The amount of electricity required to deposit 0.9 g of aluminium when the electrode reaction is Al 3++3 e- longrightarrow Al ; (atomic mass of . Al =27)

Question

The amount of electricity required to deposit 0.9 g of aluminium when the electrode reaction is Al 3++3 e- longrightarrow Al ; (atomic mass of . Al =27) (A) 9.65 × 103 C (B) 1.93 × 104 C (C) 9.65 × 104 C (D) 4.34 × 105 C

Tardigrade Admin · Accepted Answer

A l3++3 e- longrightarrow A l 27 g of Al require =3 × 96500 C 0.9 g of Al requires =(3 × 96500/27) × 0.9=9650 C =9.65 × 103 C

Q. The amount of electricity required to deposit $0.9 g$ of aluminium when the electrode reaction is $A l^{3 +} + 3 e^{-} ⟶ A l;$ (atomic mass of $A l = 27)$

$9.65 \times 1 0^{3} C$

$1.93 \times 1 0^{4} C$

$9.65 \times 1 0^{4} C$

$4.34 \times 1 0^{5} C$

$A l^{3 +} + 3 e^{-} ⟶ A l$

$27 g$ of $A l$ require $= 3 \times 96500 C$

$0.9 g$ of $A l$ requires $= \frac{3 \times 96500}{27} \times 0.9 = 9650 C = 9.65 \times 1 0^{3} C$

Q. The amount of electricity required to deposit 0.9g of aluminium when the electrode reaction is Al3++3e−⟶Al; (atomic mass of Al=27)

9.65×103C

1.93×104C

9.65×104C

4.34×105C