A quantity of electricity liberates 3.6g of Silver from its salt. What mass of aluminium Will be liberated…

To solve this problem, we use Faraday’s Laws of Electrolysis. The key principle here is that the same quantity of electricity (charge, $Q$) will liberate masses of different elements in proportion to their equivalent masses.

1. Analyze the question

• Given: • Mass of Silver ($Ag$) liberated = $3.6\text{ g}$ • Molar mass of $Ag = 108\text{ g/mol}$ • Molar mass of $Al = 27\text{ g/mol}$
• Required: Mass of Aluminium ($Al$) liberated by the same quantity of electricity.

2. Explain the concept

The relationship between the mass ($m$) of a substance deposited and the quantity of electricity ($Q$) is given by:

Where:

• $M$ = Molar mass
• $n$ = Number of electrons transferred (valency)
• $F$ = Faraday’s constant ($96500\text{ C/mol}$)

Since the quantity of electricity ($Q$) is the same for both, we can use the ratio:

Where $E$ is the equivalent mass ($E = \frac{\text{Molar Mass}}{\text{Valency}}$).

3. Show the solution

Step 1: Determine the valency ($n$) of the metals.

• Silver ($Ag$) is in Group 1, so its valency is $1$.
• Aluminium ($Al$) is in Group 3, so its valency is $3$.

Step 2: Calculate the equivalent mass for each.

• Equivalent mass of $Ag = \frac{108}{1} = 108$
• Equivalent mass of $Al = \frac{27}{3} = 9$

Step 3: Use the ratio formula.

Step 4: Solve for the Mass of Aluminium.

4. Identify the answer

Comparing our result to the options provided, the correct answer is D.

5. Exam Tip

Always remember that the equivalent mass is simply the Molar Mass divided by the valency. If you ever forget the formula, just remember: "Mass deposited is directly proportional to the equivalent mass." Also, always check the valency of the metal—common ones like $Al^{3+}$, $Mg^{2+}$, and $Ag^+$ are frequently tested in JAMB!