entropy because solvation 溶剂化 also affects particle arrangements. Compare the actual states and conditions.

Every substance has a positive absolute entropy 绝对熵 $S^\circ$. For a reaction,

Predict its sign from the change in gas moles: making more gas raises entropy ($\Delta S>0$).

Gibbs free energy 吉布斯自由能 combines enthalpy and entropy:

At constant temperature and pressure, a process is thermodynamically favorable 热力学有利 when $\Delta G<0$. So exothermic ($\Delta H<0$) and entropy-increasing ($\Delta S>0$) reactions are always favorable; when the two oppose, temperature decides.

Worked example. A reaction has $\Delta H=+40\ \text{kJ/mol}$ and $\Delta S=+120\ \text{J/(mol K)}$. It is endothermic (unfavorable enthalpy) but entropy-increasing, so it becomes favorable only when hot enough. Setting $\Delta G=\Delta H-T\Delta S<0$ and matching units ($\Delta S=0.120\ \text{kJ/(mol K)}$):

$\Delta G<0$ says a reaction can happen, not that it will happen fast. A reaction can be thermodynamically favorable yet kinetically slow because of a high activation energy 活化能 (diamond → graphite). Thermodynamics 热力学 gives the direction; kinetics 动力学 gives the speed.

Free energy links to the equilibrium constant 平衡常数:

So $\Delta G^\circ<0$ gives $K>1$ (products favored), and $\Delta G^\circ>0$ gives $K<1$. At equilibrium $\Delta G=0$.

Whether a salt dissolves depends on the free-energy change of dissolving 溶解. Dissolving often increases entropy (ordered solid → dispersed ions) but may cost enthalpy; the sign of $\Delta G$ (and thus $K_{sp}$) follows from $\Delta H - T\Delta S$.

An unfavorable reaction ($\Delta G>0$) can be driven by coupling 偶联 it to a favorable one ($\Delta G<0$) that shares a common intermediate 中间体, as long as the sum has $\Delta G<0$. This is how cells use ATP to power otherwise unfavorable processes.

Redox reactions 氧化还原反应 can move electrons through a wire:

• A galvanic (voltaic) cell 原电池 uses a favorable reaction ($\Delta G<0$) to produce electricity – a battery.
• An electrolytic cell 电解池 uses electricity to force an unfavorable reaction ($\Delta G>0$).

In both, oxidation 氧化 happens at the anode 阳极 and reduction 还原 at the cathode 阴极.

The cell potential 电池电势 $E^\circ_{\text{cell}}$ measures the driving force in volts, found from standard reduction potentials 标准还原电势 ($E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$). It links to free energy by

where $n$ is moles of electrons and $F$ the Faraday constant 法拉第常数. A positive $E^\circ_{\text{cell}}$ means a favorable (galvanic) reaction under standard conditions.

Worked example. A cell pairs a copper cathode ($\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$, $E^\circ=+0.34\ \text{V}$) with a zinc anode ($\text{Zn}^{2+}+2e^-\rightarrow\text{Zn}$, $E^\circ=-0.76\ \text{V}$). The cell potential is

It is positive, so the reaction is spontaneous 自发的 and the cell (a Daniell cell) acts as a battery.

Away from standard conditions, the potential shifts with concentration – the Nernst equation 能斯特方程 (qualitatively): as reactants are consumed, $Q$ rises and $E_{\text{cell}}$ falls, reaching zero at equilibrium (a dead battery). Changing a concentration shifts $E_{\text{cell}}$ the way Le Chatelier's principle 勒夏特列原理 predicts.

In electrolysis 电解, the charge passed determines how much substance is deposited or produced – Faraday's law 法拉第定律. Convert current × time to charge, charge to moles of electrons ($F=96{,}485$ C/mol), then use the electron ratio in the half-reaction 半反应 to get moles (and mass) of product.

Worked example. A current of $2.0\ \text{A}$ flows for $30\ \text{minutes}$ through copper(II) sulfate ($\text{Cu}^{2+}+2e^-\rightarrow\text{Cu}$). How much copper is deposited? The charge is $Q=It=2.0\times1800=3600\ \text{C}$, giving $3600/96485=0.0373\ \text{mol}$ of electrons. Since each Cu needs $2$ electrons, $0.0187\ \text{mol}$ of Cu forms, a mass of $0.0187\times63.5=1.2\ \text{g}$.

Common mistakes.

• Mixing units in $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. $\Delta S^\circ$ is usually given in J/(mol K) and $\Delta H^\circ$ in kJ/mol: convert one of them before you subtract.
• Saying a thermodynamically favorable reaction is fast. $\Delta G^\circ < 0$ says nothing about the rate; a high activation energy can keep it under kinetic control.
• Multiplying a reduction potential by the coefficient in the balanced equation. $E^\circ$ is an intensive property, so it is not multiplied; only $n$ in $\Delta G^\circ = -nFE^\circ$ changes.
• Thinking the anode is always negative. Oxidation always happens at the anode; it is negative in a galvanic cell but positive in an electrolytic cell.
• Sending electrons through the salt bridge. Electrons move only through the external wire; ions move through the salt bridge to keep each half-cell neutral.
• Getting the sign link between $K$ and $\Delta G^\circ$ wrong. $\Delta G^\circ = -RT\ln K$: $K > 1$ goes with $\Delta G^\circ < 0$ and $E^\circ_\text{cell} > 0$.

• A process is thermodynamically favourable when $\Delta G<0$; combine enthalpy and entropy with $\Delta G=\Delta H-T\Delta S$ (match units — kJ vs J).
• Entropy increases solid→liquid→gas and when more gas moles are produced.
• Favourable does not mean fast — a high activation energy can make a $\Delta G<0$ reaction extremely slow (kinetic control).
• In electrochemistry a positive $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$ means a spontaneous (galvanic) cell; oxidation is at the anode, reduction at the cathode.
• In electrolysis the charge passed ($Q=It$) fixes the amount deposited (Faraday's law).