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Three faraday of electricity is passed through aqueous solutions of \(\mathrm{AgNO}_{3}, \mathrm{NiSO}_{4}\) and \(\mathrm{CrCl}_{3}\) kept in three vessels using inert electrodes. The ratio in moles in which the metals \(\mathrm{Ag}, \mathrm{Ni}\) and \(\mathrm{Cr}\) will be deposited is (a) \(1: 2: 3\) (b) \(2: 3: 6\) (c) \(6: 3: 2\) (d) \(3: 2: 6\)

Short Answer

Expert verified
The ratio of the moles in which the metals are deposited is \(6: 3: 2\).

Step by step solution

01

Understand Faraday's Laws of Electrolysis

Faraday's First Law states that the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity that passes through the electrolyte. The mass \( m \) deposited is given by \( m = \frac{Q}{F} \times \frac{M}{n} \), where \( Q \) is the charge, \( F \) is Faraday's constant (96500 C/mol), \( M \) is the molar mass of the element, and \( n \) is the number of electrons exchanged in the half-reaction.
02

Write the Half-Reactions

Next, identify the electrochemical half-reactions for each metal: - Silver: \( \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag} \), where \( n = 1 \).- Nickel: \( \mathrm{Ni}^{2+} + 2e^{-} \rightarrow \mathrm{Ni} \), where \( n = 2 \).- Chromium: \( \mathrm{Cr}^{3+} + 3e^{-} \rightarrow \mathrm{Cr} \), where \( n = 3 \).
03

Calculate Moles of Metal Deposited

For 1 Faraday (96500 C), one mole of \( n \)-electron species will be deposited:- \( \mathrm{Ag} \): \( n = 1 \), 3 moles of electrons produce 3 moles of \( \mathrm{Ag} \).- \( \mathrm{Ni} \): \( n = 2 \), 3 moles of electrons produce \( \frac{3}{2} = 1.5 \) moles of \( \mathrm{Ni} \).- \( \mathrm{Cr} \): \( n = 3 \), 3 moles of electrons produce \( \frac{3}{3} = 1 \) mole of \( \mathrm{Cr} \).
04

Determine the Ratio of Moles

The moles of metals deposited for each element are: 3 moles of \( \mathrm{Ag} \), 1.5 moles of \( \mathrm{Ni} \), and 1 mole of \( \mathrm{Cr} \). Therefore, the ratio of \( \mathrm{Ag} : \mathrm{Ni} : \mathrm{Cr} \) is \( 3 : 1.5 : 1 \), which simplifies to \( 6 : 3 : 2 \).

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Key Concepts

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

Electrochemical Reactions
Electrochemical reactions are chemical reactions that occur through the transfer of electrons. These reactions are the basis of electrolysis and many other processes in chemistry. In essence, an electrochemical reaction involves the movement of electrons from one substance to another. This electron movement is usually facilitated by an external source of electricity, such as a battery or power supply.

In an electrochemical cell, two electrodes are immersed in an electrolyte. One electrode acts as the cathode where reduction (gain of electrons) occurs, and the other acts as the anode where oxidation (loss of electrons) occurs. The flow of electrons through the external circuit allows the electrochemical reaction to continue and results in the deposition or liberation of substances at the electrodes.
Electrolysis
Electrolysis is a process that uses electrical energy to bring about a chemical change, usually the breakdown of compounds into their elements or the deposition of a metal from a solution onto a substrate. This process is key in the fields of metallurgy and material science for producing pure elements from their ores or for plating metals onto various surfaces.

In electrolysis, the electrochemical cell setup is critical. It involves connecting two electrodes, typically inert ones like platinum or graphite, to an external power source and immersing them in a solution of the electrolyte. When electricity flows, it forces a chemical reaction at the electrodes, enabling the processing of substances in a controlled manner. Specific applications include the purification of metals like copper and aluminum and the electroplating of metals such as silver and chromium.
Metal Deposition
Metal deposition refers to the process by which metals are deposited onto a substrate during electrolysis. The underlying principle is that when an electrical current passes through an electrolyte, it causes the reduction of metal ions on the cathode, resulting in the deposition of metal atoms.

For instance, when silver ions are present in the solution, they gain electrons at the cathode forming solid silver. Similarly, nickel or chromium ions are reduced and deposited in their metallic form. The amount of metal deposited is governed by Faraday's Laws of Electrolysis, which state that the mass of metal deposited at an electrode is directly proportional to the number of electrons (or quantity of electricity) used. This proportional relationship enables precise control over the thickness and quality of the metal layer deposited.
Electrochemical Equations
Electrochemical equations are essential for understanding the reactions that occur during electrolysis. These equations represent the half-reactions that occur at the cathode and anode, showing the transfer of electrons and the resulting changes in oxidation states.

Each metal ion has a specific electrochemical equation. For example:
  • Silver: \( \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag} \)
  • Nickel: \( \mathrm{Ni}^{2+} + 2e^{-} \rightarrow \mathrm{Ni} \)
  • Chromium: \( \mathrm{Cr}^{3+} + 3e^{-} \rightarrow \mathrm{Cr} \)
The number of electrons involved in these reactions is represented by \( n \), which varies depending on the metal. These electrochemical equations not only illustrate the reduction process but also help calculate the amount of metal deposited from its ions based on the quantity of electricity used. By understanding these equations, one can determine the efficiency and outcome of the electrolysis process.

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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

For the equation \(\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{NO} .\) The number of electrons in its balanced form would be (a) 6 (b) 4 (c) 3 (d) 9

In an electrolytic cell, electrolysis is carried out. Electrical energy is converted into chemical energy. In an electrochemical cell, chemical reaction, i.e., redox reaction occurs and electricity is generated. So chemical energy is converted into electrical energy. Electrolysis is governed by Faraday's laws. The potential difference between the electrodes which is called electromotive force is responsible for the generation of electric energy in the electrochemical cells. The standard reduction potential values of three metallic cations \(\mathrm{X}, \mathrm{Y}\) and \(\mathrm{Z}\) are \(0.50 \mathrm{~V},-3.03 \mathrm{~V}\) and \(-1.2 \mathrm{~V}\) respectively. The order of reducing power of the corresponding metals is (a) \(X>Y>Z\) (b) \(\mathrm{Z}>\mathrm{Y}>\mathrm{X}\) (c) \(\mathrm{Y}>Z>\mathrm{X}\) (d) \(\mathrm{X}>\mathrm{Z}>\mathrm{Y}\)

Which of the following statements are correct? (a) \(\mathrm{KMnO}_{4}\) is a powerful oxidising agent. (b) \(\mathrm{KMnO}_{4}\) is a weaker oxidising agent than \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in acid medium. (c) \(\mathrm{KMnO}_{4}\) is a stronger oxidising agent than \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in acid medium. (d) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) oxidises a secondary alcohol to a ketone.

The values of standard oxidation potentials of following reactions are given below: \(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} ; E^{\circ}=0.762 \mathrm{~V}\) \(\mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} ; E^{\circ}=0.440 \mathrm{~V}\) \(\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}-E^{\circ}=-0.345 \mathrm{~V}\) \(\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+2 \mathrm{e}^{-} ; E^{\circ}=-0.800 \mathrm{~V}\) Which of the following is most easily reduced? (a) \(\mathrm{Fe}^{2+}\) (b) \(\mathrm{Ag}^{+}\) (c) \(\mathrm{Zn}^{2+}\) (d) \(\mathrm{Cu}^{2+}\)

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