A quantity of electricity liberates 3.6 g of silver from…

A quantity of electricity liberates 3.6 g of silver from its salt. What mass of aluminum will be liberated from its salt by the same quantity of electricity?

[Al = 27, Ag = 108].

A. 2.7 g

B. 1.2 g

C. 0.9 g

D. 0.3 g

QUICK ANSWER…

D

SOLUTION…  

Ag will form a mono-valent ion when in solution,

therefore, 1 faraday of electricity is required to liberate 1 mole of Ag

or 1 faraday of electricity is required to liberate 108g of Ag

hence, x faraday of electricity will liberate 3.6g of Ag

1F = 108g

x = 3.6g

cross multiplying, we have: 1(3.6) = x(108)

x = 3.6/108 = 0.03333333 faraday,

this means it will take 0.033333333 faraday of electricity to liberate 3.6g of Ag.

Applying this same quantity of electricity to aluminum, we have:

For aluminum, which forms tri-valent ions,

3 faradays is required to liberate 1 mole of Al

Or 3 faradays is required to liberate 27 g of Al, therefore,

0.033333 faradays will liberate y g of Al

3F = 27g

0.3333F = y

Cross multiplying, we have:

3(y) = 0.033333(27)

3y = 0.9

y = 0.9/3 = 0.3 g

Now for the right answer to the above question:

1. Option A is incorrect.
2. Option B is also incorrect.
3. C is incorrect.
4. D is the correct answer.

QUICK TIPS…

You may please note these/this:

• Elements with mono-valent ions (positive or negative) need 1 faraday of electricity to liberate 1 mole of the element.
• Elements with di-valent ions (positive or negative) need 2 faraday of electricity to liberate 1 mole of the element.
• Elements with tri-valent ions (positive or negative) need 3 faraday of electricity to liberate 1 mole of the element.
• For tetra-valent elements, electrolysis is not applicable, this is because tetravalent elements do not form ionic compounds which could easily be dissociated into ions to enable electrolysis.

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/ culled from 2018 JAMB-UTME Chemistry question 12 /