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During electrolysis of a solution of \(\mathrm{AgNO}_{3}, 9650\) coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be [2003] (a) \(1.08 \mathrm{~g}\) (b) \(10.8 \mathrm{~g}\) (c) \(21.6 \mathrm{~g}\) (d) \(108 \mathrm{~g}\) \(=\)

Short Answer

Expert verified
The mass of silver deposited is 10.8 g.

Step by step solution

01

Identify the Electrolysis Formula

In electrolysis, the mass of substance deposited at an electrode is given by Faraday's First Law of Electrolysis. The law states \( m = \frac{Q}{F} \times \frac{M}{z} \), where \( m \) is the mass of substance deposited, \( Q \) is the total charge, \( F \) is Faraday's constant \( 96500 \text{ C/mol} \), \( M \) is the molar mass of the substance, and \( z \) is the number of electrons exchanged per ion.
02

Calculate Molar Mass of Silver

Using the periodic table, we find the molar mass of silver, \( M = 107.87 \text{ g/mol} \).
03

Determine Electrons Involved

Silver ions \( \text{Ag}^+ \) gain one electron to become silver metal. Therefore, \( z = 1 \).
04

Plug in Values to Formula

Substitute the values into the formula \( m = \frac{Q}{F} \times \frac{M}{z} = \frac{9650}{96500} \times \frac{107.87}{1} \).
05

Simplify the Expression

Calculate \( m = \frac{9650}{96500} \times 107.87 = 0.1 \times 107.87 \), which simplifies to \( 10.787 \approx 10.8 \text{ g} \).

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Key Concepts

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

Faraday's First Law of Electrolysis
Faraday's First Law of Electrolysis is a fundamental principle in electrochemistry. It tells us how much substance will be deposited or dissolved during an electrochemical reaction. The law is stated mathematically as \( m = \frac{Q}{F} \times \frac{M}{z} \). Here is what each symbol in the formula means:
  • \( m \): mass of the substance (in grams) that is deposited or dissolved.
  • \( Q \): total electric charge passed through the solution, measured in coulombs.
  • \( F \): Faraday's constant, approximately \( 96500 \text{ C/mol} \).
  • \( M \): molar mass of the substance in grams per mole.
  • \( z \): number of electrons exchanged in the electrochemical reaction per ion.

Faraday's constant is a cornerstone in electrochemistry. It represents the charge of one mole of electrons. This means, if you have \( 96500 \) coulombs of charge, it equals one mole of electrons. This forms the backbone of calculating how much mass of a substance is affected during electrolysis.
Molar Mass Calculation
Molar mass is a crucial value in chemistry. It represents the mass of one mole of a substance and is expressed in grams per mole. To find the molar mass of an element, you look it up in the periodic table. For example, in our problem, the molar mass of silver (Ag) is \( 107.87 \text{ g/mol} \).
Calculating the molar mass is easy if you follow these steps:
  • Find the atomic mass of each element in the compound using the periodic table.
  • Sum the atomic masses if the substance is a compound with more than one element.

Silver is a pure element, so its molar mass is simply the atomic mass you find on the periodic table.Whenever you deal with electrolysis or any chemical reaction, determining the molar mass is often one of the first steps. Without this value, you wouldn't be able to predict or calculate the mass of products formed in the reaction.
Electrochemical Reaction
An electrochemical reaction involves chemical changes driven by electricity. Such reactions are central to electrolysis, where electrical energy causes a chemical change. These are the basics of an electrochemical reaction:
  • Ion movement: Ions move toward electrodes, causing a chemical change.
  • Cathode: Called the reduction site, where ions gain electrons.
  • Anode: The oxidation site, where ions lose electrons.

In the specific reaction where silver nitrate is electrolyzed, silver ions (\( \text{Ag}^+ \)) are reduced at the cathode. This reduction means they gain electrons to form silver metal (\( \text{Ag} \)).
The equation for what happens at the cathode is \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \). This signifies that each silver ion gains a single electron (\( z = 1 \)).Understanding this concept is key in predicting how much of a substance will be deposited in electrochemical reactions.

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Most popular questions from this chapter

In which of the following compounds the oxidation state of oxygen is other than \(-2 ?\) (a) \(\mathrm{H}_{2} \mathrm{O}_{2}\) (b) \(\mathrm{O}_{2}\) (c) \(\mathrm{O}_{2} \mathrm{~F}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{O}\)

For a \(\mathrm{Ag}-\mathrm{Zn}\) button cell, net reaction is \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \longrightarrow \mathrm{ZnO}(\mathrm{s})+2 \mathrm{Ag}(\mathrm{s})\) \(\Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left(\mathrm{Ag}_{2} \mathrm{O}\right)=-11.21 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta \mathrm{G}_{\mathrm{f}}^{\circ}(\mathrm{ZnO})=-318.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Hence \(E_{\text {cell }}^{\circ}\) of the button cell is (a) \(3.591 \mathrm{~V}\) (b) \(2.591 \mathrm{~V}\) (c) \(-1.591 \mathrm{~V}\) (d) \(1.591 \mathrm{~V}\)

A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{O}_{2}, \mathrm{Na}\) (c) \(\mathrm{O}_{2}, \mathrm{SO}_{2}\) (d) \(\mathrm{H}_{2}, \mathrm{O}_{2}\)

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) \(3: 2: 6\)

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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