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The time required to deposit one millimole of aluminum metal from Al₂O₃, by passing 9.65 amperes through aqueous solution of aluminum ion is
- a)30 s
- b)10 s
- c)30,000 s
- d)10,000 s
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The time required to deposit one millimole of aluminum metal from Al&#...
To determine the time required to deposit one millimole of aluminum metal from AlO, we can use Faraday's law of electrolysis. According to Faraday's law, the amount of substance deposited or liberated at an electrode is directly proportional to the amount of charge passed through the electrolyte.
The equation for Faraday's law is:
Mass (g) = (Current (A) x Time (s) x Molar mass (g/mol)) / (Faraday's constant (C/mol))
In this case, we want to deposit one millimole (0.001 mol) of aluminum metal. The molar mass of aluminum is 26.98 g/mol. The current is given as 9.65 amperes.
We can rearrange the equation to solve for time:
Time (s) = (Mass (g) x Faraday's constant (C/mol)) / (Current (A) x Molar mass (g/mol))
Plugging in the values:
Time (s) = (0.001 mol x 26.98 g/mol x 96485 C/mol) / (9.65 A)
Simplifying the equation:
Time (s) = 30.01 s
Therefore, the time required to deposit one millimole of aluminum metal from AlO by passing 9.65 amperes through the aqueous solution of aluminum ion is approximately 30 seconds.
The equation for Faraday's law is:
Mass (g) = (Current (A) x Time (s) x Molar mass (g/mol)) / (Faraday's constant (C/mol))
In this case, we want to deposit one millimole (0.001 mol) of aluminum metal. The molar mass of aluminum is 26.98 g/mol. The current is given as 9.65 amperes.
We can rearrange the equation to solve for time:
Time (s) = (Mass (g) x Faraday's constant (C/mol)) / (Current (A) x Molar mass (g/mol))
Plugging in the values:
Time (s) = (0.001 mol x 26.98 g/mol x 96485 C/mol) / (9.65 A)
Simplifying the equation:
Time (s) = 30.01 s
Therefore, the time required to deposit one millimole of aluminum metal from AlO by passing 9.65 amperes through the aqueous solution of aluminum ion is approximately 30 seconds.
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Community Answer
The time required to deposit one millimole of aluminum metal from Al&#...
Al2IIIO3 → Al0 + AlVIO3
so, here n=3
9.65 ×t = 10^-3 ×3 ×96500
t=30s
so, here n=3
9.65 ×t = 10^-3 ×3 ×96500
t=30s
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The time required to deposit one millimole of aluminum metal from Al₂O₃, by passing 9.65 amperes through aqueous solution of aluminum ion isa)30 sb)10 sc)30,000 sd)10,000 sCorrect answer is option 'A'. Can you explain this answer?
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