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Chapter 10: Problem 58

Which of the following are electron-transfer reactions? (a) \(2 \mathrm{CrO}_{4}^{2-}+2 \mathrm{H}^{+} \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Fe}+\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{6}+7 \mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O}\) (d) \(2 \mathrm{AgBr} \rightarrow 2 \mathrm{Ag}+\mathrm{Br}_{\dot{2}}\)

Short Answer

Expert verified
The electron-transfer reactions are: (b), (c), and (d).

Step by step solution

01

Reaction (a)

For reaction (a): \(2 \mathrm{CrO}_{4}^{2-} + 2 \mathrm{H}^{+} \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{H}_{2} \mathrm{O}\) Oxidation states for reaction (a): Cr in \(\mathrm{CrO}_{4}^{2-}\): +6 H in \(\mathrm{H}^{+}\): +1 Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\): +6 H in \(\mathrm{H}_{2} \mathrm{O}\): +1 O in \(\mathrm{H}_{2} \mathrm{O}\): -2 Since there's no change in the oxidation states of all elements involved in this reaction, no electron transfer has occurred. Therefore, reaction (a) is not an electron-transfer reaction.
02

Reaction (b)

For reaction (b): \(\mathrm{Fe} + \mathrm{NO}_{3}^{-} + 4 \mathrm{H}^{+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{NO} + 2 \mathrm{H}_{2} \mathrm{O}\) Oxidation states for reaction (b): Fe: 0 N in \(\mathrm{NO}_{3}^{-}\): +5 H in \(\mathrm{H}^{+}\): +1 Fe in \(\mathrm{Fe}^{3+}\): +3 N in \(\mathrm{NO}\): +2 H in \(\mathrm{H}_{2} \mathrm{O}\): +1 O in \(\mathrm{H}_{2} \mathrm{O}\): -2 We can observe that the oxidation state of Fe changes from 0 to +3, and the oxidation state of N changes from +5 to +2. There is a transfer of electrons in this reaction, making it an electron-transfer (redox) reaction. Therefore, reaction (b) is an electron-transfer reaction.
03

Reaction (c)

For reaction (c): \(2 \mathrm{C}_{2} \mathrm{H}_{6} + 7 \mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2} + 6 \mathrm{H}_{2} \mathrm{O}\) Oxidation states for reaction (c): C in \(\mathrm{C}_{2} \mathrm{H}_{6}\): -3 H in \(\mathrm{C}_{2} \mathrm{H}_{6}\): +1 O in \(\mathrm{O}_{2}\): 0 C in \(\mathrm{CO}_{2}\): +4 H in \(\mathrm{H}_{2} \mathrm{O}\): +1 O in \(\mathrm{CO}_{2}\): -2 O in \(\mathrm{H}_{2} \mathrm{O}\): -2 In this reaction, the oxidation state of C changes from -3 to +4, and the oxidation state of O changes from 0 to -2. This indicates a transfer of electrons between the reactants, identifying it as an electron-transfer (redox) reaction. Therefore, reaction (c) is an electron-transfer reaction.
04

Reaction (d)

For reaction (d): \(2 \mathrm{AgBr} \rightarrow 2 \mathrm{Ag} + \mathrm{Br}_{\dot{2}}\) Oxidation states for reaction (d): Ag in \(\mathrm{AgBr}\): +1 Br in \(\mathrm{AgBr}\): -1 Ag: 0 Br in \(\mathrm{Br}_{\dot{2}}\): 0 In this reaction, the oxidation state of Ag changes from +1 to 0, and the oxidation state of Br changes from -1 to 0. This indicates a transfer of electrons between the reactants, classifying this as an electron-transfer (redox) reaction. Therefore, reaction (d) is an electron-transfer reaction. In conclusion, the electron-transfer reactions are: (b), (c), and (d).

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Most popular questions from this chapter

What is the range of oxidation states possible for carbon? (Hint: Consider that carbon bonds first to atoms that are all higher in electronegativity and then to atoms all lower in electronegativity.)

Why did chemists find it necessary to invent oxidation states?

Use the shortcut rules to assign an oxidation state to each atom in: (a) \(\mathrm{O}^{2-}\) (b) \(\mathrm{Li}_{3} \mathrm{~N}\) (c) \(\mathrm{MgSO}_{4}\) (d) \(\mathrm{MnO}_{2}\)

What happens when you place a less active metal in a solution of ions of a more active metal?

Use the shortcut rules to assign an oxidation state to each atom in: (a) \(\mathrm{PCl}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{~S}\) (c) \(\mathrm{MnO}_{4}^{-}\) (d) \(\mathrm{HNO}_{3}\) (e) \(\mathrm{HCOOH}\) (f) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\)
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