Problem 90
Question
In electrolysis of \(\mathrm{NaCl}\) when Pt electrode is used, then \(\mathrm{H}_{2}\) is liberated at cathode while with \(\mathrm{Hg}\) cathode it forms sodium amalgam. The reason for this is (a) more voltage is required to reduce \(\mathrm{H}^{+}\)at \(\mathrm{Hg}\) than at \(\mathrm{Pt}\) (b) concentration of \(\mathrm{H}^{+}\)ions is larger when Pt electrode is taken. (c) Na is dissolved in \(\mathrm{Hg}\) while it does not dissolve in \(\mathrm{Pt}\) (d) \(\mathrm{Hg}\) is more inert than \(\mathrm{Pt}\)
Step-by-Step Solution
Verified Answer
(c) Na is dissolved in Hg while it does not dissolve in Pt.
1Step 1: Understanding Electrolysis with Different Cathodes
In electrolysis, different cathodes impact what substance is reduced. When a platinum (Pt) electrode is used as a cathode, the more prevalent reaction is the reduction of hydrogen ions (\(\mathrm{H}^+\)) to form hydrogen gas (\(\mathrm{H}_2\)). This reaction occurs because of the voltage preference and the nature of Pt, which does not form an amalgam with sodium (Na).
2Step 2: Analyzing the Mercury Cathode Scenario
When mercury (Hg) is used as the cathode, sodium (Na) ions are reduced and dissolve in Hg to form a sodium amalgam. This does not occur with Pt because Pt does not dissolve Na. Hence, the dissolution property of Na in Hg is key to the reaction.
3Step 3: Evaluating the Options
Let's evaluate each option:
(a) Incorrect - It's not about voltage preference but solubility of Na.
(b) Incorrect - The concentration of hydrogen ions is not the primary factor here.
(c) Correct - Na dissolves in Hg while it does not in Pt, leading to amalgamation in the Hg case.
(d) Incorrect - The inertness of Hg compared to Pt is not the cause of the observed phenomenon.
4Step 4: Conclusion
The reason why \(\mathrm{H}_2\) is liberated at the Pt cathode while sodium amalgam is formed at the Hg cathode is that Na dissolves in Hg but not in Pt. Therefore, the property of dissolution influences the electrochemical reaction.
Key Concepts
Cathode MaterialsSodium AmalgamReduction Reactions
Cathode Materials
In electrolysis, the choice of cathode material has a significant effect on the chemical processes that occur. A cathode is the electrode where reduction takes place, meaning electrons are gained by ions. The material used for the cathode plays a role in determining which ions are reduced, due to their interactions with the materials.
For example, when platinum (Pt) is used as a cathode in the electrolysis of sodium chloride ( NaCl), hydrogen ions (68^+69) from the solution are typically reduced first. Platinum does not form an amalgam with sodium and offers a stable surface for the reduction of hydrogen ions to hydrogen gas (68_269).
In contrast, mercury (Hg) as a cathode leads to a different scenario. Instead of reducing hydrogen ions, mercury reacts with sodium ions to form sodium amalgam. This difference arises from mercury's unique ability to dissolve sodium, forming a homogeneous mixture.
For example, when platinum (Pt) is used as a cathode in the electrolysis of sodium chloride ( NaCl), hydrogen ions (68^+69) from the solution are typically reduced first. Platinum does not form an amalgam with sodium and offers a stable surface for the reduction of hydrogen ions to hydrogen gas (68_269).
In contrast, mercury (Hg) as a cathode leads to a different scenario. Instead of reducing hydrogen ions, mercury reacts with sodium ions to form sodium amalgam. This difference arises from mercury's unique ability to dissolve sodium, forming a homogeneous mixture.
- Platinum: Supports hydrogen ion reduction without forming amalgams.
- Mercury: Reacts with sodium ions to create sodium amalgam.
Sodium Amalgam
Sodium amalgam is formed uniquely when mercury is used as the cathode material during the electrolysis of sodium chloride. It is essentially a type of alloy where sodium is dissolved in mercury. This mixture is crucial in various industrial applications but also illustrates a fascinating chemical behavior that occurs due to the amalgamation.
The formation of sodium amalgam happens because sodium ions (68^+69) in the solution are reduced at the mercury cathode. These sodium ions gain electrons and then easily dissolve in mercury, forming a mixture that is both stable and homogeneous.
Sodium amalgam is not only a product of electrolysis but also a useful reducing agent in organic chemistry.
The formation of sodium amalgam happens because sodium ions (68^+69) in the solution are reduced at the mercury cathode. These sodium ions gain electrons and then easily dissolve in mercury, forming a mixture that is both stable and homogeneous.
Sodium amalgam is not only a product of electrolysis but also a useful reducing agent in organic chemistry.
- Reduction at Mercury: Sodium ions gain electrons and form sodium amalgam.
- Industrial Use: Acts as a key reducing agent in various chemical processes.
Reduction Reactions
Reduction reactions are a fundamental part of electrolysis. A reduction reaction involves a gain of electrons. In the context of electrolysis, this means that ions in the solution are reduced to their elemental forms at the cathode.
During the electrolysis of 68_6696C , two common reduction reactions can occur depending on the cathode material:
During the electrolysis of 68_6696C , two common reduction reactions can occur depending on the cathode material:
- At Platinum: Hydrogen ions (68^+69) are readily reduced to form hydrogen gas (68_269) due to the non-reactivity of platinum with sodium ions.
- At Mercury: Sodium ions instead of hydrogen are reduced due to mercury's ability to dissolve sodium and form an amalgam. This leads to the formation of sodium amalgam rather than hydrogen gas.
Other exercises in this chapter
Problem 88
In the reaction: \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+\mathrm{K}_{2} \mathrm{CO}_{3} \longrightarrow\) the product formed is (a) \(\mathrm{
View solution Problem 89
Which one of the following reactions is not associated with the Solvay process of manufacture of sodium carbonate? (a) \(\mathrm{NaCl}+\mathrm{NH}_{4} \mathrm{C
View solution Problem 91
A certain metal \(\mathrm{M}\) is used to prepare an antacid, which is used as a medicine for acidity. This metal accidently catches fire, which can not be exti
View solution Problem 92
In the detection of sulphur, sodium nitroprusside solution is added to the Lassaigne's test solution, a purple coloured is obtained. It is due to (a) \(\mathrm{
View solution