Why Copper Plates Out of Copper Sulphate but Sodium Never Plates Out of Brine

Pass a current through copper sulphate solution and reddish-brown copper coats the cathode. Pass the same current through common salt solution and no sodium appears at all — bubbles of hydrogen rise from the cathode instead. Both solutions contain a metal ion and hydrogen ions from water, yet the result is opposite.

An aqueous solution always contains more than one kind of cation and more than one kind of anion. Copper sulphate solution holds ${\text{Cu}}^{2+}$ and ${\text{SO}}_4^{2-}$ from the salt, and ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ from the slight ionisation of water. At each electrode, the ions compete — and usually only one of them is discharged.

Which ion wins is not random. It depends on three things:

- How readily the ion accepts or gives up electrons, shown by its position in the activity series
- How concentrated the ion is
- What the electrode is made of — an inert electrode stands aside, but an active one can take part itself

Once those three factors are understood, the products of any electrolysis can be predicted rather than memorised.

This part then applies them to four standard electrolytic cells:

- Molten lead bromide, where there is no competition because no water is present
- Acidified water with platinum electrodes, which splits water into hydrogen and oxygen
- Copper sulphate with platinum electrodes, which deposits copper and releases oxygen
- Copper sulphate with copper electrodes, which moves copper from one electrode to the other

An everyday reason this matters. Sodium metal, used to make many chemicals, is obtained from molten sodium chloride, never from brine. Selective discharge is the reason: in water, hydrogen ions are always discharged ahead of sodium ions.

This page covers the second part of the ICSE Class 10 Chemistry chapter on electrolysis: migration and selective discharge of ions, the activity series, and the electrolysis of lead bromide, acidified water and copper sulphate.

Cations migrate to the cathode and anions to the anode, and when several compete, the one discharged depends on its position in the activity series, its concentration and the nature of the electrode.

Migration. When the circuit is switched on, the cathode becomes negative and the anode positive.

- Cations drift towards the cathode
- Anions drift towards the anode

Discharge. On reaching an electrode, an ion gains or loses electrons and becomes a neutral atom or molecule:

$${\text{H}}^{+}+e^{-}\to \text{H}\text{H}+\text{H}\to {\text{H}}_2$$

Selective or preferential discharge — when two or more ions of the same charge sign reach an electrode, one is discharged in preference to the others. Three factors decide which.

Factor 1 — position in the activity series. For cations, the lower a metal is in the activity series, the more readily its ion is discharged. The order of increasing ease of discharge at the cathode is:

$${\text{K}}^{+},{\text{Na}}^{+},{\text{Ca}}^{2+},{\text{Mg}}^{2+},{\text{Al}}^{3+},{\text{Zn}}^{2+},{\text{Fe}}^{2+},{\text{Pb}}^{2+},{\text{H}}^{+},{\text{Cu}}^{2+},{\text{Ag}}^{+}$$

For anions, sulphate and nitrate ions are the hardest to discharge; hydroxyl ions are discharged in preference to them. Halide ions such as chloride are discharged readily as well.

Factor 2 — concentration. An ion present in much greater concentration can be discharged in preference to one that would otherwise win. Concentrated brine gives chlorine at the anode, because chloride ions greatly outnumber hydroxyl ions.

Factor 3 — nature of the electrodes.

- Inert electrodes, such as platinum and graphite, do not react; they only pass electrons
- Active electrodes, such as a copper anode in copper sulphate, take part themselves — the anode metal loses electrons and dissolves, so no anion is discharged at all

Worked example — predict the cathode product. A solution contains ${\text{Na}}^{+}$ and ${\text{H}}^{+}$ ions, with inert electrodes.

Hydrogen lies below sodium in the series, so ${\text{H}}^{+}$ is discharged: hydrogen gas forms and sodium ions remain in solution.

Worked example — predict the anode product. A solution contains ${\text{SO}}_4^{2-}$ and ${\text{OH}}^{-}$ ions, with inert electrodes.

Hydroxyl ions are discharged in preference to sulphate ions:

$$4{\text{OH}}^{-}\to 2{\text{H}}_2\text{O}+{\text{O}}_2+4e^{-}$$

so oxygen is given off and sulphate ions remain.

An everyday example. The industrial electrolysis that supplies chlorine for water treatment uses concentrated brine on purpose. With dilute salt solution, far more oxygen would form at the anode — the concentration factor put to practical use.

The boundary case. The three factors can pull in different directions, and the concentration or electrode factor can override the activity series. A prediction based on the series alone is only safe when the electrodes are inert and the competing ions are at similar concentrations.

