The overall reaction for the electralytic production of aluminium may be represented as Al 2 O 3( s )+3 C ( s ) longrightarrow 2 Al ( s )+3 CO ( g ) At 1000° C , Δ G ° for the process is 594 kJ / mole. Calculate the minimum voltage required to produce 2 mole of Al at this temperature.

Question

The overall reaction for the electralytic production of aluminium may be represented as Al 2 O 3( s )+3 C ( s ) longrightarrow 2 Al ( s )+3 CO ( g ) At 1000° C , Δ G ° for the process is 594 kJ / mole. Calculate the minimum voltage required to produce 2 mole of Al at this temperature. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The relation between Δ G ° and E is Δ G °=-n FE ° In the given reaction n=6, then E °=(-Δ G °/n F )=(-594 × 103 J / mol /6 × 96500 J / V ⋅ mol )=-1.03 V

Q. The overall reaction for the electralytic production of aluminium may be represented as Al2​O3​(s)+3C(s)⟶2Al(s)+3CO(g) At 1000∘C,ΔG∘ for the process is 594kJ/ mole. Calculate the minimum voltage required to produce 2 mole of Al at this temperature.

The relation between ΔG∘ and E is ΔG∘=−nFE∘ In the given reaction n=6, then E∘=nF−ΔG∘​=6×96500J/V⋅mol−594×103J/mol​=−1.03V

The relation between $Δ G^{\circ}$ and $E$ is $Δ G^{\circ} = - n F E^{\circ}$
In the given reaction $n = 6$ , then
$E^{\circ} = \frac{- Δ G ^{\circ}}{n F} = \frac{- 594 \times 1 0 ^{3} J / m o l}{6 \times 96500 J / V \cdot m o l} = - 1.03 V$