Δ G for the reaction (4/3) Al + O 2 longrightarrow (2/3) Al 2 O 3 is -772 kJ mol -1 of O 2. Calculate the minimum EMF in volts required to carry out an electrolysis of Al 2 O 3.

Question

Δ G for the reaction (4/3) Al + O 2 longrightarrow (2/3) Al 2 O 3 is -772 kJ mol -1 of O 2. Calculate the minimum EMF in volts required to carry out an electrolysis of Al 2 O 3. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Al longrightarrow Al 3++3 e - (4/3) mol of Al =(4/3) × 3 mol e -=4 mol e - n=4 Δ G =- nFE -772 × 1000 J =-4 × 96500 × E ∴ E =(772 × 1000/4 × 96500)=2.0 V.

Q. $Δ G$ for the reaction
$\frac{4}{3} A l + O_{2} ⟶ \frac{2}{3} A l_{2} O_{3}$
is $- 772 k J m o l^{- 1}$ of $O_{2} .$
Calculate the minimum EMF in volts required to carry out an electrolysis of $A l_{2} O_{3}$ .

$A l ⟶ A l^{3 +} + 3 e^{-}$

$\frac{4}{3}$ mol of $A l = \frac{4}{3} \times 3$ mol $e^{-} = 4$ mol $e^{-}$

$n = 4$

$Δ G = - n FE$

$- 772 \times 1000 J = - 4 \times 96500 \times E$

$∴ E = \frac{772 \times 1000}{4 \times 96500} = 2.0 V .$

Q. ΔG for the reaction 34​Al+O2​⟶32​Al2​O3​ is −772kJmol−1 of O2​. Calculate the minimum EMF in volts required to carry out an electrolysis of Al2​O3​.

Al⟶Al3++3e− 34​ mol of Al=34​×3 mol e−=4 mol e− n=4 ΔG=−nFE −772×1000J=−4×96500×E ∴E=4×96500772×1000​=2.0V.

Q. $Δ G$ for the reaction
$\frac{4}{3} A l + O_{2} ⟶ \frac{2}{3} A l_{2} O_{3}$
is $- 772 k J m o l^{- 1}$ of $O_{2} .$
Calculate the minimum EMF in volts required to carry out an electrolysis of $A l_{2} O_{3}$ .

$A l ⟶ A l^{3 +} + 3 e^{-}$

$\frac{4}{3}$ mol of $A l = \frac{4}{3} \times 3$ mol $e^{-} = 4$ mol $e^{-}$

$n = 4$

$Δ G = - n FE$

$- 772 \times 1000 J = - 4 \times 96500 \times E$

$∴ E = \frac{772 \times 1000}{4 \times 96500} = 2.0 V .$