Chapter 19, Problem 66AP

Interpretation Introduction

Interpretation:

The percentage by mass of iron in the given iron ore is to be calculated.

Concept introduction:

The mass of an ion discharge during electrolysis is proportional to the quantity of the electricity passed. The deposition of substances in the form of ions on the electrode with the passage of current is called Faraday’s law of electrolysis.

Oxidation is the addition of the electronegative element and the removal of the electropositive element in a chemical reaction. Reduction is the addition of the electropositive element and the removal of the electronegative element in a chemical reaction.

Electrolysis is the process in which the reaction takes place at the electrodes. The electrode that is charged positively is called anode. The electrode that is charged negatively is called cathode. During electrolysis, the reduction takes place at the cathode and the oxidation takes place at the anode.

Answer to Problem 66AP

Solution: $45.1\%$

Explanation of Solution

Given information: The mass of $\text{iron}$ ore is $0.2792\text{g}$, the volume of ${\text{KMnO}}_{\text{4}}$

is $23.30\text{mL}$, and the concentration of ${\text{KMnO}}_{\text{4}}$ is $0.0194\text{M}$.

Consider that the galvanic cell is made up of ${\text{MnO}}_4^{-}$ half-cells and ${\text{Fe}}^{\text{2+}}$ half-cells.

The cell diagram is as follows:

${\text{Fe}}^{2+}\underset{\text{anode}}{\left(aq\right){\text{|Fe}}^{3+}}\left(aq\right)\underset{\text{salt bridge}}{\text{|| }}\underset{\text{cathode}}{{\text{MnO}}_4^{-}\left(aq\right){\text{|Mn}}^{2+}}\left(aq\right)$

The two half cell equations are as follow:

$\begin{array}{l}{\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)\\ {\text{Fe}}^{3+}\left(aq\right)+e^{-}\to {\text{Fe}}^{2+}\left(aq\right)\end{array}$

${\text{MnO}}_4^{-}$ undergoes the half-cell as reduction at the cathode as:

${\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$.

${\text{Fe}}^{2+}$ undergoes the half-cell as oxidation at the anode as:

${\text{Fe}}^{2+}\left(aq\right){\text{+ e}}^{-}\to {\text{Fe}}^{3+}\left(aq\right)$

The number of electrons take part in the two half-cells reaction should be the same.

${\text{Fe}}^{2+}$ half-cell reaction is multiplied by 5 to make the number of electrons that take part in the reaction as:

$5\times \left({\text{Fe}}^{2+}\left(aq\right){\text{+ e}}^{-}\to {\text{Fe}}^{3+}\left(aq\right)\right)$

${\text{MnO}}_4^{-}$ half-cell reaction is multiplied by 1 to make the number of electrons that take part in the reaction as:

$1\times \left({\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)\right)$

Adding the two half reactions gives the overall cell reaction, which is as follows:

$\begin{array}{l}5{\text{Fe}}^{2+}\left(aq\right)\to 5{\text{Fe}}^{3+}\left(aq\right)+5e^{-}\\ {\text{MnO}}_4^{-}\left(aq\right)+5e^{-}+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ \overline{5{\text{Fe}}^{2+}\left(aq\right)+{\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\rightleftharpoons 5{\text{Fe}}^{3+}\left(aq\right)+{\text{Mn}}^{2+}\left(aq\right)}+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

The overall balanced equation is as follows:

${\text{MnO}}_4^{-}\left(aq\right)+5{\text{Fe}}^{2+}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)+5{\text{Fe}}^{3+}\left(aq\right)$

The number of moles of ${\text{KMnO}}_4$

is calculated as follows:

$\text{Moles}\text{of}{\text{KMnO}}_{\text{4}}=\text{Concentration}\text{of}{\text{KMnO}}_{\text{4}}\times \text{Volume}\text{of}{\text{KMnO}}_{\text{4}}$.

Substitute the value of the concentration and the volume in the above expression as follows:

$\begin{array}{l}\text{Moles}\text{of}{\text{KMnO}}_{\text{4}}=\left(23.30\cancel{\text{mL}}\times \frac{0.0194\text{mol}}{1000\cancel{\text{mL}}\text{sol}}\right)\\ =4.52\times {10}^{-4}\text{mol}{\text{KMnO}}_{\text{4}}\end{array}$

The overall balanced equation is as follows:

${\text{MnO}}_4^{-}\left(aq\right)+5{\text{Fe}}^{2+}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)+5{\text{Fe}}^{3+}\left(aq\right)$

From the balance equation, 1 mole of ${\text{MnO}}_4^{-}$ is stoichiometrically equivalent to 5 moles of ${\text{Fe}}^{2+}$.

The number of moles of ${\text{Fe}}^{2+}$ oxidized is calculated as follows:

$\begin{array}{c}\text{Moles}\text{of}{\text{Fe}}^{2+}=\left(4.52\times {10}^{-4}\cancel{\text{mol}{\text{MnO}}_4^{-}}\times \frac{5\text{mol}{\text{Fe}}^{2+}}{1\cancel{\text{mol}{\text{MnO}}_4^{-}}}\right)\\ =2.26\times {10}^{-3}\text{mol}{\text{Fe}}^{2+}\end{array}$.

The mass ${\text{Fe}}^{2+}$ oxidized is calculated as follows:

$\text{Mass}\text{of}{\text{Fe}}^{2+}=\text{Moles}\text{of }{\text{Fe}}^{2+}\times \text{Molecular}\text{mass}{\text{of Fe}}^{2+}$

Substitute the value of moles and the molecular mass in the above expression as:

$\begin{array}{c}\text{Mass}\text{of}{\text{Fe}}^{2+}=\left(2.26\times {10}^{-3}\text{mol}{\text{Fe}}^{2+}\times \frac{55.85\text{g}{\text{Fe}}^{2+}}{1\text{mol}{\text{MnO}}_4^{-}}\right)\\ =0.126\text{g}{\text{Fe}}^{2+}\end{array}$

The mass percentage of iron ore is calculated by using the expression, which is as follows:

$\text{Mass}\%\text{Fe}=\frac{\text{Mass}\text{of}\text{iron}}{\text{Total}\text{mass}\text{of}\text{sample}}\times 100\%$.

Substitute the mass of iron and the mass of sample in the above expression,

$\begin{array}{c}\text{Mass}\%\text{Fe}=\left(\frac{0.126\cancel{\text{g}}}{0.2792\cancel{\text{g}}}\times 100\%\right)\\ =45.1\%\end{array}$

Conclusion

The percentage by mass of iron in the ore is $45.1\%$.