# Question #eaf9b

Aug 9, 2016

$\textsf{30 \textcolor{w h i t e}{x} m l}$

#### Explanation:

$\textsf{F {e}^{2 +} \rightarrow F {e}^{3 +} + e} \text{ } \textcolor{red}{\left(1\right)}$

$\textsf{C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 e \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O \text{ } \textcolor{red}{\left(2\right)}}$

In order to get the electrons to balance, you can see that we need to x $\textcolor{red}{\left(1\right)}$ by 6 then add to $\textcolor{red}{\left(2\right)} \Rightarrow$

$\textsf{C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + \cancel{6 e} + 6 F {e}^{2 +} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O + 6 F {e}^{3 +} + \cancel{6 e}}$

This tells us that $\textsf{1 \text{mol"color(white)(x)Cr_2O_7^(2-)-=6"mol} \textcolor{w h i t e}{x} F {e}^{2 +}}$

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{n = c \times v}$

$\therefore$$\textsf{n C r \left(V I\right) = 0.02 \times \frac{25}{1000} = 0.0005}$

$\therefore$$\textsf{n F e \left(I I\right) = 0.0005 \times 6 = 0.003}$

For the manganate(VII) titration the 1/2 equation is :

$\textsf{M n {O}_{4}^{-} + 8 {H}^{+} + 5 e \rightarrow M {n}^{2 +} + 4 {H}_{2} O \text{ } \textcolor{red}{\left(3\right)}}$

To get the electrons to balance, we need to x $\textcolor{red}{\left(1\right)}$ by 5 then add to $\textcolor{red}{\left(3\right)} \Rightarrow$

$\textsf{M n {O}_{4}^{-} + 8 {H}^{+} + \cancel{5 e} + 5 F {e}^{2 +} \rightarrow M {n}^{2 +} + 4 {H}_{2} O + 5 F {e}^{3 +} + \cancel{5 e}}$ .

This tells us that $\textsf{1 \text{mol"color(white)(x)MnO_4^(-)-=5"mol} \textcolor{w h i t e}{x} F {e}^{2 +}}$

$\therefore$$\textsf{n M n {O}_{4}^{-} = \frac{0.003}{5} = 0.0006}$

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{v = \frac{n}{c} = \frac{0.0006}{0.02} = 0.03 \textcolor{w h i t e}{x} L}$

$\textsf{= 30 \textcolor{w h i t e}{x} m l}$

Aug 9, 2016

The reduction half reaction of $C {r}_{2} {O}_{7}^{\text{2-}}$ ion in acid medium

$C {r}_{2} {O}_{7}^{\text{2-" +14H^+ +6e->2Cr^"3+}} + 7 {H}_{2} O \ldots . \left(1\right)$

And the oxidation half reaction of $F {e}^{\text{2+}}$

$F {e}^{\text{2+"-e->Fe^"3+}} \ldots \ldots \left(2\right)$

The reduction half reaction of $M n {O}_{4}^{\text{-}}$ ion in acid medium

$M n {O}_{4}^{\text{-" +8H^+ +5e->Mn^"2+}} + 4 {H}_{2} O \ldots . \left(3\right)$

Multiplying (2) by 6 and then adding with (1) the balanced equation of redox reaction by dichhromate becomes

$C {r}_{2} {O}_{7}^{\text{2-" +14H^+ +6Fe^"2+"->2Cr^"3+"+7H_2O+6Fe^"3+}} . . \left(4\right)$

This equation (4)reveals that

$1 \text{mol "Fe^"2+"equiv 1/6 "mol "Cr_2O_7^"2-} \textcolor{red}{\ldots \ldots \left(4 A\right)}$

Multiplying (2) by 5 and then adding with (3) the balanced equation of redox reaction by permanganate becomes

$M n {O}_{4}^{\text{-" +8H^+ +5Fe^"2+"->Mn^"2+"+4H_2O+5Fe^"3+}} . . \left(5\right)$

This equation (5)reveals that

$1 \text{mol "Fe^"2+"equiv 1/5 "mol "MnO_4^"-} \textcolor{red}{\ldots \ldots \left(5 A\right)}$

So by combining (4A) and (5A) we can say

$\frac{1}{6} \text{mol "Cr_2O_7^"2-"equiv1/5" mol "MnO_4^"-}$

$\therefore 1 \text{mol "Cr_2O_7^"2-"equiv6/5" mol "MnO_4^"-}$

Now
$50 \text{ml "Fe^"2+"equiv25ml" "0.02M" "Cr_2O_7^"2-}$

$= 25 \times 0.02 \times {10}^{-} 3 \text{mol "Cr_2O_7^"2-}$

$\equiv \frac{6}{5} \times 25 \times 0.02 \times {10}^{-} 3 \text{mol "MnO_4^"-}$

$\equiv 30 \times 0.02 \times {10}^{-} 3 \text{mol "MnO_4^"-}$

$\equiv 30 \text{ml "0.02M" "MnO_4^"-}$

So if titration is carried out with 0.02 M $M n {O}_{4}^{-}$ solution instead of 0.02M $C {r}_{2} {O}_{7}^{\text{2-}}$
solution then 30ml 0.02M permanganate solution will be required.