Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options given.
|
Column I (Electrolysis) |
Column II (Products at anode and cathode respectively) |
| (a) NaCl solution using inert electrodes | (i) O2 and Cu |
| (b) NaCl solution using Hg as cathode and graphite rod as anode | (ii) O2 and H2 |
| (c) Very dilute NaCl solution using inert electrodes | (iii) Cl2 and Na |
| (d) CuSO4 solution using inert electrodes | (iv) Cl2 and H2 |
Answer & explanation
Correct answer: option 3
The correct answer is option 3. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).
|
Column I (Electrolysis) |
Column II (Products at anode and cathode respectively) |
| (a) NaCl solution using inert electrodes | (iv) Cl2 and H2 |
| (b) NaCl solution using Hg as cathode and graphite rod as anode | (iii) Cl2 and Na |
| (c) Very dilute NaCl solution using inert electrodes | (ii) O2 and H2 |
| (d) CuSO4 solution using inert electrodes | (i) O2 and Cu |
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\(_2\)