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Chemistry - Electrolysis

Grade 10ICSE

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

🔑Concepts

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Electrolysis is the process of decomposition of a chemical compound in the fused (molten) or aqueous state by the passage of a direct electric current, resulting in the discharge of ions at the electrodes.

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Electrolytes are substances that allow electricity to pass through them in their molten or aqueous state and undergo chemical decomposition. Strong electrolytes like NaClNaCl, HClHCl, and NaOHNaOH dissociate completely, while weak electrolytes like CH3COOHCH_3COOH and NH4OHNH_4OH dissociate partially.

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The Electrolytic Cell consists of a Cathode (negative electrode connected to the negative terminal of the battery) and an Anode (positive electrode connected to the positive terminal).

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At the Cathode, reduction occurs as cations (Mn+M^{n+}) gain electrons: Mn++ne−ightarrowMM^{n+} + ne^- ightarrow M.

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At the Anode, oxidation occurs as anions (An−A^{n-}) lose electrons: An−−ne−ightarrowAA^{n-} - ne^- ightarrow A.

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Selective Discharge of Ions: When multiple ions are present, the ion lower in the electrochemical series is preferentially discharged. For example, in a solution containing H+H^+ and Cu2+Cu^{2+}, Cu2+Cu^{2+} is discharged first.

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Electrolysis of molten Lead Bromide (PbBr2PbBr_2): Uses inert graphite electrodes. Pb2+Pb^{2+} ions are reduced to silvery-grey Lead metal at the cathode, and Br−Br^- ions are oxidized to reddish-brown Bromine vapors at the anode.

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Electrolysis of Acidified Water: Water is acidified with dilute H2SO4H_2SO_4 to increase conductivity. Hydrogen gas is evolved at the cathode and Oxygen gas at the anode in a 2:12:1 ratio by volume.

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Electroplating: The process of depositing a superior metal (like AgAg, AuAu, NiNi) on an inferior metal object. The object to be plated is made the Cathode, and the pure metal to be deposited is made the Anode.

📐Formulae

Mn++ne−→M (Reduction at Cathode)M^{n+} + ne^- \rightarrow M \text{ (Reduction at Cathode)}

Xn−−ne−→X (Oxidation at Anode)X^{n-} - ne^- \rightarrow X \text{ (Oxidation at Anode)}

2H++2e−→H2 (Hydrogen evolution)2H^+ + 2e^- \rightarrow H_2 \text{ (Hydrogen evolution)}

4OH−−4e−→2H2O+O2 (Oxygen evolution)4OH^- - 4e^- \rightarrow 2H_2O + O_2 \text{ (Oxygen evolution)}

Pb2++2e−→Pb (Lead deposition)Pb^{2+} + 2e^- \rightarrow Pb \text{ (Lead deposition)}

2Br−−2e−→Br2 (Bromine formation)2Br^- - 2e^- \rightarrow Br_2 \text{ (Bromine formation)}

💡Examples

Problem 1:

Explain the electrolysis of aqueous copper sulphate (CuSO4CuSO_4) using copper electrodes.

Solution:

Electrolyte: Aqueous CuSO4CuSO_4 solution. Dissociation: CuSO4⇌Cu2++SO42−CuSO_4 \rightleftharpoons Cu^{2+} + SO_4^{2-} and H2O⇌H++OH−H_2O \rightleftharpoons H^+ + OH^-. At Cathode: Cu2++2e−→CuCu^{2+} + 2e^- \rightarrow Cu (Copper metal is deposited). At Anode: Cu−2e−→Cu2+Cu - 2e^- \rightarrow Cu^{2+} (Copper atoms from the anode dissolve into the solution).

Explanation:

Since active Copper electrodes are used, the anode participates in the reaction. Instead of SO42−SO_4^{2-} or OH−OH^- discharging, the copper atoms of the anode lose electrons and enter the solution as Cu2+Cu^{2+} ions. This process is used in the electro-refining of copper.

Problem 2:

Calculate the ratio of the volumes of gases collected at the cathode and anode during the electrolysis of acidified water.

Solution:

Reaction at Cathode: 4H++4e−→2H24H^+ + 4e^- \rightarrow 2H_2 Reaction at Anode: 4OH−→2H2O+O2+4e−4OH^- \rightarrow 2H_2O + O_2 + 4e^- Overall Reaction: 2H2O→Electrolysis2H2+O22H_2O \xrightarrow{\text{Electrolysis}} 2H_2 + O_2

Explanation:

According to the balanced chemical equation 2H2O→2H2+O22H_2O \rightarrow 2H_2 + O_2, for every two molecules of water decomposed, two molecules of Hydrogen gas are produced at the cathode for every one molecule of Oxygen gas at the anode. Therefore, the volume ratio H2:O2H_2 : O_2 is 2:12:1.