Grams of Al can form by passing 305 C through an electrolytic cell containing a molten aluminium salt is 0.028429 g
Given
The balanced half-reaction is,
Al⁺³ (aq) + 3e⁻ → Al(s)
Now given Coloumb of charge = 305 C
1 Mole of Al⁺³ contains charge of 3 Faraday
hence
1 Mole of Al⁺³ = 3 Faraday
= 3× 96485 C
= 289455 C
So,
289455 C = 1 Mole of Al⁺³
1 C = [tex]\frac{1}{289455}[/tex] mole of Al⁺³
= 3.454 × 10⁻⁶ moles of Al⁺³
305 C = 3.454 × 10⁻⁶ × 305 moles of Al⁺³
= 0.0010537 moles of Al⁺³
Mass of Al⁺³ = 0.0010537 × 26.981 g
= 0.028429 g
Thus from the above conclusion we can say that grams of Al that formed by passing 305 C through an electrolytic cell containing a molten aluminium salt is 0.028429 g.
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