krit.club logo

Electrochemistry - Electrolysis of aqueous solutions

Grade 11A LevelChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Electrolysis is the process by which ionic substances are decomposed (broken down) into simpler substances when an electric current is passed through them in a molten state or in an aqueous solution.

•

In aqueous solutions, besides the ions from the solute, water dissociates slightly to provide hydrogen ions (H+H^+) and hydroxide ions (OH−OH^-) according to: H2O(l)⇌H(aq)++OH(aq)−H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}.

•

At the Cathode (negative electrode), reduction occurs. If the metal in the salt is more reactive than hydrogen (e.g., NaNa, KK, MgMg), then H2H_2 gas is produced: 2H(aq)++2e−→H2(g)2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}. If the metal is less reactive (e.g., CuCu, AgAg), the metal is deposited.

•

At the Anode (positive electrode), oxidation occurs. If halide ions (Cl−Cl^-, Br−Br^-, I−I^-) are present in high concentration, the halogen gas is produced (e.g., 2Cl−→Cl2(g)+2e−2Cl^- \rightarrow Cl_{2(g)} + 2e^-). If not, or if the solution is dilute, oxygen gas is produced from OH−OH^- ions: 4OH(aq)−→O2(g)+2H2O(l)+4e−4OH^-_{(aq)} \rightarrow O_{2(g)} + 2H_2O_{(l)} + 4e^-.

•

The ease of discharge of cations follows the inverse of the reactivity series: Au3+>Ag+>Cu2+>H+>Pb2+>Sn2+>Ni2+>Fe2+>Zn2+>Al3+>Mg2+>Ca2+>Na+>K+Au^{3+} > Ag^+ > Cu^{2+} > H^+ > Pb^{2+} > Sn^{2+} > Ni^{2+} > Fe^{2+} > Zn^{2+} > Al^{3+} > Mg^{2+} > Ca^{2+} > Na^+ > K^+.

•

The ease of discharge of anions: I−>Br−>Cl−>OH−>SO42−>NO3−I^- > Br^- > Cl^- > OH^- > SO_4^{2-} > NO_3^-.

📐Formulae

Q=I×tQ = I \times t

2H(aq)++2e−→H2(g)2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}

4OH(aq)−→O2(g)+2H2O(l)+4e−4OH^-_{(aq)} \rightarrow O_{2(g)} + 2H_2O_{(l)} + 4e^-

2Cl(aq)−→Cl2(g)+2e−2Cl^-_{(aq)} \rightarrow Cl_{2(g)} + 2e^-

Cu(aq)2++2e−→Cu(s)Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}

💡Examples

Problem 1:

Predict the products at the electrodes during the electrolysis of concentrated aqueous sodium chloride (NaClNaCl) using inert carbon electrodes and write the half-equations.

Solution:

At the Cathode: H2(g)H_{2(g)} is produced. At the Anode: Cl2(g)Cl_{2(g)} is produced.

Explanation:

In aqueous NaClNaCl, the ions present are Na+Na^+, Cl−Cl^-, H+H^+, and OH−OH^-. At the cathode, H+H^+ is lower than Na+Na^+ in the reactivity series, so it is preferentially reduced: 2H++2e−→H22H^+ + 2e^- \rightarrow H_2. At the anode, because the solution is concentrated, Cl−Cl^- ions are discharged instead of OH−OH^-: 2Cl−→Cl2+2e−2Cl^- \rightarrow Cl_2 + 2e^-. The remaining ions (Na+Na^+ and OH−OH^-) form NaOHNaOH in the solution.

Problem 2:

Determine the products of the electrolysis of dilute aqueous copper(II) sulfate (CuSO4CuSO_4) using inert platinum electrodes.

Solution:

At the Cathode: Copper metal (Cu(s)Cu_{(s)}). At the Anode: Oxygen gas (O2(g)O_{2(g)}).

Explanation:

The ions present are Cu2+Cu^{2+}, SO42−SO_4^{2-}, H+H^+, and OH−OH^-. At the cathode, Cu2+Cu^{2+} is less reactive than H+H^+, so copper is deposited: Cu2++2e−→CuCu^{2+} + 2e^- \rightarrow Cu. At the anode, SO42−SO_4^{2-} is very stable and difficult to oxidize, so OH−OH^- ions from water are discharged to produce oxygen: 4OH−→O2+2H2O+4e−4OH^- \rightarrow O_2 + 2H_2O + 4e^-.