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Chapter 4: Problem 54

Classify each of the following half-reactions as oxidation or reduction. (a) \(\mathrm{TiO}_{2}(s) \longrightarrow \mathrm{Ti}^{3+}(a q)\) (b) \(\mathrm{Zn}^{2+}(a q) \longrightarrow \mathrm{Zn}(s)\) (c) \(\mathrm{NH}_{4}{ }^{+}(a q) \longrightarrow \mathrm{N}_{2}(g)\) (d) \(\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{CH}_{2} \mathrm{O}(a q)\)

Short Answer

Expert verified
Question: Classify each of the given half-reactions as either oxidation or reduction: (a) Ti in TiO2 has an oxidation state of +4 and later changes to an oxidation state of +3. (b) Zn with an initial oxidation state of +2 changes to an oxidation state of 0. (c) N is present in NH4+ with an oxidation state of -3 and later changes to an oxidation state of 0 in N2(g). (d) C is initially present in CH3OH with an oxidation state of -2 and changes to an oxidation state of 0 in CH2O. Answer: (a) Reduction (b) Reduction (c) Oxidation (d) Oxidation

Step by step solution

01

(a) Determine the oxidation state of Ti before and after reaction

Initially, Ti is present in \(\mathrm{TiO}_{2}\). In such a compound, the oxidation state of O is always -2 (except in peroxides, peroxy acis and in \(\mathrm{F}_2 \mathrm{O}\).). Therefore, the oxidation state of Ti in \(\mathrm{TiO}_{2}\) can be calculated using the following equation: $$ \mathrm{Ti} + 2(-2) = 0 $$ Solving for Ti, we get: $$ \mathrm{Ti} = +4$$ After the reaction, the oxidation state of Ti is given as +3.
02

(a) Classify the half-reaction as oxidation or reduction

Since the oxidation state of Ti decreases from +4 to +3, it gained an electron. Thus, the half-reaction is a reduction reaction.
03

(b) Determine the oxidation state of Zn before and after the reaction

Initially, Zn is present with +2 oxidation state. After the reaction, the elemental form of Zn has an oxidation state of 0.
04

(b) Classify the half-reaction as oxidation or reduction

Since the oxidation state of Zn decreases from +2 to 0, it gained two electrons. Therefore, this half-reaction is a reduction reaction.
05

(c) Determine the oxidation state of N before and after the reaction

Initially, N is present in \(\mathrm{NH}_{4}{ }^{+}\). In this compound, the oxidation state of H is +1. We can calculate the oxidation state of N using the following equation: $$ \mathrm{N} + 4(+1) = +1 $$ Solving for N, we get: $$ \mathrm{N} = -3$$ After the reaction, N is present in \(\mathrm{N}_{2}(g)\), where the oxidation state of N is 0.
06

(c) Classify the half-reaction as oxidation or reduction

Since the oxidation state of N increases from -3 to 0, it lost three electrons. Therefore, this half-reaction is an oxidation reaction.
07

(d) Determine the oxidation state of C before and after the reaction

Initially, C is present in \(\mathrm{CH}_{3} \mathrm{OH}\). In this compound, O has an oxidation state of -2, and H has an oxidation state of +1. We can calculate the oxidation state of C using the following equation: $$ \mathrm{C} + 3(+1) + (-2) + (+1) = 0 $$ Solving for C, we get: $$ \mathrm{C} = -2$$ After the reaction, C is present in \(\mathrm{CH}_{2} \mathrm{O}\). Using the same reasoning as before, the oxidation state of C can be calculated using: $$ \mathrm{C} + 2(+1) + (-2) = 0 $$ Solving for C, we get: $$ \mathrm{C} = 0$$
08

(d) Classify the half-reaction as oxidation or reduction

Since the oxidation state of C increases from -2 to 0, it lost two electrons. Therefore, this half-reaction is an oxidation reaction. In summary: (a) Reduction (b) Reduction (c) Oxidation (d) Oxidation

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Most popular questions from this chapter

What volume of \(0.285 \mathrm{M}\) strontium hydroxide is required to neutralize \(25.00 \mathrm{~mL}\) of \(0.275 \mathrm{M}\) hydrofluoric acid (HF)?

Assign oxidation numbers to each element in (a) \(\mathrm{P}_{2} \mathrm{O}_{5}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{CO}_{3}{ }^{2-}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (e) \(\mathrm{N}_{2} \mathrm{H}_{4}\)

III Classify each of the following half-reactions as oxidation or reduction. (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\)

The percentage of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), in a powder for stomach upsets is found by titrating with \(0.275 M\) hydrochloric acid. If \(15.5 \mathrm{~mL}\) of hydrochloric acid is required to react with \(0.500 \mathrm{~g}\) of the sample, what is the percentage of sodium hydrogen carbonate in the sample? The balanced equation for the reaction that takes place is $$ \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$

Consider the following generic equation: $$ \mathrm{H}^{+}(a q)+\mathrm{B}^{-}(a q) \longrightarrow \mathrm{HB}(a q) $$ For which of the following pairs would this be the correct prototype equation for the acid-base reaction in solution? If it is not correct, write the proper equation for the acid-base reaction between the pair. (a) nitric acid and calcium hydroxide (b) hydrochloric acid and \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (c) hydrobromic acid and aqueous ammonia (d) perchloric acid and barium hydroxide (e) sodium hydroxide and nitrous acid
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