A metal high in the activity series loses electrons easily and its ions hold on to their positive charge, so they are hard to discharge; a metal low in the series forms ions reluctantly, and its ions accept electrons readily.

The activity series, most reactive first:

$$\text{K}>\text{Na}>\text{Ca}>\text{Mg}>\text{Al}>\text{Zn}>\text{Fe}>\text{Pb}>\text{H}>\text{Cu}>\text{Ag}$$

The series shows the tendency of a metal to form ions.

- Sodium and magnesium, near the top, lose electrons very readily: $\text{Na}\to {\text{Na}}^{+}+e^{-}$
- Iron, in the middle, forms ions less readily
- Copper, below hydrogen, forms ions reluctantly

Discharge is the reverse process — the ion taking electrons back. So the series works upside down for discharge:

- **${\text{Na}}^{+}$ and ${\text{Mg}}^{2+}$ have little tendency to take electrons back — very hard to discharge
- ${\text{Fe}}^{2+}$** — harder to discharge than ${\text{H}}^{+}$, since iron lies above hydrogen
- **${\text{Cu}}^{2+}$ takes electrons back readily — discharged in preference to ${\text{H}}^{+}$**

$${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$$

The decisive comparison in aqueous solutions is with hydrogen, because water always supplies some ${\text{H}}^{+}$ ions.

- Metal above hydrogen (Na, Mg, Al, Zn, Fe) — hydrogen ions compete strongly, and for the most reactive metals hydrogen gas is released at the cathode instead of the metal
- Metal below hydrogen (Cu, Ag) — the metal is deposited

Worked example 1. Predict the cathode product for aqueous solutions of copper chloride and sodium chloride with inert electrodes.

- Copper chloride: copper lies below hydrogen, so copper is deposited
- Sodium chloride: sodium lies far above hydrogen, so hydrogen is released

Worked example 2. Predict the cathode product for aqueous magnesium sulphate.

Magnesium is far above hydrogen, so hydrogen gas forms at the cathode and magnesium ions stay in solution.

Worked example 3. Predict the cathode product for aqueous silver nitrate.

Silver is below hydrogen, so silver is deposited — the basis of silver plating in Part 3.

The boundary case — molten salts. When sodium chloride is molten, there is no water and therefore **no ${\text{H}}^{+}$ to compete. Sodium ions are the only cations present, so sodium metal is discharged even though it sits near the top of the series. The activity series decides only between ions that are actually present.

An everyday example. Magnesium and sodium, used in alloys and chemicals, are both produced by electrolysing molten compounds. Their ions could never be discharged from water**, so the industry has no choice but to remove the water first — at considerable cost in heat.

Molten lead bromide gives lead at the cathode and bromine vapour at the anode; acidified water gives hydrogen at the cathode and oxygen at the anode in a volume ratio of two to one.

1. Electrolysis of molten lead bromide.

- Electrolyte: molten lead bromide, in a silica crucible, which withstands strong heating
- Electrodes: inert graphite rods
- Condition: the solid is heated until it melts — solid lead bromide does not conduct

Ionisation:

$${\text{PbBr}}_2\to {\text{Pb}}^{2+}+2{\text{Br}}^{-}$$

At the cathode — reduction:

$${\text{Pb}}^{2+}+2e^{-}\to \text{Pb}$$

At the anode — oxidation:

$${\text{Br}}^{-}-e^{-}\to \text{Br}\text{Br}+\text{Br}\to {\text{Br}}_2$$

Observations:

- No current flows while the lead bromide is solid; the bulb lights only after it melts
- Silvery grey molten lead collects at the cathode
- Dense reddish-brown vapours of bromine appear at the anode

Precaution: bromine vapour is toxic, so the electrolysis is carried out in a fume cupboard.

Worked check — electrons. Each lead ion takes $2$ electrons; each bromine molecule releases $2$. Electrons gained at the cathode equal electrons lost at the anode.

