Question

Using $Cr _{2} O _{7}^{2-}$ aqueous, solution $E_{ red }^{\circ}=1.33\, V$ and $\Delta G^{\circ}= 777.07\, kJ\, mol ^{-1}$. What is the valency of the ion formed after reduction?

Answer 1

$\underset{(+12)}{Cr _{2} O _{7}^{2-}} \longrightarrow \underset{\uparrow 2 x}{2 Cr _{ x +}}$ (of two $Cr _{2}$ -atoms reduced product)

Change in oxidation number $\rightarrow(12-2 x)$

$\Delta G^{\circ}=-x F E_{\text {cell }}^{\circ}$

$-777.07 \times 10^{3}\, mol ^{-1}$

$=-(12-2 x) \times 96500 \times 1.33$

$(12-2 x)=6.00$

$\therefore 2 x=6 x=3 .$