Question

Give a clear idea about how ${MnO}_4^{-}{,CrO}_4^{2-},{Cr}_2{O}_7^{2-}$ changes its oxidation state when it is in the acidic, basic and neutral medium.

Answer 1

(1) (a) In acidic medium,

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

(b) In neutral medium,

$Mn{O}_4^{-}+2H_{2}O+3e^{-}\to Mn{O}_2+4O{H}^{-}$

(c) In basic medium,

$Mn{O}_4^{-}+1e^{-}\to Mn{O}_4^{2-}$

(2)(a) In acidic medium,

$K_{2}Cr{O}_4+8H^{+}\to 2K^{+}+C{r}^{3+}+4H_{2}O$

(3)(a) In acidic medium,

$6e^{-}+C{r}_2{O}_7^{2-}+14H_2{O}^{+}\to C{r}^{3+}+7H_{2}O$

(b) In basic medium,

$K_{2}C{r}_2{O}_7+2NaOH\to K_{2}C{r}_2{O}_4+N{a}_{2}Cr{O}_4+K_{2}O$

(c) In neutral medium,

$K_{2}C{r}_2{O}_7+H_{2}O\to 2KCr{O}_4^{2-}+2H^{+}$