Which of the statement below correctly describes the combustion ofglucose? C6H12

Question

Which of the statement below correctly describes the combustion ofglucose?
C6H12O6+6O2->6Co2+6H2o

A. Hydrogen is in C6H12O6 is being reduced
B. Oxygen in O2 is being oxidized
C. Hydrogen in C6H12O6 is the reducing agent
D. Oxygen in C6H12O6 is the oxidizing agent
E Carbon in C6H12O6 is being oxidized.

I said it was E. But now I'm uncertain because I try doing theoxidation number method but I kind of got lost from the Glucosepart.

Explanation / Answer

Use oxidation numbers to figure it out. A species is said to be reduced if the oxidation number lowers fromreactants to products (the oxidation number REDUCES). This is theoxidizing agent. A species is said to be oxidized if the oxidation number increasesfrom reactants to products. This is the reducing agent. The oxidation numbers for each atom is written above theelement. 0 +1   -2       0              +4 -2     +1   -2 C6H12O6 + 6O2 ----->6CO2 + H2O The H in glucose does not take part in the redox, neither gainingor losing electrons. The O in glucose does not take part in the redox, neither gainingor losing electrons. The O in O2 is the oxidizing agent, thus being reduced (going from0 to -2). The carbon in glucose is the reducing agent, thus it is beingoxidized (E).

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