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Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options given.
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(a)-(i), (b)-(iii), (c)-(ii), (d)-(iv) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) |
(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) |
The correct answer is option 3. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).
Let us go through each case of electrolysis to understand the products formed at the anode and cathode. (a) NaCl solution using inert electrodes Inert Electrodes: These electrodes do not participate in the reaction and are typically made of materials like platinum or graphite. At the anode (oxidation occurs): \(2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-\) Chloride ions (Cl\(^-\)) are oxidized to chlorine gas (Cl\(_2\)). At the cathode (reduction occurs): \(2 \text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2 \text{OH}^-\) Water is reduced to hydrogen gas (H\(_2\)) and hydroxide ions (OH\(^-\)). Products: Anode: Cl\(_2\) Cathode: H\(_2\) (b) NaCl solution using Hg as cathode and graphite rod as anode Mercury (Hg) Cathode and Graphite Anode: At the anode (oxidation occurs): \(2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-\) Chloride ions (Cl\(^-\)) are oxidized to chlorine gas (Cl\(_2\)). At the cathode (reduction occurs): \(\text{Na}^+ + e^- \rightarrow \text{Na (amalgam)}\) Sodium ions (Na\(^+\)) are reduced to form a sodium amalgam with mercury (Na-Hg). Products: Anode: Cl\(_2\) Cathode: Na (amalgam) (c) Very dilute NaCl solution using inert electrodes Inert Electrodes and Very Dilute NaCl Solution: At the anode (oxidation occurs): \(2 \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \text{H}^+ + 4e^-\) Water is oxidized to oxygen gas (O\(_2\)). At the cathode (reduction occurs): \(2 \text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2 \text{OH}^- \) Water is reduced to hydrogen gas (H\(_2\)) and hydroxide ions (OH\(^-\)). Products: Anode: O\(_2\) Cathode: H\(_2\) (d) CuSO\(_4\) solution using inert electrodes Inert Electrodes and CuSO\(_4\) Solution: At the anode (oxidation occurs): \(2 \text{H}_2\text{O} \rightarrow \text{O}_2 + 4 \text{H}^+ + 4e^-\) Water is oxidized to oxygen gas (O\(_2\)). At the cathode (reduction occurs): \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\) Copper ions (Cu\(^{2+}\)) are reduced to copper metal (Cu). Products: Anode: O\(_2\) |
