When electricity is passed through the solution of AlCl3, 13.5 g of Al are deposited. The number of faradays must be

Question

When electricity is passed through the solution of AlCl3, 13.5 g of Al are deposited. The number of faradays must be (A) 0.50 (B) 1.50 (C) 1.00 (D) 2.00

Tardigrade Admin · Accepted Answer

Eq. of Al =(13.5/27/3)=1.5 Thus 1.5 F is needed.

Q. When electricity is passed through the solution of $A lC l_{3}, 13.5 g$ of $A l$ are deposited. The number of faradays must be

$0.50$

$1.50$

$1.00$

$2.00$

$Eq . o f A l = \frac{13.5}{27/3} = 1.5$
Thus 1.5 F is needed.

Q. When electricity is passed through the solution of AlCl3​,13.5g of Al are deposited. The number of faradays must be

0.50

1.50

1.00

2.00