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A solution containing one mole per litre of each \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ; \mathrm{AgNO}_{3} ; \mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2} ;\) is being electrolyzed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are \(\mathrm{Ag} / \mathrm{Ag}^{+}=+0.80,2 \mathrm{Hg} / \mathrm{Hg}_{2}^{+4}=+0.79\) \(\mathrm{Cu} / \mathrm{Cu}^{++}=+0.34, \mathrm{Mg} / \mathrm{Mg}^{++}=-2.37\) With increasing voltage, the sequence of deposition of metals on the cathode will be (a) \(\mathrm{Cu}, \mathrm{Hg}, \mathrm{Ag}\) (b) \(\mathrm{Ag}, \mathrm{Hg}, \mathrm{Cu}, \mathrm{Mg}\) (c) \(\mathrm{Mg}, \mathrm{Cu}, \mathrm{Hg}, \mathrm{Ag}\) (d) \(\mathrm{Ag}, \mathrm{Hg}, \mathrm{Cu}\)

Short Answer

Expert verified
The sequence of deposition is (d) Ag, Hg, Cu.

Step by step solution

01

Identifying the Ions and Electrode Potentials

We are given a solution containing ions: Cu²⁺, Ag⁺, and Hg₂²⁺ with their respective reduction potentials as follows: - Ag⁺/Ag: +0.80 V - Hg₂²⁺/Hg: +0.79 V - Cu²⁺/Cu: +0.34 V These values will guide the sequence of metal deposition based on increasing electrode potential.
02

Understanding Deposition Sequence

In electrolysis, ions with higher reduction potentials will reduce and deposit before those with lower potentials. Therefore, metals will deposit in order from highest to lowest standard electrode potential.
03

Arranging Metals by Reduction Potential

From the list provided: 1. Ag⁺ (+0.80 V) 2. Hg₂²⁺ (+0.79 V) 3. Cu²⁺ (+0.34 V) Higher reduction potential means a metal will deposit first. Thus, Ag will deposit first, followed by Hg, then Cu.
04

Determining the Correct Sequence

The correct order of deposition based on reduction potentials is Ag, Hg, then Cu. Therefore, the sequence given in option (d) Ag, Hg, Cu matches this order.

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Key Concepts

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

Standard Electrode Potential
The standard electrode potential is a measure of the individual potential of a reversible electrode at standard state. In simpler terms, it tells us how likely it is for a chemical species, like an ion, to gain electrons and become reduced. This is a key concept in electrochemistry, helping us understand reactions that occur during electrolysis.
When electrodes are placed in a solution, ions can either gain or lose electrons. The standard electrode potential for a reaction is measured under specific conditions: a concentration of 1 molar, a pressure of 1 atmosphere, and a temperature of 25°C (298 K). Each half-cell reaction has its own standard electrode potential, which is measured in volts (V).
This potential allows us to predict the direction of electron flow. A positive standard electrode potential for a reduction reaction suggests that it occurs spontaneously. In electrolysis exercises, recognizing the standard electrode potentials helps determine the sequence in which metals will deposit on the electrode.
Reduction Potential
Reduction potential, also known as electrode potential, indicates the tendency of a chemical species to acquire electrons and be reduced. It's a measure of the readiness of an ion to gain electrons compared to hydrogen ions in solution.
The reduction potential of a half-reaction is expressed in volts, just like the standard electrode potential. In a series of experiments such as electrolysis, ions with a higher reduction potential will be reduced first. This is because they have a greater tendency to gain electrons and become a solid metal deposit on the electrode.
  • The more positive the reduction potential, the more likely the ion will gain electrons and become reduced.
  • A more negative reduction potential means the ion is less likely to gain electrons.
Using the gathered reduction potentials from the problem statement—such as Ag⁺/Ag with +0.80 V, Hg₂²⁺/Hg with +0.79 V, and Cu²⁺/Cu with +0.34 V—the sequence of reduction reactions can be predicted and thus, the sequence of metal deposition at the cathode.
Metal Deposition Sequence
During electrolysis, ions in the solution gain electrons at the cathode and deposit as metals. The sequence of this metal deposition on the electrode is crucial for understanding the competition between ions in solution.
The deposition sequence is primarily determined by the reduction potentials of the ions present in the solution. Ions with higher reduction potentials will become reduced first and deposit on the cathode more readily than ions with lower potentials.
  • In the given problem, the sequence followed is determined by the standard reduction potentials.
  • Ag⁺ has the highest reduction potential (+0.80 V), so it deposits first.
  • Hg₂²⁺ follows with a slightly lower potential of +0.79 V.
  • Finally, Cu²⁺ deposits last with +0.34 V.
This sequence, Ag -> Hg -> Cu, ensures that each ion deposits according to its ability to be reduced under the electrolytic conditions.

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Most popular questions from this chapter

What is the quantity of electricity (in coulombs) required to deposit all the silver from \(250 \mathrm{~mL}\) of \(1 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) solution? \((\mathrm{Ag}=108)\) (a) \(2412.5\) (b) 24125 (c) \(4825.0\) (d) 48250

Standard reduction electrode potentials of three metals A, B and \(C\) are \(+0.5 \mathrm{~V},-3.0 \mathrm{~V}\) and \(-1.2 \mathrm{~V}\) respectively. The reducing power of these metals are (a) \(\mathrm{B}>\mathrm{C}>\mathrm{A}\) (b) \(A>B>C\) (c) \(C>B>A\) (d) \(\mathrm{A}>\mathrm{C}>\mathrm{B}\)

The half cell reaction for the corrosion \(2 \mathrm{H}^{+}+1 / 2 \mathrm{O}_{2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}, E^{\circ}=1.23 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s}) ; E^{\circ}=-0.44 \mathrm{~V}\) Find the \(\Delta \mathrm{G}^{\circ}\) (in \(\mathrm{kJ}\) ) for the overall reaction. (a) \(-76\) (b) \(-322\) (c) \(-161\) (d) \(-152\)

When a quantity of electricity is passed through \(\mathrm{CuSO}_{4}\) solution, \(0.16 \mathrm{~g}\) of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of \(\mathrm{H}_{2}\) liberated at STP will be (At. \(\mathrm{wt}\) of \(\mathrm{Cu}=64\) ) (a) \(4.0 \mathrm{~cm}^{3}\) (b) \(56 \mathrm{~cm}^{3}\) (c) \(604 \mathrm{~cm}^{3}\) (d) \(8.0 \mathrm{~cm}^{3}\)

The hydrogen electrode is dipped in a solution of \(\mathrm{pH}=\) \(3.0\) at \(25^{\circ} \mathrm{C}\). The potential of hydrogen electrode would be (a) \(-0.177 \mathrm{~V}\) (b) \(0.177 \mathrm{~V}\) (c) \(1.77 \mathrm{~V}\) (d) \(0.277 \mathrm{~V}\)

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