Electrolysis — ICSE Class 10 Chemistry Important Questions

Answer: (a) cathode.

Cations migrate to the cathode ($-$- electrode) where they gain electrons (reduction), e.g. $Cu^{2+}+2e^-\rightarrow Cu$Cu^2++2e^-→ Cu.

Answer: (b) bromine.

Bromide ions are oxidised at the anode: $2Br^-\rightarrow Br_2+2e^-$2Br^-→ Br_2+2e^- (reddish-brown vapour). Lead is deposited at the cathode.

Answer: (c) dilute sulphuric acid.

Dilute sulphuric acid ionises almost completely into $H^+$H^+ and $SO_4^{2-}$SO_4^2- ions, making it a strong electrolyte. The others are non-electrolytes.

Answer: (a) the cathode gains mass and the anode loses mass.

At the cathode $Cu^{2+}+2e^-\rightarrow Cu$Cu^2++2e^-→ Cu (deposited), while the copper anode dissolves: $Cu-2e^-\rightarrow Cu^{2+}$Cu-2e^-→ Cu^2+. The electrolyte concentration stays constant.

Answer: (a) Water, $H_2O$H_2O, contains $H:O=2:1$H:O=2:1; on electrolysis $2$2 volumes of hydrogen form at the cathode for every $1$1 volume of oxygen at the anode, so R correctly explains A.

(a) Electrolyte: a compound which, in the molten state or in aqueous solution, allows an electric current to pass through it and is chemically decomposed by it (e.g. dilute $H_2SO_4$H_2SO_4, molten $PbBr_2$PbBr_2).

(b) Electrolysis: the chemical decomposition of an electrolyte, in the molten or aqueous state, by the passage of direct electric current.

In the solid state the ions of lead bromide are held in fixed positions in the crystal lattice by strong electrostatic forces and are not free to move, so no current flows.

On melting, these forces are overcome and the ions ($Pb^{2+}$Pb^2+ and $Br^-$Br^-) become free to move towards the electrodes, allowing the molten compound to conduct and undergo electrolysis.

Cathode (reduction): hydrogen gas is liberated.
$2H^+ + 2e^- \rightarrow H_2\uparrow$2H^+ + 2e^- → H_2

Anode (oxidation): oxygen gas is liberated.
$4OH^- \rightarrow 2H_2O + O_2\uparrow + 4e^-$4OH^- → 2H_2O + O_2 + 4e^-

The overall effect is the decomposition of water, with hydrogen and oxygen collected in the volume ratio $2:1$2:1.

Electrolyte: a solution of sodium argentocyanide, $Na[Ag(CN)_2]$Na[Ag(CN)_2] (a complex silver salt).

Anode: a pure silver plate (connected to the $+$+ terminal).

Cathode: the clean iron spoon to be plated (connected to the $-$- terminal).

When current is passed, $Ag^+$Ag^+ ions move to the cathode and are deposited on the spoon: $Ag^+ + e^- \rightarrow Ag$Ag^+ + e^- → Ag. At the anode the silver plate dissolves, $Ag - e^- \rightarrow Ag^+$Ag - e^- → Ag^+, keeping the concentration of silver ions in the electrolyte constant. A low current density with a complex salt gives a smooth, adherent silver coating.

The solution contains $Cu^{2+}$Cu^2+, $H^+$H^+, $SO_4^{2-}$SO_4^2- and $OH^-$OH^- ions. In both cases copper is deposited at the cathode ($Cu^{2+}+2e^-\rightarrow Cu$Cu^2++2e^-→ Cu).

(i) Platinum (inert) anode: $OH^-$OH^- ions are discharged in preference to $SO_4^{2-}$SO_4^2-, so oxygen gas is liberated: $4OH^-\rightarrow 2H_2O+O_2\uparrow+4e^-$4OH^-→ 2H_2O+O_2+4e^-. The blue colour fades as $Cu^{2+}$Cu^2+ is used up.

(ii) Copper (active) anode: no gas is evolved; instead the copper anode itself dissolves, $Cu-2e^-\rightarrow Cu^{2+}$Cu-2e^-→ Cu^2+, because it is easier to oxidise the metal than to discharge $OH^-$OH^-. The blue colour remains unchanged. This is the principle of electro-refining.

Impure (blister) copper is purified by electrolysis.

Electrolyte: acidified copper sulphate solution ($CuSO_4$CuSO_4 acidified with a little dilute $H_2SO_4$H_2SO_4).

Anode: a thick block of impure copper (connected to $+$+).

Cathode: a thin strip of pure copper (connected to $-$-).

At the anode (oxidation): the impure copper dissolves into the solution.
$Cu - 2e^- \rightarrow Cu^{2+}$Cu - 2e^- → Cu^2+

At the cathode (reduction): copper ions from the solution are deposited as pure copper.
$Cu^{2+} + 2e^- \rightarrow Cu$Cu^2+ + 2e^- → Cu

Impurities: more reactive metals (e.g. iron, zinc) pass into solution as ions and remain there; less reactive impurities such as silver and gold do not dissolve and collect below the anode as anode mud. The net result is transfer of pure copper to the cathode, giving copper of about $99.9\%$99.9\% purity.

(a) Ionic theory: An electrolyte, in the molten state or in aqueous solution, splits up (dissociates) into positively charged cations and negatively charged anions. These ions move freely and are attracted to the oppositely charged electrode, where they are discharged, causing electrolysis. The current is carried through the electrolyte by the movement of these ions.

(b) Ions present: from $CuSO_4$CuSO_4: $Cu^{2+}$Cu^2+ and $SO_4^{2-}$SO_4^2-; from water: $H^+$H^+ and $OH^-$OH^-.

Cathode: $Cu^{2+}$Cu^2+ is discharged in preference to $H^+$H^+ (copper is below hydrogen in the discharge/activity considerations here): $Cu^{2+}+2e^-\rightarrow Cu$Cu^2++2e^-→ Cu (reddish deposit).

Anode (platinum, inert): $OH^-$OH^- is discharged in preference to $SO_4^{2-}$SO_4^2-: $4OH^-\rightarrow 2H_2O+O_2\uparrow+4e^-$4OH^-→ 2H_2O+O_2+4e^-.

Colour: the blue colour fades as $Cu^{2+}$Cu^2+ ions are removed from solution.

pH: as $Cu^{2+}$Cu^2+ and $OH^-$OH^- are used up, $H^+$H^+ ions and $SO_4^{2-}$SO_4^2- accumulate (effectively sulphuric acid forms), so the solution becomes acidic and its pH decreases.

(i) The bangle must be the cathode (connected to the $-$- terminal). Metal cations ($Au^+$Au^+) are attracted to and deposited on the cathode, so the object to be plated is always made the cathode.

(ii) At the bangle (cathode): $Au^+ + e^- \rightarrow Au$Au^+ + e^- → Au (gold is deposited as a coating).

(iii) At the gold plate (anode): the gold dissolves, $Au - e^- \rightarrow Au^+$Au - e^- → Au^+, replenishing the ions in the electrolyte.

(iv) A complex (double) salt dissociates slowly to give a low, steady concentration of metal ions. This slow, uniform discharge produces a smooth, fine-grained and firmly adherent coating, whereas a simple salt would give a rough, loose deposit.