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
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:
- Option A is incorrect.
- Option B is also incorrect.
- C is incorrect.
- D is the correct answer.
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.
Use the questions and answers session to deal further on this topic…
You can also freely share this page with friends.
/ culled from 2018 JAMB-UTME Chemistry question 12 /