Electrolysis Worked ExamplesIGCSE Chemistry 0620

(a) Electrolysis is the decomposition of an ionic compound, when molten or in aqueous solution, by the passage of an electric current. Each bold phrase is a marking point.

(b) P is joined to the positive terminal (the long line of the cell symbol), so P is the anode. Q is joined to the negative terminal, so Q is the cathode. R, the molten substance that is decomposed, is the electrolyte.

(c) In a molten binary compound the only ions are $\text{Zn}^{2+}$ and $\text{I}^-$. The metal forms at the cathode and the non-metal at the anode, so iodine forms at P (anode) and zinc forms at Q (cathode).

(d) In the solid the ions are held in fixed positions in the lattice and cannot move. When molten, the ions are free to move to the electrodes, so the compound conducts and is decomposed.

Dilute sulfuric acid gives hydrogen at the cathode and oxygen at the anode. Twice as many molecules of hydrogen as oxygen are released, so the hydrogen volume is twice the oxygen volume.

(a) Tube X (larger volume) contains hydrogen; tube Y contains oxygen. (b) Hydrogen forms at the cathode, so the electrode under X is the cathode.

(c) The observation is bubbles (effervescence) at both electrodes, with bubbles forming faster at the cathode. "Hydrogen is produced" is a product, not an observation.

(d) Hydrogen: a lighted splint gives a 'pop'. Oxygen: a glowing splint relights.

Molten chlorides contain only the metal ion and $\text{Cl}^-$, so chlorine is always formed at the anode. Electrolyte 3 gives chlorine.

In concentrated aqueous chlorides, chloride ions are discharged in preference to hydroxide ions, so chlorine is the main anode product. Electrolyte 2 gives chlorine (with copper at the cathode).

In dilute aqueous halides, hydroxide ions from the water are discharged instead, so the main anode product is oxygen. Electrolyte 1 does not give chlorine as the main product.

The answer is C. A and B wrongly assume every chloride solution gives chlorine, whatever its concentration. D forgets that a concentrated aqueous chloride also gives chlorine.

(a) From the salt: $\text{Li}^+$ and $\text{Cl}^-$. From the water: $\text{H}^+$ and $\text{OH}^-$.

Cathode: $\text{Li}^+$ and $\text{H}^+$ are attracted. Lithium is more reactive than hydrogen, so hydrogen is formed — lithium metal cannot be formed in aqueous solution. Seen: bubbles.

Anode: $\text{Cl}^-$ and $\text{OH}^-$ are attracted. The solution is dilute, so hydroxide ions are discharged: oxygen (and water) is formed. Seen: bubbles.

(c) At the cathode, hydrogen ions gain electrons — reduction.

At the anode, hydroxide ions lose electrons — oxidation. Check: charge on the left $= 4-$; on the right $0 + 4(1-) = 4-$. Atoms: 4 O and 4 H on each side.

(a) Inert electrodes do not react with the electrolyte or the products: graphite (carbon) is the non-metal and platinum is the metal. Copper is not inert — as an anode it dissolves.

Molten: only $\text{K}^+$ and $\text{I}^-$ are present, so potassium forms at the cathode and iodine at the anode.

Any aqueous solution — cathode: potassium is more reactive than hydrogen, so hydrogen forms from $\text{H}^+$ ions (from the water).

Aqueous — anode: dilute → hydroxide ions discharged → oxygen. Concentrated → iodide ions discharged → iodine.

P is the anode. Negative ions lose electrons at the anode (oxidation). These electrons travel through the external wires, through the power supply, to the cathode Q. So electrons flow from P to Q.

Inside the electrolyte, the current is carried by ions, not electrons. Positive ions (cations) are attracted to the negative electrode, Q, where they gain electrons (reduction).

So the answer is B. A has the electrons right but sends the cations to the positive electrode. C has the cations right but reverses the electron flow — electrons are pushed out of the negative terminal of the supply into Q, having entered the supply from P.

A is connected to the positive terminal, so A is the anode; B is the cathode.

