Question

Identify whether the atom or ion in each equation shows oxidation or reduction. Cu2+ + e− → Cu+Cu2+ is Fe → Fe3+ + 3e−Fe is F + e− F−F− is 2l− → l2 + 2e−l− is 2H+ + 2e− → H2H+ is

Answer 1

Cu2+ is reduced.

Fe is oxidized.

F- is reduced.

I- is oxidized.

H+ is reduced.

Answer 2

$Cu^{2+}$ is reduced, $Fe$ is oxidized, $I^-$ is oxidized, and $H^+$ is reduced.

Oxidation/Reduction

Oxidation and reduction each have several definitions. One of the definitions of each is as follows:

• Oxidation is an increase in the oxidation number of atoms
• Reduction is a decrease in the oxidation number of atoms.

For: $Cu^{2+} + e^- --- > Cu$, the oxidation number of Cu decreased from +2 to 0. thus, it is reduced.

For: $Fe --- > Fe^{3+} + 3e^-$, the oxidation number of Fe increased from 0 to +3, it is therefore oxidized.

For: $2I^- --- > I_2 + 2e^-$, the oxidation number of I increased from -1 to 0. It is oxidized.

For: $2H^+ + 2e^- --- > H_2$, the oxidation number of H decreased from +1 to 0. It has been reduced.

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