2. Electrolysis of acidified water with platinum electrodes.

- Electrolyte: water acidified with a little dilute sulphuric acid, which supplies ions so that current can flow
- Electrodes: platinum, which is not attacked by the acid or by oxygen

Ionisation:

$${\text{H}}_2{\text{SO}}_4\to 2{\text{H}}^{+}+{\text{SO}}_4^{2-}{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}^{+}+{\text{OH}}^{-}$$

At the cathode — hydrogen ions are the only cations, so they are discharged:

$$4{\text{H}}^{+}+4e^{-}\to 2{\text{H}}_2$$

At the anode — hydroxyl ions are discharged in preference to sulphate ions:

$$4{\text{OH}}^{-}\to 2{\text{H}}_2\text{O}+{\text{O}}_2+4e^{-}$$

Overall:

$$2{\text{H}}_2\text{O}\to 2{\text{H}}_2+{\text{O}}_2$$

Observations:

- Bubbles of colourless gas at both electrodes
- The gas at the cathode has twice the volume of the gas at the anode
- The cathode gas burns with a pop when a lighted splint is brought near: hydrogen
- The anode gas rekindles a glowing splint: oxygen

Worked check — why the ratio is two to one. Four electrons discharge four hydrogen ions to give $2$ molecules of hydrogen, and the same four electrons come from four hydroxyl ions giving $1$ molecule of oxygen. Equal volumes hold equal numbers of molecules, so the volumes are in the ratio $2:1$ — Gay Lussac's and Avogadro's laws from the mole chapter, appearing in an electrolysis cell.

The boundary case about the acid. The sulphuric acid is not used up overall; sulphate ions are never discharged, and the hydrogen ions removed at the cathode are replaced by water's ionisation. Only water is decomposed, so the acid actually becomes slightly more concentrated as the electrolysis continues.

An everyday example. Hydrogen produced by splitting water with electricity from solar or wind power is being developed as a clean fuel, and the chemistry at the electrodes is exactly the pair of equations above.

With platinum electrodes, copper is deposited and oxygen released, so the solution loses its blue colour and turns acidic; with copper electrodes, the copper anode dissolves as fast as copper is deposited, so no gas forms and the solution stays unchanged.

The ions present in both cases:

$${\text{CuSO}}_4\to {\text{Cu}}^{2+}+{\text{SO}}_4^{2-}{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}^{+}+{\text{OH}}^{-}$$

1. With inert platinum electrodes.

At the cathode — copper lies below hydrogen, so ${\text{Cu}}^{2+}$ is discharged:

$${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$$

At the anode — the electrode is inert, so an anion must be discharged, and hydroxyl ions win over sulphate ions:

$$4{\text{OH}}^{-}\to 2{\text{H}}_2\text{O}+{\text{O}}_2+4e^{-}$$

Observations:

- Reddish-brown copper deposits on the cathode
- Bubbles of oxygen form at the anode
- The blue colour fades, as copper ions are removed and not replaced
- The solution becomes acidic, as hydrogen ions and sulphate ions accumulate — effectively sulphuric acid

Overall:

$$2{\text{CuSO}}_4+2{\text{H}}_2\text{O}\to 2\text{Cu}+2{\text{H}}_2{\text{SO}}_4+{\text{O}}_2$$

2. With active copper electrodes.

At the cathode — exactly as before:

$${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$$

At the anode — neither hydroxyl nor sulphate ions are discharged. The copper anode itself gives up electrons more readily and dissolves:

$$\text{Cu}\to {\text{Cu}}^{2+}+2e^{-}$$

Observations:

- Copper deposits on the cathode, which gains mass
- The anode gets thinner, losing mass
- No gas is given off at either electrode
- The blue colour stays the same, because every copper ion removed at the cathode is replaced by one formed at the anode

The electron transfer, electrode by electrode. At the anode, each copper atom hands $2$ electrons to the external circuit and enters the solution as an ion. At the cathode, each copper ion receives $2$ electrons from the circuit and becomes an atom. The net effect is copper moving from anode to cathode through the solution.

Worked example — the same charge in both cells. In each cell the cathode gains $0.635\text{ g}$ of copper, atomic mass $63.5$.

$$\text{moles of Cu}=\frac{0.635}{63.5}=0.01\text{electrons}=0.01\times 2=0.02\text{ mol}$$

- Copper electrodes: the anode loses $0.01$ mol of copper, which is $0.635\text{ g}$ — equal to the cathode's gain
- Platinum electrodes: $0.02$ mol of electrons comes from hydroxyl ions giving $\frac{0.02}{4}=0.005$ mol of oxygen, which is $0.005\times 22.4=0.112\text{ L}$ at STP

The comparison in one list:

- Cathode product: copper in both
- Anode product: oxygen with platinum; copper dissolves with copper
- Colour of solution: fades with platinum; unchanged with copper
- Nature of solution: becomes acidic with platinum; unchanged with copper
- Anode mass: unchanged with platinum; decreases with copper

The boundary case that turns this into technology. With copper electrodes, only the anode metal moves. Make the anode impure copper and the cathode a thin pure sheet, and pure copper builds up on the cathode — electrorefining. Make the cathode an object to be coated, and the process becomes electroplating. Both are the subject of Part 3.