Question

How many Faraday of electricity is required to deposit 0.8 g of calcium at cathode by the electrolysis of $\text{CaCl}_2$? (Molar mass of Ca = $40\ \text{g}\ \text{mol}^{-1}$)

• A. 4 F
• B. 0.04 F
• C. 2.5 F
• D. 2 F

Hint:

To solve electrolysis stoichiometry rapidly, write down the relationship as a single line: $\text{Molar Mass} \rightarrow (z \times F)$, where $z$ is the valency factor (charge of the ion). Then scale down to your given value using basic mental math.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the quantity of electricity in Faradays ($F$) needed to deposit a specific mass ($0.8\ \text{g}$) of calcium (Ca) metal at the cathode during the electrolysis of molten calcium chloride ($\text{CaCl}_2$).

Step 2: Key Formula or Approach:
According to Faraday's Laws of Electrolysis, the reduction reaction for a divalent metal ion like calcium ($\text{Ca}^{2+}$) at the cathode is given by: $$\text{Ca}^{2+} + 2\text{e}^- \rightarrow \text{Ca}_{(s)}$$ This stoichiometry demonstrates that 1 mole of Ca atoms requires exactly 2 moles of electrons, which is equivalent to $2\ F$ of electricity.

Step 3: Detailed Explanation:
From the molar mass, 1 mole of $\text{Ca} = 40\ \text{g}$. Therefore, depositing $40\ \text{g}$ of Ca requires $2\ F$ of electricity. We can set up a simple proportion to find the charge required for $0.8\ \text{g}$: $$\text{Electricity required} = \frac{2\ F \times 0.8\ \text{g}}{40\ \text{g}}$$ $$\text{Electricity required} = \frac{1.6}{40} = \frac{16 \times 10^{-1}}{40} = 0.4 \times 10^{-1} = 0.04\ F$$

Step 4: Final Answer:
The amount of electricity required is $0.04\ F$, which matches option (B).