Calculate the mass of Magnesium that will be liberated from its salt by the same quantity of electricity that liberated 16.0 g of Silver. [Mg = 24.0, Ag = 1...

Assessment: JAMB UTME - Chemistry - 2024 Subject: Chemistry

Question 1 Report

Calculate the mass of Magnesium that will be liberated from its salt by the same quantity of electricity that liberated 16.0 g of Silver.

[Mg = 24.0, Ag = 108 ]

Answer Details

To solve this problem, we must consider the concept of electrochemistry and Faraday's laws of electrolysis. These laws are crucial for determining the mass of a substance liberated during electrolysis.


Faraday's first law states that the mass of a substance liberated is directly proportional to the quantity of electricity that passes through the electrolyte. The mass can be calculated using the formula:


m = (Q * M) / (n * F)


Where:

  • m = mass of substance liberated
  • Q = total electric charge (in Coulombs)
  • M = molar mass of the isolated element
  • n = number of moles of electrons required to reduce 1 mole of ions of the element
  • F = Faraday's constant (approximately 96500 C/mol)

For silver (Ag), the chemical reaction at the cathode is:

Ag⁺ + e⁻ → Ag

This shows that **1 mole of electrons** is required to discharge **1 mole** of silver ions.


For magnesium (Mg), the chemical reaction at the cathode is:

Mg²⁺ + 2e⁻ → Mg

This means that **2 moles of electrons** are required to discharge **1 mole** of magnesium ions.


Given:

  • Mass of Ag liberated = **16 g**
  • Molar mass of Ag (M) = **108 g/mol**
  • Molar mass of Mg (M) = **24 g/mol**

First, find the number of moles of Ag liberated:

Number of moles of Ag = 16 g / 108 g/mol = 0.1481 mol


The same quantity of electricity will be used to liberate an equivalent in moles of electrons for Mg.

0.1481 moles of Ag require 0.1481 moles of electrons, equivalent to:

0.1481 moles of electrons for Mg. Since Mg requires 2 moles of electrons for 1 mole of Mg:

Number of moles of Mg = 0.1481 / 2 = 0.07405 mol


Finally, calculate the mass of Mg liberated:

m = 0.07405 mol * 24 g/mol = 1.7772 g


Rounding this to the closest answer provided:

The mass of magnesium that will be liberated is approximately **1.78 g**.

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