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Chapter 4: Problem 40

Use the following reaction to define redox reaction, half-reaction, oxidizing agent, and reducing agent. \(4 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Na}_{2} \mathrm{O}(s)\)

Short Answer

Expert verified
The redox reaction is \(4\, \mathrm{Na}(s) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Na}_{2}\mathrm{O}(s)\). The oxidation half-reaction is \(4\, \mathrm{Na}(s) \longrightarrow 4\, \mathrm{Na^{+}}(aq) + 4e^{-}\) and the reduction half-reaction is \(\mathrm{O_{2}}(g) + 4e^{-} \longrightarrow 2\, \mathrm{O^{2-}}(aq)\). The reducing agent is Sodium (Na) and the oxidizing agent is Oxygen (\(\mathrm{O_{2}}\)).

Step by step solution

01

Identify the Oxidation State Changes

Determine the initial and final oxidation states of the elements in the reaction:\n\nFor Sodium (Na), initial state = 0 (pure element), final state in Na2O = +1.\n\nFor Oxygen (O), initial state = 0 (pure element), final state in Na2O = -2. \nSodium's oxidation state increases, implying it has been oxidized. Oxygen's oxidation state decreases, implying it has been reduced.
02

Define the Half-Reactions and Redox Reaction

The half-reactions are the individual oxidation and reduction processes. Oxidation half-reaction: \(4\, \mathrm{Na}(s) \longrightarrow 4\, \mathrm{Na}^{+}(aq) + 4 e^{-}\). Reduction half-reaction: \( \mathrm{O}_{2}(g) + 4 e^{-} \longrightarrow 2 \mathrm{O}^{2-}(aq)\). The sum of these half-reactions is the full redox reaction: \(4\, \mathrm{Na}(s) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Na}_{2}\mathrm{O}(s)\)
03

Identify the Oxidizing and Reducing Agents

The oxidizing agent is the substance that is reduced (gains electrons), and the reducing agent is the substance that is oxidized (loses electrons). In this case, Sodium (Na) is the reducing agent as it loses electrons and gets oxidized, while Oxygen (O2) is the oxidizing agent as it gains electrons and gets reduced.

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Most popular questions from this chapter

Water is added to \(25.0 \mathrm{~mL}\) of a \(0.866 \mathrm{M} \mathrm{KNO}_{3}\) solution until the volume of the solution is exactly \(500 \mathrm{~mL}\). What is the concentration of the final solution?

Write the equation that enables us to calculate the concentration of a diluted solution. Give units for all the terms.

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with all the carbonate and then "back titrating" the remaining acid with a \(\mathrm{NaOH}\) solution. (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a 0.1022 -g sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

A quantity of \(18.68 \mathrm{~mL}\) of a KOH solution is needed to neutralize \(0.4218 \mathrm{~g}\) of \(\mathrm{KHP}\). What is the concentration (in molarity) of the KOH solution?

A \(5.00 \times 10^{2} \mathrm{~mL}\) sample of \(2.00 \mathrm{M} \mathrm{HCl}\) solution is treated with \(4.47 \mathrm{~g}\) of magnesium. Calculate the concentration of the acid solution after all the metal has reacted. Assume that the volume remains un- changed.
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