Question

${\mathrm{E}}_{\text{cell }}^0$ for reaction, $4\mathrm{Al}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{s})+6{\mathrm{H}}_2\mathrm{O}+4{\mathrm{OH}}^{-}\to 4{\left[\mathrm{Al}(\mathrm{OH}{)}_4\right]}^{-}$is $2.73\mathrm{V}$. If ${\mathrm{G}}_{\mathrm{f}}^0$ of ${\mathrm{OH}}^{-}$and ${\mathrm{H}}_2\mathrm{O}$ are $-157\mathrm{kJ}{\mathrm{mole}}^{-1}$ and $-237.2\mathrm{kJ}{\mathrm{mol}}^{-1}$ determine ${\mathrm{G}}_{\mathrm{f}}^0$ for ${\left[\mathrm{Al}(\mathrm{OH}{)}_4\right]}^{-}$

• A. -1580 kJ
• B. -1303 kJ
• C. -1260 kJ
• D. -1380 kJ

Answer 1

Answer: (B)

For given cell reaction,
$\Delta G^{\circ }=-n{E}^{\circ }F$
So, $\Delta G^{\circ }=-12\times 2.73\times 96500J$
$=-3.1613\times 10^{3}kJ$
Now for given reactions,
$\Delta {\mathrm{G}}^0=4\times {\mathrm{G}}_{\mathrm{f}}^{\mathrm{o}}{\left[\mathrm{Al}(\mathrm{OH}{)}_4\right]}^{-}-6\times {\mathrm{G}}_{\mathrm{f}}^0\left[{\mathrm{H}}_2\mathrm{O}\right]-4\times {\mathrm{G}}_{\mathrm{f}}^0\left[{\mathrm{OH}}^{-}\right]$
(Also note that ${\mathrm{G}}_{\mathrm{f}}^{\mathrm{o}}$ for elements is zero)
$-3.1613\times 10^3=4{\mathrm{G}}_{\mathrm{f}}^0{\left[\mathrm{Al}(\mathrm{OH}{)}_4\right]}^{-}-6\times (-237.2)-4\times (-157)$
So, ${\mathrm{G}}_{\mathrm{f}}^{\mathrm{O}}{\left[\mathrm{Al}(\mathrm{OH}{)}_4\right]}^{-}=-1303\mathrm{kJ}$