Question

Using $C{r}_2{O}_7^{2-}$ aqueous, solution $E_{red}^{\circ }=1.33V$ and $\Delta G^{\circ }=777.07kJmo{l}^{-1}$. What is the valency of the ion formed after reduction?

Answer 1

Answer: 3

$\underset{(+12)}{C{r}_2{O}_7^{2-}}⟶\underset{\uparrow 2x}{2C{r}_{x+}}$ (of two $C{r}_2$ -atoms reduced product)

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

$\Delta G^{\circ }=-xF{E}_{\text{cell }}^{\circ }$

$-777.07\times 10^{3}mo{l}^{-1}$

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

$(12-2x)=6.00$

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