Chapter 20, Problem 39QAP

Iron(II) can be oxidized to iron(III) by permanganate ion in acidic solution. The permanganate ion is reduced to manganese(II) ion.
(a) Write the oxidation half-reaction, the reduction half-reaction, and the overall redox equation.
(b) Calculate E° for the reaction.
(c) Calculate the percentage of Fe in an ore if a 0.3500-g sample is dissolved and the Fe²+ formed requires for titration 55.63 mL of a 0.0200 M solution of KMnO₄.

Interpretation Introduction

(a)

Interpretation:

The oxidation half, reduction half and overall redox reaction should be written for the given reaction.

Concept introduction:

The redox equation can be balanced by following a few steps:

Step 1: Split the reaction into two halves; oxidation and reduction half.

Step 2: The half reaction should be balanced.

Step 3: Number of electrons should be made equal in both the reaction halves by multiplying with suitable coefficients.

Step 4: The half reactions should be added and electrons should be cancelled to get final reaction.

Step 5: General reactions can be balanced by balancing the number of atoms of each element in the molecules of both the LHS and RHS sides.

Answer to Problem 39QAP

The oxidation half is: $F{e}^{+2}\to F{e}^{+3}+e^{-}$

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

The redox reaction is: $5F{e}^{+2}+Mn{O}_4^{-}+8H^{+}(aq)\to 5F{e}^{+3}+M{n}^{+2}+4H_{2}O$

Explanation of Solution

Given:

Iron (II) can be oxidized to iron (III) by permanganate ion in acidic medium. The permanganate ion is reduced to manganese (II) ion.

From the given reaction details first of all, the oxidation half is written as where the iron (II) is oxidized to iron (III):

Step1: Oxidation half-reaction:

$F{e}^{+2}\to F{e}^{+3}+e^{-}$

Step 2: In reduction half-reaction the permanganate ion is reduced to manganese (II) ion as:

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

Step4: Both the halves are added to get redox reaction as:

$\begin{array}{l}\left[F{e}^{+2}\to F{e}^{+3}+e^{-}\right]\times 5\\ Mn{O}_4^{-}+5e^{-}+8H^{+}\to M{n}^{+2}+4H_{2}O\\ -----------------------------\\ 5F{e}^{+2}+Mn{O}_4^{-}+8H^{+}(aq)\to 5F{e}^{+3}+M{n}^{+2}+4H_{2}O\end{array}$

Hence, the balanced redox equation is

$5F{e}^{+2}+Mn{O}_4^{-}+8H^{+}(aq)\to 5F{e}^{+3}+M{n}^{+2}+4H_{2}O$.

Interpretation Introduction

(b)

Interpretation:

The E⁰ of the reaction should be calculated.

Concept introduction:

For any redox reaction, the $E^0$ value can be determined by taking the sum of oxidation and reduction potential of the half reaction. The formula is expressed by following expression:

${\text{E}}^{\text{0}}{\text{=E}}_{\text{ox}}^{\text{0}}{\text{+E}}_{\text{red}}^{\text{0}}$

Since, for the same reaction the oxidation process will be reverse of reduction, so to obtain the oxidation potential, check for the reduction potential of the reverse reaction.

Answer to Problem 39QAP

${\text{E}}^{\text{0}}\text{= 0}\text{.743 V}$.

Explanation of Solution

The reduction potential for oxidation ands reduction half reactions are:

The oxidation half-reaction is: ${\text{Fe}}^{\text{+2}}\to {\text{Fe}}^{\text{+3}}{\text{+e}}^{\text{-}}{\text{ E}}_{\text{Ox}}^{\text{0}}\text{= - 0}\text{.769 V}$

The reduction half-reaction is: ${\text{MnO}}_{\text{4}}^{\text{-}}{\text{+5e}}^{\text{-}}{\text{+8H}}^{\text{+}}\to {\text{Mn}}^{\text{+2}}{\text{+4H}}_{\text{2}}{\text{O E}}_{\text{red}}^{\text{0}}\text{=+1}\text{.512 V}$

Hence the E° value is obtained as:

$\begin{array}{l}{\text{E}}^{\text{0}}{\text{=E}}_{\text{ox}}^{\text{0}}{\text{+E}}_{\text{red}}^{\text{0}}\text{= -0}\text{.769 V+ 1}\text{.512 V= 0}\text{.743 V}\\ {\text{E}}^{\text{0}}\text{= 0}\text{.743 V}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The percentage of Fe in an ore should be calculated.

Concept introduction:

The moles of any compound or element can be calculated by given formula:

$n=\frac{m}{M}$

Where n is the number of moles, m is the given mass and M is the molar mass of the element/compound.

From the given formula, the mass is also calculated as:

$m=n\times M$

For the calculation of mass %, the basic formula is used as:

$\text{Mass Percent=}\frac{\text{mass of element}}{\text{total mass of compound}}\times 100$

Answer to Problem 39QAP

The percentage of iron in the ore by mass is 87.7 %.

Explanation of Solution

Given:

The amount of ore dissolved is 0.3500 g and the Fe+2 formed require 55.63 mL of a 0.0200 M solution of KMnO₄ for titration.

The balanced equation of redox titration of Fe+2 is:

$5F{e}^{+2}+Mn{O}_4^{-}+8H^{+}(aq)\to 5F{e}^{+3}+M{n}^{+2}+4H_{2}O$.

Here, 5 moles of $F{e}^{+2}$ ion reacts with 1 mole of $Mn{O}_4^{-}$ ion. The ratio is 5:1.

Here, for the KMnO₄ solution, the molarity is 0.0200 M and volume taken is 55.63 mL.

The volume in L will be:

$\text{55}\text{.63 mL×}\frac{\text{1 L}}{\text{1000 mL}}\text{= 0}\text{.05563 L}$

Now, as one mole of KMnO₄ gives one mole of $Mn{O}_4^{-}$ ions, so the moles of $Mn{O}_4^{-}$ can be calculated by obtaining moles of KMnO₄ as follows:

$\begin{array}{c}\text{Molarity=}\frac{\text{Moles of solute}}{\text{Volume of solution in L}}\\ \text{Moles of solute=Molarity×Volume of solution in L}\\ {\text{Moles of MnO}}_{\text{4}}^{\text{-}}\text{=0}\text{.0200 mol/L×0}\text{.05563 L}\\ {\text{Moles of MnO}}_{\text{4}}^{\text{-}}\text{= 0}\text{.0011 mol}\end{array}$

Since, the Fe+2 reacts with $Mn{O}_4^{-}$ in 5:1 ratio, so the moles of Fe+2 produced is:

$\text{0}\text{.0011mol of }Mn{O}_4^{-}\times \frac{5{\text{ moles of Fe}}^{+2}}{1\text{mol of }Mn{O}_4^{-}}=0.0055{\text{ moles of Fe}}^{+2}$

Thus, the mass of Fe used is:

$\text{m = n×M = 0}\text{.0055 moles ×55}\text{.84 g/mol= 0}\text{.307 g}$

The mass % of Fe in ore is:

$\text{Mass Percent=}\frac{\text{mass of element}}{\text{total mass of compound}}\times 100$

$\text{Fe = }\frac{\text{0}\text{.307}}{\text{0}\text{.3500}}\text{×100 = 87}\text{.7 \%}$