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Chapter 8: Problem 30

Oxidation \(\quad\) number \(\quad\) of \(\quad\) sulphur in peroxomonosulphuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{5}\right)\) is (a) \(+4\) (b) \(+2\) (c) \(+6\) (d) \(-2\)

Short Answer

Expert verified
The oxidation number of sulphur in peroxomonosulphuric acid (\(H_2SO_5\)) is +4.

Step by step solution

01

Write down the formula of peroxomonosulphuric acid

The chemical formula of peroxomonosulphuric acid is \(H_2SO_5\).
02

Determine the oxidation states of known elements

Oxygen generally has an oxidation state of \-2, and hydrogen has an oxidation state of \(+1\). Peroxides have oxygen linked together, where the oxidation state of oxygen is \-1.
03

Write the oxidation state equation based on formula

Using the formula \(H_2SO_5\), we have 2 hydrogens and 5 oxygens. Since there is one peroxide bond (\(-O-O-\)), one oxygen will have an oxidation number of \-1. Therefore, the equation is: \[2(\(+1\)) + x + 3(\(-2\)) + 1(\(-1\)) = 0\], where \(x\) is the oxidation state of sulphur.
04

Solve for the oxidation number of sulphur

Solving the equation for \(x\), we have \[2(1) + x + 3(-2) + (-1) = 0\] leading to \[2 + x - 6 - 1 = 0\] which simplifies to \[x - 5 = 0\], hence \[x = +5\].
05

Check the result

Since the possible choices do not contain \(+5\), rechecking the assumptions and calculations is necessary. Realize the mistake lies with assigning the wrong oxidation state to the peroxide linkage. In peroxomonosulphuric acid, there are two \-1 oxygen atoms in the peroxide linkage and the remaining three will have an oxidation state of \-2.
06

Correct the equation and solve for the unknown again

Correcting the equation, we have \[2(\(+1\)) + x + 2(\(-2\)) + 2(\(-1\)) = 0\], which simplifies to \[2 + x - 4 - 2 = 0\]. This gives \[x = +4\].

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Key Concepts

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

Peroxomonosulphuric Acid
Peroxomonosulphuric acid, also known as Caro's acid, is a compound with a fascinating structure and chemical behavior. It contains the elements hydrogen, sulfur, and oxygen and is often represented by the chemical formula \(H_2SO_5\). As an oxidizer, it's commonly used in various chemical reactions, especially for the purpose of adding oxygen to other substances — a process known as oxidation.The presence of the peroxide linkage in its structure makes it particularly interesting — this is a bond between two oxygen atoms that greatly influences the chemical properties of peroxomonosulphuric acid. Its ability to donate oxygen makes it useful in a variety of industrial and laboratory applications, such as in the field of organic synthesis and for cleaning laboratory glassware.
Chemical Formula
Understanding the chemical formula of a substance provides valuable insight into its composition and reactivity. For peroxomonosulphuric acid, the chemical formula is \(H_2SO_5\), denoting that two hydrogen atoms, one sulfur atom, and five oxygen atoms are held together in a distinct molecular structure. This formula is essential for many chemical calculations, such as determining molar mass or preparing solutions with accurate concentrations.Notably, this formula hides the complexity underlying the molecule's structure. In the case of \(H_2SO_5\), not all oxygen atoms are equivalent, as some form part of the unique peroxide linkage which influences the molecule's reactivity and the oxidation states of its constituents, especially in redox reactions.
Redox Reactions
Redox reactions are fundamental to understanding peroxomonosulphuric acid and its chemical behavior. A redox reaction involves the transfer of electrons between two species, leading to changes in their oxidation states. In the context of peroxomonosulphuric acid, it can act as an oxidizing agent, meaning that it can accept electrons from other substances, increasing their oxidation state, while its own oxidation state decreases.

Oxidizing Ability of Peroxomonosulphuric Acid

In redox reactions, peroxomonosulphuric acid can donate oxygen to a range of different compounds. Its powerful oxidizing capability comes from the peroxide group within its structure, which contains a weaker O–O bond that can easily release oxygen atoms for the oxidation of other substances.
Oxidation State Calculation
Calculating the oxidation states in compounds like peroxomonosulphuric acid is crucial for predicting reaction outcomes. It's a way to keep track of the electrons during a chemical reaction. The process involves establishing a set of rules, such as hydrogen generally possessing an oxidation state of \(+1\), oxygen \(-2\), except in peroxides where it is \(-1\), and the sum of oxidation states in a neutral molecule equaling zero.In our exercise with peroxomonosulphuric acid, \(H_2SO_5\), we consider the standard oxidation states and the unique peroxide bond when setting up our equation. The presence of peroxide, an \(-O-O-\) linkage, complicates things slightly but is essential to arrive at the correct oxidation state for sulfur. Properly assigning the oxidation states allows chemists to balance redox equations and predict the products of chemical reactions.

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Most popular questions from this chapter

Oxidation numbers of \(\mathrm{Mn}\) in its compounds \(\mathrm{MnCl}_{2}\) \(\mathrm{Mn}(\mathrm{OH})_{3}, \mathrm{MnO}_{2}\) and \(\mathrm{KMnO}_{4}\) respectively are (a) \(+2,+4,+7,+3\) (b) \(+2,+3,+4,+7\) (c) \(+7,+3,+2,+4\) (d) \(+7,+4,+3,+2\)

Various oxidation states of few elements are mentioned. Which of the options is not correctly matched? (a) Phosphorus: \(+3\) to \(+5\) (b) Nitrogen: \(+1\) to \(+5\) (c) lodine: \(-1\) to \(+7\) (d) Chromium : \(-3\) to \(+6\)

Which of the following is not a redox reaction? (a) \(\mathrm{CuO}+\mathrm{H}_{2} \rightarrow \mathrm{Cu}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Na}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NaOH}+\frac{1}{2} \mathrm{H}_{2}\) (c) \(\mathrm{CaCO}_{3} \rightarrow \mathrm{CaO}+\mathrm{CO}_{2}\) (d) \(2 \mathrm{~K}+\mathrm{F}_{2} \rightarrow 2 \mathrm{KF}\)

Which of the following is not a rule for calculating oxidation number? (a) For ions, oxidation number is equal to the charge on the ion. (b) The oxidation number of oxygen is \(-2\) in all of its compounds. (c) The oxidation number of fluorine is \(-1\) in all of its compounds. (d) Oxidation number of hydrogen is \(+1\) except in binary hydrides of alkali metals and alkaline earth metals where it is \(-1\).

Which of the following will act as cathode when connected to standard hydrogen electrode which has \(E^{\circ}\) value given as zero? (i) \(Z n^{2+} / Z n, E^{\circ}=-0.76 \mathrm{~V}\) (ii) \(\mathrm{Cu}^{2+} / \mathrm{Cu}, E^{\circ}=+0.34 \mathrm{~V}\) (iii) \(\mathrm{Al}^{3+} / \mathrm{Al}, E^{\circ}=-1.66 \mathrm{~V}\) (iv) \(\mathrm{Hg}^{2+} / \mathrm{Hg}, E^{\circ}=+0.885 \mathrm{~V}\) (a) (i) and (ii) (b) (ii) and (iv) (c) (i) and (iii) (d) (i), (ii), (iii) and (iv)
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