Electrochemistry

Using electricity to force a reaction

Every redox reaction met so far ran because it wanted to: the more reactive species donated electrons to the less reactive one, spontaneously. Electrolysis is the opposite deal — electrical energy from a DC supply forces redox to happen that would never run on its own. Pass a current through molten sodium chloride and sodium metal appears at one electrode while chlorine gas bubbles off at the other: sodium and chlorine are far too reactive to coexist by choice, but the supply pushes electrons where chemistry never would.

That makes electrolysis the natural opening for this topic. Its questions ask two things — what forms at each electrode (this section) and how much (§02–§03) — and both answers fall out of electron bookkeeping you already own from AS. Later in the note (§04 onwards) the machinery runs in reverse: spontaneous reactions generating electricity instead of consuming it. Keep those two directions separate in your head from the start.

What is in the beaker, and what each electrode does

An electrolyte is a liquid that conducts because it contains mobile ions. The two states the syllabus cares about offer very different ion menus:

• Molten ionic compound (heated past its melting point): contains only its own ions. Molten lead(II) bromide contains $\text{Pb}^{2+}$ and $\text{Br}^-$ — nothing else.
• Aqueous solution (salt dissolved in water): contains the salt's own ions plus traces of $\text{H}^+\text{(aq)}$ and $\text{OH}^-\text{(aq)}$ from the water itself. Every aqueous electrolysis therefore hosts a competition at each electrode.

The electrodes are inert (platinum or carbon/graphite): they carry electrons into and out of the cell but never react themselves. The DC supply pumps electrons onto one electrode and drags them off the other:

• The cathode is flooded with electrons, so it is the negative electrode. It attracts positive ions (cations), which gain electrons there: reduction.
• The anode has electrons dragged away, so it is the positive electrode. It attracts negative ions (anions), which lose electrons there: oxidation.

The competition rules in aqueous solution

Molten electrolytes are easy: only one candidate per electrode, so the products are certain — the metal at the cathode, the non-metal at the anode. Aqueous ones need tie-break rules, because water's own $\text{H}^+$ and $\text{OH}^-$ compete with the salt's ions at both electrodes.

At the cathode (reduction) the contest is between the metal cations and hydrogen:

• For metals below hydrogen in the reactivity series (the syllabus's "redox series") — copper, silver, with lead borderline — the cations win and the metal deposits: $\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}$.
• For metals above hydrogen — K, Na, Ca, Mg, Al, Zn — the cations stay dissolved and hydrogen is released instead: $2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}$.

At the anode (oxidation) the contest is between the anions present and hydroxide/water:

• Concentrated halides ($\text{Cl}^-$, $\text{Br}^-$, $\text{I}^-$) win and the halogen is produced: $2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-$. Fluoride is the exception even when concentrated — $\text{F}_2$ is such a fierce oxidising agent that it reacts with water, so hydroxide still wins and oxygen comes off.
• Anything else — sulfates, nitrates, or a dilute halide — loses to hydroxide/water, which is oxidised to oxygen: $4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-$.

Concentration overrides position. The same salt can give different anode products depending on how concentrated it is. Electrolyse concentrated $\text{NaCl(aq)}$: chlorine bubbles off. Electrolyse dilute $\text{NaCl(aq)}$: oxygen does. In the concentrated solution the abundant $\text{Cl}^-$ outcompetes the trace $\text{OH}^-$; in the dilute solution there are too few chloride ions to win.

A clean demonstration: two predictions from scratch

Molten KBr. Molten, so the only ions are $\text{K}^+$ and $\text{Br}^-$ — no competition anywhere. Cathode: $\text{K}^+$ is the only cation available, so potassium forms, $\text{K}^+ + \text{e}^- \rightarrow \text{K}$. Anode: $\text{Br}^-$ is the only anion, so bromine forms, $2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-$. Molten-salt products are certain because each electrode is offered exactly one candidate.

Dilute CuSO₄(aq). Aqueous, so four ions compete: $\text{Cu}^{2+}$ and $\text{SO}_4^{2-}$ from the salt, plus $\text{H}^+$ and $\text{OH}^-$ from the water. Cathode: copper sits below hydrogen in the reactivity series, so $\text{Cu}^{2+}$ wins and copper coats the electrode: $\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}$. Anode: sulfate is not a halide, so it loses to hydroxide — oxygen is released: $4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-$. (As the $\text{Cu}^{2+}$ is removed the blue colour fades — and because water is consumed at the anode while $\text{H}^+$ is not, the solution slowly becomes acidic.)