Help with Oxidation-Reduction Reactions - Organic Chemistry

In an oxidation-reduction (redox) reaction, reduction and oxidation both occur. Thus, not all of the elements can be oxidized, and not all of them can be reduced. In this equation, the oxidation number of oxygen is . Multiplying that by the number of oxygen atoms (), the overall charge on oxygen is . has an overall charge, so the oxidation number on must combine with to form . Thus, the oxidation number of at the beginning of the reaction is . The iodine at the beginning of the reaction has an oxidation number of , as seen by the negative superscript.

At the end of the reaction, has an oxidation number of , as seen by the positive superscript. Iodine has an oxidation number of (there is no charge on the ).

Thus, went from having an oxidation number of to one of . Iodine went from having an oxidation number of to one of . Oxidation occurs when electrons are lost (the number becomes more positive), and reduction occurs when electrons are gained (the number becomes more negative). Because the oxidation number of iodine became more positive, iodine was oxidized. Because the oxidation number of manganese became more negative (less positive), manganese was reduced.

Lithium aluminum hydride is correct because it is a reducing agent, and is therefore not capable of oxidizing secondary alcohols. Instead, LAH could be used to perform the reverse reaction, reducing a ketone to an alcohol. The other answer choices are oxidizing agents.

In an oxidation-reduction (redox) reaction, reduction and oxidation both occur. Thus, not all of the elements can be oxidized, and not all of them can be reduced. In this equation, the oxidation number of oxygen is . Multiplying that by the number of oxygen atoms (), the overall charge on oxygen is . has an overall charge, so the oxidation number on must combine with to form . Thus, the oxidation number of at the beginning of the reaction is . The iodine at the beginning of the reaction has an oxidation number of , as seen by the negative superscript.

At the end of the reaction, has an oxidation number of , as seen by the positive superscript. Iodine has an oxidation number of (there is no charge on the ).

Thus, went from having an oxidation number of to one of . Iodine went from having an oxidation number of to one of . Oxidation occurs when electrons are lost (the number becomes more positive), and reduction occurs when electrons are gained (the number becomes more negative). Because the oxidation number of iodine became more positive, iodine was oxidized. Because the oxidation number of manganese became more negative (less positive), manganese was reduced.

Lithium aluminum hydride is correct because it is a reducing agent, and is therefore not capable of oxidizing secondary alcohols. Instead, LAH could be used to perform the reverse reaction, reducing a ketone to an alcohol. The other answer choices are oxidizing agents.

In an oxidation-reduction (redox) reaction, reduction and oxidation both occur. Thus, not all of the elements can be oxidized, and not all of them can be reduced. In this equation, the oxidation number of oxygen is . Multiplying that by the number of oxygen atoms (), the overall charge on oxygen is . has an overall charge, so the oxidation number on must combine with to form . Thus, the oxidation number of at the beginning of the reaction is . The iodine at the beginning of the reaction has an oxidation number of , as seen by the negative superscript.

At the end of the reaction, has an oxidation number of , as seen by the positive superscript. Iodine has an oxidation number of (there is no charge on the ).

Thus, went from having an oxidation number of to one of . Iodine went from having an oxidation number of to one of . Oxidation occurs when electrons are lost (the number becomes more positive), and reduction occurs when electrons are gained (the number becomes more negative). Because the oxidation number of iodine became more positive, iodine was oxidized. Because the oxidation number of manganese became more negative (less positive), manganese was reduced.

Lithium aluminum hydride is correct because it is a reducing agent, and is therefore not capable of oxidizing secondary alcohols. Instead, LAH could be used to perform the reverse reaction, reducing a ketone to an alcohol. The other answer choices are oxidizing agents.

In an oxidation-reduction (redox) reaction, reduction and oxidation both occur. Thus, not all of the elements can be oxidized, and not all of them can be reduced. In this equation, the oxidation number of oxygen is . Multiplying that by the number of oxygen atoms (), the overall charge on oxygen is . has an overall charge, so the oxidation number on must combine with to form . Thus, the oxidation number of at the beginning of the reaction is . The iodine at the beginning of the reaction has an oxidation number of , as seen by the negative superscript.

At the end of the reaction, has an oxidation number of , as seen by the positive superscript. Iodine has an oxidation number of (there is no charge on the ).

Thus, went from having an oxidation number of to one of . Iodine went from having an oxidation number of to one of . Oxidation occurs when electrons are lost (the number becomes more positive), and reduction occurs when electrons are gained (the number becomes more negative). Because the oxidation number of iodine became more positive, iodine was oxidized. Because the oxidation number of manganese became more negative (less positive), manganese was reduced.