Question

An equal volume of a reducing agent is treated separately with $1MKMn{O}_4$ in acid, neutral and alkaline media. The volume of $KMn{O}_4$ required is $20mL$ in acid, $33.4mL$ in neutral and $100mL$ in alkaline medium. Find out the oxidation state of manganese in each reaction product. Give the balanced equations for all the three half reactions. Find out the volume of $1M{K}_{2}C{r}_2{O}_7$ consumed, if the same volume of the reducing agent is treated in acid medium.

Answer 1

Let $N_1,N_2$ and $N_3$ be the normalities of $1MKMn{O}_4$ solution in acid, neutral and alkaline mediums, respectively.
$20mL{N}_1\equiv 33.4mL{N}_2\equiv 100mL{N}_3$
In acidic medium, the half reaction is:
$Mn{O}_4^{-}+8H^{+}+5e^{-}=M{n}^{2+}+4H_{2}O$
$1MKMn{O}_4=5NKMn{O}_4$
Thus, from above relation,
$N_2=\frac{20}{33.4}\times N_1=\frac{20}{33.4}\times 5N=3N$
and $N_3=\frac{20}{100}\times N_2=\frac{20}{100}\times 5N=1N$
The equations in the three media are:
$Mn{O}_4^{-}+5e^{-}\stackrel{Acid}{\to }M{n}^{2+}$
$Mn{O}_4^{-}+3e^{-}\stackrel{Neutral}{\to }M{n}^{4+}$
$Mn{O}_4^{-}+e^{-}\stackrel{Alkaline}{\to }M{n}^{6+}$
The balanced equations are:
$Mn{O}_4^{-}+8H^{+}+5e^{-}\stackrel{Acid}{\to }M{n}^{2+}+4H_{2}O$
$Mn{O}_4^{-}+2H_{2}O+3e^{-}\stackrel{Neutral}{\to }Mn{O}_2+4O{H}^{-}$
$Mn{O}_4^{-}+e^{-}\stackrel{Alkaline}{\to }Mn{O}_4^{2-}$
The balanced equation in the case of acidified $K_{2}C{r}_2{O}_7$ solution can be written as:
$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 3C{r}^{3+}+7H_{2}O$
$1M{K}_{2}C{r}_2{O}_7=6N{K}_{2}C{r}_2{O}_7$
The volume required for the titration of the same volume of reducing agent with acidified $K_{2}C{r}_2{O}_7$ solution as follows:
$20mL5NKMn{O}_4\equiv V6N{K}_{2}C{r}_2{O}_7$
$V=\frac{20\times 5}{6}=16.66mL$.