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Chapter 5: Problem 109

Consider the following redox reaction: $$\begin{array}{r}4 \mathrm{NO}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 4 \mathrm{NO}_{3}^{-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq})\end{array} $$ (a) Which species is oxidized? (b) Which species is reduced? (c) Which species is the oxidizing agent? (d) Which species is the reducing agent? (e) Which species gains electrons? (f) Which species loses electrons?

Short Answer

Expert verified
a) NO is oxidized. b) O2 is reduced. c) O2 is the oxidizing agent. d) NO is the reducing agent. e) O2 gains electrons. f) NO loses electrons.

Step by step solution

01

Identification of oxidized species

To find which species is oxidized, compare the oxidation numbers. NO gas has oxidation number of +2, and NO3- ion (nitrate) has nitrogen with oxidation number of +5. Hence, Nitrogen in NO gas is oxidized since it increases its oxidation number.
02

Identification of reduced species

To find which species is reduced, compare the oxidation numbers. The oxidation number for oxygen in O2 gas is 0, and in NO3- ion (nitrate) is -2. Hence, Oxygen in O2 gas is reduced as it decreases its oxidation number.
03

Identification of oxidizing agent

The species that is reduced is the oxidizing agent. From step 2, it can be seen that O2 was reduced, thus it is the oxidizing agent.
04

Identification of reducing agent

The species that is oxidized is the reducing agent. From step 1, it can be seen that NO was oxidized, thus it is the reducing agent.
05

Identification of the species that gains electrons

The process of gaining electrons corresponds to reduction - the species that is reduced gains electrons. From step 2, it can be seen that O2 was reduced, thus it gains electrons.
06

Identification of the species that loses electrons

The process of losing electrons corresponds to oxidation - the species that is oxidized loses electrons. From step 1, it can be seen that NO was oxidized, thus it loses electrons.

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Most popular questions from this chapter

The number of moles of hydroxide ion in 0.300 L of \(0.0050 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is (a) \(0.0015 ;(\mathrm{b}) 0.0030 ;(\mathrm{c}) 0.0050\) (d) 0.010.

Which aqueous solution has the greatest \(\left[\mathrm{H}^{+}\right]:\) (a) \(0.011 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} ;\) (b) \(0.010 \mathrm{M} \mathrm{HCl} ;\) (c) \(0.010 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(1.00 \mathrm{M} \mathrm{NH}_{3} ?\) Explain your choice.

To precipitate \(\mathrm{Zn}^{2+}\) from \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}),\) add (a) \(\mathrm{NH}_{4} \mathrm{Cl} ;\) (b) \(\mathrm{MgBr}_{2} ;\) (c) \(\mathrm{K}_{2} \mathrm{CO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

An iron ore sample weighing \(0.9132 \mathrm{g}\) is dissolved in \(\mathrm{HCl}(\mathrm{aq}),\) and the iron is obtained as \(\mathrm{Fe}^{2+}(\mathrm{aq}) .\) This solution is then titrated with \(28.72 \mathrm{mL}\) of \(0.05051 \mathrm{M}\) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .\) What is the mass percent Fe in the ore sample? \(6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow_{6 \mathrm{Fe}^{3+}}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\)

Balance these equations for disproportionation reactions. (a) $\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}^{-}$ (basic solution) (b) $\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{HSO}_{3}^{-}$ (acidic solution)
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