Chapter 17: Problem 40
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?
Chapter 17: Problem 40
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?
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Get started for freeAssign 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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