/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 In the following reactions, deci... [FREE SOLUTION] | 91Ó°ÊÓ

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In the following reactions, decide which reactant is oxidized and which is reduced. Designate the oxidizing agent and the reducing agent. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+3 \mathrm{Sn}^{2+}(\mathrm{aq})+14 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Sn}^{4+}(\mathrm{aq})+21 \mathrm{H}_{2} \mathrm{O}(\ell)\) (b) \(\mathrm{FeS}(\mathrm{s})+3 \mathrm{NO}_{3}^{-}(\mathrm{aq})+4 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{NO}(\mathrm{g})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{Fe}^{3+}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Short Answer

Expert verified
(a) \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) is the oxidizing agent; \( \mathrm{Sn}^{2+} \) is the reducing agent. (b) \( \mathrm{NO}_{3}^{-} \) is the oxidizing agent; \( \mathrm{FeS} \) is the reducing agent.

Step by step solution

01

Identify Oxidation States (Reaction a)

For the reaction \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 3 \mathrm{Sn}^{2+} + 14 \mathrm{H}_{3} \mathrm{O}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 3 \mathrm{Sn}^{4+} + 21 \mathrm{H}_{2} \mathrm{O} \), determine the change in oxidation states: - The oxidation state of \( \mathrm{Cr} \) in \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) is +6. It changes to +3 in \( \mathrm{Cr}^{3+} \), indicating reduction.- The oxidation state of \( \mathrm{Sn} \) changes from +2 in \( \mathrm{Sn}^{2+} \) to +4 in \( \mathrm{Sn}^{4+} \), indicating oxidation.
02

Identify Oxidizing and Reducing Agents (Reaction a)

The oxidizing agent is the substance that gets reduced, and the reducing agent is the one that gets oxidized:- \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) is the oxidizing agent as it gains electrons and is reduced.- \( \mathrm{Sn}^{2+} \) is the reducing agent as it loses electrons and is oxidized.
03

Identify Oxidation States (Reaction b)

For the reaction \( \mathrm{FeS} + 3 \mathrm{NO}_{3}^{-} + 4 \mathrm{H}_{3} \mathrm{O}^{+} \rightarrow 3 \mathrm{NO} + \mathrm{SO}_{4}^{2-} + \mathrm{Fe}^{3+} + 6 \mathrm{H}_{2} \mathrm{O} \), determine the change in oxidation states:- \( \mathrm{Fe} \) in \( \mathrm{FeS} \) increases from 0 to +3 in \( \mathrm{Fe}^{3+} \), indicating oxidation.- \( \mathrm{N} \) in \( \mathrm{NO}_{3}^{-} \) changes from +5 to +2 in \( \mathrm{NO} \), indicating reduction.
04

Identify Oxidizing and Reducing Agents (Reaction b)

The oxidizing agent is the substance that gets reduced, and the reducing agent is the one that gets oxidized:- \( \mathrm{NO}_{3}^{-} \) is the oxidizing agent as it is reduced.- \( \mathrm{FeS} \) is the reducing agent as it is oxidized.

