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Chapter 20: Problem 22

Hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) and dinitrogen tetroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) form a self-igniting mixture that has been used as a rocket propellant. The reaction products are \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent and which as the oxidizing agent?

Short Answer

Expert verified
(a) The balanced chemical equation for the reaction is: \(2 \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{N}_{2} \mathrm{O}_{4} \rightarrow 4 \mathrm{N}_{2} + 4 \mathrm{H}_{2} \mathrm{O}\). (b) N2H4 is being oxidized (N goes from -2 to 0) and N2O4 is being reduced (N goes from +4 to 0). (c) The reducing agent is N2H4 and the oxidizing agent is N2O4.

Step by step solution

01

Write the unbalanced chemical equation

First, we write down the given reaction: N2H4 + N2O4 -> N2 + H2O
02

Balance the chemical equation

To balance the reaction, we need to make sure that the total number of atoms of each element on the left side is equal to the total number of atoms of each element on the right side of the equation. \(2 \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{N}_{2} \mathrm{O}_{4} \rightarrow 4 \mathrm{N}_{2} + 4 \mathrm{H}_{2} \mathrm{O}\) Now the chemical reaction is balanced. The balanced chemical equation is: \(2 \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{N}_{2} \mathrm{O}_{4} \rightarrow 4 \mathrm{N}_{2} + 4 \mathrm{H}_{2} \mathrm{O}\).
03

Determine oxidation states of elements

In order to find out what is being oxidized and what is being reduced, we need to find the oxidation states of all the elements in the reactants and products. Oxidation state of N in N2H4: -2 (due to the more electronegative H) Oxidation state of N in N2O4: +4 (due to the more electronegative O) Oxidation state of N in N2: 0 Oxidation state of O in H2O: -2 (due to the more electronegative O)
04

Identify the reduction and oxidation reactions

Reduction is the process of gaining electrons, which results in a decrease in oxidation state, while oxidation is the process of losing electrons, which results in an increase in oxidation state. From the given oxidation states, we can see that: - N in N2H4 goes from -2 to 0, meaning it loses electrons and therefore, gets oxidized. - N in N2O4 goes from +4 to 0, meaning it gains electrons and therefore, gets reduced. So, N2H4 gets oxidized and N2O4 gets reduced.
05

Identify the reducing and oxidizing agents

The reducing agent is the substance that causes reduction, i.e., donates electrons, while the oxidizing agent is the substance that causes oxidation, i.e., accepts electrons. Since N2H4 is getting oxidized (loses electrons), it is the reducing agent. And since N2O4 is getting reduced (gains electrons), it is the oxidizing agent. Answer: (a) The balanced chemical equation for the reaction is: \(2 \mathrm{N}_{2} \mathrm{H}_{4} + \mathrm{N}_{2} \mathrm{O}_{4} \rightarrow 4 \mathrm{N}_{2} + 4 \mathrm{H}_{2} \mathrm{O}\). (b) N2H4 is being oxidized (N goes from -2 to 0) and N2O4 is being reduced (N goes from +4 to 0). (c) The reducing agent is N2H4 and the oxidizing agent is N2O4.

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Most popular questions from this chapter

(a) What conditions must be met for a reduction potential to be a standard reduction potential? (b) What is the standard reduction potential of a standard hydrogen electrode? (c) Why is it impossible to measure the standard reduction potential of a single half-reaction?

In the Brønsted-Lowry concept of acids and bases, acidbase reactions are viewed as proton-transfer reactions. The stronger the acid, the weaker is its conjugate base. If we were to think of redox reactions in a similar way, what particle would be analogous to the proton? Would strong oxidizing agents be analogous to strong acids or strong bases?

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. (a) \(14 \mathrm{H}^{+}(a q)+2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{NaBiO}_{3}(s)\) $\quad \longrightarrow 7 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{Bi}^{3+}(a q)+5 \mathrm{Na}^{+}(a q)$ (b) $2 \mathrm{KMnO}_{4}(a q)+3 \mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\( \)\quad \longrightarrow 2 \mathrm{MnO}_{2}(s)+3 \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{KOH}(a q)$ (c) $\mathrm{Cu}(s)+2 \mathrm{AgNO}_{3}(a q) \longrightarrow 2 \mathrm{Ag (s)+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)$

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: (a) $\mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s)$ (b) $\mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q)$ (c) $\mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q)$ (d) $2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{NO}(g)+ 4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{Cu}^{2+}(a q)$

(a) What is electrolysis? (b) Are electrolysis reactions thermodynamically spontaneous? (c) What process occurs at the anode in the electrolysis of molten \(\mathrm{NaCl}\) ? (d) Why is sodium metal not obtained when an aqueous solution of \(\mathrm{NaCl}\) undergoes electrolysis?
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