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Why must the number of electrons lost in the oxidation equal the number of electrons gained in the reduction? Is it possible to have "leftover" electrons in a reaction?

Short Answer

Expert verified
In redox reactions, the number of electrons lost in oxidation must equal the number of electrons gained in reduction to maintain charge balance and adhere to the conservation of electrons principle. This ensures that the reaction is complete and reaches its stable, lowest-energy state. Having "leftover" electrons would result in unstable radicals, which indicates an incomplete reaction. Therefore, it is not possible to have leftover electrons in a typical redox reaction.

Step by step solution

01

Understanding Redox Reactions

Redox reactions, short for "reduction-oxidation" reactions, are chemical processes in which one reactant is oxidized (loses electrons) and another reactant is reduced (gains electrons). In simpler terms, oxidation is a process where a substance loses electrons, while reduction is a process where a substance gains electrons.
02

Recognize The Role of Electrons

In redox reactions, electrons are being transferred between species. This transfer of electrons allows substances to change their oxidation states. The change in oxidation states, in turn, causes the observed chemical changes in a redox reaction.
03

Electron Balance in Redox Reactions

In a redox reaction, the number of electrons lost by the oxidized species must equal the number of electrons gained by the reduced species. This condition ensures that the reaction is balanced in terms of charge and that no "leftover" electrons are present. The conservation of electrons principle dictates that electrons cannot be created or destroyed during a chemical reaction; they can only change their location or be transferred between species.
04

No Leftover Electrons

If there were "leftover" electrons in a reaction, it would mean that the charges are not balanced and that the reaction is not complete. Incomplete redox reactions would result in reactive intermediates with unpaired electrons, called radicals, which are highly unstable and reactive. Most common redox reactions strive for a stable, lowest-energy state by ensuring that electrons are fully transferred between species, leaving no leftover electrons.
05

Conclusion

In redox reactions, the number of electrons lost in the oxidation process must equal the number of electrons gained in the reduction process, ensuring that the charges are balanced and adhering to the conservation of electrons principle. This balance prevents the formation of unstable radicals and ensures that the reaction reaches its stable, lowest-energy state with no leftover electrons.

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Most popular questions from this chapter

Assign oxidation states to all of the atoms in each of the following: a. \(\mathrm{MnCl}_{2}\) b. \(\mathrm{HBrO}_{3}\) c. \(\mathrm{B}_{2} \mathrm{H}_{6}\) d. \(\mathrm{CF}_{4}\)

Why must the sum of all the oxidation states of the atoms in a neutral molecule be zero?

An oxidizing agent causes the (oxidation/reduction) of another species, and the oxidizing agent itself is (oxidized/reduced).

Iodide ion, I\(^{-}\), is one of the most easily oxidized species. Balance each of the following oxidationreduction reactions, which take place in acidic \(50^{-}\) lution, by using the "half-reaction" method. a. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{I}_{2}(a q)\) b. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{C} \mathrm{r}^{3+}(a q)+\mathrm{I}_{2}(a q)\) c. \(\mathrm{Cu}^{2+}(a q)+\mathrm{I}^{-}(a q) \rightarrow \operatorname{CuI}(s)+\mathrm{I}_{2}(a q)\)

Potassium permanganate, \(\mathrm{KMnO}_{4}\), is one of the most widely used oxidizing agents. Balance each of the following oxidation-reduction reactions of the permanganate ion in acidic solution by using the "half-reaction" method. a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{CO}_{2}(g)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)\) c. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(g)\)

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