Electrochemistry

Electrolysis 电解 uses electricity to break down a molten or aqueous electrolyte 电解质. Positive ions move to the negative electrode and negative ions move to the positive electrode.

At each electrode 电极 a half-reaction happens:

• at the cathode 阴极 (negative): positive ions gain electrons (reduction).
• at the anode 阳极 (positive): negative ions lose electrons (oxidation).

Electrolysis: the DC supply drives positive ions to the cathode (reduction) and negative ions to the anode (oxidation)

A Hofmann voltameter for the electrolysis of water: the power supply drives current through two electrodes, and the gas made at each one collects in its side tube

Predicting the products

• a molten electrolyte gives the metal at the cathode and the non-metal at the anode.
• an aqueous electrolyte also contains water. At the cathode, a less reactive metal is released, but for a reactive metal you get hydrogen instead. At the anode you usually get oxygen, but a concentrated halide solution gives the halogen.

A molten electrolyte gives the metal and non-metal; aqueous also has water

Calculations

The charge on one mole of electrons is the Faraday constant 法拉第常量, linked to the Avogadro constant 阿伏伽德罗常量 ($L$) and the charge on one electron ($e$) by $F = Le$.

The charge passed is $Q = It$ (current $\times$ time). Then:

1. moles of electrons $= Q / F$.
2. use the half-equation to find moles of product.
3. find the mass ($\times M_r$) or the gas volume.

Measuring the mass deposited for a known charge lets you work back to a value of the Avogadro constant.

Worked example. A current of $2.0\ \text{A}$ flows for $30$ minutes through copper(II) sulfate solution, depositing copper at the cathode: $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$. Find the mass of copper deposited. ($F = 96\,500\ \text{C mol}^{-1}$, $A_r$ Cu $= 64$.)

Charge passed: $Q = It = 2.0 \times (30 \times 60) = 3600\ \text{C}$. Moles of electrons $= Q/F = 3600/96\,500 = 0.0373\ \text{mol}$. From the half-equation, $2\ \text{mol}$ of electrons give $1\ \text{mol}$ of Cu, so $n(\text{Cu}) = 0.0187\ \text{mol}$ and

Electrolysis is used to electroplate an object with a thin, shiny layer of metal such as chromium

The electrode potential 电极电势 ($E$) of a half-cell shows how easily it is reduced. We measure it against a reference, under standard conditions, to get the standard electrode potential 标准电极电势 ($E^{\ominus}$), always written as a reduction.

The reference is the standard hydrogen electrode 标准氢电极: hydrogen gas at $1\ \text{atm}$ over platinum in $1\ \text{mol dm}^{-3}$ $\text{H}^+$, defined as exactly $0.00\ \text{V}$.

The standard hydrogen electrode: H$_2$ at 1 atm over platinum in 1 mol dm$^{-3}$ H$^+$, the $0.00$ V reference for every electrode potential

To measure an $E^{\ominus}$, connect the half-cell to the standard hydrogen electrode and read the voltage. A metal sits in a solution of its ions; for two ions of the same element (such as $\text{Fe}^{3+}/\text{Fe}^{2+}$), a platinum electrode dips into a solution containing both.

Combining half-cells

The standard cell potential 标准电池电势 is the difference between the two standard electrode potentials:

A simple cell: two half-cells joined by a salt bridge, with a voltmeter. Electrons flow from the negative (Zn) electrode to the positive (Cu)

Worked example. Find the standard cell potential of a cell built from the $\text{Zn}^{2+}/\text{Zn}$ half-cell ($E^{\ominus} = -0.76\ \text{V}$) and the $\text{Cu}^{2+}/\text{Cu}$ half-cell ($E^{\ominus} = +0.34\ \text{V}$).

From $E^{\ominus}$ values you can:

• find the polarity: the more negative electrode is the negative terminal, and electrons flow from it through the external circuit to the positive electrode.
• judge reactivity: a more positive $E^{\ominus}$ means a better oxidising agent 氧化剂 (easily reduced); a more negative $E^{\ominus}$ means a better reducing agent 还原剂.

The electrochemical series: a more positive $E^{\ominus}$ (top) is a stronger oxidising agent; a more negative $E^{\ominus}$ (bottom) is a stronger reducing agent

Feasibility and redox equations

A reaction is feasible when $E^{\ominus}_{\text{cell}}$ is positive. To build the full equation, take the two half-equations 半反应方程式, reverse the one that is oxidised, and add them so the electrons cancel. This $E^{\ominus}$ test tells you the feasibility 可行性 of the reaction.

You can also link it to free energy: $\Delta G^{\ominus} = -n E^{\ominus}_{\text{cell}} F$, where $n$ is the moles of electrons.

The Nernst equation

If the concentrations are not standard, the electrode potential changes. Raising the concentration of the oxidised species makes $E$ more positive. The Nernst equation 能斯特方程 gives the exact value:

where $z$ is the number of electrons in the half-equation.

• $E^{\ominus}_{\text{cell}} = E^{\ominus}(\text{more positive}) - E^{\ominus}(\text{less positive})$; a positive $E^{\ominus}_{\text{cell}}$ means the reaction is feasible.
• Standard conditions for $E^{\ominus}$: 298 K, $1\ \text{mol dm}^{-3}$, 100 kPa, platinum electrode — state them if asked.
• A more negative electrode potential means a stronger reducing agent; use the series to predict the direction.
• For electrolysis quantities use $Q = It$ and moles of electrons $= Q/F$.