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Key Concepts

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

Oxidation States
Oxidation states, sometimes known as oxidation numbers, are used to track how electrons are transferred in redox reactions. These numbers help us determine which atoms are gaining or losing electrons. Here is a simple way to understand them: * Each element has a typical oxidation state, like +1 for hydrogen in most compounds or -2 for oxygen in H2O. * Changes in oxidation states during a reaction show electron movement. For example, if an oxidation state increases, it means the element has lost electrons, which is known as oxidation.
Conversely, if the oxidation state decreases, the element has gained electrons, known as reduction.
In the given exercises: * In reaction (a), chromium in \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) changes its oxidation state from +6 to +3, showing that it is being reduced.
On the other hand, tin in \( \mathrm{Sn}^{2+} \) changes from +2 to +4, indicating oxidation.
* For reaction (b), iron in \( \mathrm{FeS} \) is oxidized as its oxidation state goes from 0 to +3.
Whereas nitrogen in \( \mathrm{NO}_{3}^{-} \) is reduced, moving from +5 to +2.
By determining these changes, we can better understand which elements are losing or gaining electrons.
Oxidizing Agent
In a redox reaction, the oxidizing agent plays a crucial role. It accepts electrons and gets reduced in the process. To remember this, think of the oxidizing agent as the electron taker. Here’s a way to identify an oxidizing agent:* Look for the substance that has a decrease in oxidation state. This shows it gains electrons.* The element that undergoes reduction in a reaction is the oxidizing agent.
In reaction (a) of the exercise, \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) serves as an oxidizing agent. It accepts electrons from \( \mathrm{Sn}^{2+} \) and undergoes a reduction in its oxidation state from +6 in \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) to +3 in \( \mathrm{Cr}^{3+} \).
Similarly, in reaction (b), \( \mathrm{NO}_{3}^{-} \) is the oxidizing agent. It reduces its oxidation state from +5 to +2, thereby gaining electrons.
Understanding oxidizing agents is key to deciphering the electron flow in chemical reactions.
Reducing Agent
The reducing agent in a redox reaction is the opposite of the oxidizing agent. It's the species that donates electrons and, as a result, itself gets oxidized. Hence, it can be called the electron giver. Ways to identify a reducing agent include:* Observing the rise in oxidation state which indicates it loses electrons.* The element that undergoes oxidation in the reaction is the reducing agent.
In the case of reaction (a), \( \mathrm{Sn}^{2+} \) acts as the reducing agent. It loses electrons as its oxidation state rises from +2 to +4 when forming \( \mathrm{Sn}^{4+} \).
In reaction (b), \( \mathrm{FeS} \) plays the role of a reducing agent. The iron in \( \mathrm{FeS} \) loses electrons while changing from the 0 oxidation state to a +3 in \( \mathrm{Fe}^{3+} \).
Comprehending the function of reducing agents helps explain which elements in the reaction are giving up their electrons.

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Most popular questions from this chapter

The following reaction can be used to prepare iodine in the laboratory. $$\begin{aligned}2 \mathrm{NaI}(\mathrm{s}) &+2 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{s}) \rightarrow \\\& \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{MnSO}_{4}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\end{aligned}$$ (a) Determine the oxidation number of each atom in the equation. (b) What is the oxidizing agent, and what has been oxidized? What is the reducing agent, and what has been reduced? (c) Is the reaction observed product-favored or reactant-favored? (d) Name the reactants and products.

Balance the equation for the following precipitation reaction, and then write the net ionic equation. Indicate the state of each species \((\mathrm{s}, \ell, \mathrm{aq}, \text { or } \mathrm{g})\). $$\mathrm{CdCl}_{2}+\mathrm{NaOH} \rightarrow \mathrm{Cd}(\mathrm{OH})_{2}+\mathrm{NaCl}$$

A common method for analyzing for the nickel content of a sample is to use a precipitation reaction. Adding the organic compound dimethylglyoxime to a solution containing \(\mathrm{Ni}^{2+}\) ions, precipitates a red solid. Derive the empirical formula for the red solid based on the following composition: \(\mathrm{Ni}, 20.315 \% ; \mathrm{C}, 33.258 \% ; \mathrm{H}\) \(4.884 \% ; \mathbf{O}, 22.151 \% ;\) and \(\mathrm{N}, 19.392 \%\).

Complete and balance the equations for the following acid-base reactions. Name the reactants and products. (a) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s}) \rightarrow\) (b) \(\mathrm{HClO}_{4}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{aq}) \rightarrow\)

Balance the following equations: (a) for the reaction to produce "superphosphate" fertilizer \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}(\mathrm{aq})+\mathrm{CaSO}_{4}(\mathrm{s})\) (b) for the reaction to produce diborane, \(\mathrm{B}_{2} \mathrm{H}_{6}\) \(\begin{aligned} \mathrm{NaBH}_{4}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) & \rightarrow \\ & \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \end{aligned}\) (c) for the reaction to produce tungsten metal from tungsten(VI) oxide \(\mathrm{WO}_{3}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{W}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (d) for the decomposition of ammonium dichromate \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})\)

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