(a) At the cathode (B), $\text{Cu}^{2+}$ ions gain electrons: a pink-brown solid (copper) coats the electrode. At the anode (A), hydroxide ions are discharged (sulfate ions are not): bubbles of oxygen are seen.

(b) Copper is deposited on B, so its mass increases.

(c) The blue colour is caused by $\text{Cu}^{2+}\text{(aq)}$ ions. They are removed from the solution as copper metal and are not replaced (graphite does not dissolve), so the solution becomes paler blue, eventually colourless.

(d) Oxidation of hydroxide ions at A:

(a) Subtract to find each change.

J loses mass: copper atoms leave it as ions, so J is the anode. K gains copper, so K is the cathode.

(b) Anode (oxidation) and cathode (reduction):

(c) The colour does not change (stays blue). Each $\text{Cu}^{2+}$ ion removed at the cathode is replaced by one formed at the anode, so the concentration of $\text{Cu}^{2+}$ stays the same.

(d) The cathode gained $0.02$ g less than the anode lost. Sensible reasons: some copper fell off the cathode (or anode) into the beaker; copper was lost while washing the cathode; or the anode was not fully dried before the first weighing / the cathode was still damp.

The object to be plated must be the cathode (negative), because metal ions gain electrons there and form a layer of metal. The key is attached to the negative terminal, so Y is the key.

The anode (X) is made of the plating metal, nickel. It dissolves, replacing the $\text{Ni}^{2+}$ ions that are deposited on the key.

The electrolyte must contain ions of the plating metal: aqueous nickel(II) sulfate. The answer is A.

B swaps the electrodes — the key would lose metal. C uses an inert anode, so nickel ions are not replaced and the electrolyte runs out of them. D would plate the key with copper, not nickel.

(a) To improve its appearance and to increase its resistance to corrosion.

(b) Plan (one mark per point): clean the key (e.g. with sandpaper) and weigh it on a balance.

Pour aqueous nickel(II) sulfate into a beaker. Connect the key to the negative terminal (cathode) and the nickel strip to the positive terminal (anode) of the d.c. power supply, with both dipping into the solution.

Switch on for a measured time. Remove the key, wash it with distilled water, dry it and reweigh it. Mass of nickel = mass after − mass before.

(c) Mass of nickel, then moles $= \dfrac{\text{mass}}{A_r}$.

Cathode: $\text{Cu}^{2+}$ and $\text{H}^+$ are attracted. Copper is less reactive than hydrogen, so copper is deposited: a pink-brown solid forms.

Anode: $\text{Br}^-$ and $\text{OH}^-$ are attracted. The solution is concentrated, so bromide ions are discharged: bromine forms, seen as an orange-brown colour in the solution around the anode.

(b) Bromide ions lose electrons (oxidation); copper(II) ions gain electrons (reduction). Both equations balance for atoms and charge ($2- = 0 + 2-$; $2+ + 2- = 0$).

(c) Cathode: still copper — whether the metal or hydrogen forms depends on reactivity, not concentration.

Anode: the solution is now dilute, so hydroxide ions are discharged instead of bromide ions: oxygen (bubbles) forms.

(a) J is the anode (positive). The solution is concentrated, so chloride ions are discharged: chlorine forms at J. K is the cathode: sodium is more reactive than hydrogen, so hydrogen forms at K.

Test for chlorine: hold damp litmus paper at the mouth of the arm — it is bleached (turns white).

(b) Chlorine is toxic, so carry out the experiment in a fume cupboard. (Goggles alone are a general precaution, not one specific to chlorine.)

(c) At K, $\text{H}^+$ ions are removed as hydrogen gas. $\text{OH}^-$ ions from the water are left behind, together with $\text{Na}^+$ ions, so the solution near K becomes alkaline (it contains sodium hydroxide) and the indicator turns blue-purple.

(d) Chlorine is soluble in water (and reacts with it), so some of the chlorine dissolves in the solution instead of being collected. Hydrogen is almost insoluble, so all of it is collected.