Question

The number of moles of \(\mathrm{KMnO}_{4}\) reduced by one mole of KI in alkaline medium is (a) 2 (b) 1 (c) 5 (d) 6

Short answer

The number of moles of \(\mathrm{KMnO}_4\) reduced by one mole of \(\mathrm{KI}\) in alkaline medium is 2 (option a).

Step-by-step solution

Understand the Reaction

In alkaline medium, the permanganate ion (\(\mathrm{MnO}_4^-\)) is reduced to manganese dioxide (\(\mathrm{MnO}_2\)), and iodide ions (\(\mathrm{I}^-\)) are oxidized to iodate ions (\(\mathrm{IO}_3^-\)). We need to find the stoichiometric balance between these species.

Write the Reduction Half-Reaction

The reduction half-reaction for \(\mathrm{MnO}_4^-\) in an alkaline medium is: \[ \mathrm{MnO}_4^- + 2\mathrm{H}_2\mathrm{O} + 3e^- \rightarrow \mathrm{MnO}_2 + 4\mathrm{OH}^- \] Here, \(3\) electrons are involved in reducing one \(\mathrm{MnO}_4^-\) ion.

Write the Oxidation Half-Reaction

The oxidation half-reaction for \(\mathrm{I}^-\) in an alkaline medium is: \[ \mathrm{2I}^- \rightarrow \mathrm{IO}_3^- + 6e^- \] Here, \(6\) electrons are produced when two iodide ions are oxidized.

Balance Electrons in Half-Reactions

To balance the electrons, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 1 so that the electrons cancel each other out. For reduction: \[ 2(\mathrm{MnO}_4^- + 2\mathrm{H}_2\mathrm{O} + 3e^- \rightarrow \mathrm{MnO}_2 + 4\mathrm{OH}^-) \] For oxidation: \[ \mathrm{2I}^- \rightarrow \mathrm{IO}_3^- + 6e^- \]

Combine and Simplify

Combining the reactions after balancing gives us: \[ 2\mathrm{MnO}_4^- + 4\mathrm{H}_2\mathrm{O} + 6e^- \rightarrow 2\mathrm{MnO}_2 + 8\mathrm{OH}^- \] \[ \mathrm{2I}^- \rightarrow \mathrm{IO}_3^- + 6e^- \] This shows each mole of \(\mathrm{KI}\) reacts with 2 moles of \(\mathrm{KMnO}_4\).

Choose the Correct Answer

The balanced reaction indicates that 2 moles of \(\mathrm{KMnO}_4\) are reduced per 1 mole of \(\mathrm{KI}\). Thus, the correct answer is (a) 2.

Key concepts

Redox Reactions

Redox reactions are fascinating chemical processes where two chemical species exchange electrons. These reactions involve two key changes: oxidation and reduction. In a redox reaction, one species loses electrons while the other gains them.
This simultaneous process makes it possible to drive chemical reactions where energy can flow from one molecule to another.
In the case of the reduction of potassium permanganate (KMnO4) by iodide (I-) in an alkaline medium, permanganate ions gain electrons (reduction) while iodide ions lose electrons (oxidation).

• Reduction: Gain of electrons by a species.
• Oxidation: Loss of electrons by a species.

This transfer of electrons is at the heart of redox reactions.

Stoichiometry

Stoichiometry is a fancy word for the art of balancing chemical equations. It allows us to calculate the precise amounts of reactants and products involved in a chemical reaction.
It relies on the principle of mole ratios, ensuring that the number of atoms of each element is conserved during the reaction.
In our example involving \(\mathrm{KMnO}_{4}\)\ and \(\mathrm{KI}\), stoichiometry helps in determining that 2 moles of permanganate ions react with 1 mole of iodide ions.

• Mole ratios are used to balance the reaction.
• The balanced equation aligns with the conservation of mass law.

This balance is what stoichiometry strives for, ensuring precision in chemical reactions.

Half-Reaction Method

The half-reaction method is an excellent technique for balancing redox reactions. This method splits the full reaction into two half-reactions: oxidation and reduction.
Each half-reaction is balanced separately for atoms and charge before combining them into the overall balanced equation.
For example, the reduction half-reaction for PMnO4- converts it into MnO2 by accepting electrons, while the oxidation half-reaction for I- shifts to IO3- by releasing electrons.

• The reduction half involves electron gain.
• The oxidation half shows electron loss.
• Steps taken to ensure equal electrons in both halves.

This ensures that electrons lost and gained are balanced accurately.

Oxidation-Reduction

Oxidation-reduction, or redox, processes are interconnected and involve electron transfer.
Oxidation refers to the loss of electrons, and reduction signifies the gaining of electrons, occurring simultaneously in these reactions.
When referring to \(\mathrm{KMnO}_{4}\)\ and \(\mathrm{KI}\), iodide ions get oxidized while permanganate ions are reduced, demonstrating the dual nature of redox processes.
This concept is essential to understand how reactants change their oxidation state.

• Oxidation: increase in oxidation state.
• Reduction: decrease in oxidation state.
• Dependent electron exchange in reactions.

These principles underlie the balanced approach in achieving redox reactions like our exercise.

Chemical Balancing

Chemical balancing is a core chemistry skill, ensuring that the mass and charge are conserved in a reaction.
By ensuring that the number of each type of atom is the same on both sides of the equation, we can effectively balance chemical reactions.
In the \(\mathrm{KMnO}_{4}\)\ and \(\mathrm{KI}\)\ scenario, balancing involved adjusting coefficients to match the electron exchange correctly, resulting in the reaction: 2 moles of KMnO4 reacting with 1 mole of KI.

• Balances atoms and charge.
• Maintains conservation of mass.
• Adjusts coefficients accordingly.

This ensures that all quantities match up, leading to a correctly balanced chemical equation.