If mercury is used as a cathode during the electrolysis of an aqueous \(\mathrm{NaCl}\) solution, the ions discharged at cathode are (a) \(\mathrm{H}^{+}\) (b) \(\mathrm{Na}^{+}\) (c) \(\mathrm{OH}^{-}\) (d) \(\mathrm{Cl}^{-}\)

Short Answer

Expert verified
The ions discharged at the cathode are H+.

Step by step solution

01

Understand the Electrolysis Process

In electrolysis, cations (positively charged ions) move towards the cathode (negative electrode) where they can gain electrons (be reduced). We need to consider the ions present in an aqueous NaCl solution and their propensity for reduction.
02

Identify the Ions in the Solution

In an aqueous NaCl solution, the ions present are Na+, Cl-, H+ (from the self-ionization of water), and OH- (also from the self-ionization of water).
03

Determine the Discharged Ion

Mercury used as a cathode forms an amalgam with sodium, so it doesn't reduce sodium ions (Na+). Between H+ and OH-, H+ ions are preferentially reduced over OH- because they have a more positive reduction potential. Cl- ions are not reduced at the cathode; they are oxidized at the anode.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cathode and Anode in Electrolysis
Electrolysis is a chemical process used to cause a non-spontaneous reaction by passing an electric current through an electrolyte. The electrolyte here is an aqueous NaCl solution, which breaks down into ions that move towards two electrodes submerged in the solution. Each electrode plays a pivotal role in the process.

During electrolysis, two types of electrodes are used: the cathode and the anode. The cathode is the negatively charged electrode. Here, reduction (gain of electrons) occurs. Positive ions, or cations, in the solution are attracted to the cathode where they receive electrons and are reduced to their neutral state.

The anode, on the other hand, is the positively charged electrode. Oxidation (loss of electrons) happens here. Negative ions, or anions, move towards the anode, give up electrons, and are oxidized. By understanding which ions are attracted to each electrode and which processes occur (reduction or oxidation), we can predict the outcome of the electrolysis process effectively.
Reduction Potential
The concept of reduction potential is critical to understanding which ions will be reduced at the cathode during electrolysis. Reduction potential, measured in volts, indicates an element's or ion's tendency to gain electrons. In an electrolytic cell, the substance with higher reduction potential will be preferentially reduced at the cathode.

In the case of aqueous NaCl solution electrolysis, we have multiple ions to consider: Na+, H+, OH-, and Cl-. Even though Na+ is a cation and naturally moves towards the cathode, it has a lower reduction potential compared to H+. As such, H+ ions are preferentially reduced to hydrogen gas (H2) because they have a more positive reduction potential, meaning they are more likely to gain electrons. This preference guides the identification of which ion is discharged at the cathode in our exercise.
Mercury Amalgam in Electrolysis
Mercury plays a unique role when used as a cathode material in the electrolysis of aqueous NaCl. Rather than acting as a passive recipient for the deposit of reduced metal, mercury actually forms an amalgam. An amalgam is an alloy of mercury with another metal, which in the context of common electrolytic processes, usually involves the alkali and alkaline earth metals.

When sodium ions (Na+) approach the mercury cathode, they are not reduced directly to sodium metal. Instead, they form an amalgam with the mercury, because amalgamation is thermodynamically more favorable than the electrodeposition of sodium metal on the mercury surface.

This process prevents the reduction of Na+ ions at the cathode in the presence of mercury. Thus, even though Na+ is present and is a cation that could be reduced in theory, in practice, it will not be the ion that is discharged at the mercury cathode due to the formation of the mercury amalgam.

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Most popular questions from this chapter

The dissociation constant for \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(1.8 \times 10^{-5}\) at \(298 \mathrm{~K}\). The electrode potential for the half-cell: \(\mathrm{Pt} / \mathrm{H}_{2}\) \((1\) bar \() \mid 0.5 \mathrm{M}-\mathrm{CH}_{3} \mathrm{COOH}\), at \(298 \mathrm{~K}\) is \((\log 2=0.3 ; \log 3=0.48 ; 2.303 R T / F\) \(=0.06\) ) (a) \(-0.3024 \mathrm{~V}\) (b) \(-0.1512 \mathrm{~V}\) (c) \(+0.3024 \mathrm{~V}\) (d) \(+0.1512 \mathrm{~V}\)

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