Question

For the reaction \( 4 Al ( s )+3 O _{2}( g )+6 H _{2} O +4 OH ^{-} \longrightarrow 4\left[ Al ( OH )_{4}\right]^{-} ; E _{ Cell }^{\circ}=2.73 V \) If \( \Delta G _{ f }^{\circ}[ OH ]=157 kJ / mol \) and \( \Delta G _{ f }^{\circ}\left[ H _{2} O \right]=-237.2 kJ / mol \) Determine \( \Delta G _{ f }^{\circ}\left[ Al ( OH )_{4}\right]^{-} \), (free energy of formation of \( \left[ Al ( OH )_{4}\right]^{\top} \) ).

Answer 1

Given reaction,

\(4Al (s)+3O_2 (g)+6H_2O+4OH^- \rightleftharpoons 4 [Al (OH)^-_4]\)

We have \(G ^{\circ} _ {reaction} = -nFE ^{\circ}_{cell}\)

= -4 x 96500 x 2.73

= 1053780 Jmol⁻¹

= 1053.78 KJ mol⁻

and \(\triangle ^{\circ}_{reaction} = \triangle G^{\circ} _{products} - \triangle G ^{\circ}_{reactants}\)

\(1053.78 = 4 \times \triangle G ^{\circ} _f [Al (OH_4)^-]\) \(-[4\times 0)+3\times (0)+6\times (-237.2)+4\times (157)]\)

\(1053.78 = 4 \times \triangle G_f [Al (OH_4)^-]- [-795.2]\)

\(\triangle G^\circ _{f}[Al(OH_4)^-]=\frac {1053.78-795.2}{4}\)

\(\therefore \triangle G^\circ _ f [Al(OH_4)^-]= 64.645 \,KJ\,mol^{-1}\)