Question

For the reaction :
$4Al\left(s\right)+3O_2\left(g\right)+6H_{2}O+4\stackrel{\ominus }{O}H\to 4\left[Al\right(OH{)}_4{]}^{\oplus }$
$E_{cell}^{\ominus }=2.73V$
If ${\Delta }_f{G}^{\ominus }\left[\stackrel{\oplus }{O}H\right]=-157kJmo{l}^{-1}$;
${\Delta }_f{G}^{\ominus }\left(H_{2}O\right)=-237.2kJmo{l}^{-1}$.
The value of ${\Delta }_f{G}^{\ominus }\left[Al\right(OH{)}_4{]}^{\ominus }$ is:
[free energy of formation of $\left(Al\right(OH{)}_4^{\ominus }]$.

• A. ${\Delta }_f{G}^{\ominus }\left[Al\right(OH{)}_4{]}^{\ominus }=+3.50\times {10}^{3}kJmo{l}^{-1}$
• B. ${\Delta }_f{G}^{\ominus }\left[Al\right(OH{)}_4{]}^{\ominus }=+2.40\times {10}^{3}kJmo{l}^{-1}$
• C. ${\Delta }_f{G}^{\ominus }\left[Al\right(OH{)}_4{]}^{\ominus }=-1.30\times {10}^{3}kJmo{l}^{-1}$
• D. $Noneofthese$

Answer 1

The correct option is C ${\Delta }_f{G}^{\ominus }\left[Al\right(OH{)}_4{]}^{\ominus }=-1.30\times {10}^{3}kJmo{l}^{-1}$

$4A\stackrel{0}{l}\left(s\right)+3O_2\left(g\right)+6H_{2}O+4\stackrel{}{\ominus }OH$
$\to 4\left[\stackrel{+3}{Al}\right(OH{)}_4{]}^{\ominus }$
$E_{cell}^{\ominus }=2.73V$
Here, $n=4\times 3=12$ (Change in oxidation number of Al for four moles)
$\Delta G^{\ominus }=-12\times 96500\times 2.73=-3.16\times {10}^{3}kJ$
$\Delta G^{\ominus }=4{\Delta }_f{G}^{\ominus }\left[Al\right(OH{)}_4{]}^{\ominus }-\left[6{\Delta }_f{G}^{\ominus }\right(H_{2}O)+4{\Delta }_f{G}^{\ominus }(\stackrel{\ominus }{O}H\left)\right]$
$\Rightarrow -3.16\times {10}^3=4x-\left[6\right(-237.2)+4(-157\left)\right]$
$\Rightarrow 4x\Rightarrow -5212.54\Rightarrow xd=-1.30\times {10}^{3}kJmo{l}^{-1}$
${\Delta }_f{G}^{\ominus }\left[Al\right(OH{)}_4{]}^{\ominus }=-1.30\times {10}^{3}kJmo{l}^{-1}$