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Chapter 10: Problem 118

Identify the oxidizing agent and reducing agent in the reaction \(\mathrm{NO}_{3}^{-}+4 \mathrm{Zn}+7 \mathrm{OH}^{-}+6 \mathrm{H}_{2} \mathrm{O} \rightarrow\) \(4 \mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-}+\mathrm{NH}_{3}\)

Short Answer

Expert verified
In the given reaction, the oxidizing agent is \(\mathrm{NO}_{3}^{-}\) as it undergoes reduction from +5 to -3 oxidation state, causing the oxidation of Zn. The reducing agent is Zn as it undergoes oxidation from 0 to +2 oxidation state, causing the reduction of N.

Step by step solution

01

Write down the given reaction

We are given the following reaction: \[\mathrm{NO}_{3}^{-}+4 \mathrm{Zn}+7 \mathrm{OH}^{-}+6 \mathrm{H}_{2}\mathrm{O} \rightarrow 4 \mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-}+\mathrm{NH}_{3}\]
02

Assign oxidation numbers

First, let's assign oxidation numbers to each element present in the reactants and products of the reaction: Reactants: N in \(\mathrm{NO}_{3}^{-}\): +5 (since O has an oxidation number of -2) Zn: 0 (elemental form) O in \(\mathrm{OH}^{-}\): -2 H in \(\mathrm{OH}^{-}\): +1 H in \(\mathrm{H}_{2}\mathrm{O}\): +1 O in \(\mathrm{H}_{2}\mathrm{O}\): -2 Products: Zn in \(\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-}\): +2 (since 4OH has a total charge of -2 and there's one Zn) O in \(\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-}\): -2 H in \(\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-}\): +1 N in \(\mathrm{NH}_{3}\): -3 (since H has +1 and the NH3 has no charge) H in \(\mathrm{NH}_{3}\): +1
03

Identify changes in oxidation numbers

Now, we will compare the oxidation numbers of the elements in the reactants and products to identify the changes: N: +5 to -3 (reduction) Zn: 0 to +2 (oxidation) The other elements do not change their oxidation numbers during the reaction.
04

Identify the oxidizing and reducing agents

Based on the changes in the oxidation numbers of the elements, we can identify the oxidizing and reducing agents: Oxidizing agent: \(\mathrm{NO}_{3}^{-}\) (reduces itself from +5 to -3 and causes the oxidation of Zn) Reducing agent: Zn (oxidizes itself from 0 to +2 and causes the reduction of N) So, the oxidizing agent is \(\mathrm{NO}_{3}^{-}\), and the reducing agent is Zn in the given reaction.

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Most popular questions from this chapter

Hydrogen gas burns very well in the presence of oxygen to give water: \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) In principle, should it be possible to use this chemical reaction to produce electricity? Explain.

Assign an oxidation state to each atom in the amino acid glycine:

Draw a Lewis dot diagram for \(\mathrm{NO}_{3}^{-}\), and use the oxidation- state method of electron bookkeeping to determine how many electrons each atom should be assigned.

(a) Use the following results to arrange the four metals, \(\mathrm{W}, \mathrm{X}, \mathrm{Y}, \mathrm{Z}\) in the vertical column on the right from the most active (at the top) to least active (on the bottom). \(\mathrm{W}+\mathrm{X}^{+} \rightarrow \mathrm{W}^{+}+\mathrm{X}\) most active \(\mathrm{X}+\mathrm{Z}^{+} \rightarrow \mathrm{X}^{+}+\mathrm{Z}\) \(\mathrm{Y}^{+}+\mathrm{Z} \rightarrow\) no reaction \(\mathrm{X}+\mathrm{Y}^{+} \rightarrow \mathrm{X}^{+}+\mathrm{Y}\) least active (b) Based on your series in part a, circle ALL the reactions below that you would expect to occur spontaneously as written. Explain. \(\mathrm{W}^{+}+\mathrm{Y} \rightarrow \mathrm{W}+\mathrm{Y}^{+} \quad \mathrm{W}^{+}+\mathrm{Z} \rightarrow \mathrm{W}+\mathrm{Z}^{+}\)

Use the shortcut rules to assign an oxidation state to each atom in: (a) \(\mathrm{CO}\) (b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{HCOO}^{-}\) (d) \(\mathrm{PtCl}_{6}^{2-}\)
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