Consider the following redox reaction occurring in acidic solution:
Cr2O72−(aq)+14H+(aq)+6I−(aq)⟶2Cr3+(aq)+3I2(aq)+7H2O(l) \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{I}^-(\text{aq}) \longrightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{I}_2(\text{aq}) + 7\text{H}_2\text{O}(\text{l}) Cr2O72−(aq)+14H+(aq)+6I−(aq)⟶2Cr3+(aq)+3I2(aq)+7H2O(l)Which statement correctly describes the oxidation process occurring in this reaction?
I−\text{I}^-I− is oxidized because its oxidation state increases from −1-1−1 to 000 via the loss of electrons.
I−\text{I}^-I− is oxidized because its oxidation state decreases from 000 to −1-1−1 via the gain of electrons.
Cr2O72−\text{Cr}_2\text{O}_7^{2-}Cr2O72− is oxidized because the oxidation state of chromium increases from +3+3+3 to +6+6+6 via the loss of electrons.
Cr2O72−\text{Cr}_2\text{O}_7^{2-}Cr2O72− is oxidized because the oxidation state of chromium decreases from +6+6+6 to +3+3+3 via the gain of electrons.