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Chapter 10: Problem 91

Burning octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), in your car engine forms water and carbon dioxide: \(2 \mathrm{C}_{8} \mathrm{H}_{18}+25 \mathrm{O}_{2} \rightarrow 18 \mathrm{H}_{2} \mathrm{O}+16 \mathrm{CO}_{2}\) What gets oxidized and what gets reduced?

Short Answer

Expert verified
In the given reaction, the carbon in octane (\(\mathrm{C}_{8}\mathrm{H}_{18}\)) is being oxidized from an oxidation state of \(0\) to \(+4\), and oxygen is being reduced from an oxidation state of \(0\) to \(-2\).

Step by step solution

01

Assign oxidation numbers to each element in the reaction

In this step, we'll assign oxidation numbers to each element in the given balanced chemical equation: \[2 \mathrm{C}_{8} \mathrm{H}_{18} + 25 \mathrm{O}_{2} \rightarrow 18 \mathrm{H}_{2} \mathrm{O} + 16 \mathrm{CO}_{2}\] Oxidation numbers: - Carbon in octane: \(+0\) - Hydrogen in octane: \(+1\) - Oxygen in oxygen molecule: \(0\) - Hydrogen in water: \(+1\) - Oxygen in water: \(-2\) - Carbon in CO2: \(+4\) - Oxygen in CO2: \(-2\)
02

Identify the change in oxidation numbers for each element

In this step, we will compare the oxidation numbers of each element before and after the reaction and determine the change in oxidation numbers. Changes in oxidation numbers: - Carbon in octane to Carbon in CO2: \(0\) to \(+4\) (Increase of \(4\)) - Hydrogen in octane to Hydrogen in water: \(+1\) to \(+1\) (No change) - Oxygen in O2 to Oxygen in water and CO2: \(0\) to \(-2\) (Decrease of \(2\))
03

Determine which elements are oxidized and reduced

Based on the changes in the oxidation numbers, we can now determine which elements are being oxidized and reduced. - Carbon: Since the oxidation number increased from \(0\) to \(+4\), carbon is losing electrons and is being oxidized. - Oxygen: Since the oxidation number decreased from \(0\) to \(-2\), oxygen is gaining electrons and is being reduced.
04

Conclusion

In the given reaction, octane or specifically, the carbon in octane is being oxidized, and oxygen is being reduced.

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Most popular questions from this chapter

Suppose the shortcut rules can determine the oxidation state of every atom in a compound except one. How can you find the oxidation state of the remaining atom?

Using the EMF series on page 391, decide which of the following redox reactions is spontaneous. Explain your answer. (a) \(3 \mathrm{Ag}+\mathrm{Au}^{3+} \rightarrow \mathrm{Au}+3 \mathrm{Ag}^{+}\) (b) \(\mathrm{Au}+3 \mathrm{Ag}^{+} \rightarrow 3 \mathrm{Ag}+\mathrm{Au}^{3+}\)

In a nickel-cadmium battery, the relevant redox reaction is: \(2 \mathrm{NiO}(\mathrm{OH})+\mathrm{Cd}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow\) \(2 \mathrm{Ni}(\mathrm{OH})_{2}+\mathrm{Cd}(\mathrm{OH})_{2}\) Does this agree with the EMF series? What are the oxidation states of nickel before and after the reaction?

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