Question

Ratio of moles of Fe (II) oxidised by equal volumes of equimolar \(\mathrm{KMnO}_{4}\) and \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solutions in acidic medium will be: (a) \(5: 3\) (b) \(1: 1\) (c) \(1: 2\) (d) \(5: 6\)

Short answer

The ratio of moles of Fe (II) oxidised by equal volumes of equimolar \(\mathrm{KMnO}_{4}\) and \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) solutions in acidic medium will be 5:6.

Step-by-step solution

Write the balanced chemical equations

For the reaction of \(\mathrm{KMnO}_{4}\) with \(\mathrm{Fe}^{2+}\) in acidic medium: \(2 \mathrm{MnO}_{4}^{-} + 5 \mathrm{Fe}^{2+} + 16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+} + 5 \mathrm{Fe}^{3+} + 8 \mathrm{H}_{2}\mathrm{O}\). For the reaction of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) with \(\mathrm{Fe}^{2+}\) in acidic medium: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 6 \mathrm{Fe}^{2+} + 14 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2}\mathrm{O}\). Ensure that both equations are balanced.

Determine the stoichiometry of Fe (II) with both oxidizing agents

In the first equation, 5 moles of \(\mathrm{Fe}^{2+}\) react with 1 mole of \(\mathrm{MnO}_{4}^{-}\) to produce 5 moles of \(\mathrm{Fe}^{3+}\). In the second equation, 6 moles of \(\mathrm{Fe}^{2+}\) react with 1 mole of \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) to produce 6 moles of \(\mathrm{Fe}^{3+}\). As the volumes of solutions are equal and molar, the ratio is simply the number of moles of \(\mathrm{Fe}^{2+}\) required for each reaction.

Calculate the ratio of moles of Fe (II) oxidised

The ratio of moles of \(\mathrm{Fe}^{2+}\) required for oxidation by \(\mathrm{KMnO}_{4}\) to that by \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) is found from stoichiometry which is 5:6, resulting from the comparison of moles required for each reaction (5 moles for \(\mathrm{KMnO}_{4}\) and 6 moles for \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\)).

Key concepts

Balancing Redox Reactions

Balancing redox reactions is essential in understanding how substances exchange electrons during chemical processes. To balance redox reactions, assign oxidation numbers to identify the substances that get oxidized (lose electrons) and reduced (gain electrons). Then, use the half-reaction method where oxidation and reduction reactions are balanced separately.

Oxidation and reduction half-reactions are balanced by equalizing the number of electrons lost in the oxidation half-reaction with the number of electrons gained in the reduction half-reaction. In the provided example, \\(2 \mathrm{MnO}_{4}^{-}\) is reduced to \\(2 \mathrm{Mn}^{2+}\), and \\(5 \mathrm{Fe}^{2+}\) is oxidized to \\(5 \mathrm{Fe}^{3+}\). The balance of charge and mass, potentially by adding water molecules (\\(\mathrm{H}_{2}\mathrm{O}\)) and/or hydrogen ions (\\(\mathrm{H}^{+}\)), results in a balanced chemical equation.

Mole Concept

The mole concept is a cornerstone of chemistry that provides a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe. It helps chemists quantify substances in a meaningful way. One mole is defined as the amount of a substance that contains as many entities (atoms, ions, molecules, etc.) as there are atoms in 12 grams of carbon-12.

When applying the mole concept to balancing chemical reactions, it allows us to connect the volumes and concentrations of solutions to the amount of substance involved. If solutions have the same molarity, equal volumes contain an equal number of moles. For example, the stoichiometry of the redox reaction in the exercise determines the ratio of moles of \\(\mathrm{Fe}^{2+}\) required for the reactions with \\(\mathrm{KMnO}_{4}\) and \\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\).

Chemical Equation

A chemical equation is a symbolic representation of a chemical reaction where the reactants are listed on the left-hand side and the products on the right-hand side. They are separated by an arrow indicating the direction of the reaction. Coefficients preceding the chemical formulas represent the number of moles of each substance involved in the reaction and are crucial for the equation to reflect the law of conservation of mass.

In the provided exercise, we see balanced equations representing the oxidation of \\(\mathrm{Fe}^{2+}\) by two different oxidizing agents. The coefficients indicate the stoichiometric ratios needed to achieve balance. It's important to understand that these coefficients tell us not just the ratio of molecules, but the ratio of moles, allowing us to calculate quantities in actual experiments.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions are processes where one species donates electrons (oxidation) and another species receives electrons (reduction). These reactions are fundamental to many chemical processes, including metabolic pathways and industrial applications.

In any oxidative reaction, there's a corresponding reduction; both processes occur simultaneously and involve the transfer of electrons. The overall charge balance must be maintained, which is a key principle in determining the stoichiometry of a reaction. In redox reactions, the substance that gives up electrons (oxidizer) and the one that gains the electrons (reducer) are what drive the reaction forward, similar to the redox reactions seen with \\(\mathrm{KMnO}_{4}\) and \\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) oxidizing \\(\mathrm{Fe}^{2+}\) to \\(\mathrm{Fe}^{3+}\) in the exercise.