Question

Use the data given in below find out the most stable oxidised species.

${\displaystyle E^0\frac{C{r}_2{O}_1^{2-}}{C{r}_{3+}}}$ = 1.33 V   ${\displaystyle E^0\frac{C{l}_2}{C{l}^{-}}}$ = 1.36 V

${\displaystyle E^0\frac{Mn{O}_4^{-}}{M{N}^{2+}}}$ = 1.51 V   ${\displaystyle E^0\frac{C{r}^{3+}}{Cr}}$ = – 0.74 V

पर्याय

• $$\mathrm{Cr}{\phantom{A}}^{3+}$$ 

• $$\mathrm{MnO}{\phantom{A}}_4^{-}$$

• $$\mathrm{CrO}{\phantom{A}}_7^{2-}$$

• $$\mathrm{Mn}{\phantom{A}}^{2+}$$ 

Answer 1

$$\mathrm{Cr}{\phantom{A}}^{3+}$$ 

Explanation:

 $$\mathrm{Cr}{\phantom{A}}^{3+}/\mathrm{Cr}$$ has most negative value of standard reduction potential. Hence, $$\mathrm{Cr}{\phantom{A}}^{3+}$$  is the most oxidized species.