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Chapter 14: Problem 16

Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. $$ \mathrm{Sr}+\mathrm{SO}_{3} \rightarrow \mathrm{SrSO}_{3} $$

Short Answer

Expert verified
Strontium (Sr) is oxidized, and Sulfur (S) is reduced.

Step by step solution

01

Identify the Elements

Examine the reaction \( \mathrm{Sr} + \mathrm{SO}_3 \rightarrow \mathrm{SrSO}_3 \). The elements involved are Strontium (\( \mathrm{Sr} \)) and the components of \( \mathrm{SO}_3 \), which are Sulfur (\( \mathrm{S} \)) and Oxygen (\( \mathrm{O} \)).
02

Assign Oxidation Numbers to Reactants

Assign oxidation numbers for each element in the reactants.\( \mathrm{Sr} \) is a free element with an oxidation number of 0. In \( \mathrm{SO}_3 \), \( \mathrm{O} \) has an oxidation number of -2. With three oxygens, this contributes \( -6 \) in total. To balance this, \( \mathrm{S} \) must have an oxidation number of +6.
03

Assign Oxidation Numbers to Product

For the compound \( \mathrm{SrSO}_3 \), we know \( \mathrm{Sr} \) is a Group 2 element, typically having an oxidation number of +2. The \( \mathrm{SO}_3 \) group within the compound needs to counterbalance this. Since \( \mathrm{O} \) still has an oxidation number of -2, totaling \( -6 \) for three oxygens, the \( \mathrm{S} \) must still be +4 to balance the overall charge in the compound.
04

Determine Change in Oxidation Numbers

Compare the oxidation numbers in reactants and products:- \( \mathrm{Sr} \) changes from 0 to +2, thus is oxidized.- Sulfur鈥檚 oxidation number changes from +6 to +4, thus is reduced.
05

Identify Oxidized and Reduced Elements

From the oxidation number changes:- Strontium (\( \mathrm{Sr} \)) is oxidized because it loses electrons (oxidation number increases).- Sulfur (\( \mathrm{S} \)) is reduced because it gains electrons (oxidation number decreases).

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Key Concepts

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

Oxidation Number
The oxidation number is a way to keep track of electrons in a chemical reaction. It can be thought of as an 'electron bookkeeping' system. Each element in a compound or molecule is assigned an oxidation number to show the number of electrons it either gains or loses when forming a compound.
  • Free elements, like plain Strontium (\( \mathrm{Sr} \)), have an oxidation number of 0 because they are not combined with other elements.
  • In compounds, oxygen typically has an oxidation number of -2. If there are three oxygen atoms, as in \( \mathrm{SO}_{3} \), the total contribution from oxygen is \(-6\)
  • To find the oxidation number of sulfur in \( \mathrm{SO}_{3} \), consider that the total charge of neutral compound is 0; therefore, Sulfur must be \(+6\) to balance the \(-6\) from oxygen.
  • In contrast, in the compound \( \mathrm{SrSO}_{3} \), Strontium takes on its typical Group 2 oxidation number of \(+2\), while sulfur must adjust to \(+4\) so the overall charge remains neutral when combined with the \(-6\) from oxygen.
These calculations help identify the elements undergoing changes in a redox reaction. It is crucial to assign oxidation numbers correctly to determine which elements are oxidized and reduced.
Oxidation
Oxidation is a process where an element loses electrons during a chemical reaction. When an element's oxidation number increases, it signifies that the element is oxidized.
  • An easy way to remember this concept is through the mnemonic "OIL RIG" which stands for "Oxidation Is Loss" (of electrons).
  • In the redox reaction given (\( \mathrm{Sr} + \mathrm{SO}_3 \rightarrow \mathrm{SrSO}_3 \)), Strontium (\( \mathrm{Sr} \)) is oxidized. Its oxidation number changes from 0 to +2.
  • This positive change indicates that Sr loses two electrons.
Understanding oxidation is vital for grasping the broader concept of redox reactions. As one element gets oxidized, another must get reduced, balancing out the electron exchange in the reaction.
Reduction
Reduction is the gain of electrons by an element during a chemical reaction. When an element's oxidation number decreases, it signifies that the element is reduced.
  • The second part of the mnemonic "OIL RIG"鈥�"Reduction Is Gain"鈥攈elps students remember that reduction involves gaining electrons.
  • In the given reaction of \( \mathrm{Sr} + \mathrm{SO}_3 \rightarrow \mathrm{SrSO}_3 \), Sulfur (\( \mathrm{S} \)) is reduced.
  • Sulfur's oxidation number decreases from +6 in \( \mathrm{SO}_{3} \) to +4 in \( \mathrm{SrSO}_{3} \), indicating a gain of electrons.
Understanding reduction is essential for analyzing redox reactions. As one element's oxidation number increases due to oxidation, another's decreases due to reduction, creating balance in the chemical process.

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Most popular questions from this chapter

Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. $$ \mathrm{SO}_{3}+\mathrm{SCl}_{2} \rightarrow \mathrm{SOCl}_{2}+\mathrm{SO}_{2} $$

Oxidation was once defined as chemically adding oxygen to a substance. Use this reaction to argue that this definition is consistent with the modern definition of oxidation. $$ 2 \mathrm{Mg}+\mathrm{O}_{2} \rightarrow 2 \mathrm{MgO} $$

Identify what is being oxidized and reduced in this redox reaction by assigning oxidation numbers to the atoms. $$ 2 \mathrm{Rb}+\mathrm{MgCl}_{2} \rightarrow 2 \mathrm{RbCl}+\mathrm{Mg} $$

Assign oxidation numbers to the atoms in each substance. a) \(\mathrm{C}_{6} \mathrm{H}_{6}\) b) \(\mathrm{B}(\mathrm{OH})_{3}\) C) \(\mathrm{Li}_{2} \mathrm{~S}\) d) \(\mathrm{Au}\)

Assign oxidation numbers to the atoms in each substance. a) \(\mathrm{NaH}\) b) \(\mathrm{N}_{2} \mathrm{O}_{3}\) c) \(\mathrm{NO}_{2}^{-}\) d) \(\mathrm{CuNO}_{3}\)
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