Question

The number of Faradays involved in the conversion of 0.25 mol of $Al^{3+}$ to Al is x and number of Faradays involved in the conversion of 1100 mL of 0.5 M $Cu^{2+}$ to Cu is y. The values of x and y respectively are:

• A. 0.75, 1.1
• B. 0.25, 2.2
• C. 0.50, 3.3
• D. 1.00, 2.2

Hint:

Faradays = Moles of electrons transferred!

Answer 1

The Correct Option is A

Step 1: Recall Faraday's idea.
One Faraday is the charge of one mole of electrons. To deposit a metal you need as many Faradays as the total moles of electrons taken up.

Step 2: Find x for aluminium.
The half reaction is $Al^{3+} + 3e^- \rightarrow Al$. Each $Al^{3+}$ needs 3 electrons. For 0.25 mol: $x = 3 \times 0.25 = 0.75$ F.

Step 3: First get moles of copper.
Volume is 1100 mL, which is 1.1 L. Moles $= 1.1 \times 0.5 = 0.55$ mol of $Cu^{2+}$.

Step 4: Find y for copper.
The half reaction is $Cu^{2+} + 2e^- \rightarrow Cu$, so each ion needs 2 electrons. $y = 2 \times 0.55 = 1.1$ F.

Step 5: Collect the answers.
So $x = 0.75$ and $y = 1.1$.

Step 6: Conclusion.
So the answer is 0.75, 1.1. \[ \boxed{x=0.75,\ y=1.1} \]