Test Index

JAMB Chemistry 2010 Paper

Question : 24

Total: 49

[Al = 27, F = 96500 C mol -1]

Solution:

Concept:

Aluminum is deposited by electrolysis according to Faraday's law of electrolysis.

Explanation:

First, calculate the total charge passed,

Q = It

Q = 10 \, \text{A} \times 1930 \, \text{s} = 19300 \, \text{C}

Aluminum is trivalent, so the ionic equation is

\mathrm{Al}^{3+} + 3e^- \rightarrow \mathrm{Al}

Thus,

1

mole of Al (

27 \, \text{g}

) requires

3

moles of electrons =

3 \times 96500 \, \text{C}

Let the mass deposited be

x

Using proportion:

27 \, \text{g} \rightarrow 3 \times 96500 \, \text{C}

x \rightarrow 19300 \, \text{C}

x = \frac{27 \times 19300}{3 \times 96500}

x = \frac{27 \times 19300}{289500} = 1.8 \, \text{g}

Answer:

1.8 \, \text{g}

Go to Question:

Prev Question

Next Question