Selective discharge of ions

Every aqueous electrolyte is really at least a four-ion system: the salt or acid's own cation and anion, plus H⁺ and OH⁻ contributed by the slight self-ionisation of water. Selective (preferential) discharge is the outcome of a three-way competition among whichever ions are drawn to a given electrode.

Factor 1 — position in the electrochemical series. At the cathode, cations compete, and the less reactive metal is discharged first — a less reactive metal holds onto its electrons less strongly demanded, so its ion is reduced back to atoms more readily. The qualitative order (most to least reactive) used at this level is:

where: ions lower in this list (nearer Au) are discharged in preference to ions higher up (nearer K). So Cu²⁺ beats H⁺, and H⁺ beats Na⁺ or K⁺, whenever they compete at the same cathode. At the anode, anions compete, and the accepted order of ease of discharge is: simple halides (I⁻, Br⁻, Cl⁻) discharge most easily, then hydroxide (OH⁻), and last the "stable" oxy-anions such as sulphate (SO₄²⁻) and nitrate (NO₃⁻), which are essentially never discharged from dilute aqueous solution — water decomposes instead, releasing oxygen from OH⁻.

Factor 2 — concentration of the ion. A higher local concentration raises an ion's chance of discharge even against series preference. This is exactly why concentrated brine gives chlorine at the anode instead of the "expected" oxygen: high Cl⁻ concentration overrides the series' mild preference for OH⁻.

Factor 3 — nature of the electrode. Inert electrodes (platinum, graphite) are passive surfaces where discharge occurs without the electrode itself reacting. Active electrodes (for example, copper) can participate directly — a copper anode dissolves as Cu²⁺ instead of any anion being discharged at all. This third factor is the working principle behind electroplating and electro-refining.

Selective discharge has no single numeric formula at this level — it is a ranking problem, decided by weighing three qualitative factors against each other. The reasoning chain is: identify every ion present at a given electrode, rank them by the electrochemical series, then check whether concentration or electrode reactivity is strong enough to overturn that ranking.

Step 1. List every ion actually present. Any aqueous solution supplies H⁺ and OH⁻ from water's own slight ionisation, on top of whatever the dissolved electrolyte contributes. Students who skip this step often forget that H⁺ or OH⁻ is even in the race.

Step 2. Sort the cations reaching the cathode and the anions reaching the anode using the electrochemical series order given above. The default winner is the less reactive cation at the cathode and the more easily oxidised anion (simple halide, then OH⁻, then never sulphate or nitrate) at the anode.

Step 3. Check whether concentration changes the default answer. A large excess of one ion — as with concentrated brine's Cl⁻ — can push that ion into discharge even if the series alone would have favoured a different ion. Dilute the same solution and the series-preferred ion (OH⁻, giving oxygen) can take over again.

Step 4. Check the electrode material last. If the electrode itself is reactive (an active anode like copper), it can dissolve into solution directly, short-circuiting the anion competition entirely — no anion needs to be discharged because the electrode supplies the reaction instead. Applying these four steps in order, rather than jumping straight to the series list, is what lets you predict products correctly even for electrolytes you have never seen before.

Question: Dilute sulphuric acid is electrolysed between platinum electrodes, and then, in a second cell, concentrated aqueous sodium chloride (brine) is electrolysed between graphite electrodes. Identify the ions present in each case and predict the products at both electrodes, with reasoning.

Step 1. Dilute sulphuric acid ionises as $\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}$, and water contributes a little extra H⁺ and OH⁻. So the ions present are H⁺ (cation) and SO₄²⁻, OH⁻ (anions). At the cathode only H⁺ is available to be discharged, since there is no metal cation: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2\uparrow$.

Step 2. At the anode, SO₄²⁻ competes with OH⁻. Sulphate sits well below hydroxide in ease of discharge and is essentially never discharged from dilute solution, so OH⁻ is discharged instead: $4\text{OH}^- \rightarrow \text{O}_2\uparrow + 2\text{H}_2\text{O} + 4e^-$. Overall the cell liberates hydrogen and oxygen in a 2:1 volume ratio — effectively electrolysis of water, with the acid only making the solution conducting.

Step 3. Brine contains Na⁺ and Cl⁻ from the salt, plus H⁺ and OH⁻ from water. At the cathode, sodium is far above hydrogen in the electrochemical series, so H⁺ is discharged instead of Na⁺: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2\uparrow$, leaving Na⁺ and OH⁻ to accumulate as sodium hydroxide in solution.

Step 4. At the anode, Cl⁻ and OH⁻ compete. Series position alone would slightly favour OH⁻, but because the brine is concentrated, the high Cl⁻ concentration overrides this and chlorine is discharged instead: $2\text{Cl}^- \rightarrow \text{Cl}_2\uparrow + 2e^-$. Answer: sulphuric acid gives hydrogen at the cathode and oxygen at the anode; concentrated brine gives hydrogen at the cathode and chlorine at the anode, with sodium hydroxide building up in solution.

Competitive stretch

If the same sodium chloride solution were made very dilute instead of concentrated, the low Cl⁻ concentration would no longer be enough to override the series preference for OH⁻, so the anode product would shift back to oxygen — the same electrolyte can give two different anode products purely because of concentration.

• Assuming the "obvious" ion named in the electrolyte (the metal ion or the acid radical) is automatically the one discharged, forgetting the ever-present competition from H⁺ and OH⁻ supplied by water.
• Predicting a "sulphate gas" or "nitrate gas" at the anode — stable oxy-anions like SO₄²⁻ and NO₃⁻ are practically never discharged in ordinary dilute aqueous electrolysis.
• Applying the electrochemical series rigidly in every case and ignoring concentration, which is exactly why concentrated brine gives Cl₂ rather than the series-preferred O₂.
• Forgetting that an active (reactive) electrode, such as a copper anode, can dissolve directly into solution and change the outcome entirely, instead of any anion being discharged.

Ask the tutor: Want to work through what happens if you electrolyse a solution of copper sulphate with copper electrodes instead of platinum — ask the AI tutor to walk you through how the active electrode changes both the cathode and anode reactions.