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Chapter 9: Problem 3

In the following rection \(4 \mathrm{P}+3 \mathrm{KOH}+3 \mathrm{H}_{2} \mathrm{O} \longrightarrow 3 \mathrm{KH}_{2} \mathrm{PO}_{2}+\mathrm{PH}_{3}\) (a) \(\mathrm{P}\) is only oxidized (b) \(\mathrm{P}\) is only reduced (c) P is both oxidized as well as reduced (d) none of these

Short Answer

Expert verified
Option (c): P is both oxidized as well as reduced.

Step by step solution

01

Determine Oxidation State of P in Reactants

In the reactants, phosphorus (P) is in its elemental form, which means it has an oxidation state of 0.
02

Determine Oxidation State of P in Products

- In the product \( KH_2PO_2 \), phosphorus is bonded to two oxygen atoms and one hydrogen atom. The oxidation state of each oxygen is \(-2\), and hydrogen is \(+1\). Since the sum of oxidation states in the ion \( H_2PO_2^- \) must equal \(-1\), solve for P: \( x + 2(+1) + 2(-2) = -1 \), resulting in \(x = +1\). - In the product \( PH_3 \), phosphorus is bonded to three hydrogen atoms (each with an oxidation state of \(+1\)). Solve for P: \( x + 3(+1) = 0 \), giving \(x = -3\).
03

Analyze Changes in Oxidation State

- Since P goes from an oxidation state of \(0\) to \(+1\) in \(KH_2PO_2\), it is oxidized.- P also goes from \(0\) to \(-3\) in \(PH_3\), indicating that it is reduced.
04

Conclusion on P's Oxidation and Reduction

Since phosphorus undergoes both oxidation (increasing from 0 to +1) and reduction (decreasing from 0 to -3) in the reaction, it is both oxidized and reduced. Therefore, option (c) is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation State
The oxidation state is an essential concept in understanding redox reactions. It helps us keep track of electron transfer among different atoms. To determine the oxidation state, follow these basic rules:
  • For a pure element, the oxidation state is always 0.
  • In compounds, hydrogen is typically +1, and oxygen is -2.
  • The sum of oxidation states for all atoms in a molecule or polyatomic ion equals the overall charge of the molecule or ion.
In the provided reaction, elemental phosphorus (P) starts with an oxidation state of 0. We check its new oxidation states in each of the products. In the compound \( KH_2PO_2 \), phosphorus has an oxidation state of +1. We solve for this by setting up the equation: \( x + 2(+1) + 2(-2) = -1 \), which simplifies to \( x = +1 \). In \( PH_3 \), phosphorus has an oxidation state of -3, calculated as \( x + 3(+1) = 0 \), resulting in \( x = -3 \). Tracking these changes in oxidation state enables us to identify the processes of oxidation and reduction.
Elemental Phosphorus
Elemental phosphorus refers to phosphorus in its pure form, where it exists as P atoms bonded together. In this state, phosphorus has an oxidation state of 0 because it is not bonded to any other types of elements that could alter its electronic distribution.
Elemental phosphorus is unique in redox reactions as it can exhibit versatility. It can either donate electrons, leading to oxidation, or accept electrons, resulting in reduction.
In the examined reaction, we began with elemental phosphorus, which undergoes changes indicating both oxidation and reduction. The key takeaway is that elemental forms like this are starting points in redox reactions and can lead to various outcomes due to their inherent ability to change their oxidation state.
Disproportionation Reaction
A disproportionation reaction is a specific type of redox reaction where a single substance is both oxidized and reduced. This occurs when an element goes from a single oxidation state to two different oxidation states within the same reaction.
In the given reaction with elemental phosphorus, we see a classic example of disproportionation. Initially, phosphorus has an oxidation state of 0. It then moves to an oxidation state of +1 in \( KH_2PO_2 \) and -3 in \( PH_3 \).
This shift demonstrates both oxidation, where the oxidation state increases, and reduction, where it decreases. Because phosphorus becomes both more and less oxidized, the reaction serves as a prime illustration of how one element can undergo two opposing processes simultaneously. Understanding disproportionation is crucial as it showcases the flexibility of elements in chemical reactions and the dynamic nature of redox processes.

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Most popular questions from this chapter

Electrolysis of dilute aqueous \(\mathrm{NaCl}\) solution was carried out by passing 10 mili ampere current. The time required to librate \(0.01\) mole of \(\mathrm{H}_{2}\) gas at the cathode? (a) \(9.65 \times 10^{4} \mathrm{Sec}\) (b) \(19.3 \times 10^{4} \mathrm{Sec}\) (c) \(28.95 \times 10^{4} \mathrm{Sec}\) (d) \(38.6 \times 10^{4} \mathrm{Sec}\)

A dilute aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is electrolyzed using platinum electrodes. The product at the anode and cathode are (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{Na}\) (c) \(\mathrm{O}_{2}, \mathrm{Na}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{H}_{2}\)

The standard reduction potentials of \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) and \(\mathrm{Cu}^{2+} /\) \(\mathrm{Cu}^{+}\)are \(0.337 \mathrm{~V}\) and \(0.153 \mathrm{~V}\) respectively. The standard electrode potential of \(\mathrm{Cu}^{+} / \mathrm{Cu}\) half cell is (a) \(0.184 \mathrm{~V}\) (b) \(0.827 \mathrm{~V}\) (c) \(0.521 \mathrm{~V}\) (d) \(0.490 \mathrm{~V}\)

For a \(\mathrm{Ag}-\mathrm{Zn}\) button cell, net reaction is \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \longrightarrow \mathrm{ZnO}(\mathrm{s})+2 \mathrm{Ag}(\mathrm{s})\) \(\Delta \mathrm{G}_{\mathrm{f}}^{\circ}\left(\mathrm{Ag}_{2} \mathrm{O}\right)=-11.21 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta \mathrm{G}_{\mathrm{f}}^{\circ}(\mathrm{ZnO})=-318.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Hence \(E_{\text {cell }}^{\circ}\) of the button cell is (a) \(3.591 \mathrm{~V}\) (b) \(2.591 \mathrm{~V}\) (c) \(-1.591 \mathrm{~V}\) (d) \(1.591 \mathrm{~V}\)

Given the data at \(25^{\circ} \mathrm{C}\) \(\mathrm{Ag}+\mathrm{I}^{-} \longrightarrow \mathrm{AgI}+\mathrm{e}^{-} ; E^{\circ}=0.152 \mathrm{~V}\) \(\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; E^{\circ}=-0.800 \mathrm{~V}\) What is the value of \(\log \mathrm{K}_{\text {sp }}\) for \(\mathrm{AgI}\) ? \((2.303 \mathrm{RT} / F=0.059 \mathrm{~V})\) (a) \(-8.12\) (b) \(+8.612\) (c) \(-37.83\) (d) \(-16.13\)
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